Compensators and Controllers

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|

For phase lead compensator
\begin{aligned} G_{c}(s)&=\frac{(1+s \tau)}{(1+\alpha s \tau)} \quad ; \quad \alpha \lt 1\\ \text{Here,}\quad \tau&=\frac{1}{a}\\ \text{and}\quad \alpha \tau&=\frac{1}{b}\\ \text{or,}\quad \alpha&=\frac{a}{b} \lt 1\\ \text{or,}\quad a& \lt b \end{aligned}

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

Given open loop transfer function is
\begin{aligned} G(s)&=\frac{1}{(s+1)(s+2)} \\ \therefore T(s)&=\frac{G(s)}{1+G(s)}=\frac{1}{(s+1)(s+2)+1} \\ T(s)&=\frac{1}{s^{2}+3 s+3} &\ldots(i) \end{aligned}
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

For phase to be maximum,
\begin{aligned} \frac{\partial P}{\partial \omega}&=0 \\ \frac{1}{1+\frac{\omega^{2}}{a^{2}}}- \frac{1/b}{1+\frac{\omega^{2}}{b^{2}}}&=0 \\ \frac{1}{1+\omega^{2}}- \frac{1/b}{1+\frac{\omega^{2}}{4}}&=0\\ \frac{1}{1+\omega^{2}}-\frac{2}{\omega^{2}+4} &=0 \\ \omega^{2}+4-2-2 \omega^{2} &=0 \\ \omega^{2} &=2 \\ \omega &=\sqrt{2} \mathrm{rad} / \mathrm{sec} \end{aligned}

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

Steady state error,
\begin{aligned} e_{s s} &=\lim _{s \rightarrow 0} \frac{s R(s)}{1+G(s) G_{c}(s)} \\ r(t) &=u(t) \\ R(s) &=\frac{1}{s} \\ e_{ss} &=\lim _{s \rightarrow 0} \frac{s \cdot \frac{1}{s}}{1+G(s) G_{c}(s)} \\ e_{ss}&=\lim _{s \rightarrow 0} \frac{1}{1+G(s) G_{c}(s)} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+1}{s+2}, e_{s s}=\frac{2}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+2}{s+1}, e_{s s}=\frac{1}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{(s+1)(s+4)}{(s+2)(s+3)}, e_{s s}=\frac{3}{5} \\ \text { Taking, } \quad & G_{c}(s)=1+\frac{2}{s}+3 s, e_{s s}=0 \end{aligned}

\begin{array}{c} \frac{V_{i}\left(R_{1} C_{1} S+1\right) 1}{R_{1}}=-\frac{V_{0}}{Z} \\ \Rightarrow \frac{V_{0}}{V_{i}}=-\frac{Z\left(R_{1} C_{1} S+1\right)}{R_{1}} \end{array}
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.