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Problem 4.29PP
The transfer functions of speed control for a magnetic tape-drive system are shown in Fig. The 
speed sensor is fast enough that its dynamics can be neglected and the diagram shows the 
equivalent unity feedback system.
(a) Assuming the reference is zero, what is the steady-state error due to a step disturbance 
torque of 1 N m? What must the amplifier gain K be in order to make the steady-state error ess ̂ s 44.94, ^ s0.13- 
Hence, the roots are undesirable because, the damping is too low and there is high overshoot.
The roots of the closed loop are plotted and sketched.
4>
• 3
I
Figure 2
The time response of the system for a step input is sketched.
1 I
Figure 3
(c)
Given that there is 1 % of settling time of ^ 0.1 sec and an over shoot of 
We know that for i 0.1, we have ^ ^ 46- 
Similarly for S 0.05, ^ ^ 0.7
Step 8 of 11 ^
The region of acceptable closed loop poles is sketched in the complex plane.
Lines ofdam ping and
( d )
For a PD controller larger values of and ̂ are required.
This can be achieved by increasing and adding derivative fed back.
y ( i)finri ' ^
lOAT,_L
y (» ) 0 . 5 i - f l J s * b
w k , ( k ^ + \) 
(o .5 j + i) ( y * + ft)
y ( j ) 200X.
n , ( s ) + ( 1 2 + 2 0 O * :,* r„ ) i + 20(1 + 10*:,)
Hence, the transfer function is
r(s ) 200* : ,
a , ( s ) s* + 12 + 2 0 0 * :,* : .)s + 20(1 + 10* : , )
By choosing the suitable values for Kp and KD any values for ojj,, and ̂ can be achieved.
Step 10 of 11
(e)
Find the transfer function — for the disturbance.
1
J s + b
IT 1 io * : , ( * : „ f + i )
J s + ft (0.5» + 1 )
20(0.5t +1)
ly ” j ’ + ( i2 + 200* : , * : . ) j + 20(1 + 10* : , )
The error when a step input is given is e,(stepm W) = —
10*:,
Step 11 of 11
The derivative feedback affects the transient response only.
To eliminate the steady state error an integrator is added to the loop..
This can be represented by replacing AT^with AT̂ + in the fon/vard loop and still keeping PD 
control in the feedback loop.
Thus, we get
y 20(0 .Ss + l ) i
ly " s’ + (12 + 200* : , * : . )s ’ + (2 0 + 200* : , + 200* : , * : . )s + 200* :,
Hence, the steady state error to the disturbance will be [c , (step in W) = O] - 
Thus, the steady state error to the disturbance torque is eliminated entirely.