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Problem 7.1 OPP
Use the transformation matrix in Eq. to explicitly multiply out the equations at the end of Example. 
Eq.
T = [ - t ? ] ’ ' ] •
Example
EXAM PLE 7.9 Tran^ormation o f The mud Systemfrom Control to Modal Form
s ia Eq. (7.1 ̂ into theRod thee
modal fonn of Eq. (7.14).
StdvUon. Aoconfing to Eqi. (7 J4) and (7.35), we need fint to find die 
eigenveclois and eigenvalues of the Af maliu. Wb take the eigenvectofi to be
[S]
aaUL'Wei
[s]
es oi me Af maanL >ve nice me eigenvecton lo He
i5I
-7 f |i -12^1 —pliu 
'll
Suhstiniting Eq. (7.36c) into Eq. (7.36b) results in 
-7 p ll i- l2 U i= p * * 2 l.
^ t i i + 7 ^ 1 + 12f2i = 0,
p ^ + lp + l l-O ,
- 3 , - 4 .
a J 6 a )
036b)
(7J6c)
Wb have found (again!) that the eigenvalues (poles) a e 
fartfaennore. Eq. (7.36c) tdls os that the two eigeuvecton are
a J 7 a ) 
(7.37b) 
(737c) 
(7.37d) 
- 3 and- 4 ;
r - ^ i
L ft. ]
where til and tn a e arbitrary nonzero scale focton. We arant to select the 
two scale focton such tha both elements of Bg| in Eq. (7.14a) are mhy. Hie 
equation for Bn in terms of Br is TB* » Br. and its solution b I2i = ^1 
and l22 — !• Therefore, the traMforinatioo matrix and its iovene^ are
"■[-t 1]- ""-[I *]■ a.38)
Elementary matrix muhqiUcation abows that, using T as defined fay 
Eq. (738), the matrices of Eqs. (7.12) and (7.14) are related as follows:
Am^T~'A€T, B in = T “*Bc.
C , = CfT. D»=Dc. a » )
*1b fied the iavMse e f a 2 X 2 ■
M l '22.'* d w ^ die s ^ o f * e ** i r H d d ie ‘7 P I 
[slisE4.(7JB)l.
These computations can be carried out by using the following Matiab
T -[4-3; -11];
Am-inv(T)*Ac*T;
Bm«1nv(T)*Bc
Cm-CcT;
Dm-Dc;
Step-by-step solution
Step 1 of 5
According to equation (7.30) 
Transformation Matrix
And die equations are
A . = r % T
C . = C.T
- = [ ; a
B , = T ^B 
D , = D,
Step 2 of 5
and
A = 0'
•7 -12 
1 0 
■28+12 
4+10
1 3 
1 4 
1 3'
1 4 
1 3‘
1 4 
-16+12 9 
-16+16 9
][-. -.1
21-121
- 3 - l o J
•16 9 ]
4 - 3 J
. - 9-1 
- I 2 J
4 =
- 4 o ' 
0 - 3
Ans.
Step 3 of 5
a T %
1 3 
1 4 
'l+O ' 
1+0
a
B .= Ans.
Step 4 of 5
c . - c ^ r
■" 4 - . V]
=[4 - 2 - 3 + 2 ] 
= [ 2 - 1]
IC, = [ 2 -1]| A?is.
Step 5 of 5
= 0| Ans