Solution_manual_Introduction_to_Robotics_Analysis_Contro

Prévia do material em texto

Introduction to Robotics 
Analysis, Control, Applications 
 
 
 
Solution Manual 
 
 
 
 
Saeed B. Niku 
 
 
 
 
 
 
 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on 
BlackBoard or any other electronic media system and the Internet without prior expressed 
consent of the copyright owner.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
2
CHAPTER ONE 
 
Problem 1.1 
Draw the approximate workspace for the following robot. Assume the dimensions of the 
base and other parts of the structure of the robot are as shown. 
 
Estimated student time to complete: 15-25 minutes 
Prerequisite knowledge required: Text Section(s) 1.14 
 
Solution: 
 
The workspace shown is approximate. 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
3
 
Problem 1.2 
Draw the approximate workspace for the following robot. Assume the dimensions of the 
base and other parts of the structure of the robot are as shown. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 1.14 
 
Solution: 
 
The workspace shown is approximate. 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
4
Problem 1.3 
Draw the approximate workspace for the following robot. Assume the dimensions of the 
base and other parts of the structure of the robot are as shown. 
 
Estimated student time to complete: 10-15 minutes 
Prerequisite knowledge required: Text Section(s) 1.14 
 
Solution: 
 
The workspace shown is approximate. 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
5
CHAPTER TWO 
 
Problem 2.1 
Write a unit vector in matrix form that describes the direction of the cross product of 
5 3= +p i k and 3 4 5= + +q i j k . 
 
Estimated student time to complete: 5-10 minutes 
Prerequisite knowledge required: Text Section(s) 2.4 
 
Solution: 
 
( ) ( ) ( )
2 2 2
5 0 3 0 12 25 9 20 0 12 16 20
3 4 5
144 256 400 28.28
12
28.28 0.424
16 0.566
28.28
0.70720
28.28
x y zr r rλ
⎡ ⎤
⎢ ⎥= × = = − − − + − = − − +⎢ ⎥
⎢ ⎥⎣ ⎦
= + + = + + =
−⎡ ⎤
⎢ ⎥
−⎡ ⎤⎢ ⎥
− ⎢ ⎥⎢ ⎥= = −⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥ ⎣ ⎦
⎢ ⎥
⎢ ⎥⎣ ⎦
i j k
r p q i j k i j k
r
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
6
Problem 2.2 
A vector p is 8 units long and is perpendicular to vectors q and r described below. 
Express the vector in matrix form. 
 
0.3
0.4
0
y
unit
q
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
q 
0.5
0.4
0
x
unit
r⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
r 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 2.4 
 
Solution: 
 
The two vectors given are unit vectors. Therefore, each missing component can be found 
as: 
 
1 0.09 0.16 0.866
1 0.25 0.16 0.768
y
x
q
r
= − − =
= − − =
 
 
Since p is perpendicular to the other two vectors, it is in the direction of the cross product 
of the two. Therefore: 
 
( ) ( ) ( )
( ) ( ) ( )
0.3 0.866 0.4 0.346 0.2 0.12 0.307 0.15 0.665
0.768 0.5 0.4
0.146 0.187 0.515
pλ
⎡ ⎤
⎢ ⎥= = − − − + −⎢ ⎥
⎢ ⎥⎣ ⎦
= + −
i j k
i j k
i j k
 
 
Since q and r are not perpendicular to each other, the resulting p is not a unit vector. 
Vector p can be found as: 
 
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )( )
( ) ( ) ( )
2 2 2
0.146 0.187 0.515
0.146 0.187 0.515 0.567
8 14.1
0.567
0.146 0.187 0.515
2.06 2.64 7.27
p
p
w
w
λ
λ
= + −
= + + =
= =
= + −
= + −
i j k
p i j k
p i j k
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
7
Problem 2.3 
Will the three vectors p, q, and r in Problem 2.2 form a traditional frame? If not, find the 
necessary unit vector s to form a frame between p, q, and s. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 2.4 
 
Solution: 
 
As we saw in Problem 2.2, since q×r is not a unit vector, it means that q and r and not 
perpendicular to each other, and therefore, they cannot form a frame. However, p and q 
are perpendicular to each other, and we can select s to be perpendicular to those two. Of 
course, p is not a unit length, therefore we use the unit vector representing it. 
 
( ) ( ) ( )
( ) ( ) ( )( )
( ) ( ) ( )
2 2 20.146 0.187 0.515 0.567
1 1.764
0.567
0.146 0.187 0.515
0.257 0.33 0.908
p
w
w
λ = + + =
= =
= + −
= + −
p i j k
p i j k
 
( ) ( ) ( )
( ) ( ) ( )
0.257 0.33 0.908
0.257 0.33 0.908 0.918 0.375 0.124
0.3 0.866 0.4
= + −
⎡ ⎤
⎢ ⎥= − = − +⎢ ⎥
⎢ ⎥⎣ ⎦
p i j k
i j k
s i j k
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
8
Problem 2.4 
Suppose that instead of a frame, a point ( )3,5,7 TP = in space was translated a distance 
of ( )2,3,4 Td = . Find the new location of the point relative to the reference frame. 
 
Estimated student time to complete: 5 minutes 
Prerequisite knowledge required: Text Section(s) 2.6 
 
Solution: 
 
As for a frame, 
 
1 0 0 2 3 5
0 1 0 3 5 8
0 0 1 4 7 11
0 0 0 1 1 1
newP
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
9
Problem 2.5 
The following frame B was moved a distance of ( )5,2,6 Td = . Find the new location of 
the frame relative to the reference frame. 
 
0 1 0 2
1 0 0 4
0 0 1 6
0 0 0 1
B
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 5-10 minutes 
Prerequisite knowledge required: Text Section(s) 2.6 
 
Solution: 
 
The transformation matrix representing the translation is used to find the new location as: 
 
1 0 0 5 0 1 0 2 0 1 0 7
0 1 0 2 1 0 0 4 1 0 0 6
0 0 1 6 0 0 1 6 0 0 1 12
0 0 0 1 0 0 0 1 0 0 0 1
newB
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
10
Problem 2.6 
For frame F, find the values of the missing elements and complete the matrix 
representation of the frame. 
 
? 0 1 5
? 0 0 3
? 1 0 2
0 0 0 1
F
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 2.4 
 
Solution: 
 
0 1 5
0 0 3
1 0 2
0 0 0 1
x
y
z
n
n
F
n
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
From × =n o a 
0 0 1
x y zn n n
⎡ ⎤
⎢ ⎥ = −⎢ ⎥
⎢ ⎥−⎣ ⎦
i j k
i 
Or: ( ) ( ) ( )0y xn n− − − + = −i j k i , and therefore: 1, 0, 0y x zn n n= = = 
 
0 0 1 5
1 0 0 3
0 1 0 2
0 0 0 1
F
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied,posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
11
Problem 2.7 
Find the values of the missing elements of frame B and complete the matrix 
representation of the frame. 
 
 
0.707 ? 0 2
? 0 1 4
? 0.707 0 5
0 0 0 1
B
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 2.4 
 
Solution: 
 
0.707 0 2
0 1 4
0.707 0 5
0 0 0 1
x
y
z
o
n
B
n
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
From × =n o a 0.707
0 0.707
y z
x
n n
o
⎡ ⎤
⎢ ⎥ =⎢ ⎥
⎢ ⎥⎣ ⎦
i j k
j 
 
Therefore: 0.707
0 0.707
y z
x
n n
o
⎡ ⎤
⎢ ⎥ =⎢ ⎥
⎢ ⎥⎣ ⎦
i j k
j 
And ( ) ( ) ( )0.707 0.5 0y z x y x yn n o n o n− − + − = → =i j k j 
 
From length equations: 1=n or 
2 2 2
2
0.707 1 0.707
0.5 1 0.707
y z z
x x
n n n
o o
+ + = → = ±
+ = → = ±
 
 
Therefore, there are two possible acceptable solutions: 
 
 
0.707 0.707 0 2
0 0 1 4
0.707 0.707 0 5
0 0 0 1
B
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 and 
0.707 0.707 0 2
0 0 1 4
0.707 0.707 0 5
0 0 0 1
B
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
12
Problem 2.8 
Derive the matrix that represents a pure rotation about the y-axis of the reference frame. 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.2. 
 
Solution: 
 
From the figure: 
cos sin
sin cos
x n a
y o
z n a
p p p
p p
p p p
θ θ
θ θ
= +
=
= − +
 and 
0
0 1 0
0
x n
y o
z a
p C S p
p p
p S C p
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
13
Problem 2.9 
Derive the matrix that represents a pure rotation about the z-axis of the reference frame. 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.2. 
 
Solution: 
From the Figure: 
 
cos sin
sin cos
x n o
y n o
z a
p p p
p p p
p p
θ θ
θ θ
= −
= +
=
 and 
0
0
0 0 1
x n
y o
z a
p C S p
p S C p
p p
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
14
Problem 2.10 
Verify that the rotation matrices about the reference frame axes follow the required 
constraint equations set by orthogonality and length requirements of directional unit 
vectors. 
 
Estimated student time to complete: 10 minutes. 
Prerequisite knowledge required: Text Section(s) 2.4.5 and 2.6.2 
 
Solution: 
 
For 
1 0 0
( , ) 0
0
Rot x C S
S C
θ
⎡ ⎤
⎢ ⎥= −⎢ ⎥
⎢ ⎥⎣ ⎦
 we have: 
 
0=n oi or nx ox + ny oy + nz oz = 0 
0=n ai 
0=a oi 
1=n 
2 2 1S C= + =o 
( )2 2 1S C= − + =a 
 
( , )Rot y θ and ( , )Rot z θ will be the same.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
15
Problem 2.11 
Find the coordinates of point (2,3,4)TP relative to the reference frame after a rotation of 
45 about the x-axis. 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.3. 
 
Solution: 
 
2 1 0 0 2 2
( , 45) 3 0 0.707 0.707 3 0.707
4 0 0.707 0.707 4 4.95
U P Rot x
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Note that the rotation is written in a 3 3× , not homogeneous form, because we are only 
concerned about the rotation part.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
16
Problem 2.12 
Find the coordinates of point (3,5,7)TP relative to the reference frame after a rotation of 
30 about the z-axis. 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.3. 
 
Solution: 
 
3 0.866 0.5 0 3 0.1
( ,30) 5 0.5 0.866 0 5 5.83
7 0 0 1 7 7
U P Rot z
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Note that the rotation is written in a 3 3× form not homogeneous form, because we are 
only concerned about the rotation part.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
17
Problem 2.13 
Find the new location of point (1, 2,3)TP relative to the reference frame after a rotation of 
30 about the z-axis followed by a rotation of 60 about the y-axis. 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.3. 
 
Solution: 
 
( ,60) ( ,30)
0.5 0 0.866 0 0.866 0.5 0 0 1 2.531
0 1 0 0 0.5 0.866 0 0 2 2.232
0.866 0 0.5 0 0 0 1 0 3 1.616
0 0 0 1 0 0 0 1 1 1
U
U
P Rot y Rot z P
P
=
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
18
Problem 2.14 
A point P in space is defined as (5,3,4)B TP = relative to frame B which is attached to the 
origin of the reference frame A and is parallel to it. Apply the following transformations 
to frame B and find AP. Using the 3-D grid, plot the transformations and the result and 
verify it. Also verify graphically that you would not get the same results if you apply the 
transformations relative to the current frame: 
• Rotate 90 about x-axis, then 
• Translate 3 units about y-axis, 6 units about z-axis, and 5 units about x-axis. Then, 
• Rotate 90 about z-axis. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.3. 
 
Solution: 
 
( ,90) (5,3,6) ( ,90)
0 1 0 0 1 0 0 5 1 0 0 0 5 1
1 0 0 0 0 1 0 3 0 0 1 0 3 10
0 0 1 0 0 0 1 6 0 1 0 0 4 9
0 0 0 1 0 0 0 1 0 0 0 1 1 1
A B
A
P Rot z Trans Rot x P
P
=
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
19
Problem 2.15 
A point P in space is defined as (2,3,5)B TP = relative to frame B which is attached to the 
origin of the reference frame A and is parallel to it. Apply the following transformations 
to frame B and find AP. Using the 3-D grid, plot the transformations and the result and 
verify it: 
• Rotate 90 about x-axis, then 
• Rotate 90 about local a-axis, then 
• Translate 3 units about y-, 6 units about z-, and 5 units about x-axes. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.4. 
 
Solution: 
 
(5,3,6) ( ,90) ( ,90)
1 0 0 5 1 0 0 0 0 1 0 0 2 2
0 1 0 3 0 0 1 0 1 0 0 0 3 2
0 0 1 6 0 1 0 0 0 0 1 0 5 8
0 0 0 1 0 0 0 1 0 0 0 1 1 1
A B
A
P Trans Rot x Rot a P
P
=
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
20
Problem 2.16 
A frame B is rotated 90 about the z-axis, then translated 3 and 5 units relative to the n- 
and o-axes respectively, then rotated another 90 about the n-axis, and finally, 90 about 
the y-axis. Find the new location and orientation of the frame. 
 
 
0 1 0 1
1 0 0 1
0 0 1 1
0 0 0 1
B
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.4. 
 
Solution: 
 
( ,90) ( ,90) (3,5,0) ( ,90)
0 0 1 0 0 1 0 0 0 1 0 1 1 0 0 3 1 0 0 0 0 1 0 1
0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 5 0 0 1 0
1 0 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 1 0 0
0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
new
new
B Rot y Rot z BTrans Rot n
B
=
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
0 0 1 6
1 0 0 4
0 0 0 1
⎡ ⎤
⎢ ⎥−⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
21
Problem 2.17 
The frame B of problem 2.16 is rotated 90 about the a-axis, 90 about the y-axis, then 
translated 2 and 4 units relative to the x- and y-axes respectively, then rotated another 
90 about the n-axis. Find the new location and orientation of the frame. 
 
 
0 1 0 1
1 0 0 1
0 0 1 1
0 0 0 1
B
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 2.6.4. 
 
Solution: 
 
(2,4,0) ( ,90) ( ,90) ( ,90)
1 0 0 2 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 0 0 1 0 3
0 1 0 4 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 0
0 0 1 0 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0
0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
new
new
B Trans Rot y B Rot a Rot n
B
=
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
0 0 1 5
1 0 0 1
0 0 0 1
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
22
Problem 2.18 
Show that rotation matrices about the y- and the z-axes are unitary. 
 
Estimated student time to complete: 
Prerequisite knowledge required: Text Section(s) 2.7. 
 
Solution: 
 
0
( , ) 0 1 0
0
C S
A Rot y
S C
θ
⎡ ⎤
⎢ ⎥= = ⎢ ⎥
⎢ ⎥−⎣ ⎦
 
 
2 2det 1( ) 0 1A C Sθ θ= + + = 
1 0 0
0
0
TA C S
S C
θ θ
θ θ
⎡ ⎤
⎢ ⎥= ⎢ ⎥
⎢ ⎥−⎣ ⎦
 
 
1
0
adj 0 1 0 /1
0
T
minor
C S
A A A
S C
−
−⎡ ⎤
⎢ ⎥= = =⎢ ⎥
⎢ ⎥⎣ ⎦
 
 
1 TA A− = 
 
Exact same is true for rotation about the z-axis.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
23
Problem 2.19 
Calculate the inverse of the following transformation matrices: 
 
 1 2
0.527 0.574 0.628 2 0.92 0 0.39 5
0.369 0.819 0.439 5 0 1 0 6
and
0.766 0 0.643 3 0.39 0 0.92 2
0 0 0 1 0 0 0 1
T T
−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
 
 
Estimated student time to complete: 15-20 minutes. 
Prerequisite knowledge required: Text Section(s) 2.7. 
 
Solution: 
 
Since transformation matrices are unitary, we calculate the inverses simply by 
transposing the rotation part and calculating the position part by: 
 
0.527 2 0.369 5 0.766 3 0.601
0.574 2 0.819 5 0 3 2.947
0.628 2 0.439 5 0.643 3 5.38
× + × − × = −
− × + × − × = −
× + × + × = −
 
 
Therefore: 
 
1
1
0.527 0.369 0.766 0.601
0.574 0.819 0 2.947
0.628 0.439 0.643 5.38
0 0 0 1
T −
− −⎡ ⎤
⎢ ⎥− −⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
And similarly, 
 
1
2
0.92 0 0.39 3.82
0 1 0 6
0.39 0 0.92 3.79
0 0 0 1
T −
− −⎡ ⎤
⎢ ⎥−⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
24
Problem 2.20 
Calculate the inverse of the matrix B of Problem 2.17. 
 
Estimated student time to complete: 5 minutes 
Prerequisite knowledge required: Text Section(s) 2.7. 
 
Solution: 
 
0 1 0 1
1 0 0 1
0 0 1 1
0 0 0 1
B
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
1
0 1 0 1
1 0 0 1
0 0 1 1
0 0 0 1
B−
−⎡ ⎤
⎢ ⎥−⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
25
Problem 2.21 
Write the correct sequence of movements that must be made in order to restore the 
original orientation of the spherical coordinates and make it parallel to the reference 
frame. About what axes are these rotations supposed to be? 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 2.9.3. 
 
Solution: 
 
1 0 0
0 1 0
( , , ) ( , ) ( , )
0 0 1
0 0 0 1
sph
rS C
rS S
T r Rot o Rot a
rC
β γ
β γ
β γ β γ
β
⎡ ⎤
⎢ ⎥
⎢ ⎥− − =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
As shown, the rotations are relative to o- and a-axes in the sequence shown. These 
rotations must be relative to the current moving frame in order to prevent changing the 
position of the frame.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
26
Problem 2.22 
A spherical coordinate system is used to position the hand of a robot. In a certain 
situation, the hand orientation of the frame is later restored in order to be parallel to the 
reference frame, and the matrix representing it is described as: 
 
1 0 0 3.1375
0 1 0 2.195
0 0 1 3.214
0 0 0 1
sphT
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
• Find the necessary values of , ,r β γ to achieve this location. 
• Find the components of the original matrix , ,n o a vectors for the hand before the 
orientation was restored. 
 
Estimated student time to complete: 20 minutes 
Prerequisite knowledge required: Text Section(s) 2.9.3. 
 
Solution: 
 
a) The equations representing position in a spherical coordinates may be set equal to the 
given values as: 
i. 3.1375
ii. 2.195
iii. 3.214
rC S
rS S
rC
γ β
γ β
β
=
=
=
 
I. Assuming Sβ is positive, from i and ii 
 35γ = 
 from ii and iii 50β = 
 from iii r=5 units. 
 
II. If Sβ were negative, then 215γ = 
 50β = − 
 r=5 units. 
 
Since orientation is not specified, no more information is available to check the results. 
 
b) For case I, substitute corresponding values of Sβ , Cβ , Sγ ,Cγ and r in spherical 
coordinates to get: 
 
0.5265 0.5735 0.6275 3.1375
0.3687 0.819 0.439 2.195
( , , ) (35,50,5)
0.766 0 0.6428 3.214
0 0 0 1
sph sphT r Tβ γ
−⎡ ⎤
⎢ ⎥
⎢ ⎥= =
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
27
Problem 2.23 
Suppose that a robot is made of a Cartesian and RPY combination of joints. Find the 
necessary RPY angles to achieve the following: 
 
 
0.527 0.574 0.628 4
0.369 0.819 0.439 6
0.766 0 0.643 9
0 0 0 1
T
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 20 minutes 
Prerequisite knowledge required: Text Section(s) 2.10.1 
 
Solution: 
 
The position isachieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units 
along the x-, y-, and z-axes. The orientation is achieved by RPY rotations, therefore: 
 
 2( , ) 2(0.369,0.527) 35a y xATAN n n ATANφ = = = 
or 2( , ) 2( 0.369, 0.527) 215a y xATAN n n ATANφ = − − = − − = 
 
For 35aφ = 
( )( )2( , ( )) 2 0.766, 0.527 0.819 0.369 0.574 50o z x a y aATAN n n C n S ATANφ φ φ= − + = × + × =
 
For 215aφ = 
( )2 0.766, 0.643 130o ATANφ = − = 
 
For 35aφ = 
( ) ( )
( )
2(( ), ( ))
2
2
0.439 0.819 0.628 0.574 , 0.819 0.819 0.574 0.574
0,1 0
n y a x a y a x aATAN a C a S o C o S
ATAN
ATAN
φ φ φ φ φ= − + −
=
=
− × + × × + ×⎡ ⎤⎣ ⎦
=
 
 
For 215aφ = 
( )2 0, 1 180n ATANφ = − = 
 
Either solution is acceptable: 
35 215
50 130
0 180
a a
o o
n n
φ φ
φ φ
φ φ
⎧ ⎧= =
⎪ ⎪= =⎨ ⎨
⎪ ⎪= =⎩ ⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
28
Problem 2.24 
Suppose that a robot is made of a Cartesian and Euler combination of joints. Find the 
necessary Euler angles to achieve the following: 
 
 
 
0.527 0.574 0.628 4
0.369 0.819 0.439 6
0.766 0 0.643 9
0 0 0 1
T
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 20 minutes 
Prerequisite knowledge required: Text Section(s) 2.10.2 
 
Solution: 
 
The position is achieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units 
along the x-, y-, and z-axes. The orientation is achieved by Euler rotations, therefore: 
 
2( , ) 2( , )y x y xATAN a a or ATAN a aφ φ= = − − 
or 
2(0.439,0.628) 35
2( 0.439, 0.628) 215
ATAN
ATAN
φ
φ
⎧ = =⎪
⎨
= − − =⎪⎩
 
 
2[( ), ( )]x y x yATAN n S n C o S o Cψ φ φ φ φ= − + − + 
or 
2[0,1] 0
2[0, 1] 180
ATAN
ATAN
ψ
ψ
⎧ = =⎪
⎨
= − =⎪⎩
 
 
2[( ), )]x y zATAN a C a S aθ φ φ= + 
or 
( ) ( )
2[(0.628 0.819 0.439 0.573),0.643)] 50
2[(0.628 0.819 0.439 0.573 ),0.643)] 50
ATAN
ATAN
θ
θ
⎧ = × + × =⎪
⎨
= × − + × − = −⎪⎩
 
 
Then 
35 215
0 180
50 50
φ φ
ψ ψ
θ θ
⎧ ⎧= =
⎪ ⎪= =⎨ ⎨
⎪ ⎪= = −⎩ ⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
29
Problem 2.25 
Assume that the three Euler angles used with a robot are 30 ,40 ,50 respectively. 
Determine what angles should be used to achieve the same result if RPY is used instead. 
 
