ISRO EC 2025 Q.71: solution and answer

Official answer: ASource: ISRO's final answer key of 15 May 2026

1. Recall the standard DTFT pair: $a^{n}u[n]⟷\frac{1}{1-a{e}^{-j\omega }}$ for $|a|<1$.

2. Compare with $H(e^{j\omega })=\frac{1}{1-0.5e^{-j\omega }}$: here $a=0.5$ and $|a|<1$, so the series converges on the unit circle and the frequency response exists.

3. Check with the geometric series (valid since $|0.5e^{-j\omega }|=0.5<1$): $$\frac{1}{1-0.5e^{-j\omega }}=\sum _{n=0}^{\infty }(0.5{)}^n{e}^{-j\omega n}.$$ Reading off the coefficients of $e^{-j\omega n}$ gives $h[n]=(0.5{)}^n$ for $n\ge 0$ and $0$ for $n<0$.

4. Hence $h[n]=(0.5{)}^{n}u[n]$, option (A).

5. Why the others fail. Option (B) is the same sequence delayed by one sample; a delay multiplies the transform by $e^{-j\omega }$, giving $\frac{e^{-j\omega }}{1-0.5e^{-j\omega }}$. Option (D) is a finite sequence with transform $1-0.5e^{-j\omega }$, the reciprocal of the given $H$. Option (C), $u[n]-0.5u[n-1]$, has transform $\frac{1-0.5e^{-j\omega }}{1-e^{-j\omega }}$ (the step has transform $\frac{1}{1-e^{-j\omega }}$ apart from an impulse term at $\omega =0$), which does not match.