Frequency Response Analysis Practice Questions

From the given Bode plot, it is clear that corner frequencies, $\omega _{c1}=\frac{1}{3}$ and $\omega _{c2}=1$ $\therefore$ Transfer function of given system is given by $T(s)=\frac{C(s)}{R(s)}=\left[ \frac{1+3s}{1+s} \right]=\left[ \frac{1+Ts}{1+\alpha Ts} \right]$ Here, $T=3$ and $\alpha T=1,$ $\alpha =\frac{1}{3}$ As $\alpha \lt 1$ , therefore the transfer function T(s) represents a lead compensator having $\alpha =\frac{1}{3}$ $\therefore$ Maximum phase shift, $\phi _{m}=tan^{-1}\left[ \frac{1-\alpha }{2\sqrt{\alpha }} \right]$ $=tan^{-1}\left[ \frac{1-\frac{1}{3}}{\frac{2}{\sqrt{3}}} \right]$ $=tan^{-1}\left[ \frac{2}{3}\times \frac{\sqrt{3}}{2} \right]$ $=tan^{-1}\left[ \frac{1}{\sqrt{3}} \right]=30^{\circ}$ $\therefore \;\; \phi _m=30^{\circ}$ Also, $T(j\omega )=\frac{1+j3\omega }{1+j\omega }$ $\therefore \;\; |T(j\omega )|=\frac{\sqrt{1+9\omega ^2}}{\sqrt{1+\omega ^2}}$ $\therefore \;\; Gain G_m$ in dB $=20log |T(j\omega )|$ $=20 log_{10}\left[ \frac{\sqrt{1+9\omega ^2_m}}{1+\omega ^2_m} \right]$ Now, $\omega _m=\frac{1}{T\sqrt{\alpha }}=\frac{1}{\sqrt{3}}$ $\therefore \;\; G_m=20 log_{10}\left[ \sqrt{\frac{1+9\times \left ( \frac{1}{\sqrt{3}} \right )^2}{1+\left ( \frac{1}{\sqrt{3}} \right )^2}} \right]$ $=20 log_{10}\left[ \sqrt{\frac{1+3}{1+\frac{1}{3}}} \right]=20 log_{10}\sqrt{3}$ $G_m=4.77dB$

Given, $G(s)=\frac{5(s+4)}{s(s+0.25)(s^{2+}4s+25)}$ Reducing G(s) in time constant form, we have: G(s)=\frac{5 \times 4\left ( 1+\frac{s}{4} \right )}{s\left[ 0.25\left ( 1+\frac{s}{0.25} \right ) \right]\left[ 25\left ( 1+\frac{4}{25}s+\frac{1}{25}s^2 \right ) \right]\}$$=\frac{20(1+0.25s)}{s[0.25(1+4s)]$[25(1+0.16s+0.04s^2)]$}$$G(s)=\frac{3.2(1+0.25s)}{s(1+4s)(1+0.16s+0.04s^2)}$$\therefore$The value of constant gain term =3.2 Also,$G(j\omega )=\frac{3.2\left ( 1+\frac{j\omega }{4} \right )}{(j\omega)\left ( 1+\frac{j\omega}{0.25} \right )\left ( 1+j0.16\omega -\frac{\omega ^2}{25} \right )}$Corner frequencies are:$\omega _1=4$rad/sec$\omega _2=0.25$rad/sec$\omega _3=5$rad/sec$\therefore$Highest corner frequency$=\omega _3=5$ rad/sec

Let, $G(s)=\frac{K}{s^n}$ $G(j\omega )=\frac{K}{(j\omega )^n}$ $\because$ Gain=32 dB at $\omega =1$ rad/sec $\therefore \; 20log\left | \frac{K}{1^n} \right |=32$ $\Rightarrow \; 20 log K=32$ $\Rightarrow \; K=39.8$ $Now,$ $\because \;$ Gain=-8 dB at $\omega =10$ rad/sec $\therefore \; 20log\left | \frac{39.8}{10^n} \right |=-8$ $\Rightarrow \; 10^n=100$ $\Rightarrow \; n=2$

Fpr sinusoidal excitation, $s=j\omega$ $\therefore \;\;G(j\omega )=\frac{(-\omega ^{2+}9)(j\omega +2)}{(j\omega +1)(j\omega +3)(j\omega +4)}$ For zero steady state output $|G(j\omega )|=0$ $\;\;\;=\frac{(-\omega ^{2+}9)\sqrt{\omega ^{2+}4}}{(\omega ^{2+}1)(\sqrt{\omega ^{2+}9})(\sqrt{\omega ^{2+}16})}$ For zero steady state output, $\Rightarrow \; \omega ^2=9$ $\Rightarrow \; \omega =3 \; rad/sec$

