Frequency Response Analysis Practice Questions

The open loop transfer function of the system $G(s) H(s)=\frac{K\left(\frac{s}{2}+1\right)}{s\left(\frac{s}{10}+1\right)}\qquad\ldots(i)$ where, $\left.26.02\right|_{\omega=0.1}=20 \log K-20 \log \omega\qquad\ldots(ii)$ Or, $\quad K=1.99 \simeq 2$ From equation (i) and (ii), we get $G(s)=\frac{20(s+2)}{2 s(s+10)}\qquad\ldots(iii)$ Steady state error for ramp input is $e_{s s}=\frac{A}{K_{V}} \text { where } K_{V}=\lim _{s \rightarrow 0} s G(s)$ From equation (iii), $K_{V}=\lim _{s \rightarrow 0} s \times \frac{10(s+2)}{s(s+10)}=2$ $\therefore \quad e_{s s}=\frac{1}{2}=0.5$

From the given transfer function,

$\therefore$ From equation(ii) and (iii) we get, $\text { Phase margin } \mathrm{PM}=180^{\circ}-135^{\circ}=45^{\circ}$

A $4^{\text {th }}$ order all-pole system means that the system must be having no zero or s-terms in numerator or and $s^{4}$ terms in denominator or 4 -poles. i.e. $H(s) \propto \frac{1}{s^{4}}$ One pole exhibits -20 dB/decade slope, so 4 pole exhibits a slope of -80 dB/decade.

$10 \mathrm{rad} / \mathrm{s}$ to $1 \mathrm{rad} / \mathrm{s}$ is 1 decade $32-(-8)=40 d B$ So, the slope is 40dB/decade it means there are two poles at origin, it means either option (B) or option (D) is correct put $\omega=1$ rad/sec in both the options. 20 \log \left[\frac{39.8}{(1)^{2}}\right]\=32 \mathrm{dB}$$20 \log \left[\frac{32}{(1)^{2}}\right]$=30.1 \mathrm{dB}$So, option (B) is correct option$=\frac{39.8}{s^{2}}$

Transfer function is

So output $Y(s)=\frac{\left(s^{2}+9\right)(s+2)}{(s+1)(s+3)(s+4)} \cdot \frac{\omega}{\left(s^{2}+\omega^{2}\right)}$ Now $\lim _{t \rightarrow \infty } y(t)=\lim _{s \rightarrow 0} S Y(s)=\lim _{s \rightarrow 0} s \frac{\left(s^{2}+9\right)(s+2)}{(s+1)(s+3)(s+4)} \cdot \frac{\omega}{\left(s^{2}+\omega^{2}\right)}$ this will be zero if $s^{2}+\omega^{2}$ will cancel $\left(s^{2}+9\right)$ term as only then final value theorem will be applicable. So at $\omega^{2}=9 \Rightarrow \omega=3 \mathrm{rad} / \mathrm{sec},$ steady state output will be zero.

For phase cross-over frequency, $\phi=-180^{\circ}$

G.M. in $\mathrm{dB}=20 \log 20=26 \mathrm{dB}$

From the transfer function,

Only option (A) satisfies the conditions.

$G(s)=\frac{K}{s(s+1)(s+20)}=\frac{K / 20}{s(1+s)\left(1+\frac{s}{20}\right)}$ Bode plot is in (1+sT) form

when $s=0, A=5$

For 2nd order system G.M. $=\infty$

Let $\frac{1}{s+1}$ be a lag network At $\omega=0, \text{Mag}=\infty , \angle=-\tan ^{-1} 0=0$ $\omega=\infty , \mathrm{Mag}=0, \angle=-\tan ^{-1} \infty =-90^{\circ}$ If in the direction of $\omega$ increasing phase shift is decreasing system is lag network.

since, G.M. is negative system is unstable. $\therefore$ P.M. is also negative. $\mathrm{P} \mathrm{M} .=-87.5^{\circ}$