GATE 2022 IN Q.53: solution and answer

GATE 2022 · INQ.53 · 2 marks · NAT

In a unity-gain feedback control system, the plant
$P (s) = \frac{0.001}{s ( 2 s + 1 ) ( 0.01 s + 1 )}$
is controlled by a lag compensator
$C (s) = \frac{s + 10}{s + 0.1}$
The slope (in dB/decade) of the asymptotic Bode magnitude plot of the loop gain at $ω = 3$ rad/s is ________ (in integer)

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Answer and solution

ANSWER

The asymptotic Bode magnitude slope of the loop gain at $ω = 3 rad/s$ is $- 60 dB/decade$ .

Official answer: -60Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Loop gain with unity feedback: $L (s) = C (s) P (s) = \frac{0.001 ( s + 10 )}{s ( s + 0.1 ) ( 2 s + 1 ) ( 0.01 s + 1 )}$
Put the compensator in time-constant form so every corner can be read directly: $C (s) = \frac{s + 10}{s + 0.1} = \frac{10 ( 0.1 s + 1 )}{0.1 ( 10 s + 1 )} = \frac{100 ( 0.1 s + 1 )}{10 s + 1}$ so $L (s) = \frac{0.1 ( 0.1 s + 1 )}{s ( 10 s + 1 ) ( 2 s + 1 ) ( 0.01 s + 1 )}$
Corner frequencies, in increasing order: pole at $ω = \frac{1}{10} = 0.1 rad/s$ (compensator), pole at $ω = \frac{1}{2} = 0.5 rad/s$ from $(2 s + 1)$ , zero at $ω = \frac{1}{0.1} = 10 rad/s$ (compensator), pole at $ω = \frac{1}{0.01} = 100 rad/s$ .
Starting slope: there is one pole at the origin (type 1), so the asymptote starts at $- 20 dB/decade$ .
Asymptotic Bode rule: each simple real pole adds $- 20 dB/decade$ and each simple real zero adds $+ 20 dB/decade$ beyond its corner frequency.
Slope sequence: $- 20$ for $ω < 0.1$ ; $- 40$ on $0.1 < ω < 0.5$ ; $- 60$ on $0.5 < ω < 10$ ; $- 40$ on $10 < ω < 100$ ; $- 60$ for $ω > 100$ (all in $dB/decade$ ).
$ω = 3 rad/s$ lies between $0.5$ and $10 rad/s$ , where the slope is $- 60 dB/decade$ . The constant gain only shifts the whole plot up or down, so it has no effect on the slope.

KEY IDEA

An asymptotic Bode slope at a given frequency depends only on how many poles and zeros have corner frequencies below it: start from $- 20 N$ for $N$ poles at the origin, then add $- 20$ for each pole corner and $+ 20$ for each zero corner already passed. Here, by $ω = 3$ , the integrator and the poles at $0.1$ and $0.5$ have all acted, and the zero at $10$ has not yet, so $- 20 - 20 - 20 = - 60 dB/decade$ .

WATCH OUT FOR

Reading the compensator corners wrongly: in $\frac{s + 10}{s + 0.1}$ the pole is at $0.1 rad/s$ and the zero is at $10 rad/s$ (a lag network puts its pole first), not the other way round. Swapping them would place the zero below $ω = 3$ and give $- 20 dB/decade$ . Another slip is taking the $(2 s + 1)$ corner to be $2 rad/s$ instead of $0.5 rad/s$ . That does not change this answer, because both values are below 3, but it gives wrong slopes elsewhere. Also, the constant $0.001$ has no effect on slope.

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GATE 2022 IN Q.53: solution and answer

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