Estimated student time to complete: 20-25 minutes 
Prerequisite knowledge required: Text Section(s) 2.10 
 
Solution: 
 
For Euler angles: 
 
Euler( , , ) ( , ) ( , ), ( , ) ( ,30) ( , 40), ( ,50)Rot a Rot o Rot a Rot a Rot o Rot aφ θ ψ φ θ ψ= = 
=
0.0435 0.8296 0.5568 0
0.9096 0.2635 0.3215 0
0.4134 0.4925 0.766 0
0 0 0 1
−⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
2( , ) 2(0.9096,0.0435) 87.26a y xATAN n n ATANφ = = = 
or 2( , ) 2( 0.9096, 0.0435) 267.26a y xATAN n n ATANφ = − − = − − = 
 
( )2( , ( )) 2 0.4134,0.9106 24.5o z x a y aATAN n n C n S ATANφ φ φ= − + = = 
or ( )2 0.4134, 0.9106 155.5o ATANφ = − = 
 
( )
2(( ), ( ))
2 0.5408,0.8412 32.7
n y a x a y a x aATAN a C a S o C o S
ATAN
φ φ φ φ φ= − + −
= =
 
 
or 2( 0.5408, 0.8412) 212.7n ATANφ = − − = 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
30
Problem 2.26 
A frame UB was moved along its own o-axis a distance of 6 units, then rotated about its 
n-axis an angle of 60 , then translated about the z-axis for 3 units, followed by a rotation 
of 60 about the z-axis, and finally rotated about x-axis for 45 . 
• Calculate the total transformation performed. 
• What angles and movements would we have to make if we were to create the same 
location and orientation using Cartesian and Euler configurations? 
 
Estimated student time to complete: 25-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.9- 2.10 
 
Solution: 
 
( , 45) ( ,60) (0,0,3) (0,6,0) ( ,60)
0.5 0.433 0.75 5.196
0.6123 0.4355 0.6596 0
0.6123 0.789 0.0474 4.242
0 0 0 1
U
B
new
T Rot x Rot z Trans Trans Rot n
B
=
− −⎡ ⎤
⎢ ⎥− −⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
The Cartesian translations are: 5.196, 0, 4.242x x xp p p= − = = . Euler angles are: 
 
2( , ) 2( , )y x y xATAN a a or ATAN a aφ φ= = − − 
or 
2( 0.6596,0.75) 41.33
2(0.6596, 0.75) 138.67
ATAN
ATAN
φ
φ
⎧ = − = −⎪
⎨
= − =⎪⎩
 
 
2[( ), ( )]x y x yATAN n S n C o S o Cψ φ φ φ φ= − + − + 
or 
2[0.79, 0.613] 127.8
2[ 0.79,0.613] 52.2
ATAN
ATAN
ψ
ψ
⎧ = − =⎪
⎨
= − = −⎪⎩
 
 
2[( ), )]x y zATAN a C a S aθ φ φ= + 
or 
2[0.9988,0.0474] 87.28
2[ 0.9988,0.0474] 87.28
ATAN
ATAN
θ
θ
⎧ = =⎪
⎨
= − = −⎪⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
31
Problem 2.27 
A frame UF was moved along its own n-axis a distance of 5 units, then rotated about its 
o-axis an angle of 60 , followed by a rotation of 60 about the z-axis, then translated 
about its a-axis for 3 units, and finally rotated 45 about the x-axis. 
• Calculate the total transformation performed. 
• What angles and movements would we have to make if we were to create the same 
location and orientation using Cartesian and RPY configurations? 
 
Estimated student time to complete: 20-25 minutes 
Prerequisite knowledge required: Text Section(s) 2.11 
 
Solution: 
 
( , 45) ( ,60) (5,0,0) ( ,60) (0,0,3)
0.25 0.866 0.433 3.8
0.918 0.354 0.177 3.59
0.306 0.354 0.884 5.71
0 0 0 1
U
B
new
T Rot x Rot z Trans Rot o Trans
B
=
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
For Cartesian coordinates: 3.8, 3.59, 5.71x y zp p p= = = 
 
For RPY angles: 
2( , ) 2(0.918,0.25) 74.8a y xATAN n n ATANφ = = = 
or 2( , ) 254.8a y xATAN n nφ = − − = 
 
( )2( , ( )) 2 0.306,0.951 17.8o z x a y aATAN n n C n S ATANφ φ φ= − + = = 
( )2 0.306, 0.951 162.2o ATANφ = − = 
 
[ ]
2(( ), ( ))
2 0.372,0.928 21.8
n y a x a y a x aATAN a C a S o C o S
ATAN
φ φ φ φ φ= − + −
= =
 
 
( )2 0.372, 0.928 201.8n ATANφ = − − = 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
32
Problem 2.28 
Frames describing the base of a robot and an object are given relative to the Universe 
frame. 
• Find a transformation RTH of the robot configuration if the hand of the robot is to be 
placed on the object. 
• By inspection, show whether this robot can be a 3-axis spherical robot, and if so, find 
, , rα β . 
• Assuming that the robot is a 6-axis robot with Cartesian and Euler coordinates, find 
, , , , ,x y zp p p φ θ ψ . 
 
 
1 0 0 1 0 1 0 2
0 0 1 4 1 0 0 1
0 1 0 0 0 0 1 0
0 0 0 1 0 0 0 1
U U
obj RT T
−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
 
 
Estimated student time to complete: 20-25 minutes 
Prerequisite knowledge required: Text Section(s) 2.11 
 
Solution: 
 
a. We want to place the hand on the part. Therefore, 1R U UH H objT T T
−= 
 
1U U R U R R U U
obj R H R obj H R objT T T T T T T T
−= = → = 
 
Substitute: 
0 1 0 1 1 0 0 1 0 0 1 5
1 0 0 2 0 0 1 4 1 0 0 1
0 0 1 0 0 1 0 0 0 1 0 0
0 0 0 1 0 0 0 1 0 0 0 1
R
HT
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
b. No. The 3,2 element of spherical coordinate transformation matrix should be zero. This 
is not. 
 
c. 5, 1, 0x y zp p p= = = 
 
2[0, 1] 180 or 2[0,1] 0ATAN ATANφ φ= − = = = 
2[ 1( 1),0] 90 or 2[ 1(1),0] 270ATAN ATANψ ψ= − − = = − = 
2[ 1( 1),0] 90 or 2[ 1(1),0] 270ATAN ATANθ θ= − − = = − = 
Then 
180 0
90 27090 270
φ φ
ψ ψ
θ θ
⎧ ⎧= =
⎪ ⎪= =⎨ ⎨
⎪ ⎪= =⎩ ⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
33
Problem 2.29 
A 3-DOF robot arm has been designed for applying paint on flat walls, as shown. 
• Assign coordinate frames as necessary based on the D-H representation. 
• Fill out the parameters table. 
• Find the UTH matrix. 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. Please also note that no particular reset position is specified. 
 
# θ d a α 
0-1 1θ l4 0 0 
1-2 0 0 l5 -90 
2-H 90 l6 0 0 
 
1 2 3
1 1 1 5
2 1 1
3 4 6
1 1 1 6 1 5 1
1 1 1 6 1 5 2
4 3
1 0 0 0 0 1 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0 1 0 0 0
0 0 1 0 0 1 0 1 0 0 0 0 1
0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
0
0
1 0 0
0 0 0 1
U U R U
H R H RT T T T A A A
l C S l
l S C
l l l
C S S l C l l
S C C l S l l
l l
= =
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥=
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
− − − + +⎡
⎢ + +
=
− +
⎣
⎤
⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
34
Problem 2.30 
In the 2-DOF robot shown, the transformation matrix 0TH is given in symbolic form, as 
well as in numerical form for a specific location. The length of each link l1 and l2 is 1 ft. 
Calculate the values of 1 2 and θ θ for the given location. 
 
12 12 2 12 1 1
12 12 2 12 1 10
0 0.2924 0.9563 0 0.6978
0 0.9563 0.2924 0 0.8172
0 0 1 0 0 0 1 0
0 0 0 1 0 0 0 1
H
C S l C l C
S C l S l S
T
− + − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥+ −⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
 
z
x
y
x0
z1
x1
zH
xH2θ
1θ
z0
 
Figure P.2.30 
 
Estimated student time to complete: 20 minutes 
Prerequisite knowledge required: Text Section(s) 2.13 
 
Solution: 
 
This example is very simple, with only two degrees of freedom. Therefore, there is really 
no need to use the traditional method of inverse kinematic mentioned in Section 2.13. 
The joint angles can be determined as follows: 
 
( )
( ) ( ) ( )( )
1 2 12 12
1 2 12 12 12 1 1
1 2 12 1 2 12 1
2 12 1 1 1 2 12 1
2( , ) 2(0.9563, 0.2924) 107
/
2 / , /
/
xx
y x
y y
ATAN S C ATAN
C p l C ll C l C p
ATAN p l S l p l C l
l S l S p S p l S l
θ θ
θ
+ = = − =
= −⎧+ =⎧ ⎪→ → = − −⎨ ⎨+ = = −⎩ ⎪⎩
 
Substitute: 
( ) ( )( )
( ) ( )( ) ( )
1 2 12 1 2 12 1
1
2
2 / , /
2 0.8172 0.9563 , 0.6978 0.2924 2 0.1391,0.9902 8
107 ( 8) 115
y xATAN p l S l p l C l
ATAN ATAN
θ
θ
θ
= − −
= − + = − = −
= − − =
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
35
Problem 2.31 
For the following SCARA-type robot: 
• Assign the coordinate frames based on the D-H representation. 
• Fill out the parameters table. 
• Write all the A matrices. 
• Write the UTH matrix in terms of the A matrices. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. Please also note that no particular reset position is specified. 
 
# θ d a α 
0-1 1θ 0 d3 0 
1-2 2θ 0 d4 180 
2-H 3θ d5 0 0 
 
0
0 0 1 2 3
1 1 3 1 2 2 4 2 3 3
1 1 3 1 2 2 4 2 3 3
1 2 5
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 1 0 0 0 1 0 0 0 1
0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
U U U
H HT T T T A A A
C S d C C S d C C S
S C d S S C d S S C
d d d
= =
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥=
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Multiplying these matrices will yield the total transformation.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
36
Problem 2.32 
A special 3-DOF spraying robot has been designed as shown: 
• Assign the coordinate frames based on the D-H representation. 
• Fill out the parameters table. 
• Write all the A matrices. 
• Write the UTH matrix in terms of the A matrices. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
Please note that no particular reset position is specified for this robot. 
 
# θ d a α 
0-1 190 θ+ 0 0 90 
1-2 0 l2 0 -90 
2-3 3θ 0 0 90 
3-H 0 l3 0 0 
 
1 1 1
1 1
0 1
3 3
3 3
2 3 4
2 3
1 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 1 0 0
0 0 0 1 0 0 1 1
1 0 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 1 0 0
0 1 0 0 1 0 0 0 0 1
0 0 0 1 0 0 0 1 0 0 0 1
U
l S C
C S
T A
C S
S C
A A A
l l
−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
0
0 0 1 2 3 4
U U U
H HT T T T A A A A= = 
The total transformation can be summarized by multiplying these matrices. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
37
Problem 2.33 
For the Unimation Puma 562, 6-axis robot shown, 
• Assign the coordinate frames based on the D-H representation. 
• Fill out the parameters table. 
• Write all the A matrices. 
• Find the RTH matrix for the following values: 
 Base height=27 in, d2=6 in, a2=15 in, a3=1 in, d4=18 in
 1 2 3 4 5 60 , 45 , 0 , 0 , 45 , 0θ θ θ θ θ θ= = = = = − = 
 
Estimated student time to complete: 30-40 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. To express the height from the table, the 27-in base height must be added to 
the zp value. Please also note that no particular reset position is specified. Therefore, the 
results relate to a position and orientation that would correspond to the given 
transformations from a reset position. We assume at reset, there is 90 degrees between 0x 
and 1x . Please note how frames 2, 3, and 4 relate to each other by referring to the side 
view of the arm. 
 
 
(Reprinted with permission from Staubli Robotics.) 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
38
 
 
1 2 3
4 5
0 0 1 0 0.707 0.707 0 10.61 1 0 0 0
1 0 0 0 0.707 0.707 0 10.61 0 0 1 0
0 1 0 0 0 0 1 6 0 1 0 0
0 0 0 1 0 0 0 1 0 0 0 1
1 0 0 0 0.707 0 0.707 0
0 0 1 0 0.707 0 0.707 0
0 1 0 18 0 1 0 0
0 0 0 1 0 0 0 1
A A A
A A
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢= =
⎢ ⎥ ⎢ −
⎢ ⎥ ⎢
⎣ ⎦ ⎣ ⎦
6
0
6
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
0 1 0 6
1 0 0 1.41
0 0 1 24
0 0 0 1
A
T
⎡ ⎤
⎢ ⎥
⎥ ⎢ ⎥=
⎥ ⎢ ⎥
⎥ ⎢ ⎥
⎣ ⎦
−⎡ ⎤
⎢ ⎥−⎢ ⎥=
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
39
Problem 2.34 
For the given4-DOF robot: 
• Assign appropriate frames for the Denavit-Hartenberg representation. 
• Fill out the parameters table. 
• Write an equation in terms of A-matrices that shows how U HT can be calculated. 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. Please also note that no particular reset position is specified. 
 
 
 
 
1
20
0 0 1 2 3 4 1 2 3 4
3
1 0 0
0 1 0
0 0 1
0 0 0 1
U U U
H H
l
l
T T T T A A A A A A A A
l
⎡ ⎤
⎢ ⎥
⎢ ⎥= = =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
40
Problem 2.35 
For the given 4-DOF robot designed for a specific operation: 
 
• Assign appropriate frames for the Denavit-Hartenberg representation. 
• Fill out the parameters table. 
• Write an equation in terms of A-matrices that shows how U HT can be calculated. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. Please also note that no particular reset position is specified. 
 
 
 
* ( )4 5 / sinL l l β= + 
The total transformation for the robot can be determined from: 
1
0
0 0 1 2 3 4 1 2 3 4
1 0 0
0 1 0 0
0 0 1 0
0 0 0 1
U U U
H H
l
T T T T A A A A A A A A
⎡ ⎤
⎢ ⎥
⎢ ⎥= = =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
41
Problem 2.36 
For the given specialty-designed 4-DOF robot: 
• Assign appropriate frames for the Denavit-Hartenberg representation. 
• Fill out the parameters table. 
• Write an equation in terms of A-matrices that shows how U HT can be calculated. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. Please also note that no particular reset position is specified. 
 
 
1
20
0 0 1 2 3 1 2 3
1 0 0
0 1 0
0 0 1 0
0 0 0 1
U U U
H H
l
l
T T T T A A A A A A
⎡ ⎤
⎢ ⎥
⎢ ⎥= = =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
42
Problem 2.37 
For the given 3-DOF robot: 
• Assign appropriate frames for the Denavit-Hartenberg representation. 
• Fill out the parameters table. 
• Write an equation in terms of A-matrices that shows how U HT can be calculated. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. Please also note that no particular reset position is specified. Please also note 
that there can be an additional frame between frames 1 and H. In that case, the two joint 
variables will be represented by two separate matrices. 
 
 
# θ d a α 
0-1 190 θ− + 3 4l l− − 0 -90 
1-H 2θ 5 6 7l l l+ + 0 0 
 
1
20
0 0 1 2 1 2
1 0 0
0 1 0
0 0 1 0
0 0 0 1
U U U
H H
l
l
T T T T A A A A
⎡ ⎤
⎢ ⎥−⎢ ⎥= = =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
43
Problem 2.38 
For the given 4-DOF robot: 
• Assign appropriate frames for the Denavit-Hartenberg representation. 
• Fill out the parameters table. 
• Write an equation in terms of A matrices that shows how U HT can be calculated. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 2.12 
 
Solution: 
 
In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, 
intermediate matrices might be different for each case. However, the final answer should 
be the same. Please also note that no particular reset position is specified. 
 
 
# θ d a α 
0-1 1θ 3l 4l 90 
1-2 290 θ+ 5l− 0 90 
2-H 3θ 6l 0 0 
 
1
20
0 0 1 2 3 1 2 3
1 0 0
0 1 0
0 0 1 0
0 0 0 1
U U U
H H
l
l
T T T T A A A A A A
⎡ ⎤
⎢ ⎥
⎢ ⎥= = =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
44
Problem 2.39 
 
Derive the inverse kinematic equations for the robot of Problem 2.36. 
 
Estimated student time to complete: 20-30 minutes if Problem 2.36 is already solved 
Prerequisite knowledge required: Text Section(s) 2.13 
 
Solution: 
 
We assume all positions are made relative to the base of the robot, and therefore, 0
UT is 
not included in the solution. Using the Denavit-Hartenberg representation and the joint 
parameters shown, we get: 
 
 
1 1 2 2 5 2 3 3
1 1 2 2 5 2 3 3
1 2 3
3 4 6 7
0 1 0
1 2 3 1 2 3
1 1
3 4
1 1
0 0 0 0 0
0 0 0 0 0
0 1 0 0 1 0 0 0 0 1
0 0 0 1 0 0 0 1 0 0 0 1
0 0
0 0 1
0 0
0 0 0 1
H H
x x x x
y y y
C S C S l C C S
S C S C l S S C
A A A
l l l l
T A A A A T A A
C S n o a p
l l n o a
S C
−
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ − +
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
= → =
⎡ ⎤
⎢ ⎥− −⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎣ ⎦
2 2 5 2 3 3
2 2 5 2 3 3
6 7
0 0 0
0 0 0
0 1 0 0 0 0 1
0 0 0 1 0 0 0 1 0 0 0 1
y
z z z z
C S l C C S
p S C l S S C
n o a p l l
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥=
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− +
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Multiplying these matrices we get: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
45
 
1 1 1 1 1 1 1 1
3 4
1 1 1 1 1 1 1 1
2 3 2 3 2 2 6 7 5 2
2 3 2 3 2 2 6 7 5 2
3 3
0 0 0 1
( )
( )
0 0
0 0 0 1
x y x y x y x y
z z z z
x y x y x y x y
C n S n C o S o C a S a C p S p
n o a p l l
LHS
S n C n S o C o S a C a S p C p
C C C S S S l l l C
S C S S C C l l l S
RHS
S C
+ + + +⎡ ⎤
⎢ ⎥− −⎢ ⎥=
⎢ ⎥− − − −
⎢ ⎥
⎣ ⎦
− − − + +⎡ ⎤
⎢ ⎥− + +⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
From 3,3 or 3,4 elements we get: 
1
1 1 1
1
1 1 1
0 tan
0 tan
y
x y
x
y
x y
x
a
S a C a
a
p
S p C p
p
θ
θ
−
−
⎛ ⎞
− = → = ⎜ ⎟
⎝ ⎠
⎛ ⎞
− = → = ⎜ ⎟
⎝ ⎠
 
From 1,3 and 2,3 elements we 
get: ( )2 1 1 2 1 1
2
( )
2 ,x y x y z
z
S C a S a
ATAN C a S a a
C a
θ
= − +⎧
→ = − −⎨
=⎩
 
 
From 2,1 and 2,2 elements we get: 
 
 ( )3 1 1 3 1 1 1 1
3 1 1
2 ,x y x y x y
x y
S S n C n
ATAN S n C n S o C o
C S o C o
θ
= −⎧⎪ → = − −⎨ = −⎪⎩
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
46
Problem 2.40 
Derive the inverse kinematic equations for the robot of Problem 2.37. 
 
Estimated student time to complete: 20-30 minutes if Problem 2.37 is already solved 
Prerequisiteknowledge required: Text Section(s) 2.13 
 
Solution: 
 
We assume all positions are made relative to the base of the robot, and therefore, 0
UT is 
not included in the solution. Using the Denavit-Hartenberg representation and the joint 
parameters shown, we get: 
 
 
 
# θ d a α 
0-1 190 θ− + 3 4l l− − 0 -90 
1-H 2θ 5 6 7l l l+ + 0 0 
 
1 1 2 2
1 1 2 2
1 2
3 4 6 6 7
0
1 2
0 0 0 0
0 0 0 0
0 1 0 0 0 1
0 0 0 1 0 0 0 1
H
S C C S
C S S C
A A
l l l l l
T A A
−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥− − − + +
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
=
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
47
1 1 2 2
1 1 2 2
3 4 5 6 7
1 2 1 2 1 1 5 6 7
1 2 2 1 1 1 5 6 7
2
0 0 0 0
0 0 0 0
0 1 0 0 0 1
0 0 0 1 0 0 0 1 0 0 0 1
( )
( )
 
x x x x
y y y y
z z z z
n o a p S C C S
n o a p C S S C
n o a p l l l l l
S C S S C C l l l
C C S C S S l l l
S
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥=
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − + +
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
− + +
− + +
=
− 2 3 40
0 0 0 1
C l l
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥− − −
⎢ ⎥
⎣ ⎦
 
 
From 1,3 and 2,3 elements we get: ( )1 1
1
2 ,y y x
x
S a
ATAN a a
C a
θ
=⎧⎪ → =⎨
=⎪⎩
 
 
From 3,1 and 3,2 elements we get: ( )2 2
2
2 ,z z z
z
S n
ATAN n o
C o
θ
= −⎧
→ = − −⎨ = −⎩
 
 
From 1,4 elements we get: 1 5 6 7 5 6 7
1
( ) xx
pC l l l p l l l
C
+ + = → + + = 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
48
CHAPTER THREE 
 
Problem 3.1 
Suppose the location and orientation of a hand frame is expressed by the following 
matrix. What is the effect of a differential rotation of 0.15 radians about the z-axis, 
followed by a differential translation of [0.1, 0.1, 0.3]? Find the new location of the hand. 
 
