GATE 2024 EE Q.22: solution and answer

GATE 2024 · EEQ.22 · 1 mark · MCQ

Consider the cascaded system as shown in the figure. Neglecting the faster component of the transient response, which one of the following options is a first-order pole-only approximation such that the steady-state values of the unit step responses of the original and the approximated systems are same?

GATE 2024 EE question 22: Two blocks in cascade (series), no feedback. — Figure from the GATE 2024 EE paper. It is part of the question.

(A)
$\frac{1}{s + 1}$
(B)
$\frac{2}{s + 1}$
(C)
$\frac{1}{s + 20}$
(D)
$\frac{2}{s + 20}$

Answer and solution

ANSWER

The first-order pole-only approximation is $\frac{2}{s + 1}$ , option (B). It keeps the dominant pole at $s = - 1$ , and its DC gain is 2, the same as the original system's $G (0) = \frac{40}{20} = 2$ .

Official answer: BSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

Blocks in series multiply, so the overall transfer function is $G (s) = \frac{1}{s + 1} \cdot \frac{s + 40}{s + 20} = \frac{s + 40}{( s + 1 ) ( s + 20 )}$
The poles are at $s = - 1$ and $s = - 20$ . The pole at $- 20$ produces an $e^{- 20 t}$ term, which decays 20 times faster than the $e^{- t}$ term (time constant $0.05 s$ against $1 s$ ). So $s = - 1$ is the dominant pole and the one to keep. The zero at $- 40$ is even further from the origin, and a pole-only model drops it as well.
The approximation must therefore have the form $G_{a} (s) = \frac{K}{s + 1}$ . That rules out (C) and (D), which keep the fast pole and drop the slow one.
Final value theorem for a unit step input: $y (\infty) = lim_{s \to 0} s G (s) \frac{1}{s} = G (0)$ . Matching the steady-state values is the same as matching the DC gains.
DC gain of the original system: $G (0) = \frac{0 + 40}{( 0 + 1 ) ( 0 + 20 )} = \frac{40}{20} = 2$
DC gain of the approximation: $G_{a} (0) = \frac{K}{0 + 1} = K$ . Setting $K = 2$ gives $G_{a} (s) = \frac{2}{s + 1}$ which is option (B).

KEY IDEA

To reduce the order, keep the dominant pole (the one closest to the origin, here $s = - 1$ ) and drop the far pole and zero. Then pick the gain so the DC gains match, because by the final value theorem the unit-step steady state equals $G (0)$ .

WATCH OUT FOR

There are three common traps. (1) Dropping the second block $\frac{s + 40}{s + 20}$ entirely and choosing (A) $\frac{1}{s + 1}$ . This ignores the block's DC gain of $\frac{40}{20} = 2$ and halves the steady-state value. (2) Keeping the pole at $- 20$ because it 'looks bigger', which gives (C) or (D). The pole to keep is the slow one closest to the origin, not the fast one. (3) Expanding in partial fractions, $G (s) = \frac{39/19}{s + 1} - \frac{20/19}{s + 20}$ , and simply deleting the fast term. That leaves $\frac{39/19}{s + 1}$ , whose DC gain of $\frac{39}{19} \approx 2.05$ does not equal 2. The question asks for the steady states to match, so the gain has to be reset to $G (0) = 2$ .

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 22 Time response questions, with notes

GATE 2024 EE Q.22: solution and answer

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