GATE 2022 IN Q.38: solution and answer

GATE 2022 · INQ.38 · 2 marks · MCQ

Consider the transfer function

H_{c} (s) = \frac{1}{( s + 1 ) ( s + 3 )}

Bilinear transformation with a sampling period of 0.1 s is employed to obtain the discrete-time transfer function $H_{d} (z)$ . Then $H_{d} (z)$ is ________

(A)
$\frac{( 1 + z ^{- 1} ) ^{2}}{( 19 - 21 z ^{- 1} ) ( 23 - 17 z ^{- 1} )}$
(B)
$\frac{( 1 - z ^{- 1} ) ^{2}}{( 21 - 19 z ^{- 1} ) ( 17 - 23 z ^{- 1} )}$
(C)
$\frac{( 1 + z ^{- 1} ) ^{2}}{( 21 - 19 z ^{- 1} ) ( 23 - 17 z ^{- 1} )}$
(D)
$\frac{( 1 + z ^{- 1} ) ^{2}}{( 21 - 19 z ^{- 1} ) ( 17 - 23 z ^{- 1} )}$

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Answer and solution

ANSWER

The answer is (C): $H_{d} (z) = \frac{( 1 + z ^{- 1} ) ^{2}}{( 21 - 19 z ^{- 1} ) ( 23 - 17 z ^{- 1} )}$ .

Official answer: CSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Bilinear transformation: $s = \frac{2}{T} \frac{1 - z ^{- 1}}{1 + z ^{- 1}}$ . With $T = 0.1 s$ , $\frac{2}{T} = 20$ , so $s = 20 \frac{1 - z ^{- 1}}{1 + z ^{- 1}}$ .
First factor, over the common denominator $1 + z^{- 1}$ : $s + 1 = \frac{20 ( 1 - z ^{- 1} ) + ( 1 + z ^{- 1} )}{1 + z ^{- 1}} = \frac{21 - 19 z ^{- 1}}{1 + z ^{- 1}}$
Second factor: $s + 3 = \frac{20 ( 1 - z ^{- 1} ) + 3 ( 1 + z ^{- 1} )}{1 + z ^{- 1}} = \frac{23 - 17 z ^{- 1}}{1 + z ^{- 1}}$
Substitute into $H_{c} (s)$ . Each denominator factor contributes one $(1 + z^{- 1})$ in its own denominator, so both move up to the numerator: $H_{d} (z) = \frac{( 1 + z ^{- 1} ) ^{2}}{( 21 - 19 z ^{- 1} ) ( 23 - 17 z ^{- 1} )}$
DC-gain check: the bilinear map sends $s = 0$ to $z = 1$ , so $H_{d} (1)$ must equal $H_{c} (0) = \frac{1}{3}$ . Here $H_{d} (1) = \frac{2 ^{2}}{( 2 ) ( 6 )} = \frac{1}{3}$ , which confirms the result.
Pole check: the bilinear map sends $s = p$ to $z = \frac{20 + p}{20 - p}$ . The poles $s = - 1$ and $s = - 3$ therefore go to $z = \frac{19}{21}$ and $z = \frac{17}{23}$ , both inside the unit circle as they must be for a stable $H_{c} (s)$ . These are exactly the roots of $21 - 19 z^{- 1}$ and $23 - 17 z^{- 1}$ .
This matches option (C). Option (A) swaps the coefficients of the first factor to $19 - 21 z^{- 1}$ (pole at $z = \frac{21}{19}$ , outside the unit circle; DC gain $- \frac{1}{3}$ ). Option (D) flips the second factor to $17 - 23 z^{- 1}$ (pole at $z = \frac{23}{17}$ ; DC gain $- \frac{1}{3}$ ). Option (B) has $(1 - z^{- 1})^{2}$ in the numerator, which would make the DC gain zero.

KEY IDEA

Replace $s$ by $\frac{2}{T} \frac{1 - z ^{- 1}}{1 + z ^{- 1}}$ (here $\frac{2}{T} = 20$ ) and write each factor $(s + a)$ over the common denominator $1 + z^{- 1}$ , which gives $\frac{( 20 + a ) - ( 20 - a ) z ^{- 1}}{1 + z ^{- 1}}$ . Each first-order pole factor puts one $(1 + z^{- 1})$ in the numerator. You can check the result two quick ways: the DC gain must survive ( $z = 1$ matches $s = 0$ ), and a stable pole $s = p$ must land inside the unit circle at $z = \frac{20 + p}{20 - p}$ .

WATCH OUT FOR

Writing the factor as $(20 - a) - (20 + a) z^{- 1}$ , i.e. swapping the two coefficients. That gives $19 - 21 z^{- 1}$ or $17 - 23 z^{- 1}$ , which put a pole outside the unit circle and flip the sign of the DC gain (the traps in options A and D). Another mistake is using $s = \frac{1}{T} \frac{1 - z ^{- 1}}{1 + z ^{- 1}}$ without the factor 2, or putting $(1 - z^{- 1})$ in the numerator (option B).

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