At gain crossover frequency $(\omega_{gc})$ , magnitude of $G(j\omega)$ is 1. $|G(j\omega_{gc})|=1$ Phase of $G(j\omega )=\angle G(j\omega_{gc})=-150^{\circ}$ Phase margin $=180^{\circ}+\angle G(j\omega_{gc})$ $\;\;\; =180^{\circ}-150^{\circ}=30^{\circ}$ At phase cross frequency $(\omega_{pc})$ , phase of $G(j\omega )$ is $-180^{\circ}$ , $\angle G(j\omega_{pc})=-180^{\circ}$ M=magnitude of $G(j\omega )$ at $\omega _{pc}$ $=|G(j\omega_{pc})|=0.5$ Gain margin $=20log\frac{1}{M}$ $=20 log \frac{1}{0.5}=6 dB$

Currently no explanation available

Two clockwise encirclement of 1+j0. N=-2 Open-loop system is stable $\Rightarrow \; P=0$ N=P-Z -2=0-Z Z= Number of closed loop poles in RHS of 8-plane. Hence the system is unstable.

Initial slope =-40 dB/dec. Hence the system is type-2. So the corresponding term of the transfer function is $1/s^2$ . At $\omega =2$ rad/sec, slope changes by -20 dB/dec from -40 dB/dec to -60 dB/dec. Hence the corresponding term of the transfer function is $\frac{1}{\left ( 1+\frac{s}{2} \right )}$ . At $\omega =3$ rad/sec, slope changes by 20 dB/dec from -60 dB/dec to -40 dB/dec. Hence the corresponding term of the transfer function is $\left ( 1+\frac{s}{5} \right )$ . At $\omega =25$ rad/sec, slope changes by -20 dB/dec from -40 dB/dec to -60 dB/dec. Hence the corresponding term of the transfer function is $\frac{1}{\left ( 1+\frac{s}{25} \right )}$ . $G(s)=\frac{k\left ( 1+\frac{s}{5} \right )}{s^2\left ( 1+\frac{s}{2} \right )\left ( 1+\frac{s}{25} \right )}$ At $\omega =0.1(\omega \leq 2 \;rad/sec)$ $Gain(dB)=20 \log k-40 \log \omega$ $80=20 \log k-40 \log 0.1$ $20 \log k=40$ $k=100$ $G(s)=\frac{100 \left ( 1+\frac{s}{5} \right )}{s^2\left ( 1+\frac{s}{2} \right )\left ( 1+\frac{s}{25} \right )} =\frac{1000(s+5)}{s^2(s+2)(s+25)}$

Open loop transfer function, $G(s)=\frac{e^{-0.1s}}{s}$ Put, $s=j\omega$ $G(j\omega)=\frac{e^{-j0.1\omega}}{j\omega}$ At phase crossover frequency $(\omega_{pc})$ , phase of OLTF is $-180^{\circ}$ $\angle G(j\omega)|_{\omega/\omega_{pc}}=-180^{\circ}=-\pi$ $-\frac{\pi}{2}-0.1\omega_{pc}=-\pi$ $-0.1\omega_{pc}=-\frac{\pi}{2}$ $\omega_{pc}=5 \pi$ $|e^{-0.1\omega}|$ is always 1 for any value of $\omega$ $|G(j\omega)|=\frac{1}{\omega }$ Gain at $\omega_{pc}$ ( phase-cross frequency) $|G(j\omega)|_{\omega_{pc}=5 \pi}=\frac{1}{5 \pi }$ Gain margin $=20 log \frac{1}{|G(j\omega_{pc}|}$ $\;\;\;=20 log 5\pi=23.9 dB$

Initial slope is -40 dB/decade, it means there are double pole at origin. Slope changes from -40 dB/decade to -20 dB/decade. It means there is a zero. Slope changes from -20 dB/decade to -0 dB/decade at some other frequency.eans ther is one more zero. Therefore transfer function has two poles and two zeros.

