Compensators And Controllers Practice Questions

$G(S) C(s)=\frac{1}{S^{2}} \frac{k(S+\alpha)}{(S+4)}$ Characteristic equation is $S^{3}+4 S^{2}+k S+\alpha=0$ Poles of the system present at $=-1 \pm j \sqrt{3}$ Characteristic equation is

Compare equation (I) and (II)

Sufficient:

,

$\therefore$ For $K_{p}=4, K_{I}$ is maximum, which is $K_{I}=\frac{36+32 \times 4-64}{32}=3.125$ For $K_{p}=4, K_{I} \lt 3.125$ for stability $\therefore K_{I \max }=3.125$

In_ case of high type systems Lag compensator fails to give stability.

Phase lag controller transfer function is $G_{C}(s)=\frac{s+Z}{s+P} ;|Z|\gt|P|$

$G_{C}(s)=\frac{s+2}{s+4} \quad\dots(i)$ For lead compensator, Transfer function $=\frac{1+s \tau}{1+\alpha s \tau} \quad\dots(ii)$ Where, $\tau=$ Lead time constant $=R_{1} C$ and $\alpha=\frac{R_{2}}{R_{1}+R_{2}}$ Comparing equation (i) and (ii), we get $\tau=\frac{1}{2} \quad$ and $\quad \alpha \tau=\frac{1}{4}$ or, $\alpha=\frac{1}{2}$ $\therefore$ RC time constant =0.5

For phase lead compensator

Given open loop transfer function is

Comparing equation (i) with standard transfer function $\xi \omega_{n}=\frac{3}{2}=1.5$ $\therefore$ 2% settling time $\tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{1.5}=2.67>2 \mathrm{sec}$ Thus, in order to make settling time $\left(\tau_{s}\right)$ less than 2 sec, PD controller should be used. Hence, option (C) is the correct choice. Where, $C(s)=2(s+4)$ $\therefore$ New transfer function $=T^{\prime}(G)$ $=\frac{C(s) G(s)}{1+C(s) G(s)}=\frac{2(s+4)}{s^{2}+3 s+2+2 s+8}$ or $T^{\prime}(s)=\frac{2(s+4)}{s^{2}+5 s+10}$ $\therefore \quad \tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{5 / 2}=\frac{4}{2.5}=1.6$

$G_{c}(s)=\left(\frac{s+a}{s+b}\right)$ Phase $P=\tan ^{-1}(\omega / a)-\tan ^{-1} \omega / b$ for lead comparators phase must be + ve. for this $\frac{\omega}{a} \gt \frac{\omega}{b} \Rightarrow a \lt b$ So $a=1, b=2$

For phase to be maximum,

Steady state error,

Currently no explanation available

In case of Q, $\quad Z=\frac{R_{2} C_{2} s+1}{C_{2} S+1}$ In case of R, $\quad Z=\frac{R_{2}}{R_{2} C_{2} s+1}$ Considering Q $\frac{V_{0}}{V_{i}}=-\frac{\left(R_{1} C_{1} S+1\right)}{R_{1}} \cdot \frac{\left(R_{2} C_{2} S+1\right)}{C_{2} S}$ Considering R , $\frac{V_{0}}{V_{i}}=-\frac{\left(R_{1} C_{1} S+1\right)}{R_{1}} \cdot \frac{R_{2}}{\left(R_{2} C_{2} s+1\right)}$ $\because$ Given that $R_{2} C_{2}>R_{1} C_{1}$ $\therefore \quad$ Considering R, controller is lag compensator. and Considering Q , Controller is PID controller.

Now characteristics eq. $(1+G(s) H(s))=0$

Comparing with standard second order eq.

Lead compensator is required for stability. Controllers given in option (A) and (B) are not lead compensator. From option (C) and (d), with option (D) system becomes unstable, hence option (C) is correct.

Maximum phase shift

For maximum phase shift

$\therefore$ For compensated system $\angle=180-120; \; 60^{\circ}$ (b) & (d) are lag network & for compensating lag network $K /s^{2}$ , a lead network is required $\therefore$ Putting $s=-2.5+4.33 j$ in (a) gives $\frac{K(s+3)}{s^{2}(s+9.9)}=\frac{0.5+j 4.33}{7.4+j 4.33}=53^{\circ} ; 60^{\circ}$ $\therefore$ Option (A) is the correct answer.

A PD controller does not effect the type of the system. It reduces the overshoot and increases the damping. It increases the bandwidth therefor SNR decreases, System becomes more prone to noise