GATE 2018 EE Q.25: solution and answer

GATE 2018 · EEQ.25 · 1 mark · NAT

Consider a unity feedback system with forward transfer function given by

G (s) = \frac{1}{( s + 1 ) ( s + 2 )}

The steady-state error in the output of the system for a unit-step input is ________ (up to 2 decimal places).

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Answer and solution

ANSWER

The steady-state error for a unit-step input is $e_{ss} = \frac{1}{1 + K _{p}} = \frac{2}{3} \approx 0.67$ .

Official answer: 0.65 to 0.69Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Check stability first, because the final value theorem only applies to a stable closed loop. The characteristic equation is $1 + G (s) = 0$ , which gives $(s + 1) (s + 2) + 1 = s^{2} + 3 s + 3 = 0$ . For a second-order polynomial, all coefficients positive is necessary and sufficient for both roots to lie in the left half-plane. So the closed loop is stable.
Find the system type. $G (s)$ has no pole at $s = 0$ , so the system is type 0. A type-0 system tracks a step input with a finite, non-zero steady-state error.
Position error constant: $K_{p} = s \to 0 lim G (s) = \frac{1}{( 0 + 1 ) ( 0 + 2 )} = \frac{1}{2} = 0.5$
For unity feedback the error is $E (s) = \frac{R ( s )}{1 + G ( s )}$ , with $R (s) = \frac{1}{s}$ for a unit step. Apply the final value theorem: $e_{ss} = s \to 0 lim s \cdot \frac{1}{s} \cdot \frac{1}{1 + G ( s )} = \frac{1}{1 + K _{p}} = \frac{1}{1 + 0.5} = \frac{2}{3}$
$e_{ss} = 0.6667 \approx 0.67$ . Cross-check: the closed-loop DC gain is $\frac{G ( 0 )}{1 + G ( 0 )} = \frac{0.5}{1.5} = \frac{1}{3}$ , so the output settles at $\frac{1}{3}$ and the error is $1 - \frac{1}{3} = \frac{2}{3}$ .

KEY IDEA

In a stable unity-feedback type-0 system, the unit-step steady-state error is $e_{ss} = \frac{1}{1 + K _{p}}$ , where $K_{p} = G (0)$ is the DC gain of the open-loop transfer function. You can check it quickly: the closed-loop DC gain is $\frac{G ( 0 )}{1 + G ( 0 )}$ , and the error is 1 minus that.

WATCH OUT FOR

Writing $e_{ss} = \frac{1}{K _{p}} = 2$ (the formula for ramp error with $K_{v}$ , or a slip in the step formula). Another error is to give the steady-state output, $\frac{1}{3} \approx 0.33$ , as the error. A third is to skip the stability check. The final value theorem only holds when the closed loop is stable, and here that means confirming $s^{2} + 3 s + 3$ has both roots in the left half-plane.

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GATE 2018 EE Q.25: solution and answer

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