 
0 0 1 2
1 0 0 7
0 1 0 5
0 0 0 1
R
HT
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 3.6, 3.7. 
 
Solution: 
 
For 0.15, 0.1, 0.1, 0.3,z dx dy dzδ = = = = we get: 
 
0 0.15 0 0.1
0.15 0 0 0.1
0 0 0 0.3
0 0 0 0
−⎡ ⎤
⎢ ⎥
⎢ ⎥Δ =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
[ ]
0 0.15 0 0.1 0 0 1 2 0.15 0 0 0.95
0.15 0 0 0.1 1 0 0 7 0 0 0.15 0.4
0 0 0 0.3 0 1 0 5 0 0 0 0.3
0 0 0 0 0 0 0 1 0 0 0 0
0.15 0 1 1.05
1 0 0.15 7.4
0 1 0 5.3
0 0 0 1
R R
H H
R R R
H H Hnew old
d T T
T T d T
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤= Δ = =⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−⎡ ⎤
⎢
⎢⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢
⎢
⎣ ⎦
⎥
⎥
⎥
⎥
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
49
Problem 3.2 
As a result of applying a set of differential motions to frame T shown, it has changed an 
amount dT as shown. Find the magnitude of the differential changes made 
( , , , , ,dx dy dz x y zδ δ δ ) and the differential operator with respect to frame T. 
 
 
1 0 0 5 0 0.1 0.1 0.6
0 0 1 3 0.1 0 0 0.5
0 1 0 8 0.1 0 0 0.5
0 0 0 1 0 0 0 0
T dT
− −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥− − −
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.8. 
 
Solution: 
 
[ ] [ ][ ] [ ] [ ][ ]
[ ]
1
0 0.1 0.1 0.6 1 0 0 5 0 0.1 0.1 0.1
0.1 0 0 0.5 0 0 1 8 0.1 0 0 0
0.1 0 0 0.5 0 1 0 3 0.1 0 0 0
0 0 0 0 0 0 0 1 0 0 0 0
dT T dT T −= Δ → Δ =
− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥Δ = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
By inspection: d = [0.1,0,0], δ = [0,0.1,0.1]. 
 
[ ] [ ] [ ] [ ]1
1 0 0 5 0 0.1 0.1 0.6 0 0.1 0.1 0.6
0 0 1 8 0.1 0 0 0.5 0.1 0 0 0.5
0 1 0 3 0.1 0 0 0.5 0.1 0 0 0.5
0 0 0 1 0 0 0 0 0 0 0 0
T T
T
dT T T dT−⎡ ⎤ ⎡ ⎤= Δ → Δ =⎣ ⎦ ⎣ ⎦
− − − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤Δ = =⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
50
Problem 3.3 
Suppose the following frame was subjected to a differential translation of [1 0 0.5]d = 
units and a differential rotation of [0 0.1 0]δ = . 
a. What is the differential operator relative to the reference frame? 
b. What is the differential operator relative to the frame A? 
 
 
0 0 1 10
1 0 0 5
0 1 0 0
0 0 0 1
A
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.8 
 
Solution: 
a. 
0 0 0.1 1
0 0 0 0
0.1 0 0 0.5
0 0 0 0
⎡ ⎤
⎢ ⎥
⎢ ⎥Δ =
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
b. 
[ ] [ ][ ]1
0 1 0 5 0 0 0.1 1 0 0 1 10
0 0 1 0 0 0 0 0 1 0 0 5
1 0 0 10 0.1 0 0 0.5 0 1 0 0
0 0 0 1 0 0 0 0 0 0 0 1
0 0 0 0
0 0 0.1 0.5
0 0.1 0 1
0 0 0 0
A A A−
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤Δ = Δ =⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤
⎢ ⎥− −⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
51
Problem 3.4 
The initial location and orientation of a robot’s hand are given by T1, and its new location 
and orientation after a change are given by T2. 
a. Find a transformation matrix Q that will accomplish this transform (in the Universe 
frame). 
b. Assuming the change is small, find a differential operator Δ that will do the same. 
c. By inspection, find a differential translation and a differential rotation that 
constitute this operator. 
 
 1 2
1 0 0 5 1 0 0.1 4.8
0 0 1 3 0.1 0 1 3.5
0 1 0 6 0 1 0 6.2
0 0 0 1 0 0 0 1
T T
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.8 
 
Solution: 
 
a. 
1
2 1 2 1
1 0 0.1 4.8 1 0 0 5 1 0.1 0 0.1
0.1 0 1 3.5 0 0 1 6 0.1 1 0 0
0 1 0 6.2 0 1 0 3 0 0 1 0.2
0 0 0 1 0 0 0 1 0 0 0 1
T QT Q T T −
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥= → = = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
b. 
( )2 1 1 1 1 1T T dT T T I T QT= + = + Δ = Δ + = 
Therefore: Q IΔ = − 
or 
0 0.1 0 0.1
0.1 0 0 0
0 0 0 0.2
0 0 0 0
−⎡ ⎤
⎢ ⎥
⎢ ⎥Δ =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
c. 
By inspection: d = [0.1, 0, 0.2] and δ = [0, 0, 0.1].
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
52
Problem 3.5 
The hand frame of a robot and the corresponding Jacobian are given. For the given 
differential changes of the joints, compute the change in the hand frame, its new location, 
and correspondingΔ . 
 
6
6
8 0 0 0 0 0 0
0 1 0 10 3 0 1 0 0 0 0.1
1 0 0 5 0 10 0 0 0 0 0.1
0 0 1 0 0 1 0 0 1 0 0.2
0 0 0 1 0 0 0 1 0 0 0.2
1 0 0 0 0 1 0
TT J Dθ
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥−⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥
−⎣ ⎦ ⎣ ⎦
 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.11 
 
Solution: 
[ ]
6
6
6
6 6
6
6
6
8 0 0 0 0 0 0 0
3 0 1 0 0 0 0.1 0.1
0 10 0 0 0 0 0.1 1
0 1 0 0 1 0 0.2 0.3
0 0 0 1 0 0 0.2 0.2
1 0 0 0 0 1 0 0
T
T
T
T T
T
T
T
dx
dy
dz
D J D
x
y
z
θ δ
δ
δ
⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−
⎡ ⎤ ⎡ ⎤= = = = ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
− ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Substitute in 6T Δ to get: 6
0 0 0.2 0
0 0 0.3 0.1
0.2 0.3 0 1
0 0 0 0
T
⎡ ⎤
⎢ ⎥− −⎢ ⎥Δ =
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
Then 
[ ] 66 6
0 1 0 10 0 0 0.2 0 0 0 0.3 0.1
1 0 0 5 0 0 0.3 0.1 00 0.2 0
0 0 1 0 0.2 0.3 0 1 0.2 0.3 0 1
0 0 0 1 0 0 0 0 0 0 0 0
TdT T
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤= Δ = =⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
[ ] [ ] [ ]6 6 6
0 1 0.3 9.9
1 0 0.2 5
0.2 0.3 1 1
0 0 0 1
new old
T T dT
−⎡ ⎤
⎢ ⎥
⎢ ⎥= + =
⎢ ⎥− − −
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
53
[ ] [ ]6 16 6
0 1 0 10 0 0 0.2 0 0 1 0 5
1 0 0 5 0 0 0.3 0.1 1 0 0 10
0 0 1 0 0.2 0.3 0 1 0 0 1 0
0 0 0 1 0 0 0 0 0 0 0 1
0 0 0.3 0.1
0 0 0.2 0
0.3 0.2 0 1
0 0 0 0
TT T −
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤Δ = Δ =⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−⎡ ⎤
⎢ ⎥−⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
54
Problem 3.6 
Two consecutive frames describe the old (T1) and new (T2) positions and orientations of 
the end of a 3-DOF robot. The corresponding Jacobian relative to T1, relating to 
1 1 1, ,T T Tdz x zδ δ is also given. Find values of joint differential motions 1 2 3, ,ds d dθ θ of the 
robot that caused the given frame change. 
1
1 2
0 0 1 8 0 0.01 1 8.1
5 10 0
1 0 0 5 1 0.05 0 5
3 0 0
0 1 0 2 0.05 1 0.01 2
0 1 1
0 0 0 1 0 0 0 1
TT T J
⎡ ⎤ ⎡ ⎤
⎡ ⎤⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥= = = ⎢ ⎥⎢ ⎥ ⎢ ⎥−
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦
⎣ ⎦ ⎣ ⎦
 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.11 
 
Solution: 
 
[ ][ ] [ ]
[ ]
1
1
2 1 1 1
1
1
0 0.01 0 0.1
0 0.05 0 0
0.05 0 0.01 0
0 0 0 0
0 1 0 5 0 0.01 0 0.1 0 0.05 0 0
0 0 1 2 0 0.05 0 0 0.05 0 0.01 0
1 0 0 8 0.05 0 0.01 0 0 0.01 0 0.1
0 0 0 1 0 0 0 0 0 0 0 0
T
T
T T dT T T
T dT−
⎡ ⎤
⎢ ⎥−⎢ ⎥ ⎡ ⎤− = = = Δ = Δ⎣ ⎦⎢ ⎥−
⎢ ⎥
⎣ ⎦
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤Δ = = =⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢− −
⎢ ⎥ ⎢ ⎥ ⎢
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎥
⎥
 
Therefore, 
1
1 1
1
0.05 0.1
0.01 0.01
0.050.1
T
z
T T
x
T
z
D
d
δ
δ
⎧ = ⎡ ⎤
⎪ ⎢ ⎥= → =⎨ ⎢ ⎥
⎪ ⎢ ⎥= ⎣ ⎦⎩
 
And: [ ] [ ]1 1 1 11T T T TD J D D J Dθ θ
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= → =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 
 
Please see Appendix A for methods to calculate the inverse of a non-unitary matrix. 
Substitute to get: 
 
[ ] 1 1
1
1
5 10 0 0.1 0.0033
3 0 0 0.01 0.00833
0 1 1 0.05 0.0417
T TD J Dθ
−
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
55
Problem 3.7 
Two consecutive frames describe the old (T1) and new (T2) positions and orientations of 
the end of a 3-DOF robot. The corresponding Jacobian, relating to , ,dz x zδ δ is also 
given. Find values of joint differential motions 1 2 3, ,ds d dθ θ of the robot that caused the 
given frame change. 
 
1 2
0 0 1 10 0.05 0 1 9.75
5 10 0
1 0 0 5 1 0.1 0.05 5.2
3 0 0
0 1 0 3 0.1 1 0 3.7
0 1 1
0 0 0 1 0 0 0 1
T T J
−⎡ ⎤ ⎡ ⎤
⎡ ⎤⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥= = = ⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦
⎣ ⎦ ⎣ ⎦
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.11 
 
Solution: 
 
2 1
0.05 0 0 0.25
0 0.1 0.05 0.2
0.1 0 0 0.7
0 0 0 0
T T dT
− −⎡ ⎤
⎢ ⎥−⎢ ⎥− = =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
[ ] [ ][ ] [ ] [ ][ ] 11 1
0.05 0 0 0.25 0 1 0 5
0 0.1 0.05 0.2 0 0 1 3
0.1 0 0 0.7 1 0 0 10
0 0 0 0 0 0 0 1
0 0.05 0 0
0.05 0 0.1 0
0 0.1 0 0.2
0 0 0 0
dT T dT T −
− − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥= Δ → Δ = =
⎢ ⎥ ⎢ ⎥−
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
−⎡ ⎤
⎢ ⎥−⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
Therefore, 
0.2 0.2
0.1 0.1
0.050.05
z
x
z
d
Dδ
δ
=⎧ ⎡ ⎤
⎪ ⎢ ⎥= → =⎨ ⎢ ⎥
⎪ ⎢ ⎥= ⎣ ⎦⎩
 
 
Substitute to get: [ ] [ ] [ ]
1
1
5 10 0 0.2 0.0334
3 0 0 0.1 0.00334
0 1 1 0.05 0.0467
D J Dθ
−
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
56
Problem 3.8 
A camera is attached to the hand frame T of a robot as given. The corresponding inverse 
Jacobian of the robot at this location is also given. The robot makes a differential motion, 
as a result of which, the change in the frame dT is recorded as given. 
a. Find the new location of the camera after the differential motion. 
b. Find the differential operator. 
c. Find the joint differential motion values associated with this move. 
d. Find how much the differential motions of the hand-frame ( T D ) should have been 
instead, if measured relative to frame T, to move the robot to the same new location 
as in part a. 
1
1 0 0 0 0 0
0 1 0 3 2 0 1 0 0 0
1 0 0 2 0 0.2 0 0 0 0
0 0 1 8 0 1 0 0 1 0
0 0 0 1 0 0 0 1 0 0
1 0 0 0 0 1
T J −
⎡ ⎤
⎢ ⎥−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥= = ⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥
⎣ ⎦
 
0.03 0 0.1 0.79
0 0.03 0 0.09
0 0.1 0 0.4
0 0 0 0
dT
− −⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 30-40 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.11 
 
Solution: 
a. 
0.03 1 0.1 3.79
1 0.03 0 2.09
0 0.1 1 7.6
0 0 0 1
new oldT T dT
− −⎡ ⎤
⎢ ⎥
⎢ ⎥= + =
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
 
b. 
1
0.03 1 0.1 3.79 0 1 0 2 0 0.03 0.1 0.05
1 0.03 0 2.09 1 0 0 3 0.03 0 0 0
0 0.1 1 7.6 0 0 1 8 0.1 0 0 0.1
0 0 0 1 0 0 0 1 0 0 0 0
dT T −
− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥Δ = ⋅ = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
c. [ ]0.05 0 0.1 0 0.1 0.03 TD = − 
 
Therefore: 1
1 0 0 0 0 0 0.05 0.05
2 0 1 0 0 0 0 0.2
0 0.2 0 0 0 0 0.1 0
0 1 0 0 1 0 0 0.1
0 0 0 1 0 0 0.1 0
1 0 0 0 0 1 0.03 0.08
D J Dθ
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
= ⋅ = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
57
 
d. 
 
1
0 1 0 2 0.03 0 0.1 0.79
1 0 0 3 0 0.03 0 0.09
0 0 1 8 0 0.1 0 0.4
0 0 0 1 0 0 0 0
0 0.03 0 0.09
0.03 0 0.1 0.79
0 0.1 0 0.4
0 0 0 0
T TdT T T dT−
− − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= ⋅ Δ → Δ = ⋅ =
⎢ ⎥ ⎢ ⎥− − −
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
⎡ ⎤
⎢ ⎥− −⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
[ ]0.09 0.79 0.4 0.1 0 0.03 TT D = −
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
58
Problem 3.9 
A camera is attached to the hand frame T of a robot as given. The corresponding inverse 
Jacobian of the robot relative to the frame at this location is also given. The robot makes 
a differential motion, as a result of which, the change dT in the frame is recorded as 
given. 
a. Find the new location of the camera after the differential motion. 
b. Find the differential operator. 
c. Find the joint differential motion values Dθ associated with this move. 
1
1 0 0 0 0 0
0 1 0 3 2 0 1 0 0 0
1 0 0 2 0 0.1 0 0 0 0
0 0 1 8 0 1 0 0 1 0
0 0 0 1 0 0 0 1 0 0
1 0 0 0 0 1
TT J −
⎡ ⎤
⎢ ⎥−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥= = ⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥
⎣ ⎦
0.02 0 0.1 0.7
0 0.02 0 0.08
0 0.1 0 0.3
0 0 0 0
dT
− −⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 30-40 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.11 
 
Solution: 
 
a. 
 
0.02 1 0.1 3.7
1 0.02 0 2.08
0 0.1 1 7.7
0 0 0 1
new oldT T dT
− −⎡ ⎤
⎢ ⎥
⎢ ⎥= + =
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
 
b. 
 
1
0.02 0 0.1 0.7 0 1 0 2 0 0.02 0.1 0.06
0 0.02 0 0.08 1 0 0 3 0.02 0 0 0.02
0 0.1 0 0.3 0 0 1 8 0.1 0 0 0
0 0 0 0 0 0 0 1 0 0 0 0
dT T −
− − − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥Δ = ⋅ = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −
⎢ ⎥ ⎢ ⎥ ⎢⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
c. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
59
[ ][ ] [ ]1 1
0 1 0 2 0.02 0 0.1 0.7 0 0.02 0 0.08
1 0 0 3 0 0.02 0 0.08 0.02 0 0.1 0.7
0 0 1 8 0 0.1 0 0.3 0 0.1 0 0.3
0 0 0 1 0 0 0 0 0 0 0 0
T T T T dT− −⎡ ⎤ ⎡ ⎤Δ = Δ = =⎣ ⎦ ⎣ ⎦
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Therefore: [ ]0.08 0.7 0.3 0.1 0 0.02 TT D = − 
 
1
1 0 0 0 0 0 0.08 0.08
2 0 1 0 0 0 0.7 0.14
0 0.1 0 0 0 0 0.3 0.07
0 1 0 0 1 0 0.1 0.7
0 0 0 1 0 0 0 0.1
1 0 0 0 0 1 0.02 0.06
T T T TD J D D J Dθ θ
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
= ⋅ → = ⋅ = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
60
Problem 3.10 
The hand frame TH of a robot is given. The corresponding inverse Jacobian of the robot at 
this location relative to this frame is also shown. The robot makes a differential motion 
relative to this frame described as 
[ ]0.05 0 0.1 0 0.1 0.1H TT D = − . 
a. Find which joints must make a differential motion, and by how much, in order to 
create the indicated differential motions. 
b. Find the change in the frame. 
c. Find the new location of the frame after the differential motion. 
d. Find how much the differential motions (given above) should have been if 
measured relative to the Universe, to move the robot to the same new location as in Part 
c. 
1
5 0 0 0 0 0
0 1 0 3 2 0 1 0 0 0
1 0 0 3 0 0.2 0 0 0 0
0 0 1 8 0 1 0 0 1 0
0 0 0 1 0 0 0 1 0 0
1 0 0 0 0 1
HT
HT J
−
⎡ ⎤
⎢ ⎥−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥= = ⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.11 
 
Solution: 
a. 
 
[ ] 1
5 0 0 0 0 0 0.05 0.25
2 0 1 0 0 0 0 0.2
0 0.2 0 0 0 0 0.1 0
0 1 0 0 1 0 0 0.1
0 0 0 1 0 0 0.1 0
1 0 0 0 0 1 0.1 0.15
H HT TD J Dθ
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ −⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
Therefore, joints 1, 2, 4, and 6 must move as shown. 
 
b. Substituting values from HT D to form HT Δ , we get: 
 
0 1 0 3 0 0.1 0.1 0.05 0.1 0 0 0
1 0 0 3 0.1 0 0 0 0 0.1 0.1 0.05
0 0 1 8 0.1 0 0 0.1 0.1 0 0 0.1
0 0 0 1 0 0 0 0 0 0 0 0
HT
HdT T
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⋅ Δ = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
c. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
61
0 1 0 3 0.1 0 0 0 0.1 1 0 3
1 0 0 3 0 0.1 0.1 0.05 1 0.1 0.1 3.05
0 0 1 8 0.1 0 0 0.1 0.1 0 1 8.1
0 0 0 1 0 0 0 0 01 0 0 1
new oldH H
T T dT
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + = + =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
d. 
 
1
0.1 0 0 0 0 1 0 3 0 0.1 0 0.3
0 0.1 0.1 0.05 1 0 0 3 0.1 0 0.1 1.15
0.1 0 0 0.1 0 0 1 8 0 0.1 0 0.2
0 0 0 0 0 0 0 1 0 0 0 0
HT
H H
H
dT T T
dT T −
= ⋅ Δ = Δ ⋅
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥Δ = ⋅ = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Therefore, [ ]0.3 1.15 0.2 0.1 0 0.1 TD = − − −
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
62
Problem 3.11 
The hand frame T of a robot is given. The corresponding inverse Jacobian of the robot at 
this location is also shown. The robot makes a differential motion described as 
[ ]0.05 0 0.1 0 0.1 0.1 TD = − . 
a. Find which joints must make a differential motion, and by how much, in order to 
create the indicated differential motions. 
b. Find the change in the frame. 
c. Find the new location of the frame after the differential motion. 
d. Find how much the differential motions (given above) should have been, if 
measured relative to Frame T, to move the robot to the same new location as in Part c. 
 
1
5 0 0 0 0 0
0 1 0 3 2 0 1 0 0 0
1 0 0 3 0 0.2 0 0 0 0
0 0 1 8 0 1 0 0 1 0
0 0 0 1 0 0 0 1 0 0
1 0 0 0 0 1
T J −
⎡ ⎤
⎢ ⎥−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥= = ⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 3.6-3.11 
 
Solution: 
 
a. 
 
