Question 10.7.2: Design of a PD Control System For the system shown in Figure...

a. From the figure we obtain the following output, error, and actuator equations.
\Theta(s) = \frac{K_{P}  +  K_{D} s}{10 s^{2}  +  (2  +  K_{D})s  +  K_{P}} \Theta_{r}(s)  −  \frac{1}{10s^{2}  +  (2  +  K_{D})s  +  K_{P}} T_{d} (s)             (1)
E(s) = \frac{10s^{2}  +  2s}{10 s^{2}  +  (2  +  K_{D})s  +  K_{P}} \Theta_{r}(s)  + \frac{1}{10 s^{2}  +  (2  +  K_{D})s  +  K_{P}} T_{d} (s)                 (2)
T (s) = \frac{(10s^{2}  +  2s)(K_{P}  +  K_{D}s)}{10 s^{2}  +  (2  +  K_{D})s  +  K_{P}} \Theta_{r}(s) + \frac{K_{D} s  +  K_{P}}{10 s^{2}  +  (2  +  K_{D})s  +  K_{P}} T_{d} (s)            (3)
The characteristic equation is
10 s^{2} + (2 + K_{D})s + K_{P} = 0
and the damping ratio is
ζ = \frac{2  +  K_{D}}{2 \sqrt{10K_{P}}} = 1                 (4)

Since ζ = 1, the expression for the time constant is
\tau = \frac{20}{2  +  K_{D}} = 2  sec
which gives K_{D} = 8. Using this value in equation (4) gives
\frac{2  +  8}{2 \sqrt{10K_{P}}} = 1
which gives K_{P} = 2.5 .
Applying the final value theorem to the error equation (2) with \Theta_{r}(s) = 1/s and T_{d} (s) = 0 gives
e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{10s^{2}  +  2s}{10s^{2}  +  (2  +  K_{D})s  +  K_{P}} \frac{1}{s} = 0
Thus the system has zero command error for a step command.
Applying the final value theorem to the error equation (2) with \Theta_{r}(s) = 0 and T_{d} (s) = 1/s gives
e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{1}{10s^{2}  +  (2  +  K_{D})s  +  K_{P}} \frac{1}{s} = \frac{1}{K_{P}} = \frac{1}{2.5} = 0.4  rad
Thus the system has a nonzero disturbance error for a step disturbance.
b. Applying the final value theorem to the error equation (2) with \Theta_{r}(s) = 1/s^{2} and T_{d} (s) = 0 gives
e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{s 10s^{2}  +  2s}{10s^{2}  +  (2  +  K_{D})s  +  K_{P}} \frac{1}{s^{2}} = \frac{2}{K_{P}} = 0.8  rad
Thus the ramp command error is nonzero.
c. The disturbance transfer function is
\frac{\Theta_{r}(s)}{T_{d} (s)} = − \frac{1}{10s^{2}  +  (2  +  K_{D})s  +  K_{P}} = − \frac{1}{10s^{2}  +  10s  +  2.5}
Its bandwidth is the frequency range 0 ≤ ω ≤ 0.32 rad/s. Its low-frequency gain is 1/2.5 = 0.4 or 20 log 0.4 = −7.86 dB. Thus the system will not respond very much to disturbances that have a frequency content higher than 0.32 rad/s.
d. Substituting the values of K_{P} and K_{D} into the actuator equation (3) and using \Theta_{r}(s) = 1/s and T_{d} (s) = 0, we obtain
T (s) = \frac{(10s^{2}  +  2s)(8s  +  2.5)}{10s^{2}  +  10s  +  2.5} \frac{1}{s} = \frac{80s^{2}  +  41s  +  5}{10s^{2}  +  10s  +  2.5}       (5)
Because the orders of the numerator and denominator are equal, this can be expressed as
T (s) = C_{1} + \frac{C_{2}s  +  C_{3}}{10s^{2}  +  10s  +  2.5}
which may be arranged as a single fraction as follows.
T (s) = \frac{10C_{1}s^{2}  +  (10C_{1}  +  C_{2})s  +  2.5C_{1}  +  C_{3}}{10s^{2}  +  10s  +  2.5}          (6)
Comparing the numerators of equations (5) and (6), we find that C_{1} = 8, C_{2} = −39, and C_{3} = −15, so that
T (s) = 8  −  \frac{39s  +  15}{10s^{2}  +  10s  +  2.5} = 8  −  \frac{3.9s  +  1.5}{(s  +  0.5)^{2}}

This could also have been obtained with synthetic division. Using the inverse transform we obtain
T (t) = 8δ(t)  −  3.9e^{−0.5t} + 0.45te^{−0.5t }
where δ(t) is the unit impulse function. Therefore, the actuator output predicted by the model will contain an impulse of strength 8 at t = 0. Of course, this is impossible physically, and so we must view these  results with skepticism. The impulse is caused by the derivative term K_{D^{s}} in the control algorithm.