ISRO EC 2023 Q.26: solution and answer

ISRO 2023 · ECQ.26 · 1 mark · MCQ

Two systems $h_{1} [n] = A (b_{1})^{n} u [n]$ and $h_{2} [n] = A (b_{2})^{n} u [n]$ are cascaded. If the effective $H (z) = \frac{4}{4 - z ^{- 2}}$ , find a possible value of $(A, b_{1}, b_{2})$ .

(A)
$A = 1/2, b_{1} = 1/2, b_{2} = - 1/2$
(B)
$A = 2, b_{1} = 2, b_{2} = - 2$
(C)
$A = 1, b_{1} = 1/2, b_{2} = - 1/2$
(D)
$A = 1, b_{1} = 1, b_{2} = - 1$

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Answer and solution

ANSWER

Option (C): $A = 1, b_{1} = 1/2, b_{2} = - 1/2$ .

Official answer: CSource: ISRO's revised and final answer key of 16 February 2024

Take the $z$ -transform of each system, using $a^{n} u [n] \leftrightarrow \frac{1}{1 - a z ^{- 1}}$ : $H_{1} (z) = \frac{A}{1 - b _{1} z ^{- 1}}$ and $H_{2} (z) = \frac{A}{1 - b _{2} z ^{- 1}}$ .
A cascade multiplies the transfer functions: $H (z) = \frac{A ^{2}}{( 1 - b _{1} z ^{- 1} ) ( 1 - b _{2} z ^{- 1} )} = \frac{A ^{2}}{1 - ( b _{1} + b _{2} ) z ^{- 1} + b _{1} b _{2} z ^{- 2}} .$
Divide the numerator and denominator of the given function by 4: $\frac{4}{4 - z ^{- 2}} = \frac{1}{1 - \frac{1}{4} z ^{- 2}} .$
Match coefficients of the two forms: $A^{2} = 1$ , the $z^{- 1}$ coefficient gives $b_{1} + b_{2} = 0$ , and the $z^{- 2}$ coefficient gives $b_{1} b_{2} = - \frac{1}{4}$ .
From $b_{2} = - b_{1}$ we get $- b_{1}^{2} = - \frac{1}{4}$ , so $b_{1} = \frac{1}{2}$ and $b_{2} = - \frac{1}{2}$ (or the same pair swapped), with $A = \pm 1$ .
Check the options. (a) $A = 1/2$ gives $A^{2} = 1/4 \neq = 1$ . (b) $A = 2$ gives $A^{2} = 4 \neq = 1$ , and $b_{1} b_{2} = - 4 \neq = - \frac{1}{4}$ . (d) $b_{1} b_{2} = - 1 \neq = - \frac{1}{4}$ . (c) $A = 1$ , $b_{1} = \frac{1}{2}$ , $b_{2} = - \frac{1}{2}$ gives $A^{2} = 1$ , $b_{1} + b_{2} = 0$ and $b_{1} b_{2} = - \frac{1}{4}$ , so $H (z) = \frac{1}{( 1 - \frac{1}{2} z ^{- 1} ) ( 1 + \frac{1}{2} z ^{- 1} )} = \frac{1}{1 - \frac{1}{4} z ^{- 2}}$ , as required.

KEY IDEA

Cascaded systems multiply their $z$ -transforms. Write the target $H (z)$ with a leading 1 in the denominator, then match the coefficients of $z^{- 1}$ and $z^{- 2}$ and the numerator constant.

WATCH OUT FOR

Forgetting that the gain $A$ appears in both systems, so the cascade has $A^{2}$ in the numerator, not $A$ . Also forgetting to divide by 4 first, which hides the $1 - \frac{1}{4} z^{- 2}$ form and the product $b_{1} b_{2} = - \frac{1}{4}$ .

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 5 ISRO z-transform questions, with notes

ISRO EC 2023 Q.26: solution and answer

Answer and solution

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