1
5 0 0 0 0 0 0.05 0.25
2 0 1 0 0 0 0 0.2
0 0.2 0 0 0 0 0.1 0
0 1 0 0 1 0 0 0.1
0 0 0 1 0 0 0.1 0
1 0 0 0 0 1 0.1 0.15
D J Dθ
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −
= ⋅ = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
Therefore, joints 1, 2, 4, and 6 must move as shown. 
 
b. Using D given above to form Δ , 
 
0 0.1 0.1 0.05 0 1 0 3 0.1 0 0.1 0.55
0.1 0 0 0 1 0 0 3 0 0.1 0 0.3
0.1 0 0 0.1 0 0 1 8 0 0.1 0 0.4
0 0 0 0 0 0 0 1 0 0 0 0
dT T
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥= Δ ⋅ = =
⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − − −
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
63
c. 
0.1 1 0.1 3.55
1 0.1 0 3.3
0 0.1 1 7.6
0 0 0 1
new oldT T dT
− −⎡ ⎤
⎢ ⎥
⎢ ⎥= + =
⎢ ⎥− −
⎢ ⎥
⎣ ⎦
 
 
d. 
 
[ ][ ] [ ]1 1
0 1 0 3 0.1 0 0.1 0.55
1 0 0 3 0 0.1 0 0.3
0 0 1 8 0 0.1 0 0.4
0 0 0 1 0 0 0 0
0 0.1 0 0.3
0.1 0 0.1 0.55
0 0.1 0 0.4
0 0 0 0
T T T T dT− −
− − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤Δ = Δ = =⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥− − −
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
⎡ ⎤
⎢ ⎥− −⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
The differential motions should have been: 
 
[ ]0.3 0.55 0.4 0.1 0 0.1T D = −
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
64
Problem 3.12 
Calculate the 6 21
T J element of the Jacobian for the revolute robot of Example 2.25. 
 
Estimated student time to complete: 20 minutes 
Prerequisite knowledge required: Text Section(s) 3.10 
 
Solution: 
 
The robot has 6 revolute joints. From Equation 3.26: 
 
( )6 2T i x y y xJ o p o p= − + where for i=1 we use 6 1 2 3 4 5 6oT A A A A A A= given in Equation 2.59. 
Substitute to get: 
 
[ ] [ ]
[ ] [ ]
6
21 1 234 5 6 234 6 1 5 6 1 234 4 23 3 2 2
1 234 5 6 234 6 1 5 6 1 234 4 23 3 2 2
( ) ( )
( ) ( )
T J C C C C S C S S S S C a C a C a
S C C C S C C S S C C a C a C a
= − − − + × + +
+ − − − × + +
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
65
Problem 3.13 
Calculate the 6 16
T J element of the Jacobian for the revolute robot of Example 2.25. 
 
Estimated student time to complete: 20 minutes 
Prerequisite knowledge required: Text Section(s) 3.10 
 
Solution: 
 
The robot has 6 revolute joints. From Equation 3.26: 
 
( )6 1T i x y y xJ n p n p= − + where for i=6 we use 5 6 6T A= given in Equation 2.57. Substitute 
to get: 
 
( )( ) ( )( )6 21 6 60 0 0T J C S= − + =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
66
Problem 3.14 
Using Equation (2.34), differentiate proper elements of the matrix to develop a set of 
symbolic equations for joint differential motions of a cylindrical robot and write the 
corresponding Jacobian. 
 
Estimated student timeto complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 3.10-3.11 
 
Solution: 
 
For a cylindrical coordinate (without additional rotation) there can only be 3 variables. 
The Jacobian will be a 3 3× matrix. 
 
0
0
( , , )
0 0 1
0 0 0 1
cyl
C S rC
S C rS
T r l
l
α α α
α α α
α
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
Then 
x
y
z
p rC
p rS
p l
α
α
⎧ =
⎪ =⎨
⎪ =⎩
 and 
x
y
z
dp drC rS d
dp drS rC d
dp dl
α α α
α α α
⎧ = −
⎪ = +⎨
⎪ =⎩
 
 
Therefore, 
0
0
0 0 1
x
y
z
dp C rS dr
dp S rC d
dp dl
α α
α α α
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
67
Problem 3.15 
Using Equation (2.36), differentiate proper elements of the matrix to develop a set of 
symbolic equations for joint differential motions of a spherical robot and write the 
corresponding Jacobian. 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 3.10-3.11 
 
Solution: 
 
For a spherical coordinate (without additional rotation) there can only be 3 variables. The 
Jacobian will be a 3 3× matrix. 
 
( , , )
0
0 0 0 1
sph
C C S S C rS C
C S C S S rS S
T r
S C rC
β γ γ β γ β γ
β γ γ β γ β γ
β γ
β β β
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Then 
.
.
x
y
z
p rS C
p rS S
p rC
β γ
β γ
β
⎧ =
⎪ =⎨
⎪ =⎩
 and 
. . .
. . .
x
y
z
dp drS C rC C d rS S d
dp drS S rC S d rS C d
dp drC rS d
β γ β γ β β γ γ
β γ β γ β β γ γ
β β β
⎧ = + −
⎪ = + +⎨
⎪ = −⎩
 
 
Therefore, 
. . .
. . .
0
x
y
z
dp S C rC C rS S dr
dp S S rC S rS C d
dp C rS d
β γ β γ β γ
β γ β γ β γ β
β β γ
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
68
Problem 3.16 
For a cylindrical robot, the three joint velocities are given for a corresponding location. 
Find the three components of the velocity of the hand frame. 
 r =0.1 in/sec, α =0.05 rad/sec, l =0.2 in/sec, r=15 in, 30α = , l=10 in. 
 
Estimated student time to complete: 10 or 20 minutes 
Prerequisite knowledge required: Text Section(s) 3.10-3.11 
 
Solution: 
 
Referring to the solution of Problem 3.14, and dividing both sides by dt, substitute the 
given values into the equation to get: 
 
0 0.866 7.5 0 0.1 0.288
0 0.5 13 0 0.05 0.7
0 0 1 0 0 1 0.2 0.2
x
y
z
p C rS r
p S rC
p l
α α
α α α
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
in/sec
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
69
Problem 3.17 
For a spherical robot, the three joint velocities are given for a corresponding location. 
Find the three components of the velocity of the hand frame. 
 r =2 in/sec, β =0.05 rad/sec, γ =0.1 rad/sec, r=20 in, 60β = , 30γ = . 
 
Estimated student time to complete: 10 or 20 minutes 
Prerequisite knowledge required: Text Section(s) 3.10-3.11 
 
Solution: 
 
Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the 
given values into the equation to get: 
 
0.75 8.66 8.66 2 1.067
0.433 5 15 0.05 2.616
0 0.5 17.32 0 0.1 0.134
x
y
z
p S C rC C rS S r
p S S rC S rS C
p C rS
β γ β γ β γ
β γ β γ β γ β
β β γ
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
70
Problem 3.18 
For a spherical robot, the three joint velocities are given for a corresponding location. 
Find the three components of the velocity of the hand frame. 
 r =1 unit/sec, β =1 rad/sec, γ =1 rad/sec, r=5 units, 45β = , 45γ = . 
 
Estimated student time to complete: 10 or 20 minutes 
Prerequisite knowledge required: Text Section(s) 3.10-3.11 
 
Solution: 
 
Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the 
given values into the equation to get: 
 
0.5 2.5 3.535 1 0.535
0.5 2.5 3.535 1 6.535
0 0.707 3.535 0 1 2.828
x
y
z
p S C rC C rS S r
p S S rC S rS C
p C rS
β γ β γ β γ
β γ β γ β γ β
β β γ
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
71
Problem 3.19 
For a cylindrical robot, the three components of the velocity of the hand frame are given 
for a corresponding location. Find the required three joint velocities that will generate the 
given hand frame velocity. 
 1 in/sec, 3 in/sec, 5 in/sec, 45 , 20 in, 25 inx y z r lα= = = = = = 
Estimated student time to complete: 10 or 25 minutes 
Prerequisite knowledge required: Text Section(s) 3.12 
 
Solution: 
 
Referring to the solution of Problem 3.16, substitute to get: 
 
0 1 0.707 14.14 0
0 2 0.707 14.14 0
0 0 1 3 0 0 1
x
y
z
p C rS r r
p S rC
p l l
α α
α α α α
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= → =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
2.83 in/sec0.707 14.14 1
0.707 14.14 3 =0.071 /sec
5 5 in/sec
rr
r
l l
α
α α
=⎧ ⎧− =
⎪ ⎪+ = →⎨ ⎨
⎪ ⎪= =⎩ ⎩
 
 
Please note that this problem could have been solved by calculating the inverse of the 
Jacobian as well. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
72
Problem 3.20 
For a spherical robot, the three components of the velocity of the hand frame are given 
for a corresponding location. Find the required three joint velocities that will generate the 
given hand frame velocity. 
 
 5 in/sec, 9 in/sec, 6 in/sec, 60 , 20 in, =30x y z rβ γ= = = = = 
 
Estimated student time to complete: 15 or 30 minutes 
Prerequisite knowledge required: Text Section(s) 3.12 
 
Solution: 
 
Referring to the solution of Problem 3.17, substitute to get: 
 
5 0.75 8.66 8.66
9 0.433 5 15
0 6 0.5 17.32 0
x
y
z
p S C rC C rS S r r
p S S rC S rS C
p C rS l
β γ β γ β γ
β γ β γ β γ β α
β β γ
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= → =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
 
 
0.75 8.66 8.66 5 10.647 in/sec
0.433 5 15 9 = 0.039 /sec
0.306 /sec0.5 17.32 6
r r
r
r
β γ
β γ β
γβ
⎧ + − = =⎧
⎪ ⎪+ + = → −⎨ ⎨
⎪ ⎪ =− = ⎩⎩
 
 
Please note that this problem could have been solved by calculating the inverse of the 
Jacobian as well. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
73
CHAPTER FOUR 
 
Problem 4.1 
Using Lagrangian mechanics derive the equations of motion of a cart with two tires under 
the cart as shown: 
k
m
F
x
r, I r, I
 
Figure P.4.1 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 4.2 
 
Solution: 
 
I. Kinematics: 
Since the rollers are in contact with the block at the top, the contact-point velocity is x . 
Therefore: 
2
2
xx r
rω ω= → = 
x
vc 
 
II. Dynamics: 
 
2 2 2 2 2 2 2
2
1 1 1 12
2 2 2 2 4 4
w c
w c c eff
m IK mx m v I mx x x m x
r
ω⎛ ⎞= + + = + + =⎜ ⎟
⎝ ⎠
 
where 2
1
2 4 4
w c
eff
m Im m
r
⎛ ⎞+ + =⎜ ⎟
⎝ ⎠
 
21
2
P kx= 
Therefore, 2 21
2eff
L K P m x kx= − = − 
2 2eff eff
L d Lm x m x
x dt x
∂ ∂⎛ ⎞= → =⎜ ⎟∂ ∂⎝ ⎠
 
L kx
x
∂
= −
∂
 
2 effF m x kx= +
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
74
Problem 4.2 
Calculate the total kinetic energy of the link AB, attached to a roller with negligible mass, 
as shown. 
l, I
m
Vo
θ
A
B
C
x
y
 
Figure P.4.2 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 4.2 
 
Solution: 
 
First, find the velocity of point C, assuming that the roller has negligible mass too: 
 
/ ( ) cos sin2 2 2
sin cos( ) cos sin
2 2 2 2
C A C A o o
o o
l lv
l l l lv j v
θ θ θ
θ θθ θ θ θ θ θ
⎛ ⎞= + = + × = + × +⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − = − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
lv v v v θ i k i j
i i i j
 
 
2 2 2 2 2
2 2 2 2 2 2sin cossin sin
4 4 4C o o o o
l l lv v v l v v lθ θθ θθ θ θ θθ= + − + = + − 
2
2 2 2 2 21 1 1 1sin
2 2 2 4 2C o o
lK mv I m v v l Iθ θ θθ θ
⎛ ⎞
= + = + − +⎜ ⎟
⎝ ⎠
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
75
Problem 4.3 
Derive the equations of motion for the 2-link mechanism with distributed mass, as 
shown. 
x0
y0
m1
m2
x1
y1
l1 , I
1
l
2 , I
2
2θ
1θ
O
C
A
D
B
 
Figure P.4.3 
 
Estimated student time to complete: 1.5-2 hours 
Prerequisite knowledge required: Text Section(s) 4.2 
 
Solution: 
 
I. Kinematic: First we write equations describing the kinematic relationships. 
 
( )
( )
2 2
1 1 12 1 1 1 1 2 12
2 2
1 1 12 1 1 1 1 2 12
2 2
2 2
D D
D D
l lx l C C x l S S
l ly l S S y l C C
θ θ θ
θ θ θ
⎧ ⎧= + = − − +⎪ ⎪⎪ ⎪→⎨ ⎨
⎪ ⎪= + = + +
⎪ ⎪⎩ ⎩
 
 
The velocity of point D will be: 
( ) ( )
( ) ( )
( ) ( )
2
2 2 2 2 2 2 2 2 22
1 1 1 1 2 1 2 12 1 2 1 1 2 1 12
2
2 2 2 2 2 22
1 1 1 1 2 1 2 12 1 2 1 1 2 1 12
2
2 2 2 22
1 1 1 2 1 2 1 2 1 1 2 2
2
4
2
4
2
4
D D D
lv x y l S S l l S S
ll C C l l C C
ll l l C
θ θ θ θ θ θ θ θ
θ θ θ θ θ θ θ θ
θ θ θ θ θ θ θ θ
= + = + + + + +
+ + + + + +
= + + + + +
 
Rearrange: 
2 2 2
2 2 2 22 2 2
1 1 1 2 2 2 1 2 1 2 24 4 2D
l l lv l l l C l l Cθ θ θ θ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
76
 
II: Kinetics: 
 
Kinetic Energy: 
 
( )
( )
22 2
1 2 1 1 2 2
22 2 2 2
1 1 1 2 2 1 2 2
2 2 2 2 2 2 2
1 1 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 1 2 2
1 1 1
2 2 2
1 1 1 1 1
2 3 2 12 2
1 1 1 1 1 1 1
6 6 2 2 6 3 2
O D D
D
K K K I I m v
m l m l m v
m l m l m l m l l C m l m l m l l C
θ θ θ
θ θ θ
θ θ θ θ
= + = + + +
⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
 
 
Potential Energy: 
 
( ) ( ) ( )1 21 2 1 1 2 1 1 2 121 1 12 2
l lP P P m g C m gl C m g C= + = − + − + − 
 
Lagrangian: 
 
2 2 2 2 2 2 2
1 1 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 1 2 2
1 2 1 2
1 2 1 1 2 12 1 2 1 2
1 1 1 1 1 1 1
6 6 2 2 6 3 2
2 2 2 2
L K P m l m l m l m l l C m l m l m l l C
l l l lm g m gl C m g C m g m gl m g
θ θ θ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞+ + + − − −⎜ ⎟
⎝ ⎠
 
 
Derivatives and equations of motion: 
 
2 2 2 2
1 1 1 2 2 2 1 2 1 2 2 2 2 2 2 1 2 2
1
2 2 2 2
1 1 1 2 2 2 1 2 1 2 2 2 2 2 2 1 2 2
1
2
1 2 2 1 2 2 2 2 1 2 2
1 1 1 1 1 12
6 6 2 2 3 2
1 1 1 1 1 12
6 6 2 2 3 2
1
2
L m l m l m l m l l C m l m l l C
d L m l m l m l m l l C m l m l l C
dt
m l l S m l l S
θ θ
θ
θ θ
θ
θ θ θ
∂ ⎛ ⎞ ⎛ ⎞= + + + + +⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ⎛ ⎞ ⎛ ⎞= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠⎝ ⎠
− −
 
1 2
1 2 1 1 2 12
1 2 2
l lL m g m gl S m g S
θ
∂ ⎛ ⎞= − + −⎜ ⎟∂ ⎝ ⎠
 
 
Substitute to get: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
77
2 2 2 2
1 1 1 2 2 2 1 2 1 2 2 1 2 2 2 1 2 2 2
2
2 1 2 2 1 2 2 1 2 2 2 1 2 1 1 2 2 12
1 1 1 1
3 3 3 2
1 1 1
2 2 2
T m l m l m l m l l C m l m l l C
m l l S m l l S m m gl S m gl S
θ θ
θ θ θ
⎛ ⎞ ⎛ ⎞= + + + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞− − + + +⎜ ⎟
⎝ ⎠
 
 
For the second variable: 
 
2 2
2 2 2 1 2 2 2 1 2 2
2
2 2
2 2 2 1 2 2 2 1 2 2 2 1 2 2 1 2
2
1 1 1
3 3 2
1 1 1 1
3 3 2 2
L m l m l m l l C
d L m l m l m l l C m l l S
dt
θ θ
θ
θ θ θ θ
θ
∂ ⎛ ⎞= + +⎜ ⎟∂ ⎝ ⎠
⎛ ⎞∂ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠⎝ ⎠
 
2
2 1 2 2 1 2 1 2 2 1 2 2 2 12
2
1 1 1
2 2 2
L m l l S m l l S m gl Sθ θ θ
θ
∂
= − − −
∂
 
 
Substitute to get: 
 
2 2 2
2 2 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 2 12
1 1 1 1 1
3 2 3 2 2
T m l m l l C m l m l l S m gl Sθ θ θ⎛ ⎞= + + + +⎜ ⎟
⎝ ⎠
 
The result can be written in a matrix form too.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
78
Problem 4.4 
Write the equations that express U62, U35, U53, U623, and U534 for a 6-axis cylindrical-RPY 
robot in terms of the A and Q matrices. 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 4.4.1. 
 
Solution: 
 
For joints 1 and 3: ( )
0 0 0 0
0 0 0 0
0 0 0 1
0 0 0 0
prismatic PQ Q
⎡ ⎤
⎢ ⎥
⎢ ⎥= =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
For joints 2,4,5,6: ( )
0 1 0 0
1 0 0 0
0 0 0 0
0 0 0 0
revolute RQ Q
−⎡ ⎤
⎢ ⎥
⎢ ⎥= =
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
 
0
1 2.. ..iij j j i
j
TU A A Q A A j i
q
∂
= = ≤
∂
 
 
 ijk ij kU U q= ∂ ∂ 
 
( )
( )
( )
( )
( )
0
1 2 3 4 5 66
62 1 2 3 4 5 6
0
1 2 33
35
0 0
0
1 2 3 4 55
53 1 2 3 4 5
1 2 3 4 5 662
623 1 2 3 4 5 6
1 2 3 4 553
534 1 2 3 4 5
0 ( )
R
P
R
R P
P
P R
a a
A A A A A ATU AQ A A A A A
A A ATU i j
A A A A ATU A A Q A A A
l l
AQ A A A A AUU AQ A Q A A A A
l l
A A Q A A AUU A A Q A Q A A
q q
α α
φ φ
∂∂
= = =
∂ ∂
∂∂
= = = ≥
∂ ∂
∂∂
= = =
∂ ∂
∂∂
= = =
∂ ∂
∂∂
= = =
∂ ∂
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
79
Problem 4.5 
Using Equations 4.49 to 4.54, write the equations of motion for a 3-DOF revolute robot 
and describe each term. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 4.4.4. 
 
Solution: 
 
We have: 
 
2 2 2
1 11 1 12 2 13 3 1( ) 1 111 1 122 2 133 3
112 1 2 113 1 3 123 2 3 1
2 2 2
2 21 1 22 2 23 3 2( ) 2 211 1 222 2 233 3
212 1 2 213 1 3 223
2 2 2
2 2 2
act
act
T D D D I D D D
D D D D
T D D D I D D D
D D D
θ θ θ θ θ θ θ
θ θ θ θ θ θ
θ θ θ θ θ θ θ
θ θ θ θ
= + + + + + +
+ + + +
= + + + + + +
+ + + 2 3 2
2 2 2
3 31 1 32 2 33 3 3( ) 3 311 1 322 2 333 3
312 1 2 313 1 3 323 2 3 32 2 2
act
D
T D D D I D D D
D D D D
θ θ
θ θ θ θ θ θ θ
θ θ θ θ θ θ
+
= + + + + + +
+ + + +
 
 
( ) iD θ terms are caused by angular accelerations. 
( ) 2iD θ terms represent centripetal accelerations. 
( ) i jD θ θ terms are related to the Coriolis accelerations. 
( )iD terms are gravity terms.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronicmedia system or the Internet without prior expressed consent of the copyright owner. 
80
Problem 4.6 
Expand the D134 and D15 terms of Equation 4.49 in terms of their constituent matrices. 
 
