Frequency Response Analysis Practice Questions

Initial slppe $=-20 \mathrm{~dB} / \mathrm{sec}$ Open loop transfer function $=\frac{k}{s}$

$e_{s s}$ for type -1 system for ramp input $e_{s S}=\frac{1}{k_{v}}=\frac{1}{10}=0.1$

Nyquist Contour : Given:

$K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta}}$ Max. phase lead occur at, $\omega_{m}=\sqrt{a(a \beta)}$ $=\sqrt{\beta}$ Given : $\quad \omega_{\mathrm{m}}=4 \mathrm{rad} / \mathrm{sec}$ $a \sqrt{\beta}=4 \quad ...(1)$ Now, $\quad K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}$ Put, $s=j \omega$ $\mathrm{K}\left(\mathrm{j} \omega_{\mathrm{n}}\right)=\frac{1+\frac{\mathrm{j} \omega_{m}}{a}}{1+\frac{j \omega_{m}}{\beta a}}=\frac{j w_{m}}{a}$ Given : $M=6$

From eqn. (1), we get

We knaow, $y(t)=A|G(j\omega ) \sin (\omega t+\phi )$ Here, $G(j\omega )=\frac{100}{-\omega ^{2+}j0.1\omega +100}$ Put $\omega =0$ $|G(j\omega )|=1$ Now, put $\omega =10$ rad/sec $|G(j\omega )|=\left | \frac{100}{-100+i1+100} \right |=100$ Therefore, $y(t)=1+0.1 \times 100 \sin(10t+\theta ) =1+10 \sin (10t+\theta )$ On comparision $: a=1 ,b=10$

Given: P = 1 (Because, Nyquist contour encircle one pole i.e s = 0) We have, N = P - Z N = 1-Z Characteristic equation

R-H criteria:

Hence, z = 0 (because no sign change in first column of R-H criteria) (where, z = closed loop pole on RHS side of s-plane) Therefore, N = 1

The given system is non-minimum phase system Therefore, transfer function, $T.F=\frac{s-1}{s+1}$ Hence, one LHP pole and one RHP zero at the same frequency.

From response plot

From electrical network

$G(s)=\frac{k}{s\left ( 1+\frac{s}{1} \right )\left ( 1+\frac{s}{20} \right )}$ Transfer function shows 2 poles and no zeros. So statement I is false. $\angle G(j\omega )=-90-tan^{-1}\omega -tan^{-1}\frac{\omega }{20}$ $\angle G(j\omega )|_{\omega \rightarrow \infty }=-270^{\circ}=-\frac{3\pi}{2} \;rad$ So statement II is true.

$\angle G(j\omega )=-0.25 \times \frac{180\omega _{pc}}{\pi}-90^{\circ}=-180^{\circ}$ $-0.25 \times \frac{180\omega _{pc}}{\pi}=-90^{\circ}$ $\omega _{pc}=\frac{4\pi}{2}=2\pi$ $|G(j\omega )|_{\omega =\omega _{pc}}=\left | \frac{\pi e^{-0.25s}}{s} \right |_{\omega =\omega _{pc}}$ $=\frac{\pi}{2\pi}=0.5$ $\therefore$ Point is $(-0.5,j0)$

$\frac{V_o(s)}{V_i(s)}=\frac{1-s}{1+s}=H(s)$ For the minimum and maximum values of $'\phi '$ $H(j\omega )=\frac{1-j\omega }{1+j\omega }$ $\angle H(j\omega )=-2 tan^{-1}\omega$ At $\omega =0; \;\; \; \angle H(j\omega )=0^{\circ}$ At $\omega =\infty ; \;\;\; \angle H(j\omega )=-\pi$

Forward path transfer function, $G(s)=\frac{Ke^{-s}}{s}$ Given, Phase margin $=30^{\circ}$ Phase margin $=180^{\circ}+\phi$ $30^{\circ}=180^{\circ}+\phi$ $\phi =-150^{\circ}$ $\phi =\angle G(j\omega )_{\omega =\omega _{gc}}$ [where $\omega _{gc}$ is gain crossover frequency ] $\angle G(j\omega )=-90^{\circ}-57.3\omega ^{\circ}$ At $\omega =\omega _{gc}$ $|G(j\omega )|=1$ $\frac{k \times 1}{\omega }=1$ $\Rightarrow \omega =\omega _{gc}=K \; rad/sec.$ $\therefore \;\; \angle G(j\omega )_{\omega =\omega _{gc}}=-90^{\circ}-57.3K^{\circ}$ $-90^{\circ}-57.3K^{\circ}=-150^{\circ}$ $-57.3K^{\circ}=60^{\circ}$ $K=\frac{60}{57.3}=1.047$

Nyquist plot of $G(s)H(s)=\frac{s+3}{s^2(s-3)}$ is as shown below From the Nyquist plot G(s)H(s) encircle -1+j0 once in clockwise direction. Alternate Solution: Characteristic equation, $1+G(s)H(s)=0$ $s^2(s-3)+(s+3)=0$ $s^{3-}3s^{2+}s+3=0$ using Routh's array

There are two sign changes, hence two poles in right side of s-plane exist. $Z=2, \; P=1$ $N=P-Z=-1$ One encirclement in clockwise direction.

$G(s)=\frac{100}{(s+1)^3}$ $G(j\omega )=\frac{100}{(1+j\omega )^3}$ $\;\;=\frac{100}{1+(j\omega )^{3+}3(j\omega )^{2+}3j\omega }$ $\;\;=\frac{100}{(1-3\omega^2 )+j(3\omega-\omega^3)}$ \;\;=\frac{100\[(1-3\omega^2 )+j\omega(3-\omega^2)^2]$}{[ (1-3\omega^2 )+\omega^2(3-\omega^2)^2]$}$For phase corssover frequency$\omega_{ph} Img[G(j \omega )]=0;$Hence,$\omega (3-\omega ^2)=0$$\omega =0; \pm \sqrt{3}$Therefore,$\omega _{ph}=\sqrt{3}$ rad/sec

From the given Bode plot, it is evident that there are 3(three) poles in the transfer function, out of which there are double poles at corner frequency near but less than $\omega=8$ rad/sec and one pole is near but greater than $\omega=0.5$ rad/sec. The initial slope is +20 dB/dec. Therefore one zero exist at s=0. So from all the given options, option (A) satisfies all the conditions. Therefore Option (A) is correct.

$G_1(s)=\frac{1}{s}$ $G_2(s)=s$ $G_1(s) \times G_2(s)=\frac{1}{s}\cdot s=1$

From initial line equation, $20 dB=20log_{10}K$ as we know, $20 dB=20log_{10}K$ $K=10$ now, from the given plot we know type of system is one. Hence, $G(s)=\frac{K(1+sT_1)}{(1+sT_2)}$ $T_1=\frac{1}{10}; \;\;T_2=\frac{1}{1000}$ $G(s)=\frac{10(1+0.001s)}{(1+0.1s)} =\frac{(1000+s)}{10(s+10)}$