GATE 2020 EC Q.55: solution and answer

GATE 2020 · ECQ.55 · 2 marks · NAT

Consider the following closed loop control system

[Figure]

where $G (s) = \frac{1}{s ( s + 1 )}$ and $C (s) = K \frac{s + 1}{s + 3}$ . If the steady state error for a unit ramp input is 0.1, then the value of $K$ is ________.

GATE 2020 EC question 55: Unity negative feedback loop. — Figure from the GATE 2020 EC paper. It is part of the question.

Answer and solution

ANSWER

$K = 30$

Official answer: 30Source: Official answer key, GATE 2020 EC (IIT Delhi)

Forward path. The feedback is unity (the output returns to the summer directly, with no block), so the open-loop transfer function is $C (s) G (s) = K \frac{s + 1}{s + 3} \cdot \frac{1}{s ( s + 1 )} = \frac{K}{s ( s + 3 )}$ The controller zero at $s = - 1$ cancels the plant pole at $s = - 1$ .
Stability comes first, because the final-value theorem only holds for a stable closed loop. Without the cancellation, the characteristic equation $1 + C (s) G (s) = 0$ is $s (s + 1) (s + 3) + K (s + 1) = (s + 1) (s^{2} + 3 s + K) = 0$ The cancelled mode at $s = - 1$ is in the left half-plane, and $s^{2} + 3 s + K$ has both roots in the left half-plane for every $K > 0$ . So the loop is stable for $K > 0$ , and $K = 30$ satisfies this.
System type. $C (s) G (s)$ has one pole at the origin, so the loop is type 1. A unit ramp therefore leaves a finite steady-state error $e_{ss} = \frac{1}{K _{v}}$ .
Velocity error constant: $K_{v} = s \to 0 lim s C (s) G (s) = s \to 0 lim \frac{K}{s + 3} = \frac{K}{3}$
Cross-check straight from the error signal: $E (s) = \frac{R ( s )}{1 + C ( s ) G ( s )} = \frac{1}{s ^{2}} \cdot \frac{s ( s + 3 )}{s ^{2} + 3 s + K}, e_{ss} = s \to 0 lim s E (s) = s \to 0 lim \frac{s + 3}{s ^{2} + 3 s + K} = \frac{3}{K}$ This agrees with $1/ K_{v}$ .
Apply the given condition: $e_{ss} = \frac{3}{K} = 0.1 \Rightarrow K = 30$

KEY IDEA

With unity feedback, the ramp error of a type-1 loop is $e_{ss} = 1/ K_{v}$ , where $K_{v} = lim_{s \to 0} s C (s) G (s)$ . The controller zero cancels the plant pole at $s = - 1$ , which leaves $C (s) G (s) = K / (s (s + 3))$ . That gives $K_{v} = K /3$ , so $e_{ss} = 3/ K$ . The cancelled mode sits at $s = - 1$ , in the left half-plane, so it cannot make the loop unstable.

WATCH OUT FOR

A common slip is to ignore the controller's pole at $s = - 3$ and take $K_{v} = K$ , which gives $K = 10$ . Another is to use the $(s + 1)$ factor after cancelling it, for example writing $K_{v} = K / (1 \cdot 3)$ with the wrong factor or keeping $s + 1$ in the denominator, which gives a different value. Some students also apply the final-value theorem without checking stability first. Here the loop is stable for every $K > 0$ , but you should state that check before using the theorem.

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GATE 2020 EC Q.55: solution and answer

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