Estimated student time to complete: 1 hour 
Prerequisite knowledge required: Text Section(s) 4.4 
 
Solution: 
 
a. From Equation 4.44: 
max( , , )
( )
n
T
ijk pjk p pi
p i j k
D Trace U J U
=
= ∑ 
For D134 we have i =1, j =3, k =4, p =4, n =6. Then: 
( ) ( ) ( )
6
134 34 1 434 4 41 534 5 51 634 6 61
4
( )T T T Tp p pD Trace U J U Trace U J U Trace U J U Trace U J U= = + +∑ 
( )
( )
( )
0
1 2 3 44
43 1 2 3 3 4
3 3
1 2 3 3 443
434 1 2 3 3 4 4
4 4
0
1 2 3 44
41 1 1 2 3 4
1 1
A A A ATU A A Q A A
q q
A A Q A AUU A A Q A Q A
q q
A A A ATU Q A A A A
q q
∂∂
= = =
∂ ∂
∂∂
= = =
∂ ∂
∂∂
= = =
∂ ∂
 
From Equation A.11 of Appendix A: 41 4 3 2 1 1
T T T T T TU A A A A Q= 
( )
( )
( )
( )
1 2 3 4 5
53 1 2 3 3 4 5
3
1 2 3 3 4 5
534 1 2 3 3 4 4 5
4
1 2 3 4 5
51 1 1 2 3 4 5 51 5 4 3 2 1 1
1
1 2 3 4 5 6
63 1 2 3 3 4 5 6
3
63
634 1 2 3 3 4 4 5 6
4
61 6 5 4
T T T T T T T
T T T T
A A A A A
U A A Q A A A
q
A A Q A A A
U A A Q A Q A A
q
A A A A A
U Q A A A A A U A A A A A Q
q
A A A A A A
U A A Q A A A A
q
UU A A Q A Q A A A
q
U A A A
∂
= =
∂
∂
= =
∂
∂
= = → =
∂
∂
= =
∂
∂
= =
∂
= 3 2 1 1
T T T TA A A Q
 
 
Substitute to get: 
 
( )
( )
( )
134 1 2 3 3 4 4 4 4 3 2 1 1
1 2 3 3 4 4 5 5 5 4 3 2 1 1
1 2 3 3 4 4 5 6 6 6 5 4 3 2 1 1
T T T T T
T T T T T T
T T T T T T T
D Trace A A Q A Q A J A A A A Q
Trace A A Q A Q A A J A A A A A Q
Trace A A Q A Q A A A J A A A A A A Q
=
+
+
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
81
b. From Equation 4.43: 
max( , )
( )
n
T
ij pj p pi
p i j
D Trace U J U
=
= ∑ 
For D15 we have i =1, j =5, p =5, n =6. Then: 
 
( ) ( )
6
15 5 1 55 5 51 65 6 61
5
( )T T Tp p pD Trace U J U Trace U J U Trace U J U= = +∑ 
55 1 2 3 4 5 5
51 1 1 2 3 4 5 51 5 4 3 2 1 1
65 1 2 3 4 5 5 6
61 1 1 2 3 4 5 6 61 6 5 4 3 2 1 1
T T T T T T T
T T T T T T T T
U A A A A Q A
U Q A A A A A U A A A A A Q
U A A A A Q A A
U Q A A A A A A U A A A A A A Q
=
= → =
=
= → =
 
 
Substitute to get: 
 
( )
( )
15 1 2 3 4 5 5 5 5 4 3 2 1 1
1 2 3 4 5 5 6 6 6 5 4 3 2 1 1
T T T T T T
T T T T T T T
D Trace A A A A Q A J A A A A A Q
Trace A A A A Q A A J A A A A A A Q
=
+
 
 
Although a large number of matrices are involved, they are all known.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
82
Problem 4.7 
An object is subjected to the following forces and moments relative to the reference 
frame. Attached to the object is a frame, which describes the orientation and the location 
of the object. Find the equivalent forces and torques acting on the object relative to the 
current frame. 
 
0.707 0.707 0 2
0 0 1 5
[10,0,5,12,20,0] N, N.m
0.707 0.707 0 3
0 0 0 1
TB F
⎡ ⎤
⎢ ⎥
⎢ ⎥= =
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 4.6. 
 
Solution: 
 
Substitute: 
 
[10,0,5] [12, 20,0] [2,5,3]
[0.707,0,0.707] [0.707,0, 0.707] [0,1,0]
T T T
T T T
= = =
= = − =
f m p
n o a
 
 
10 0 5 ( 25) (20) (50)
2 5 3
( ) 13 50
× = = − − +
× + = − +
i j k
f p i j k
f p m i k
 
 
7.07 3.54 10.61
7.07 3.54 3.53
0
[( ) ] 9.19 35.35 26.16
[( ) ] 9.19 35.35 44.54
[( ) ] 0
B
x
B
y
B
z
B
x
B
y
B
z
f N
f N
f
m Nm
m Nm
m
= ⋅ = + =
= ⋅ = − =
= ⋅ =
= ⋅ × + = − + =
= ⋅ × + = − − = −
= ⋅ × + =
n f
o f
a f
n f p m
o f p m
a f p m
 and 
10.61
3.53
0
26.16
44.54
0
F
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
83
Problem 4.8 
In order to assemble two parts together, one part must be pushed into the other with a 
force of 10 lb in the x-axis direction, 5 lb in the y-direction, and be turned with a torque 
of 5 lb in⋅ along the x-axis direction. The object’s location relative to the base frame of a 
robot is described by RT0. Assuming that the two parts must be aligned together for this 
purpose, find the necessary forces and moments that the robot must apply to the part 
relative to its hand frame. 
 0
0 0.707 0.707 4
1 0 0 6
0 0.707 0.707 3
0 0 0 1
RT
−⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 4.6. 
 
Solution: 
 
Since the robot must be aligned with the object for this operation, then R RH OT T= . 
Therefore we have: 
 
[10,5,0] [5,0,0] [4,6,3]
[0,1,0] [ 0.707,0,0.707] [0.707,0,0.707]
T T T
T T T
= = =
= = − =
f m p
n o a
 
 
10 5 0 (15) (30) (40)
4 6 3
( ) 20 30 40
× = = − +
× + = − +
i j k
f p i j k
f p m i j k
 
 
5
7.07
7.07
[( ) ] 30
[( ) ] 14.14
[( ) ] 42.42
H
x
H
y
H
z
H
x
H
y
H
z
f units
f units
f units
m units
m units
m units
= ⋅ =
= ⋅ = −
= ⋅ =
= ⋅ × + = −
= ⋅ × + =
= ⋅ × + =
n f
o f
a f
n f p m
o f p m
a f p m
 and 
5
7.07
7.07
30
14.14
42.42
F
⎡ ⎤
⎢ ⎥−⎢ ⎥
⎢ ⎥
= ⎢ ⎥−⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
84
CHAPTER FIVE 
 
 
Problem 5.1 
It is desired to have the first joint of a 6-axis robot go from an initial angle of 50 to a 
final angle of 80 in 3 seconds. Calculate the coefficients for a third-order polynomial 
joint-space trajectory. Determine the joint angles, velocities, and accelerations at 1, 2, and 
3 seconds. It is assumed that the robot starts from rest, and stops at its destination. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 5.5.1. 
 
Solution: 
 
For a 3rd-order polynomial: 
 
2 3
0 1 2 3
2
1 2 3
( )
( ) 2 3
t c c t c t c t
t c c t c t
θ
θ
= + + +
= + +
 
 
Substitute boundary conditions to get: 
 
0
1
2 3
2 3
50 0
0 0
80 50 9 27
0 6 27
i
i
f
f
c
c
c c
c c
θ
θ
θ
θ
= = +
= = +
= = + +
= = +
 
 
Solve to get: 
 
0 1 2 3
2 3
2
5 0 10 2.222
( ) 50 10 2.222
( ) 20 6.666
( ) 20 13.332
c c c c
t t t
t t t
t t
θ
θ
θ
= = = = −
= + −
= −
= −
 
 
@t=1 sec sec
sec
57.78
13.334
6.668
θ
θ
θ
⎧ =
⎪ =⎨
⎪ =⎩
 @t=2 sec sec
sec
72.22
13.34
6.664
θ
θ
θ
⎧ =
⎪ =⎨
⎪ = −⎩
 @t=3 sec sec
sec
80
0
20
θ
θ
θ
⎧ =
⎪ =⎨
⎪ = −⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
85
Problem 5.2 
It is desired to have the third joint of a 6-axis robot go from an initial angle of 20 to a 
final angle of 80 in 4 seconds. Calculate the coefficients for a third-order polynomial 
joint-space trajectory and plot the joint angles, velocities, and accelerations. The robot 
starts from rest, but should have a final velocity of /sec5 . 
 
Estimated student time to complete: 20-30 minutes (with plotting) 
Prerequisite knowledge required: Text Section(s) 5.5.1. 
 
Solution: 
 
For a 3rd-order polynomial: 
 
2 3
0 1 2 3
2
1 2 3
( )
( ) 2 3
t c c t c t c t
t c c t c t
θ
θ
= + + +
= + +
 
 
Substitute boundary conditions to get: 
 
0
1
2 3
2 3
20 0
0 0
80 20 16 64
5 8 48
i
i
f
f
c
c
c c
c c
θ
θ
θ
θ
= = +
= = +
== + +
= = +
 
 
Solve to get: 
 
0 1 2 3
2 3
2
20 0 10 1.5625
( ) 20 10 1.5625
( ) 20 4.6875
( ) 20 9.375
c c c c
t t t
t t t
t t
θ
θ
θ
= = = = −
= + −
= −
= −
 
 
The plot is shown: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
86
Problem 5.3 
The second joint of a 6-axis robot is to go from initial angle of 20 to an intermediate 
angle of 80 in 5 seconds and continue to its destination of 25 in another 5 seconds. 
Calculate the coefficients for third-order polynomials in joint-space. Plot the joint angles, 
velocities, and accelerations. Assume the joint stops at intermediate points. 
 
Estimated student time to complete: 30-40 minutes (with plotting) 
Prerequisite knowledge required: Text Section(s) 5.5.1 
 
Solution: 
 
There are two segments to this motion. Therefore: 
 
 Segment 1 Segment 2 
2 3
0 1 2 3
2
1 2 3
( )
( ) 2 3
t c c t c t c t
t c c t c t
θ
θ
⎧ = + + +⎪
⎨
= + +⎪⎩
 
2 3
0 1 2 3
2
1 2 3
( )
( ) 2 3
t b b t b t b t
t b b t b t
θ
θ
⎧ = + + +⎪
⎨
= + +⎪⎩
 
 
There are eight unknowns; eight equations are needed. Known initial and final conditions 
are: 1 1 2 2 1 2 1 2, , , , , , andi f i f i f f iθ θ θ θ θ θ θ θ= . This results in seven equations. The eighth 
equation can be generated by making assumptions such as a maximum allowable 
acceleration or an intermediate velocity. 
 
For this problem we will assume that the joint will come to a stop at the intermediate 
point. Therefore, 
 
0 0
1 1
2 3 2
32 3
20 0 20
0 0 0
1
80 20 25 125 7.2
0.960 10 75
c c
c c
c c c
cc c
= + =⎧ ⎧
⎪ ⎪= + =⎪ ⎪→⎨ ⎨= + + =⎪ ⎪
⎪ ⎪ = −= + ⎩⎩
 
0 0
1 1
2 3 2
32 3
80 0 80
0 0 0
2
25 80 25 125 6.6
0.880 10 75
b b
b b
b b b
bb b
= + =⎧ ⎧
⎪ ⎪= + =⎪ ⎪→⎨ ⎨= + + = −⎪ ⎪
⎪ ⎪ == + ⎩⎩
 
 
Substitute to get: 
 
2 3
2
( ) 20 7.2 0.96
1 ( ) 14.4 2.88
( ) 14.4 5.76
t t t
t t t
t t
θ
θ
θ
⎧ = + −
⎪
= −⎨
⎪ = −⎩
 and 
2 3
2
( ) 80 6.6 0.88
2 ( ) 13.2 2.64
( ) 13.2 5.28
t t t
t t t
t t
θ
θ
θ
⎧ = − +
⎪
= − +⎨
⎪ = − +⎩
 
 
The position, velocities, and accelerations are shown: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
87
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
88
Problem 5.4 
A fifth-order polynomial is to be used to control the motions of the joints of a robot in 
joint-space. Find the coefficients of a fifth-order polynomial that will allow a joint to go 
from an initial angle of 0 to a final joint angle of 75 in 3 seconds, while the initial and 
final velocities are zero and initial acceleration and final decelerations are 10 2/ sec . 
 
Estimated student time to complete: 20-30 minutes (with plotting) 
Prerequisite knowledge required: Text Section(s) 5.5.2 
 
Solution: 
 
For a 5th-order polynomial: 
 
2 3 4 5
0 1 2 3 4 5
2 3 4
1 2 3 4 5
2 3
2 3 4 5
( )
( ) 2 3 4 5
( ) 2 6 12 20
t c c t c t c t c t c t
t c c t c t c t c t
t c c t c t c t
θ
θ
θ
⎧ = + + + + +
⎪
= + + + +⎨
⎪ = + + +⎩
 
 
Substitute the given initial and final conditions and get: 
 
0
1 2 3 4 5
1
2 3 4 5
2
2 3 4 5
0
75 3 9 27 81 243
0
0 6 27 108 405
10 2
10 2 18 108 540
c
c c c c c
c
c c c c
c
c c c c
=⎧
⎪ = + + + +⎪
⎪ =⎪
⎨ = + + +⎪
⎪ =
⎪
= + + +⎪⎩
 
 
Solve to get: 0 1 2 3 4 50, 0, 5, 24.24, 13.33, 1.85c c c c c c= = = = = − = . Substitute to 
get: 
 
2 3 4 5
2 3 4
2 3
( ) 5 24.44 13.33 1.85
( ) 10 73.33 53.33 9.26
( ) 10 146.66 160 37.04
t t t t t
t t t t t
t t t t
θ
θ
θ
⎧ = + − +
⎪
= + − +⎨
⎪ = + − +⎩
 
 
The position, velocity, and acceleration plots are shown: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
89
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
90
Problem 5.5 
Joint 1 of a 6-axis robot is to go from an initial angle of 30iθ = to the final angle of 
120fθ = in 4 seconds with a cruising velocity of 
sec
1 30ω = . Find the necessary blending 
time for a trajectory with linear segments and parabolic blends and plot the joint 
positions, velocities, and accelerations. 
 
Estimated student time to complete: 15-30 minutes (with plotting) 
Prerequisite knowledge required: Text Section(s) 5.5.3 
 
Solution: 
 
( )30 120 30 4 1
30
i f f
b
t
t
θ θ ω
ω
− + − +
= = = sec 
 
( ) ( ) ( )
2 2
For to For to For to 
1 130 30 120 30 (4 )30 12 2
30 30 30(4 )
30 0 30
i A A B B f
A
t tt
t t
θ θ θ θ θ θ θ θ θ
θ θθ θ
θ θ θ
θ θ θ
= = =
⎧ ⎧= + = − −⎪ ⎪= + −⎧
⎪ ⎪⎪⎪ ⎪= = = −⎨ ⎨ ⎨
⎪ ⎪ ⎪= = = −⎪ ⎩ ⎪
⎪ ⎪⎩ ⎩
 
-40
-20
0
20
40
60
80
100
120
140
1 21 41
Seconds
Position
Velocity
Acceleration
42
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
91
Problem 5.6 
A robot is to be driven from an initial position through two via points before it reaches its 
final destination using a 4-3-4 trajectory. The positions, velocities, and time duration for 
the three segments for one of the joints are given below. Determine the trajectory 
equations and plot the position, velocity, and acceleration curves for the joint. 
 
 
1 1 1 1 1
2 2 2
3 3 3
4 4 4
20 0 0 0 1
60 0 2
100 0 1
40 0 0
i f
i f
i f
θ θ θ τ τ
θ τ τ
θ τ τ
θ θ θ
= = = = =
= = =
= = =
= = =
 
 
Estimated student time to complete: 50-60 minutes (with plotting) 
Prerequisite knowledge required: Text Section(s) 5.5.5 
 
Solution: 
 
Using Equations 5.19 to 5.32 we get the following equations. The same can be found 
using the matrix equation of Equation 5.33: 
 
0
1
2
3
0
4
0
1
3 4
2
3 4 1
3 4 2
1 2 3
1 2 3 1
2 3 2
1 2 3 4
1 2 3 4
2 3 4
20
0
0
80.95
60
40.95
100
79.05
60 20
3 4
6 12 2
100 60 2 4 8
4 12
2 12 2
40 100
0 2 3 4
0 2 6 12
a
a
a
a
b
a
c
b
a a
b
a a b
a a b
b b b
b b b c
b b c
c c c c
c c c c
c c c
=⎧
⎪ =⎪
⎪ =
=⎪
=⎪ = −⎪ =
⎪ =
= + +⎪
⎪ + =⎪ →⎨ + =⎪
⎪ = + + +
⎪
+ + =⎪
⎪ + =⎪
⎪ = + + + +
⎪ = + + +⎪
⎪ = + +⎩
3
1
2
3
4
2.85
13.34
92.4
82.875
202.9
87.62
b
c
c
c
c
⎧
⎪
⎪
⎪
⎪
= −⎪
⎪ = −⎨
⎪ = −⎪
⎪ = −
⎪ =⎪
⎪ = −⎩
 
 
These equations can be solved by elimination or an equation solver. Substituting the 
coefficients into each segment we get: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
92
3 4
1
2 3
2
2 3 4
3
( ) 20 80.95 40.95
( ) 60 79.05 2.85 13.34
( ) 100 92.4 82.875 202.9 87.62
t t t
t t t t
t t t t t
θ
θ
θ
= + −
= + − −
= − − + −
 
 
The results are plotted below. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
93
Problem 5.7 
A 2-DOF planar robot is to follow a straight line in Cartesian-space between the start 
(2,6) and the end (12,3) points of the motion segment. Find the joint variables for the 
robot if the path is divided into 10 sections. Eachlink is 9 inches long. 
 
Estimated student time to complete: 60-75 minutes (with plotting) 
Prerequisite knowledge required: Text Section(s) 5.6 
 
Solution: 
 
The inverse kinematic equations can be found in different ways. We will use a simple 
Denavit-Hartenberg representation as follows: 
 
x0z0
y0
x1
x2z2
z1
1θ
2θ
# θ d a α 
0-1 1θ 0 9 0 
1-2 2θ 0 9 0 
 
 
1 1 1 2 2 2
1 1 1 2 2 2
1 2
0
2 1 2
0 9 0 9
0 9 0 9
0 0 1 0 0 0 1 0
0 0 0 1 0 0 0 1
C S C C S C
S C S S C S
A A
T A A
− −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
=
 
 
Multiplying A1 by A2 and setting px and py equal to the components of position for 0T2, we 
get: 
 
1 2 1 2 1 12 1
1 2 1 2 1 12 1
9 9 9 9 9 (1)
9 9 9 9 9 (2)
x
y
p C C S S C C C
p S S C C S S S
= − + = +⎧⎪
⎨ = + + = +⎪⎩
 
 
Squaring these equations and adding them together, and then regrouping, we get: 
 
2 2 2 2
1
2 21 cos 1162 162
x y x yp p p pC θ −
⎛ ⎞+ +
= − → = −⎜ ⎟⎜ ⎟
⎝ ⎠
 
Substitute into (1) to get: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
94
1 2
1
2
9 (3)
1
xp S S
C
C
+
=
+
 
Substitute (3) into (2) to get: 
 
( ) ( )
( )
( )
( )
( )
2 22
1 2 1 2 2
2 2 2 21
1 1
2 2
1 1
9 9
1 1
sin
18 1 18 1
yx
y x y x
pS pS C S S C
p C S p p C S p
S
C C
θ −
+ + + = +
⎛ ⎞+ − + −
= → = ⎜ ⎟⎜ ⎟+ +⎝ ⎠
 
 
Next, calculate the slope of the straight line motion: 
 
3 3 6 0.3
12 12 2
y
x
− −
= = −
− −
 
 
The following table shows the coordinates of the intermediate points and the joint angles 
from the above equations. The joint angles are shown on the graph as well: 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
95
Problem 5.8 
The 3-DOF robot of Example 5.7, as shown in Figure 5.18, is to move from point (3, 5, 
5) to point (3, -5, 5) along a straight line, divided into 10 sections. Find the angles of the 
three joints for each intermediate point and plot the results. 
 
