Question

5, (29%) Consider the feedback control system in Figure-5 in block diagram form. The reference input R(s), system output Y(s), and disturbance D(s) are denoted along with the error E(s) and control effort F(s). You will design the control law Gc(s) to achieve certain performance criteria. Answer the following questions (assume D(s)0 in all parts except part(ü) (a) [396] Show that the transfer function relating the reference R(s) to the output Y(s) is given by (b) [3%) Assuming a proportional control law, Gc(s)-Kp, choose Kp to limit the steady state error (c) [396] For the proportional control law, Gc(s)-Kp choose Kp so that 63% of the steady state Y(syR(s) GsIs+0.1+Gc(s) between the reference and the output to 5% for a unit step reference. (d) [296] Explain why you cannot achieve both criteria listed in (b) and (c) using a proportional control (e) [396] For an integral control law, Gc(s)-KUS, find the steady state error between the reference (f) [3%] Choose Ki in the integral control law from part (e) to limit the peak overshoot to 10%. response for a unit step input is reached in exactly 1 second law. What control law would you use (and why) to achieve both of these criteria? and the output for a unit step reference (g)(5%) Assuming a proportional-integral control law, Gc(s)-Kp+Ki/s, choose Ki and Kp to simultaneously limit the peak overshoot to 10% and settling time (for response to be within 2% of the final value) to 1 second (h) [396] Show that the transfer function relating the disturbance D(s) to the error E(s) is given by, E(SVD(s)1s+0.1+Gc(s)

Answer 1

(a):

Let the plant transfer function be Gp.

E(s) = R(s)-Y(s);

F(s) = Gc(s)*E(s)

Y(s) = Gp(s)*(D(s)+F(s)) = Gp(s)*(D(s)+ Gc(s)*E(s))  = Gp(s)*(D(s)+ Gc(s)*R(s)- Gc(s)*Y(s))

= Gp(s)*D(s)+ Gp(s)*Gc(s)*R(s)- Gp(s)*Gc(s)*Y(s)

=>Y(s)(1+Gp(s)*Gc(s)) = Gp(s)*D(s)+ Gp(s)*Gc(s)*R(s)

$=>Y(s) = \frac{Gp(s)}{1+Gp(s)*Gc(s)}*D(s)+ \frac{Gp(s)*Gc(s)}{1+Gp(s)*Gc(s)}*R(s)$

Transfer function from r to y is:

$=>\frac{Y(s) }{R(s)}= \frac{Gp(s)*Gc(s)}{1+Gp(s)*Gc(s)}$

$Gp(s) = \frac{1}{s+0.1}$

$=>\frac{Y(s) }{R(s)}= \frac{Gc(s)}{s+0.1+Gc(s)}$

(b):

$=>\frac{Y(s) }{R(s)}= \frac{Kp}{s+0.1+Kp}$

Steady state value for a unit step input can be obtained by setting the parameter 's' in the transfer function to 0.

Therefore, the steady state values is:

$\frac{Kp}{0.1+Kp}$

Steady state error is:

$1-\frac{Kp}{0.1+Kp}=\frac{0.1}{0.1+Kp}$

For this value to be with in 5% and for the system to be stable,

$\frac{0.1}{0.1+Kp}<0.05$

=> Kp>1.9

(c):

$=>\frac{Y(s) }{R(s)}= \frac{\frac{Kp}{0.1+Kp}}{\frac{s}{0.1+Kp}+1}$

So, the time constant is $\frac{1}{0.1+Kp}$

which is the time required to reach 63%. For this to be equal to 1,

Kp = 0.9.

(d):

For the steady state error to be less, the value of the proportional controller should be high to have enough influence and reduce the error. But this would also lead to faster response of the system. This means the time constant would be small. Therefore, it is not possible to satisfy both the conditions in this case.

An integral controller in addition to the proportional part would solve the issue. The proportional part which has a major influence on time constant can be reduced, where as the integral part can reduce the steady state error.

(e):

$=>\frac{Y(s) }{R(s)}= \frac{Ki}{s^2+0.1s+Ki}$

the steady state values is:

$\frac{Ki}{Ki} = 1$

Therefore, the steady state error is 0.

(f):

The characteristic equation is s²+0.1s+Ki = 0. Comparing it to the standard equation gives,

$\zeta = \frac{0.1}{2\sqrt{Ki}} = \frac{0.05}{\sqrt{Ki}}$

The overshoot is given by:

$M_P(\%) = e^{\frac{-\pi \zeta}{\sqrt{1-\zeta^2}}}*100$

$=> \zeta = 0.591$

=> Ki = 0.00715

(g):

$=>\frac{Y(s) }{R(s)}= \frac{Kp*s+Ki}{s^2+0.1s+Kp*s+Ki}= \frac{Kp*s+Ki}{s^2+(0.1+Kp)s+Ki}$

$=>\zeta = \frac{0.1+Kp}{2\sqrt{Ki}}$

Settling time for 2% range is approximately given by:

$T_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.5913*\sqrt{Ki}}$

T_s = 1 => Ki = 45.762

(h):

$=> E(s) = \frac{1}{1+Gp*G}R(s) - \frac{Gp}{1+Gp*Gc}*D(s)$

The transfer function between E and D is:

$=> \frac{E(s)}{D(s)} = - \frac{Gp}{1+Gp*Gc}= \frac{-1}{s+0.1+Gc}$