Q41GATE 2005MCQ

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

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📖 Explanation

System is stable in region -0.2 to -2 & on the left side of -8 as number of encirclement there is zero.

\begin{aligned} 0.2 K \lt 1 &\Rightarrow K \lt 5 \\ 2 K \gt 1 &\Rightarrow K \gt 0.5 \\ \therefore 0.5 \lt K& \lt 5\\ 1 \gt 8 &K \end{aligned}

$K \lt \frac{1}{8}$ (negative sign only shows that it is on negative axis)

Q43GATE 2004MCQ

Consider the Bode magnitude plot shown in the fig. The transfer function H(s) is

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📖 Explanation

\begin{aligned} 20 \log K &=-20 \mathrm{dB} \\ \Rightarrow \quad K &=10^{-1}=0.1 \\ \text { Zero at } \omega=1 & \& \text { poles at } \omega=10,100 \\ H(s) &=\frac{K(s+1)}{\left(\frac{s}{10}+1\right)\left(\frac{s}{100}+1\right)} \\ &=\frac{0.1(s+1) \times 10 \times 100}{(s+10)(s+100)} \\ &=\frac{10^2 (s+1)}{(s+10)(s+100)} \end{aligned}

Q44GATE 2003MCQ

Fig. shows the Nyquist plot of the open-loop transfer function G(s)H(s) of a system. If G(s)H(s) has one right-hand pole, the closed-loop system is

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📖 Explanation

The encirclement of critical point (-1,0) in A.C direction is once.

\begin{aligned} \therefore \quad N&=1, P=1 &\text{(given)}\\ Z&=P-N=O \end{aligned}

No zero in RH of s-plane So system is stable.

Q45GATE 2003MCQ

The gain margin and the phase margin of feedback system with $G(s)H(s)=\frac{s}{(s+100)^{3}}$ are

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📖 Explanation

\begin{aligned} G(s) H(s)&=\frac{s}{(s+100)^{3}}\\ \text{Calculating } \omega_{\mathrm{pc}}, \\ -180^{\circ}&=+90-3 \tan ^{-1} \frac{\omega_{p c}}{100} \\ -270^{\circ}&=-3 \tan ^{-1} \frac{\omega_{p c}}{100}\\ \text{or, }\quad\omega_{p c}&=\infty \\ \therefore \quad\mathrm{G} \cdot \mathrm{M} \cdot&=\infty \end{aligned}

Hence cannot be determined. Calculating $\omega_{g c}$ $\text { From, }|G(s) H(s)|=1$ $\omega_{g c}=$ negative value G.M. and P.M. of the system cannot be determined.

Q46GATE 2003MCQ

The approximate Bode magnitude plot of a minimum phase system is shown in Fig. below. The transfer function of the system is

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📖 Explanation

$\omega=0.1 to 10,+120 \mathrm{dB}$ change $\therefore$ 3 zeroes at 0.1 $\omega=10 \text { to } 100,-40 \mathrm{dB}$ [i.e. +60 to +20] change So two poles at $\omega=10$ $\omega=100,-20 \mathrm{dB}$ change. $\therefore$ one pole at $\omega=100$ $\therefore T(s)=\frac{K(s+0.1)^{3}}{(s+10)^{2}(s+100)}$ $\left.20 \log (K / \omega)\right|_{\omega=10}=140$ $\left.\frac{K}{\omega}\right|_{\omega=10}=10^{7}$ $\therefore \quad K=10^{8}$

Q47GATE 2002MCQ

The phase margin of a system with the open-loop transfer function $G(s)H(s)=\frac{(1-s)}{(s+1)(s+2)}$ is

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📖 Explanation

\begin{aligned} \mathrm{PM}&=180^{\circ}+\left.\angle \mathrm{GH}(j \omega)\right|_{\omega_{9 c}} \\ \text { For } \omega_{\mathrm{gc}^{\prime}} &|G(s) H(s)|_{\omega=\omega_{\infty }}=1\\ \left|\frac{1-s}{(1+s)(2+s)}\right|&=\frac{\sqrt{1+\omega^{2}}}{\sqrt{1+\omega^{2} \sqrt{4+\omega^{2}}}}&=1 \\ \sqrt{4+\omega^{2}}&=1 \\ \Rightarrow \quad 4+\omega^{2}&=1 \\ \omega^{2}&=-3 \text{ (imaginary)}\\ \end{aligned}

So no gain crossover frequency $\therefore PM=\infty$

Q48GATE 2002MCQ

The system with the open loop transfer function $G(s)H(s)=\frac{1}{s(s^{2}+s+1)}$ has a gain margin of

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📖 Explanation

$G(s) H(s)=\frac{1}{s\left(s^{2}+s+1\right)}$ $\omega$ when $\angle G(s) H(s)=-180^{\circ}$

\begin{aligned} -180 &=-90-\tan ^{-1} \frac{\omega}{1-\omega^{2}} \\ -90 &=-\tan ^{-1} \frac{\omega}{1-\omega^{2}} \\ 1-\omega^{2} &=0 \\ \omega_{\alpha c} &=1 \mathrm{rad} / \mathrm{sec} \end{aligned}

Value of gain at $\omega_{\rho c}=1$

\begin{aligned} |G(s) H(s)| &=\frac{1}{\sqrt{\left(\left(1-\omega^{2}\right)^{2}+|j \omega|^{2}\right.}} \\ \therefore \quad G \cdot M \cdot &=-20 \log 1=0 \end{aligned}

Q49GATE 2001MCQ

The Nyquist plot for the open-loop transfer function G(s) of a unity negative feedback system is shown in figure. If G(s) has no pole in the right half of splane, the number of roots of the system characteristic equation in the right half of s-plane is

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📖 Explanation

$N=0$ (1 encirclement in CW direction and other in ACW) $P=0$ (no pole in right half) So, $N=P-Z$ $Z=P-N=0$ $\therefore \quad$ No roots on RH of s-plane.