Estimated student time to complete: 30 minutes, depending on programming expertise 
Prerequisite knowledge required: Text Section(s) 5.6 
 
Solution: 
 
Using the inverse kinematic equations of the robot from Example 5.7, the joint angles can 
be found as shown. The values are also plotted: 
 
# Px Py Pz Theta 1 Theta 2 Theta 3 
1 3 5 5 30.96 48.59 137.27 
2 3 4 5 36.87 49.87 142.2 
3 3 3 5 45 52.04 146.45 
4 3 2 5 56.31 54.87 149.8 
5 3 1 5 71.57 57.51 151.98 
6 3 0 5 90 99.6 160.82 
7 3 -1 5 108.4 122.5 151.98 
8 3 -2 5 123.7 125.1 149.8 
9 3 -3 5 135 128 146.45 
10 3 -4 5 143.1 130.1 142.2 
11 3 -5 5 149 131.4 137.27 
 
0
20
40
60
80
100
120
140
160
180
1 2 3 4 5 6 7 8 9 10 11
Joint 1
Joint 2
Joint 3
 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
96
CHAPTER SIX 
 
Problem 6.1 
Derive the inverse Laplace transform of the following equation: 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 6.5 and 6.6 
 
Solution: 
( )( )
( ) ( )( )
( ) ( )( )
1 2
2
1
1
2
4
1 1 4
3 3( )
( 5 4) 1 4 1 4
3 31 1
1 4 3
3 34 1
1 4 3
1 1( )
1 4
1 1( )
1 4
s
s
t t
a aF s
s s s s s s
a s
s s
a s
s s
F s
s s
f t e e
s s
=−
=−
− − − −
= = = +
+ + + + + +
⎡ ⎤
= + = =⎢ ⎥+ +⎣ ⎦
⎡ ⎤
= + = = −⎢ ⎥+ + −⎣ ⎦
= −
+ +
−
= + = −
+ +
L L
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
97
Problem 6.2 
Derive the inverse Laplace transform of the following equation: 
2
( 6)( )
( 5 6)
sF s
s s s
+
=
+ +
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 6.5, 6.6 
 
Solution: 
 
( )
( ) ( )( )
( ) ( )( )
( ) ( )( ) ( )( )
2
31 2
1
0
2
2
3
3
1 2
( 6) ( 6)( )
( 5 6) ( 5) 3
2 3
6 6 1
2 3 6
6 42 2
2 3 2
6 33 1
2 3 3 1
1 2 1( )
2 3
1 2 1( ) ( )
2 3
s
s
s
s sF s
s s s s s s
aa a
s s s
sa s
s s s
sa s
s s s
sa s
s s s
F s
s s s
f t u t e
s s s
=
=−
=−
− −
+ +
= =
+ + + +
= + +
+ +
⎡ ⎤+
= = =⎢ ⎥+ +⎣ ⎦
⎡ ⎤+
= + = = −⎢ ⎥+ + −⎣ ⎦
⎡ ⎤+
= + = =⎢ ⎥+ + − −⎣ ⎦
−
= + +
+ +
−⎛ ⎞= + + = −⎜ ⎟+ +⎝ ⎠
L 3t te−+
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
98
Problem 6.3 
Derive the inverse Laplace transform of the following equation: 
2
1( )
( 1) ( 2)
F s
s s
=
+ +
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 6.5, 6.6 
 
Solution: 
 
( )
( )
2 1 1
2 2
2
2 2
1
1 2
1
1 2
2
2
1
2
1( )
( 1) ( 2) ( 1) ( 1) ( 2)
11 1
( 1) ( 2)
1 1 1
( 2) ( 2)
12 1
( 1) ( 2)
1 1 1( )
( 1) ( 1) ( 2)
1 1 1( )
( 1) ( 1) ( 2)
s
s
s
b b aF s
s s s s s
b s
s s
db
ds s s
a s
s s
F s
s s s
f t t
s s s
=−
=−
=−
−
= = + +
+ + + + +
⎡ ⎤
= + =⎢ ⎥+ +⎣ ⎦
⎡ ⎤ −
= = = −⎢ ⎥+ +⎣ ⎦
⎡ ⎤
= + =⎢ ⎥+ +⎣ ⎦
= − +
+ + +
⎛ ⎞
= − + =⎜ ⎟+ + +⎝ ⎠
L ( ) 21 t te e− −− +
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
99
Problem 6.4 
Derive the inverse Laplace transform of the following equation: 
3
10( )
( 4)( 2)
F s
s s
=
+ +
 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 6.5-6.6 
 
Solution: 
 
( )
( )
( )
( )
3 2 1 1
3 3 2
3
3 3
2
3
2 23
22 2
2
1 22
2 2
10( )
( 4)( 2) ( 2) ( 2) ( 2) ( 4)
102 5
( 4)( 2)
10 10 10 102 2.5
( 4)( 2) 4 44
1 10 1 10
2 4 2 4
s
ss s
s s
b b b aF s
s s s s s s
b s
s s
d db s
ds s s ds s s
d db
ds s ds s
=−
=−=− =−
=− =−
= = + + +
+ + + + + +
⎡ ⎤
= + =⎢ ⎥+ +⎣ ⎦
⎡ ⎤ − −⎛ ⎞= + = = = = −⎜ ⎟⎢ ⎥+ + +⎝ ⎠ +⎣ ⎦
−⎛ ⎞= =⎜ ⎟+⎝ ⎠ +
( )
3
2
1 3
4
3 2
1 2 2 4
3 2
1 20 5
2 ( 4) 4
10 54
( 4)( 2) 4
5 5 5 5( )
( 2) 2( 2) 4( 2) 4( 4)
5 5 5 5 5 5 5 5( )
( 2) 2( 2) 4( 2) 4( 4) 2 2 4 4
s
s
t t
s
a s
s s
F s
s s s s
f t t t e e
s s s s
=−
=−
− − −
= =
+
⎡ ⎤ −
= + =⎢ ⎥+ +⎣ ⎦
= − + −
+ + + +
⎛ ⎞ ⎛ ⎞= − + − = − + −⎜ ⎟ ⎜ ⎟+ + + + ⎝ ⎠⎝ ⎠
L
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
100
Problem 6.5 
Simplify the block diagram of Figure P.6.5. 
+-A G1
H3
H1
G2+-
H2
+-
B
 
Figure P.6.5 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 6.8. 
 
Solution: 
 
The following sequence is used to simplify the block diagram: 
 
+-A G1
H3
H1
G2
H2 /G1
+-
B
+-
 
 
+-A
H3
G2
H2 /G1
+-
B
1
1 11
G
G H+
 
 
+-A
H3
H2 /G1
+-
B
1 2
1 11
G G
G H+
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
101
 
A
H3
+-
B
1 2
1 1 2 21
G G
G H G H+ +
 
 
A
B
1 2
1 1 2 2 1 2 31
G G
G H G H G G H+ + +
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internetwithout prior expressed consent of the copyright owner. 
102
Problem 6.6 
Simplify the block diagram of Figure P.6.6. 
+- G1
H1
+- G2
H2
+-
H3
BA
 
Figure P.6.6 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 6.8. 
 
Solution: 
 
The following sequence is used to simplify the block diagram: 
 
+-
H3
BA 1
1 11
G
G H+
2
2 21
G
G H+
 
 
+-
H3
BA
( )( )
1 2
1 1 2 21 1
G G
G H G H+ +
 
 
+-
BA K
 
 
where K is calculated as follows: 
 
( )( )
( )( )
( )( )
1 2
1 1 2 2 1 2
1 2 3 1 1 2 2 1 2 3
1 1 2 2
1 1
1 11
1 1
G G
G H G H G GK G G H G H G H G G H
G H G H
+ +
= =
+ + ++
+ +
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
103
Problem 6.7 
Simplify the block diagram of Figure P.6.7. 
+- G1
H1
BA +-G2 G3
G4
H3
+-
 
Figure P.6.7 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 6.8. 
 
Solution: 
 
The following sequence is used to simplify the block diagram: 
 
BA
H3
+-
1 2
1 2 11
G G
G G H+ 3 4
G G+
 
 
BA
H3
+-
1 2
1 2 11
G G
G G H+ 3 4
G G+
 
 
+-
BA K
 
where 
( )
( )
( )
( )
1 2 3 4
1 2 3 41 2 1
1 2 3 4 3 1 2 1 1 2 3 4 3
1 2 1
1
11
1
G G G G
G G G GG G HK
G G G G H G G H G G G G H
G G H
+
++
= =
+ + + +
+
+
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
104
Problem 6.8 
Write an equation that describes the output of the system of Figure P.6.8. 
G1
H1
++ G2+-
YU1
U2
 
Figure P.6.8 
 
Estimated student time to complete: 10-15 minutes 
Prerequisite knowledge required: Text Section(s) 6.8. 
 
Solution: 
 
( )( )
( )
( )
1 1 1 2 2
1 1 1 1 2 2
1 1 2 1 2 1 2 2
1 1 2 2 2 1 2 1
1 1 2 2 2
1 2 1
1
1
U YH G U G Y
U G YG H U G Y
U G G YG G H U G Y
U G G U G Y G G H
U G G U GY
G G H
− + =
− + =
− + =
+ = +
+
=
+
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
105
Problem 6.9 
Write the equations that describe the input-output relationships for Figure P.6.9. 
G1 +
+
G2+
+
Y1
U1
U2
+
U3 Y2
H1 H2
 
Figure P.6.9 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 6.8. 
 
Solution: 
 
( )
1 1 2 2 2 1
3 1 1 1 2 2
1 1 2 2 2 1
3 2 1 1 2 1 2
U G U Y H Y
U U G H G Y
U G U Y H Y
U G U G G H Y
+ + =⎧⎪
⎨ + =⎪⎩
+ + =⎧
⎨ + =⎩
 
 
Substitute Y2 into Y1 to get: 
 
1 1 1 2 3 2 2 1 1 2 1 2
2 3 2 1 1 2 1
1
1 1 1 2 1 2 2 2
2
2 1 2 1 2
3
1
0
Y U G U U G H U G G H H
Y U G U G G H
U
Y G G G H H G H
U
Y G G H G
U
= + + +⎧
⎨ = +⎩
⎡ ⎤
+⎡ ⎤ ⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
106
Problem 6.10 
Sketch the root locus for the following: 
 
( 1)( 3)( 4)
KGH
s s s s
=
+ + +
 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 6.13. 
 
Solution: 
 
Poles are at 0, 1, 3, 4p p p p= = − = − = − 
 Number of asymptotes: 4 0 4α = − = , angles of asymptotes: 45, 135, 225, 315. 
Asymptotic center: ( )0 1 3 4 0 2
4A
p Zσ σσ
α
− − − − −
= = = −∑ ∑ 
 
The root locus, as drawn by MATLAB, is shown: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
107
Problem 6.11 
Sketch the root locus for the following: 
 ( 6)
( 4)
K sGH
s s
+
=
+
 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 6.13. 
 
Solution: 
 
Roots are at 6, 0, 4z p p= − = = − 
 Number of asymptotes: 2 1 1α = − = , angles of asymptote: 180. 
 
The root locus, as drawn by MATLAB, is shown: 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
108
Problem 6.12 
Sketch the root locus for the following: 
 
( 6)
( 10 10)( 10 10)( 12)
K sGH
s s j s j s
+
=
+ − + + +
 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 6.13. 
 
Solution: 
 
Roots are at 6, 0, 10 10, 12z p p j p= − = = − ± = − 
 Number of asymptotes: 4 1 3α = − = , angles of asymptotes: 60, 180, 300. 
Asymptotic center: ( )0 10 10 12 ( 6) 6.5
4A
p Zσ σσ
α
− − − − − −
= = = −∑ ∑ 
 
The root locus, as drawn by MATLAB, is shown: 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
109
Problem 6.13 
For the system of Problem 6.10, assume that two of the roots are chosen at 
5 2.55s j= − ± . Find the system’s gain, damping ratio, and natural frequency. Show that 
the angle criterion is met. Can you determine from the root locus whether or not the 
system is stable? 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 6.13, 6.14. 
 
Solution: 
2 2 2 2 2 2 2 2
1
1
1
2
1
3
1
4
5 2.55 4 2.55 2 2.55 1 2.55 236
2.55180 tan 180 27 153
5
2.55180 tan 180 32.52 147.48
4
2.55180 tan 180 51.9 128.1
2
2.55180 tan 180 68.59 111.4
1
t
i
i
p
z
M
K
M
θ
θ
θ
θ
θ
−
−
−
−
= = + + + + =
⎛ ⎞= − = − =⎜ ⎟
⎝ ⎠
⎛ ⎞= − = − =⎜ ⎟
⎝ ⎠
⎛ ⎞= − = − =⎜ ⎟
⎝ ⎠
⎛ ⎞= − = − =⎜ ⎟
⎝ ⎠
∏
∏
153 147.48 128.1 111.4 540 3 180otal = + + + = = ×∑
 
It meets the criterion. 
 
1 2.55cos cos tan 0.891
5
ζ θ −⎛ ⎞= = =⎜ ⎟
⎝ ⎠
 
 
2 21 0.891 1 0.891
5 2.55
5.61 / sec
n n n n
n
s j j
j
rad
ζω ω ζ ω ω
ω
= − ± − = − ± −
= − ±
=
 
 
The root have to potential to fall on the right side, and therefore, may become unstable.. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
110
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
111
Problem 6.14 
For the system of Problem 6.10, assume that two of the roots are chosen at 
4 1.24s j= − ± . Find the system’s gain, damping ratio, and natural frequency. 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 6.13,6.14 
 
Solution: 
2 2 2 2 2 2 24 1.24 3 1.24 1 1.24 1.24 26.85i
i
p
z
M
K
M
= = + + + =∏
∏
 
 
1 1.24cos cos tan 0.955
4
ζ θ −⎛ ⎞= = =⎜ ⎟
⎝ ⎠
 
 
2 21 0.891 1 0.955
4 1.24
4.19 / sec
n n n n
n
s j j
j
rad
ζω ω ζ ω ω
ω
= − ± − = − ± −
= − ±
=
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
112
Problem 6.15 
For the system of Problem 6.11, assume that the roots are chosen at 3 1.73s j= − ± . Find 
the system’s gain, damping ratio, and natural frequency. Show that the angle criterion is 
met. Can you determine whetheror not the system may become unstable as the gain 
changes? 
 
Estimated student time to complete: 30 minutes 
Prerequisite knowledge required: Text Section(s) 6.13, 6.14. 
 
Solution: 
 
2 2 2 2
2 2
3 1.73 1 1.73 2
3 1.73
i
i
p
z
M
K
M
+ +
= = =
+
∏
∏
 
1 1.73cos cos tan 0.866
3
ζ θ −⎛ ⎞= = =⎜ ⎟
⎝ ⎠
 
 
2 21 0.866 1 0.866
3 1.73
3.46 / sec
n n n n
n
s j j
j
rad
ζω ω ζ ω ω
ω
= − ± − = − ± −
= − ±
=
 
 
0
4
6
1
1
1
1.73180 tan 180 30 150
3
1.73180 tan 60
1
1.73tan 30
3
150 60 30 180
p
p
z
total
θ
θ
θ
θ
−
−
−
−
⎛ ⎞= − = − =⎜ ⎟
⎝ ⎠
⎛ ⎞= − =⎜ ⎟
⎝ ⎠
⎛ ⎞= =⎜ ⎟
⎝ ⎠
= + − =∑
 
 
It meets the criterion. 
 
The roots are always on the left side, and therefore, the system is always stable. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
113
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
114
Problem 6.16 
For the system of Problem 6.11, find the roots, the gain, and the steady-state error for a 
settling time of less that 1 second and overshoot of 4% or less. 
 
Estimated student time to complete: 30-40 minutes 
Prerequisite knowledge required: Text Section(s) 6.14 
 
Solution: 
 
( 6)
( 4)
K sGH
s s
+
=
+
 
 
By trial and error, 
 
( )2/ 1
1
% 100% 4% 0.715,
cos 44.4
OS e
ζπ ζ
ζ
θ ζ
− −
−
= × = → ≥
= =
 
 
The chosen roots must be below this value, therefore choose 4 2.83s j= − ± 
4( ) 4
1n
approximateζω = = 
System type: 1 0ssE→ = 
( )( )2 2 2
2 2
4.9 2.834 2.83 2.83 4
3.4652 2.83
i
i
p
z
M
K
M
− +
= = = =
− +
∏
∏
 
1 2.83tan 35.28
4
θ −= = 
cos 0.816actualζ θ= = 
 
The root locus, as drawn by MATLAB is shown: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
115
Problem 6.17 
For the following system, find the roots, the gain, and steady-state error for the fastest 
response and a settling time of less than 2 seconds and an overshoot of less than 4%. 
 
( 1)( 3)
KGH
s s
=
+ +
 
 
Estimated student time to complete: 30-40 minutes 
Prerequisite knowledge required: Text Section(s) 6.14 
 
Solution: 
For root locus: 1 3 2 and 90
2A A
σ θ− −= = − = ± 
For overshoot of 4%: ( )
2/ 1
% 100% 4% 0.72OS e
ζπ ζ
ζ
− −
= × = → ≅ 
 1 4 4cos 44 and 2
2s n
Tθ ζ
ζω
−= = = = = 
The root locus and the limitations are shown: 
 
 
Choose poles for fastest response: tan 44 1.93
2
l l= → = 
Therefore: 2 1.93s j= − ± 
 
2 2 2 2(3 2) 1.93 (1 2) 1.93
4.72
1
i
i
p
z
M
K
M
− + − +
= = =∏
∏
 
System type: 0 
( )( )0 0
4.72lim ( ) lim 1.57
1 3P s s
K G s
s s→ →
→ = = =
+ +
 
 1 0.39
1 1.57ss
E = =
+
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
116
Problem 6.18 
For the system of problem 6.17, select the locations and the proportional and integral 
gains to change it to a proportional-plus-integral system with zero steady-state error. 
 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 6.15 
 
Solution: 
 
We add 0s = at the origin and a zero near it (location to be found). Therefore, 
 
( ) ( )
( )( )
0.1
( 1)( 3) 1 3
K s z K s
GH
s s s s s s
+ +
= =
+ + + +
 
Assume 0.73ζ = for 4% OS even though the system is not second-order (this is close 
enough). Therefore, 
 
43θ = approximately at 2 1.87s j= − ± . 
 
2 2 2 2 2 2
2 2
2 1.87 1 1.87 1 1.87 4.6
1.9 1.87
i
i
p
z
M
K
M
+ + +
= = =
+
∏
∏
 
 
The gain can be calculated as: 
 
( )4.6 0.1
4.6 and 0.46
I
P
PP I
P I
KK s
sKK s KK
s s s
K K
⎛ ⎞
+⎜ ⎟ ++ ⎝ ⎠= = =
= =
 
 
Since this is Type-1, Ess = 0.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
117
Problem 6.19 
For the system of Problem 6.17, we would like to improve the settling time to 1 second 
by adding a zero to the system, therefore a proportional-plus-derivative system. Find a 
proper location for the zero and the loop gain. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 6.16 
 
Solution: 
 
Although not accurate, assume that 4 4
1
σ = = . Therefore, we arbitrarily choose the poles 
to be at 4 3.85s j= − ± . 
 
Angle deficiency: 
 
1
1
1
3
3.85180 tan 127.9
3
3.85180 tan 104.6
1
127.9 104.6 180 52.5
3.85tan 52.5 6.95
4
z
z
z
θ
θ
θ
−
−
−
−
= − =
= − =
= + − =
= → =
−
 
The loop gain is: 
 
2 2 2 2
2 2
3 3.85 1 3.85 4
2.95 3.85
i
i
p
z
M
K
M
+ +
= = =
+
∏
∏
 
 
( )
( )( )
4 6.95
1 3
z
GH
s s
+
=
+ +
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
118
Problem 6.20 
For the system of problem 6.19, add an integrator to the system to make it into a PID 
system in order to achieve a zero steady-state error. Find the location of an additional 
zero and proportional, derivative, and integral gains. 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 6.17 
 
Solution: 
 
Assume: ( )( )
( )( )
6.95
1 3
K s s z
GH
s s s
+ +
=
+ +
 
Magnitude criterion: 
 
2 2 2 2 2 2
2 2 2 2
4 3.85 3 3.85 1 3.85 4.06
2.95 3.85 3.9 3.85
i
i
p
z
M
K
M
+ + +
= = =
+ +
∏
∏
 
2
24.06( 6.95)( 0.1) 4.06( 7.05 0.695)
P I
D
D Da I
P D
K KK s s
K KV KG K K s
E s s
s s s s
s s
⎛ ⎞
+ +⎜ ⎟
⎝ ⎠= = + + =
+ + + +
= =
 
4.06 4.06
7.05 28.6
2.82
0.695
D
D
P
P
D
I
I
D
K K
K K
K
K
K
K
⎧
⎪ =⎪ =⎧
⎪ ⎪= → =⎨ ⎨
⎪ ⎪ =⎩⎪
=⎪
⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
119
CHAPTER SEVEN 
 
 
Problem 7.1 
A motor with a rotor inertia of 0.030 Kgm2 and maximum torque of 12 Nm is connected 
to a uniformly distributed arm with a concentrated mass at its end, as shown in Figure 
P.7.1. Ignoring the inertia of a pair of reduction gears and viscous friction in the system, 
calculate the total inertia felt by the motor and the maximum angular acceleration it can 
develop if the gear ratio is a) 5, b) 50, c) 100. Compare the results. 
 
Im
Tm
N
Tl
x
y
z
m=3Kg
0.5 m
m=3Kg
 
Figure P.7.1. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 7.2 
 
Solution: 
2 2 2 2 21 1 (3)(0.5) (3)(0.5) 1 Kgm
3 3load arm mass
I m l m l= + = + = 
 
a) For N = 5, 
2
2
2max
max
1 10.03 =0.07 Kgm
25
12 171 /
0.07
total motor load
total
I I I
N
T r s
I
θ
⎧ ⎛ ⎞= + = +⎜ ⎟⎪ ⎝ ⎠⎪
⎨
⎪ = = =
⎪⎩
 
 
b) For N = 50, 
2
2
2max
max
1 10.03 =0.0304 Kgm
2500
12 395 /
0.0304
total motor load
total
I I I
N
T r s
I
θ
⎧ ⎛ ⎞= + = +⎜ ⎟⎪ ⎝ ⎠⎪
⎨
⎪ = = =
⎪⎩
 
c) For N = 100, 
2
2
2max
max
1 10.03 =0.0301 Kgm
10000
12 399 /
0.0301
total motor load
total
I I I
N
T r s
I
θ
⎧ ⎛ ⎞= + = +⎜ ⎟⎪ ⎝ ⎠⎪
⎨
⎪ = = =
⎪⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronicmedia system or the Internet without prior expressed consent of the copyright owner. 
120
Problem 7.2 
Repeat Problem 1, but assume that the two gears have 0.002 Kgm2 and 0.005 Kgm2 
inertias respectively. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 7.2 
 
Solution: 
 
2 2 2 2 21 1 (3)(0.5) (3)(0.5) 1 Kgm
3 3load arm mass
I m l m l= + = + = 
2
2 2 2 2
0.0050.002
1 0.005 1 1.0050.03 0.002 0.032
gears
total motor gears load
I
N
I I I I
N N N N
= +
⎛ ⎞= + + = + + + = +⎜ ⎟
⎝ ⎠
 
 
a) For N = 5, 
2
2max
max
1.0050.032 =0.0722 Kgm
25
12 166 /
0.0722
total
total
I
T r s
I
θ
⎧ = +⎪⎪
⎨
⎪ = = =
⎪⎩
 
 
b) For N = 50, 
2
2max
max
1.0050.032 0.0324 Kgm
2500
12 370 /
0.0324
total
total
I
T r s
I
θ
⎧ = + =⎪⎪
⎨
⎪ = = =
⎪⎩
 
 
c) For N = 100, 
2
2max
max
1.0050.032 =0.0321 Kgm
10,000
12 374 /
0.0321
total
total
I
T r s
I
θ
⎧ = +⎪⎪
⎨
⎪ = = =
⎪⎩
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
121
Problem 7.3 
The three-axis robot shown in Figure P.7.3. is powered by geared servomotors attached to 
the joints by worm gears. Each link is 22 cm long, made of hollow aluminum bars, each 
weighing 0.5 Kg. The center of mass of the second motor is 20 cm from the center of 
rotation. The gear ratio is 1/3 in the servomotor and 1/5 in the worm gear set. The worst 
case scenario for the elbow joint is when the arm is fully extended, as shown. Calculate 
the torque needed to accelerate both arms together, fully extended, at a rate of 90 rad/s2. 
Assume the inertias of the worm gears are negligible. 
 
