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Problem 7.53PP
A simplified model for the control of a flexible robotic arm is shown in Fig., where 
k/M = 900 rad/sec2, 
y = output, the mass position, 
u = input, the position of the end of the spring.
(a) Write the equations of motion in state-space form.
(b) Design an estimator with roots at s = -100 ± 100/.
(c) Could both state-variables of the system be estimated if only a measurement of y was 
available?
(c) Could both state-variables of the system be estimated if only a measurement of y was 
available?
(d) Design a full-state feedback controller with roots at s = -20 ± 20/.
(e) Would it be reasonable to design a control law for the system with roots at s = -200 ± 200/. 
State your reasons.
(f) Write equations for the compensator, including a command input for y. Draw a Bode plot for 
the closed-loop system and give the gain and phase margins for the design.
Figure Simple robotic ann
Step-by-step solution
step 1 of 12
Refer to the Figure 7.98 from the text book.
It is clear that the value of — is 900 ra^sfc * -
(a)
The equations of motion in state space form are as follows. 
Consider that x^m y and Xj ■ y
Thus, the equations are,
X ,= X i
k k
X, ss— X + — tt
Thus, the state matrix representation is as follows.
0 1 O'
= - - i 0 f c *. M M .
y= 1 0]jt
(1)
Step 2 of 12
(b)
It is clear that roots are —100±100y- 
Use the roots and calculate the equation.
( i + 100+100y)(j + 1 0 0 -1 0 0 y)= (i+ 1 0 0 )^+ (1 0 0 )'
= « ’ + 10‘ +200s + 10‘
= « ‘ + 200«+(20000)
It is clear that * '+200 «+(20000 ) represents the characteristic equation.
Calculate the characteristic equation for the robotic arm using the matrices shown In equation (1). 
The formula to calculate the characteristic equation using the state space fonn is as follows. 
d e t[ jI - (F -L H )] = 0 
It is clear from the equation (1) that,
0 1'
F*
w = - K x
Write the following code in MATLAB and obtain the transfer function of the compensator.
» F=[ 0 1 :-900 0];
» G=[0;900]:
» H=[1 0];
» L=[200;19100]:
» K=[-0.111 0.044]:
» syms s 
» l=[1 0; 0 1];
» D=-K*inv{(s*l-F+G*K+L*H))*L 
D =
(222*(5*s + 198))/{5*(10*s''2 + 2396*s + 278201)) + 8862049/(5*{10‘ s'^2 + 2396*s + 278201)) - 
(8404*(s + 200))/{10*s'‘2 + 2396*s + 278201)
» simplify(D) 
ans =
-(8182*s -100401 )/(10*s'^2 + 2396*s + 278201)
Step 11 of 12
Draw the bode plot for the obtained transfer function of the compensator.
Bode D ig ra m
Step 12 of 12
Thus, bode plot is obtained for the compensator and the gain margin and phase margins are 
1-10.7 dB and respectiveiy.