$G(s)=\frac{1}{s(s+1)(s+2)}$ Put $s=j\omega,$ $G(j\omega)=\frac{1}{j\omega(j\omega+1)(j \omega+2)}$ $\;\;\;=\frac{-j(1-j\omega)(2-j\omega)}{\omega (1+\omega ^2)(4+\omega ^2)}$ $\;\;\;=\frac{-j(2-3j\omega -\omega ^2)}{\omega (1+\omega ^2)(4+\omega ^2)}$ $\;\;\;=\frac{-3\omega -j(2-\omega ^2)}{\omega (1+\omega ^2)(4+\omega ^2)}$ $\;\;\;=\frac{-3}{(1+\omega ^2)(4+\omega ^2)}+\frac{ -j(\omega ^{2-}2)}{\omega (1+\omega ^2)(4+\omega ^2)}$ $=x+iy$ At $\omega =0, \; x\rightarrow -3/4,\; y\rightarrow -\infty$

$H(j\omega )=\frac{10^4(1+j\omega )}{(10+j\omega)(100+j\omega)^2}$ $H(j\omega )=\frac{(1+j\omega)}{10\times \left ( 1+\frac{j\omega}{10} \right )\times \left ( 1+\frac{j\omega}{100} \right )^2}$ $H(j\omega )=\frac{k(1+j\omega)}{ \left ( 1+\frac{j\omega}{10} \right ) \left ( 1+\frac{j\omega}{100} \right )^2}$ $where \;\; k=\frac{1}{10}=0.1$ Corner frequencys $\omega _1= 1$ rad/sec; $\omega _2= 10$ rad/sec and $\omega _3=100$ rad/sec For frequency less than $\omega _1 \; i.e. \; \omega \lt \omega _1$ Gain of the system is constant aas there is no pole at origin. Gain $=20 log k=20 log 0.1=-20 dB$ At $\omega =\omega _1= 1$ rad/sec or $\omega _1= log 1 =0$ There is zero, so system gain increases with slope +20dB/decades and system gain become 0 dB at $\omega = 10$ rad/sec or $log\omega |_{\omega =10}=log 10=1$ At $\omega _2 =10$ or $log \omega |_{\omega _2=10}=log 10=1$ there is pole, so slope is -20 dB/decade. Overall slope $\omega _2 \lt \omega \lt \omega _3$ = 20 dB/decade - 20 dB/decade = 0 dB/decade So, gain remain constant between $\omega _2 \lt \omega \lt\omega _3$ or $1 \lt log \omega \lt 2$ At $\omega =\omega _3 =100$ rad/sec or $log\omega |_{\omega _3=100}=log 100=2$ the double pole are present. So, system gain decrease with -40 dB/decade.

Currently no explanation available

$T(s)=\frac{s^{2+}4}{(s+1)(s+4)}$ For frequency response, put $s=j\omega$ $T(j\omega )=\frac{(j\omega )^{2+}4}{(j\omega +1)(j\omega +4)}=\frac{4-\omega ^2}{(1+j\omega )(4+j\omega )}$ Magnitude of $T(j\omega )=|T(j\omega )|=\frac{4-\omega ^2}{\sqrt{(1+\omega ^2)(16+\omega )^2}}=0$ $\omega =2$ rad/sec The system output is zero at 2 rad/sec.

$-1+j0$ is not enclosed , system is always stable. $\therefore \;\; G.M.=\infty$

$G(s)=\frac{k+0.366s}{s(s+1)}$ and $H(s)=1$ $OLTF=G(s)H(s)=\frac{k+0.366s}{s(s+1)}$ Phase margin= $60^{\circ}$ at $\omega _{gc}$ (gain crossover frequency)

$G(s)H(s)=\frac{\pi e^{-0.25s}}{s}$ Putting $s=j \omega$ , for frequency response $G(j\omega )H(j\omega )=\frac{\pi e^{-j0.25 \omega }}{j\omega }$ Phase angle $=\angle G(j\omega )H(j\omega )$ $\;\;\;=-90-0.25\omega$ at $\omega _{pc}$ nyquist plot cuts negative real axis and this frequency $\angle G(j\omega _{pc} )H(j\omega _{pc})$ is $-180^{\circ}$

$x(t)=\sin t =1 \lt 0$ $\omega =1 \;rad/sec$ Transfer function $=G(s)=\frac{s}{s+1}$ For sinusoidal input, $s=j\omega =j$ $G(j\omega )|_{\omega =1}=\frac{j\omega }{j\omega +1}=\frac{j}{j+1}=\frac{1}{\sqrt{2}}\angle 45^{\circ}$ $y(t)=\frac{1}{\sqrt{2}}\angle 45^{\circ}.1 \angle 0^{\circ}=\frac{1}{\sqrt{2}}\angle 45^{\circ}$ $y(t)=\frac{1}{\sqrt{2}} \sin (t+45^{\circ})$

$\omega _g$ is gain cross over frequency at which the gain $|G(j\omega) H(j\omega)|$ becomes unity. In this case, phase $\angle G(j\omega) H(j\omega)$ is $-180^{\circ}$ at $\omega =\omega _g$ . $\phi =-180^{\circ}$ So phase margin $=180^{\circ}+\phi =0^{\circ}$