Worm gear
N=5
Motors
m2=0.5 Kg
m4=0.5 Kg
m3=0.5 Kg
I1=0.01 Kgm2
N=3
 
Figure P.7.3. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 7.2 
 
Solution: 
 
Total gear ratio 5 3 15N = × = 
( )
( )( ) ( )
( )( ) ( )( )
1 2 21 2
2 2
2 2
2
1
10.01 2 2
3
1 10.01 1 0.44 0.5 0.2 0.01038
3 15
0.01038 90 0.934
total rotor arm motor
arm motor motor
I I I I
N
m l m l
T I Nmθ
+
= + +
⎛ ⎞= + +⎜ ⎟
⎝ ⎠
⎡ ⎤= + + =⎢ ⎥⎣ ⎦
= = × =
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
122
Problem 7.4 
Repeat Problem 3, but suppose the maximum torque this motor can provide is 0.9 Nm. 
Therefore, a new motor must be picked. Two other motors are available, one with the 
inertia of 0.009 Kgm2 and torque of 0.85 Nm, one with inertia of 0.012 Kgm2 and torque 
of 1 Nm. Which one would you use? 
 
Estimated student time to complete:15-20 minutes 
Prerequisite knowledge required: Text Section(s) 7.2. 
 
Solution: 
 
For 5 3 15N = × = , 
 
For case 1: 
( )
( )( ) ( )
( )( ) ( )( )
1 2 21 2
2 2
2 2
2
1
10.009 2 2
3
1 10.009 1 0.44 0.5 0.2 0.00938
3 15
0.00938 90 0.844
total rotor arm motor
arm motor motor
I I I I
N
m l m l
T I Nmθ
+
= + +
⎛ ⎞= + +⎜ ⎟
⎝ ⎠
⎡ ⎤= + + =⎢ ⎥⎣ ⎦
= = × =
 
 
For case 2: 
 
( )
( )( ) ( )
( )( ) ( )( )
1 2 21 2
2 2
2 2
2
1
10.012 2 2
3
1 10.009 1 0.44 0.5 0.2 0.01238
3 15
0.01238 90 1.114
total rotor arm motor
arm motor motor
I I I I
N
m l m l
T I Nmθ
+
= + +
⎛ ⎞= + +⎜ ⎟
⎝ ⎠
⎡ ⎤= + + =⎢ ⎥⎣ ⎦
= = × =
 
 
The smaller motor is better.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
123
Problem 7.5 
Estimate how much the torque/inertia ratio of a disk motor might be if it can go from zero 
to 2000 rpm in one millisecond, and compare it to the motor of Problem 1. 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 7.2 
 
Solution: 
 
2
22 1
2 1
2000 2 209.44 / sec
60
209.44 0 209440 / sec
0.001
rad
rad
t t
πθ
θ θθ
×
= =
− −
= = =
−
 
The torque/inertia ratio is: 
 
209440TT I
I
θ θ= → = = 
For the motor of Problem 1: 
 
212 400 / sec
0.03
TT I rad
I
θ= → = =
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
124
Problem 7.6 
Using a timer circuit, design a pulse generating circuit that will deliver a range of 5-500 
pulses per second to a stepper motor driver. 
 
Estimated student time to complete:15-20 minutes 
Prerequisite knowledge required: Text Section(s) 7.6.7. 
 
Solution: 
 
We will pick a 50% duty cycle for each signal. Therefore: 
 
1
2low high total
t t t= = 
 
totalt ranges between 0.002 and 0.2 seconds. Therefore 0.001lowt = sec or 0.1lowt = sec. 
 
Pick 2 0.1C Fμ= . Therefore: 
 
( )
( )
2 2
6
2 2
6
2 2
0.693 ( )
0.001 0.693 0.1 10 14, 430
0.1 0.693 0.1 10 1, 443,000
lowt C R
R R
R R
−
−
=
= × → = Ω
= × → = Ω
 
Therefore, a potentiometer with 2 1.5R M= Ω enables us to achieve the desired values. 
For 50% duty cycle, as shown above, 
 
2 2 2 2 1 3 40.693 ( ) 0.693 ( ) 0
low high
H H
t t
C R C R R R R R R
=
= + → = + + =
 
 
Pick 1 3 4 1R R R K= = = Ω . Therefore: 
 
2
1
2
3
4
0.1
1
1.5
1
1
C F
R K
R M
R K
R K
μ=⎧
⎪ = Ω⎪⎪ = Ω⎨
⎪ = Ω⎪
⎪ = Ω⎩
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
125
Problem 7.7 
Calculate the gear ratio for a Harmonic drive if NL=100, NF=95, N2=90, N3=95. 
 
Estimated student time to complete: 10-15 minutes 
Prerequisite knowledge required: Text Section(s) 7.11 
 
Solution: 
 
From Equation 7.25: 
 
( )( ) ( )( )
( )( )
2 3
2
90 100 95 95 1
90 100 360
L FL
A L
N N N Ne
N N
ω
ω
−−
= = = = − 
 
The gear ratio between the Arm and the Last gear (with Arm as input) is: 
 
360A
L
ω
ω
= − 
 
The minus sign indicates that the input (Arm) and the output (Last) gears rotate in 
opposite directions.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
126
Problem 7.8 
Write a program to generate a variable pulse stream to drive a motor with pulse-width-
modulated voltages of 1, 2, 3, 4, and 5 volts for a 5 volt input. 
 
Estimated student time to complete: Variable depending on expertise. 
Prerequisite knowledge required: Text Section(s) 7.7. 
 
Solution: 
 
The accuracy with which this program will work depends on the processing speed of the 
microprocessor. It also depends on the level of programming language and how 
efficiently it is written. We assume that the processor can accept and run a compiled “C” 
program, and that it is fast enough such that the execution of commands will not 
negatively affect the accuracy of the output timing. Since only five distinct output 
voltages of 1, 2, 3, 4, and 5 volts are desired, we only need to generate 5 variations in the 
timing loops. We assume the output level will be indicated by an input port. Therefore, if 
input port 1 is high, the desired output level is 1 volts, etc. 
 
The following program is written in “C” language for execution by a Mini-Board™ 
microprocessor. It is the essentials of the program that are important here, not the actual 
program and the way it is written. This program must be modified for use for any other 
specific microprocessor with its own unique programming syntax and requirements. 
 
In this program, the command “motor(1,n)” indicates that the output port1 is turned on at 
n
15CC
V . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command 
“msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port 
1 is turned off. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
127
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
128
Problem 7.9 
Write a program to generate a sinusoidal pulse-width-modulated output for a constant 
input voltage. 
 
Estimated student time to complete: Variable depending on expertise. 
Prerequisite knowledge required: Text Section(s) 7.7. 
 
Solution: 
 
The accuracy with which this program will work depends on the processing speed of the 
microprocessor. It also depends on the level of programming language and how 
efficiently it is written. We assume that the processor can accept and run a compiled “C” 
program, and that it is fast enough such that the execution of commands will not 
negatively affect the accuracy of the output timing. Since a sinusoidal out put is desired 
for a 5 volt input voltage, we need to generate a variable control timing for the pulse 
width modulation. We calculate the sine value for each degree between zero and 90 
degrees. This allows for approximately 10 pulses per second. 
 
The following program is written in “C” language for execution by a Mini-Board™ 
microprocessor. It is the essentials of the program that are important here, not the actual 
program and the way it is written. This program must be modified for use for any other 
specific microprocessor with its own unique programming syntax and requirements. 
 
Since Mini-Board can generate voltages in both polarities, the program can generate a 
full sine wave. The program must be modified accordingly if the microprocessor cannot 
do this. 
 
In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at 
n
15CC
V . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command 
“msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port 
1 is turned off. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
129
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
130
Problem 7.10 
If you have access to a microprocessor and electronic components such as transistors, 
make an H-bridge and write a control program to drive a motor in either direction or to 
brake it. Be mindful of the problems associated with an H-bridge’s transistors turning on 
and off at inappropriate times. 
 
Estimated student time to complete: 
Prerequisite knowledge required: Text Section(s) 
 
Solution: 
 
The following program is written in “C” language for execution by a Mini-Board™ 
microprocessor. It is the essentials of the program that are important here, not the actual 
program and the way it is written. This program must be modified for use for any other 
specific microprocessor with its own unique programming syntax and requirements. 
 
In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at 
n
15CC
V . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command 
“msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port 
1 is turned off. 
 
It is assumed that an input to the microprocessor equal to 1 indicates rotation in the CCW 
direction, 2 indicates CW rotation, and a 0 means the motor should be braked. 
 
/************************************************************/ 
/* H-Bridge Controller */ 
/************************************************************/ 
. 
. 
main( ) 
{ 
int direction 
while(1) 
{direction = digital(0); /* motor rotation direction input */ 
if (direction = = 1) {msleep (1); /* CCW rotation */ 
motor (1,15); motor (4,15);} 
off (2); off (3); 
 
if (direction = = 2) {msleep (1); /* CW rotation */ 
motor (2,15); motor (3,15);} 
off (1); off (4); 
 
if (direction = = 0) {msleep (1); /* brake the motor */ 
motor (3,15); motor (4,15);} 
off (1); off (2); }}
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
131
CHAPTER EIGHT 
 
There are no problems in Chapter 8.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
132
CHAPTER NINE 
 
Problem 9.1 
Calculate the necessary memory requirement for a still color image from a camera with 
10 megapixels at: 
• 8-bits per pixel (256 levels) 
• 16-bits per pixel (65536 levels) 
 
Estimated student time to complete: 5 minutes 
Prerequisite knowledge required: Text Section(s) 9.5 
 
Solution: 
 
For an 8-bit per pixel system, each pixel requires 1 byte. The total memory needed is 10 
megabytes. 
 
For a 16-bit per channel system, each pixel requires 2 bytes. The total memory needed is 
20 megabytes. 
 
The assumption here is that the pixel count given is the total number for all three colors. 
The actual memory is greatly influenced by the format in which the image is saved. A 
bitmap image needs much more memory than JPEG.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
133
Problem 9.2 
Consider the pixels of an image, with values as shown in Figure P.9.2, as well as a 
convolution mask with the given values. Calculate the new values for the given pixels. 
 
4 7 1 8 
3 2 6 6 1 0 1 
5 3 2 3 1 1 1 
8 6 5 9 1 0 1 
 
Figure P.9.2 
 
Estimated student time to complete: 5-10 minutes 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeter will 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
4 7 1 8 
3 3.3 5 6 
5 4.6 4.4 3 
8 6 5 9 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
134
Problem 9.3 
Consider the pixels of an image, with values as shown in Figure P.9.3, as well as a 
convolution mask with the given values. Calculate the new values for the given pixels. 
Substitute 0 for negative grey levels. What conclusion can you make from the result? 
 
5 5 5 7 7 8 
5 5 5 8 8 9 1 0 1 
10 10 10 10 10 10 0 -4 0 
5 5 5 9 9 8 1 0 1 
8 8 8 9 9 10 
 
Figure P.9.3 
 
Estimated student time to complete: 5-10 minutes 
Prerequisite knowledge required: Text Section(s) 9.10. 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeterwill 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
5 5 5 7 7 8 
5 10 12 0 3 9 
10 0 0 0 0 10 
5 16 17 1 3 8 
8 8 8 9 9 10 
 
 
This can be an indication of an edge. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
135
Problem 9.4 
Repeat Problem 9.3, but substitute the absolute value of negative grey levels. What 
conclusion can you make from the result? 
 
Estimated student time to complete: 5 -10 minutes 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeter will 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
5 5 5 7 7 8 
5 10 12 0 3 9 
10 20 13 13 6 10 
5 16 17 1 3 8 
8 8 8 9 9 10 
 
When absolute values of negative pixels are shown, edges are no longer discernable.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
136
Problem 9.5 
Repeat Problem 9.3, but apply the mask of Figure P.9.5 and compare your results with 
Problem 9.3. Which one is better? 
 
-1 0 -1 
0 4 0 
-1 0 -1 
Figure P.9.5 
 
Estimated student time to complete: 5 -10 minutes 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeter will 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
 
5 5 5 7 7 8 
5 0 0 0 0 9 
10 20 13 13 6 10 
5 0 0 0 0 8 
8 8 8 9 9 10 
 
This mask also shows the edge, but unlike the one in problem 9.3., the pixels are brighter 
than the neighboring pixels.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
137
Problem 9.6 
Repeat Problem 9.3, but apply the mask of Figure P.9.6 and compare your results with 
Problem 9.3. Which one is better? 
 
0 1 0 
1 -4 1 
0 1 0 
Figure P.9.6 
 
Estimated student time to complete: 5 -10 minutes 
Prerequisite knowledge required: Text Section(s)9.10 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeter will 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
 
5 5 5 7 7 8 
5 5 8 0 2 9 
10 0 0 0 0 10 
5 8 12 0 0 8 
8 8 8 9 9 10 
 
At least for this image, the effect is similar to the mask in problem 9.3.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
138
Problem 9.7 
Repeat Problem 9.3, but apply the mask of Figure P.9.7 and compare your results with 
Problem 9.3. Which one is better? 
 
1 1 1 
1 -8 1 
1 1 1 
Figure P.9.7 
 
Estimated student time to complete: 5 -10 minutes 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeter will 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
 
5 5 5 7 7 8 
5 15 20 0 5 9 
10 0 0 0 0 10 
5 24 29 0 3 8 
8 8 8 9 9 10 
 
At least for this image, the effect is similar to Problem 9.3, but darker.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
139
 
Problem 9.8 
An image is represented by values shown below. 
a) Find the value of pixel 2c when mask 1 is applied. 
b) Find the value of pixel 3b when mask 2 is applied. 
c) Find the values of pixels 2b and 3c when a 3 3× median filter is applied. 
d) Find the area of the major object that results when a threshold of 4.5 is applied 
 based on a +4-connectivity (start at the first on-pixel). 
 
 a b c d e 
1 4 9 6 2 6 1 1 1 0 1 0 
2 4 3 7 4 6 1 1 1 1 -4 1 
3 6 2 6 5 3 1 1 1 0 1 0 
4 1 6 5 9 2 Mask 1 Mask 2 
5 2 8 4 4 7 
 
Figure P.9.8. 
 
Estimated student time to complete: 10 -15 minutes 
Prerequisite knowledge required: Text Section(s) 9.10 and 9.17. 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeter will 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
a) For 2c: 
R = (9×1 + 6×1 +2×1 +3×1 +7×1 +4×1 +2×1 +6×1 +5×1) / 9 = 4.9 (or 5) 
 
b) For 3b: 
R = (4×0 + 3×1 +7×0 +6×1 +2× -4 +6×1 +1×0 +6×1 +5×0) = 13 
 
c) For 2b: 2,3,4,4,6,6,6,7,9. The median is 6. 
 For 3c: 2,3,4,5,5,6,6,7,9. The median is 5. 
 
d) The resulting image after the application of the threshold is: 
 
 a b c d e 
1 X X X
2 X X
3 X X X 
4 X X X 
5 X X
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
140
 
Applying the +4 connectivity rule starting at the first on-pixel results in: 1b, 1c, 2c, 3c, 
3d, 4c, 4d, 4b, 5b, for a total of 9 pixels.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
141
 
Problem 9.9 
An image is represented by values shown below. 
a. Find the value of pixel 3b when mask 1 is applied. 
b. Find the values of pixels 2b, 2c, 2d when mask 2 is applied. 
c. Find the value of pixel 3c when a 5 5× median filter is applied. 
d. Find the area of the major object that results when a threshold of 4.5 is applied 
 based on a ×4-connectivity (start at the first on-pixel). 
 
 a b c d e 
1 8 9 6 2 5 0 1 0 1 1 1 
2 4 6 2 4 6 1 4 1 -2 -2 -2 
3 6 7 5 6 5 0 1 0 1 1 1 
4 1 10 5 9 4 Mask 1 Mask 2 
5 2 8 4 3 2 
 
Figure P.9.9. 
 
Estimated student time to complete: 10 -15 minutes 
Prerequisite knowledge required: Text Section(s) 9.10 and 9.17. 
 
Solution: 
 
The mask will only affect the middle pixels. The remaining pixels on the perimeter will 
not be affected. Depending on how a new image is formatted, the pixels on the perimeter 
may be set to zero or the current values. 
 
The new values for the image, assuming that the old file is copied into a new file, are: 
 
a) For 3b: 
R = (4×0 + 6×1 +2×0 +6×1 +7×4 +5×1 +1×0 +10×1 +5×0)/8 = 6.875 (or 7) 
 
b) For 2b: 
R = (8×1 + 9×1 +6×1 +4× -2 +6× -2 +2× -2 +6×1 +7×1 +5×1) = 17 
 
 For 2c: 
R = (9×1 + 6×1 +2×1 +6× -2 +2× -2 +4× -2 +7×1 +5×1 +6×1) = 11 
 
 For 2d: 
R = (6×1 +2×1 +5×1 +2× -2 +4× -2 +6× -2 +5×1 +6×1 +5×1) = 5 
 
c) For 3c: 1,2,2,2,2,3,4,4,4,4,5,5,5,5,6,6,6,6,6,7,8,8,9,9,10. The median is 5. 
 
 
d) The resulting image after the application of the threshold is: 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
142
 a b c d e 
1 X X X X
2 X X
3 X X X X X
4 X X X 
5 X 
 
Applying the ×4 connectivity rule starting at the first on-pixel results in: 1a, 2b, 3c, 1c, 
3a, 4d, 4b, 3e, for a total of 8 pixels.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
143
Problem 9.10 
Write a computer program for the application of a 3×3 averaging convolution mask unto 
a 15×15 image. Refer to the note on page 415 for more information. 
 
Estimated student time to complete: Variable depending on the expertise 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The following program is written in “C” language. It is the essentials of the programming 
that are important here, not the syntax. You may use the proper syntax for the program 
you are using. This program may be written in countless other languages, both graphical 
and scientific. 
 
In this program, the assumption is that the image input file is read from a data file or it is 
entered after initializing the array. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
144
Problem 9.11 
Write a computer program for the application of a 5×5 averaging convolution mask unto 
a 15×15 image. Refer to the note on page 415 for more information. 
 
Estimated student time to complete: Variable depending on the expertise 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The following program is written in “C” language. It is the essentials of the programming 
that are important here, not the syntax. You may use the proper syntax for the program 
you are using. This program may be written in countless other languages, both graphical 
and scientific. 
 
In this program, the assumption is that the image input file is read from a data file or it is 
entered after initializing the array. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
145
Problem 9.12 
Write a computer program for the application of a 3×3 high pass convolution mask unto 
a 15×15 image for edge detection. Refer to the note on page 415 for more information. 
 
Estimated student time to complete: Variable depending on expertise 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The following program is written in “C” language. It is the essentials of the programming 
that are important here, not the syntax. You may use the proper syntax for the program 
you are using. This program may be written in countless other languages, both graphical 
and scientific. 
 
In this program, the assumption is that the image input file is read from a data file or it is 
entered after initializing the array. 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
146
Problem 9.13 
Write a computer program for the application of an n×n convolution mask unto a k×k 
image. Refer to the note on page 415 for more information. You should write the routine 
such that the user can choose the size of the mask and the values of each mask cell 
individually. 
 
Estimated student time to complete: Variable depending on expertise 
Prerequisite knowledge required: Text Section(s) 9.10 
 
Solution: 
 
The following program is written in “C” language. It is the essentials of the programming 
that are important here, not the syntax. You may use the proper syntax for the program 
you are using. This program may be written in countless other languages, both graphical 
and scientific. 
 
In this program, the assumption is that the image input file is read from a data file or it is 
entered after initializing the array. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
147
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
148
Problem 9.14 
Write a computer program that will perform the Left-Right search routine for a 15×15 
image. Refer to the note on page 415 for more information. 
 
Estimated student time to complete: Variable depending on expertise 
Prerequisite knowledge required: Text Section(s) 9.13 
 
Solution: 
 
The following program is written in “C” language. It is the essentials of the programming 
that are important here, not the syntax. You may use the proper syntax for the program 
you are using. This program may be written in countless other languages, both graphical 
and scientific. 
 
In this program, the assumption is that the image input file is read from a data file or it is 
entered after initializing the array. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
149
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
150
Problem 9.15 
Using the Left-Right search technique, find the outer edge of the object in Figure P.9.15: 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
Figure P.9.15 
 
Estimated student time to complete: 5-10 minutes 
Prerequisite knowledge required: Text Section(s) 9.13 
 
Solution: 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
The edge consists of: 2c, 2d, 2e, 2f, 2g, 2h, 2i, 2j, 2k, 3k, 4k, 5j, 6j, 7k, 7j, 8i, 9h, 10h, 
11h, 12h, 12g, 12f, 11f, 10g, 9g, 8f, 8e, 7e, 7d, 7c, 6b, 5b, 4b, 4c, 3c.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
151
Problem 9.16 
The x and y coordinates of 5 points are given as (2.5, 0), (4,2), (5,4), (7,6), and (8.5,8). 
Using the Hough transform, determine which of these points form a line and find its slope 
and intercept. 
 
Estimated student time to complete: 20-25 minutes 
Prerequisite knowledge required: Text Section(s) 9.15. 
 
Solution: 
 
y x x,y m,c 
0 2.5 0=2.5m+c c= -2.5m+0 
2 4 2=4m+c c= -4m+2 
4 5 4=5m+c c= -5m+4 
6 7 6=7m+c c= -7m+6 
8 8.5 8=8.5m+c c= -8.5m+8 
The following graph shows the lines and where they intersect. Line 3 (for point 5,4) does 
not intersect the remaining lines. 
 
The approximate slope and intercept of the line on which the other four points lay are 1.4 
and -3.5. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system orthe Internet without prior expressed consent of the copyright owner. 
152
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
153
Problem 9.17 
Write a computer program that will perform a region growing operation based on +4 
connectivity. The routine should start at the 1,1 corner pixel, search for a nucleus, grow a 
region with a chosen index number, and after finishing that region, must continue 
searching for another nuclei until all object pixels have been checked. Refer to the note 
on page 415 for more information. 
 
Estimated student time to complete: Variable depending on expertise. 
Prerequisite knowledge required: Text Section(s) 9.17 
 
Solution: 
 
The following program is written in “C” language. It is the essentials of the programming 
that are important here, not the syntax. You may use the proper syntax for the program 
you are using. This program may be written in countless other languages, both graphical 
and scientific. 
 
In this program, the assumption is that the image input file is read from a data file or it is 
entered after initializing the array. 
 
In this program, when a nucleus is found, all the pixels connected to it based on a +4-
connectivity that are “on” (have a value in them and therefore, are part of the image not 
the background) are placed in a stack by changing the value of the cell form zero to 1 in 
an array of 256x256. The cell’s value is later checked until all are found and the cell’s are 
all zero. Each corresponding pixel’s value is changed from 1 to the region number. At the 
end of execution, the original image is changed to as many regions as necessary. Since 
there will be no cells (pixels) left with a value of 1, execution ends. All segments will be 
2 and larger in value. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
154
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
155
Problem 9.18 
Write a computer program that will perform a region growing operation based on ×4 
connectivity. The routine should start at the 1,1 corner pixel, search for a nucleus, grow a 
region with a chosen index number, and after finishing that region, must continue 
searching for another nuclei until all object pixels have been checked. Refer to the note 
on page 415 for more information. 
 
Estimated student time to complete: Variable depending on expertise. 
Prerequisite knowledge required: Text Section(s) 9.17 
 
Solution: 
 
Please see the solution for Problem 9.17. The program for growing a region based on ×4 
connectivity is very similar to that program, except that the search will follow the ×4 
sequence given in Equation 9.7.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
156
Problem 9.19 
Using +4 connectivity logic and starting from 1a pixel, write the sequence of pixels in 
correct order that will be detected by a region growing routine for the object in Figure 
P.9.19: 
1
3
4
+4-connectivity
2
9
8
7
6
5
4
3
2
1
f g ha b c d e
 
Figure P.9.19 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 9.17 
 
Solution: 
 
The sequence of pixels detected is: 2d, 3d, 3e, 3f, 4e, 3g, 4f, 5e, 5f, 5d, 5g, 6d, 6c, 7c, 7b, 
8b. 
 
The remaining pixels are not connected to this region and will not be detected.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
157
 
Problem 9.20 
Using ×4 connectivity logic and starting from 1a pixel, write the sequence of pixels in 
correct order that will be detected by a region growing routine for the object in Figure 
P.9.20: 
1
3
4
x4-connectivity
2
9
8
7
6
5
4
3
2
1
f g ha b c d e
 
Figure P.9.20 
 
Estimated student time to complete: 10 minutes 
Prerequisite knowledge required: Text Section(s) 9.17 
 
Solution: 
 
The sequence of pixels detected is: 2c, 3d, 4e, 4c, 5f, 5d, 3f, 4b, 6g, 6c, 7b. 
 
The remaining pixels are not connected to this region and will not be detected. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
158
 
Problem 9.21 
Find the union between the two objects in Figure P.9.21. 
 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 1 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 
Figure P.9.21 
 
Estimated student time to complete: 10-15 minutes 
Prerequisite knowledge required: Text Section(s) 9.18 
 
Solution: 
 
The union is shown: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
159
 
Problem 9.22 
Apply a single-pixel erosion based on 8-connectivity on the image of Figure P.9.22. 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
Figure P.9.22 
 
Estimated student time to complete: 10-20 minutes 
Prerequisite knowledge required: Text Section(s) 9.18 
 
Solution: 
 
The result is shown: 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
160
 
Problem 9.23 
Apply a one-pixel dilation to the result of Problem 9.22 and compare your result to 
Figure P.9.22. 
 
Estimated student time to complete: 15-25 minutes 
Prerequisite knowledge required: Text Section(s)9.18 
 
Solution: 
 
The result is shown. As you notice, the image is different from the original. 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
161
 
Problem 9.24 
Apply an open operation to Figure P.9.24. 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
Figure P.9.24 
 
Estimated student time to complete: 10-20 minutes 
Prerequisite knowledge required: Text Section(s) 9.18 
 
Solution: 
 
The result is shown: 
2
3
4
5
6
7
8
9
10
11
12
13
1
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g h i j k l ma b c d e f g
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
162
 
Problem 9.25 
Apply a close operation to Figure P.9.24. 
 
Estimated student time to complete: 10-20 minutesPrerequisite knowledge required: Text Section(s) 9.18 
 
Solution: 
 
The result is shown: 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
163
 
Problem 9.26 
Apply a skeletonization operation to Figure P.9.26. 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
Figure P.9.26 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 9.18 
 
Solution: 
 
The result is shown: 
2
3
4
5
6
7
8
9
10
11
12
13
1
h i j k l ma b c d e f g
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
164
 
Problem 9.27 
Write a computer program in which different moments of an object in an image can be 
calculated. The program should ask you for moment indices. The results may be reported 
to you in a new file, or may be stored in memory. Refer to the note on page 415 for more 
information. 
 
Estimated student time to complete: Variable depending on expertise 
Prerequisite knowledge required: Text Section(s) 9.21. 
 
Solution: 
 
The following program is written in “C” language. It is the essentials of the programming 
that are important here, not the syntax. You may use the proper syntax for the program 
you are using. This program may be written in countless other languages, both graphical 
and scientific. 
 
In this program, the assumption is that the image input file is read from a data file or it is 
entered after initializing the array. 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
165
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
166
Problem 9.28 
Calculate the 0,2M moment for the result of Problem 9.8d based on +4-connectivity. 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 9.21. 
 
Solution: 
 
The resulting image after the application of the threshold is: 
 
 a b c d e 
1 X X X
2 X X
3 X X X 
4 X X X 
5 X X
 
Applying the +4 connectivity rule starting at the first on-pixel results in the following 
area: 
 
 
 a b c d e 
1 X X 
2 X 
3 X X 
4 X X X 
5 X 
 
x
y
 
The moment, calculated relative to the first row and column of pixels with whole-number 
distances is: 
 
0 2 2 2 2 2 2
0,2 1(1) 2(2) 3(3) 1(4) 52M x y y= = = + + + =∑ ∑
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
167
Problem 9.29 
For the binary image of a key in Figure P.9.29, calculate the following: 
• Perimeter, based on the Left-Right search technique. 
• Thinness, based on
2P
Area . 
• Center of gravity 
• Moment 0,1M about the origin (pixel 1,1) and about the lowest pixel of a rectangular 
box around the key (2,2). 
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x
y 
Figure P.9.29 
 
Estimated student time to complete: 15-20 minutes 
Prerequisite knowledge required: Text Section(s) 9.13, 9.21. 
 
Solution: 
 
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x
y 
The left-right search result is shown. The perimeter based on the pixels on the search line 
is P = 23. 
 
The area may be calculated by 0,0M , by region growing or by counting. 
 
Thinness: 
2 223 17
31
P
Area
= = 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
168
The 1st moments are: 
 
1 0
1,0
0 1
0,1
2(1) 3(2) 4(3) 4(4) 5(5) 5(6) 5(7) 3(8) 150
2(1) 4(2) 4(3) 5(4) 5(5) 4(6) 4(7) 3(8) 143
M x y x
M x y y
= = = + + + + + + + =
= = = + + + + + + + =
∑ ∑
∑ ∑
 
 
Center of gravity is: 1,0 0,1
0,0 0,0
150 1434.84 and 4.61
31 31
M M
x y
M M
= = = = = = 
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x
y
x'
y'
 
 
0 1
0,1
0 1
0,1
@(1,1) 143
@(2,2) 4(1) 4(2) 5(3) 5(4) 4(5) 4(6) 3(7) 112
M x y
M x y
= =
= = + + + + + + =
∑
∑
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
169
Problem 9.30 
Using moment equations, calculate 0,2M and 2,0M about the centroidal axes of the part in 
Figure P.9.30. 
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x
y 
Figure P.9.30 
 
Estimated student time to complete: 20-30 minutes 
Prerequisite knowledge required: Text Section(s) 9.21 
 
Solution: 
 
The moment values for the image are: 
 
0 0
0,0
1 0
1,0
0 1
0,1
1,0 0,1
0,0 0,0
22 0 2 2 2 2
2,0
16(1) 16
1(2) 3(3) 3(4) 4(5) 2(6) 3(7) 76
3(2) 2(3) 4(4) 3(5) 3(6) 1(7) 68
76 684.75 and 4.25
16 16
1(2) 3(3) 3(4) 4(5) 2(
M x y
M x y x
M x y y
M M
x y
M M
M x y x
= = =
= = = + + + + + =
= = = + + + + + =
= = = = = =
= = = + + + +
∑
∑ ∑
∑ ∑
∑ 2 2
0 2 2 2 2 2 2 2 2
0,2
6) 3(7) 398
3(2) 2(3) 4(4) 3(5) 3(6) 1(7) 326M x y y
+ =
= = = + + + + + =
∑
∑ ∑
 
 
The second moments about the centroids can be calculates as follows: 
 
( )2 22,0 2,0 0,0@ 398 16(4.75) 37M x M M x= − = − = 
2 2
0,2 0,2 0,0@ ( ) 326 16(4.25) 37M y M M y= − = − =
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
170
Problem 9.31 
Using moment equations, calculate 0,2M and 2,0M about the centroidal axes of the part in 
Figure P.9.31. 
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
x
y 
Figure P.9.31 
 
Estimated student time to complete:20 -30 minutes 
Prerequisite knowledge required: Text Section(s) 9.21 
 
Solution: 
 
The moment values for the image are: 
 
0 0
0,0
1 0
1,0
0 1
0,1
1,0 0,1
0,0 0,0
22 0 2 2 2 2 2 2
2,0
18(1) 18
3(2) 2(3) 4(4) 4(5) 2(6) 3(7) 81
4(2) 5(3) 5(4) 4(5) 63
81 634.5 and 3.5
18 18
3(2) 2(3) 4(4) 4(5) 2(6) 3(7) 41
M x y
M x y x
M x y y
M M
x y
M M
M x y x
= = =
= = = + + + + + =
= = = + + + =
= = = = = =
= = = + + + + + =
∑
∑ ∑
∑ ∑
∑
0 2 2 2 2 2 2
0,2
3
4(2) 5(3) 5(4) 4(5) 241M x y y= = = + + + =
∑
∑ ∑
 
 
The second moments about the centroids can be calculates as follows: 
 
( )2 22,0 2,0 0,0@ 413 18(4.5) 48.5M x M M x= − = − = 
2 2
0,2 0,2 0,0@ ( ) 241 18(3.5) 20.5M y M M y= − = − = 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
171
CHAPTER TEN 
 
Problem 10.1 
Develop a fuzzy inference system for a robot, where the force exerted at the hand and the 
velocity of the hand are the inputs and the % power to the actuators is the output. 
 
Estimated student time to complete:-------- 
Prerequisite knowledge required: Text Section(s) 10.3-10.9 
 
Solution: 
 
As in any other design problem, the solution is not unique. For example, different inputs 
and outputs may be selected,different fuzzy sets may be chosen, the ranges selected may 
be different, and alternate rules may be assigned. Therefore, not only the solution is not 
unique, the result will also differ. This suggested solution may only be used as a guide. 
 
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any 
other available system. 
 
In this solution, the inputs, outputs, the sets, and the rules are as follows: 
 
Inputs: Force; Small, Average, Large 
 Velocity; Slow, Fast, Fast, Very-Fast 
 
Outputs: Power; Low, Medium, High, Very-High 
 
The fuzzy sets are (arbitrary numbers): 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
172
 
 
 
 
The rules are: 
 
The resulting output is: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
173
 
 
Changing the rules or the membership functions will change this result. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
174
Problem 10.2 
Develop a fuzzy inference system for a washing machine. The inputs are how dirty the 
fabrics are and how much clothes are being washed, and the output is the wash time. 
 
Estimated student time to complete:-------- 
Prerequisite knowledge required: Text Section(s) 10.3-10.9 
 
Solution: 
 
As in any other design problem, the solution is not unique. For example, different inputs 
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may 
be different, and alternate rules may be assigned. Therefore, not only the solution is not 
unique, the result will also differ. This suggested solution may only be used as a guide. 
 
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any 
other available system. 
 
In this solution, the inputs, outputs, the sets, and the rules are as follows: 
 
Inputs: Dirty; Not-Much, Some, Very. 
 Clothes; Small, Med, Much, A-Lot. 
 
Outputs: Wash-Time; Five, Ten, Twenty, Thirty 
 
The fuzzy sets are: 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
175
 
 
 
 
The rules are: 
 
The resulting output is: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
176
 
 
Changing the rules or the membership functions will change this result. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
177
Problem 10.3 
Develop a fuzzy inference system for a barbecue. The inputs may be the thickness of the 
steak and how cooked or rare it is desired to be. The output may be the temperature of the 
flame and/or the time of cooking. 
 
Estimated student time to complete: 
Prerequisite knowledge required: Text Section(s) 10.3-10.9 
 
Solution: 
 
As in any other design problem, the solution is not unique. For example, different inputs 
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may 
be different, and alternate rules may be assigned. Therefore, not only the solution is not 
unique, the result will also differ. This suggested solution may only be used as a guide. 
 
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any 
other available system. 
 
In this solution, the inputs, outputs, the sets, and the rules are as follows: 
 
Inputs: Thickness; See-Through, Average, Hefty 
 How Cooked; Rare, Medium-Rare, Medium, Well-Done 
 
Outputs: Temp; Low, Medium, High 
 Time; Three, Five, Eight, Ten 
 
The fuzzy sets are: 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
178
 
 
 
 
 
 
The rules are: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
179
 
The resulting out puts are: 
 
And 
 
 
Changing the rules or the membership functions will change this result. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
180
Problem 10.4 
Develop a fuzzy inference system for an automatic gearbox. The inputs are the speed of 
the car and the load on the engine, and the output is the gear ratio of the transmission. 
 
Estimated student time to complete: 
Prerequisite knowledge required: Text Section(s) 10.3-10.9 
 
Solution: 
 
As in any other design problem, the solution is not unique. For example, different inputs 
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may 
be different, and alternate rules may be assigned. Therefore, not only the solution is not 
unique, the result will also differ. This suggested solution may only be used as a guide. 
 
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any 
other available system. 
 
In this solution, the inputs, outputs, the sets, and the rules are as follows: 
 
Inputs: Speed; Slow, Medium, Cruise, Fast, Maximum 
 Load; Low, Average, High 
 
Outputs: Gear-Ratio; Fourth, Third, Second, First 
 
The fuzzy sets are: 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
181
 
 
 
 
 
The rules are: 
 
The resulting out put is: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
182
 
Changing the rules or the membership functions will change this result. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
183
Problem 10.5 
Develop a fuzzy logic system for a vision system in which the inputs are the intensities of 
the three colors of red, green, and blue (RGB) in a color image and the output is the 
relationship of the combination to the colors of the rainbow. 
 
Estimated student time to complete: 
Prerequisite knowledge required: Text Section(s) 10.3-10.9 
 
Solution: 
 
As in any other design problem, the solution is not unique. For example, different inputs 
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may 
be different, and alternate rules may be assigned. Therefore, not only the solution is not 
unique, the result will also differ. This suggested solution may only be used as a guide. 
 
This solution is implemented on the MATLAB Fuzzy Toolbox.You may use this or any 
other available system. 
 
In this solution, the inputs, outputs, the sets, and the rules are as follows: 
 
Inputs: Color-Red; Red-Off, Red-Medium, Red-High 
 Color-Green; Green-Off, Green-Medium, Green-High 
 Color-Blue; Blue-Off, Blue-Medium, Blue-High 
 
Outputs: Color; Black, Red, Orange, Yellow, Green, Cyan, Blue, Violet, White 
 
The fuzzy sets are: 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
184
 
 
 
The rules are: 
 
 
 
The resulting out put is: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
185
 
 
 
 
 
 
Changing the rules or the membership functions will change this result. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
186
Problem 10.6 
Develop a fuzzy inference system for grading a robotics course. The inputs are your 
effort level in the course and your exam grade, and the output is your letter grade. 
 
Estimated student time to complete: 
Prerequisite knowledge required: Text Section(s) 10.3-10.9 
 
Solution: 
 
As in any other design problem, the solution is not unique. For example, different inputs 
and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may 
be different, and alternate rules may be assigned. Therefore, not only the solution is not 
unique, the result will also differ. This suggested solution may only be used as a guide. 
 
This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any 
other available system. 
 
In this solution, the inputs, outputs, the sets, and the rules are as follows: 
 
Inputs: Effort; Low, Medium, High 
 Exam-Grade; eA, eB, eC, eD, eF 
 
Outputs: Course-Grade; A, B, C, D, F 
 
The fuzzy sets are: 
 
 
 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
187
 
 
 
The rules are: 
 
The resulting out put is: 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
188
 
 
Changing the rules or the membership functions will change this result. 
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
189
APPENDICES 
 
Problem A.1 
Show that the determinant of a matrix can be calculated by picking any row or column. 
 
Estimated student time to complete: 10-15 minutes 
Prerequisite knowledge required: Text Section(s) A.1 
 
Solution: 
 
For an arbitrary matrix: 
 
a b c
A d e f
g h i
⎡ ⎤
⎢ ⎥= ⎢ ⎥
⎢ ⎥⎣ ⎦
 
 
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
det
det
det
A a ei fh b di fg c dh eg
aei afh bdi bfg cdh ceg
A d bi ch e ai cg f ah bg
dbi dch eai ecg fah fbg
A b di fg e ai cg h af cd
bdi bfg eai ecg haf hcd
= − − − + −
= − − + + −
= − − + − − −
= − + + − − +
= − − + − − −
= − + + − − +
 
 
They are all the same.
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
190
 
Problem A.2 
Calculate the determinant of the following (4 × 4) matrix. 
 
 
1 1 0 0
0 1 2 0
3 0 1 1
1 0 0 1
A
⎡ ⎤
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
 
 
Estimated student time to complete: 5 minutes 
Prerequisite knowledge required: Text Section(s) A.1. 
 
Solution: 
 
( )( ) ( )( )det 1 1 1 0 0 0 1 0 2 3 1 0 0 0 1 4 5A = − − + − − − − + − = + =
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
191
Problem A.3 
Calculate the inverse of the following matrix using method 1: 
 
 
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
302
010
211
B 
 
Estimated student time to complete: 5-10 minutes 
Prerequisite knowledge required: Text Section(s) A.1. 
 
Solution: 
 
1
det 1(3 0) 0 2( 2) 3 4 1
1 0 2
1 1 0
2 0 3
3 3 2
0 1 0
2 2 1
3 3 2
0 1 0
2 2 1
T
B
B
adjB
B−
= − − + − = − = −
⎡ ⎤
⎢ ⎥= ⎢ ⎥
⎢ ⎥⎣ ⎦
− −⎡ ⎤
⎢ ⎥= −⎢ ⎥
⎢ ⎥−⎣ ⎦
−⎡ ⎤
⎢ ⎥= ⎢ ⎥
⎢ ⎥− −⎣ ⎦
 
© Copyrighted 2010. 
This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other 
electronic media system or the Internet without prior expressed consent of the copyright owner. 
192
Problem A.4 
Calculate the inverse of the following matrix using method 2: 
 
 
1 0 1
0 2 1
3 1 0
C
⎡ ⎤
⎢ ⎥= ⎢ ⎥
⎢ ⎥⎣ ⎦
 
 
Estimated student time to complete: 10-15 minutes 
Prerequisite knowledge required: Text Section(s) A.1. 
 
Solution: 
 
11 12 13
21 22 23
31 32 33
11 31 11
21 31 31
11 21 21
12 32 12
22 32 22
3212 22
1 0 1 1 0 0
0 2 1 0 1 0
3 1 0 0 0 1
1 1/ 7
2 0 6 / 7
3 0 3/ 7
0 1/ 7
2 1 3/ 7
1/ 73 0
x x x
x x x
x x x
x x x
x x x
x x x
x x x
x x x
xx x
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
+ = =⎧ ⎧
⎪ ⎪+ = → =⎨ ⎨
⎪ ⎪+ = = −⎩⎩
+ = = −⎧ ⎧
⎪ ⎪+ = → =⎨ ⎨
⎪ ⎪ =+ = ⎩⎩
31 33 13
23 33 23
13 23 33
1
0 2 / 7
2 0 1/ 7
3 1 2 / 7
1 1 2
7 7 7
3 3 1
7 7 7
6 1 2
7 7 7
x x x
x x x
x x x
C−
+ = =⎧ ⎧
⎪ ⎪+ = → =⎨ ⎨
⎪ ⎪+ = = −⎩ ⎩
⎡ ⎤−⎢ ⎥
⎢ ⎥
−⎢ ⎥= ⎢ ⎥
⎢ ⎥−⎢ ⎥
⎢ ⎥⎣ ⎦