GATE 2025 EC Q.18: solution and answer

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

1. Triangle inequality for integrals. The magnitude of an integral is never more than the integral of the magnitude: $$|F(\omega )|=\left|{\int }_{-\infty }^{\infty }f(t)e^{-j\omega t}dt\right|\le {\int }_{-\infty }^{\infty }\left|f(t)\right|\left|e^{-j\omega t}\right|dt$$

2. Unit magnitude of the kernel. For real $\omega$ and $t$, $\left|e^{-j\omega t}\right|=1$, so $$|F(\omega )|\le {\int }_{-\infty }^{\infty }|f(t)|dt\text{for every }\omega$$ This is statement (A). It holds for any $f(t)$ whose transform exists: if $f(t)$ is not absolutely integrable, the right-hand side is infinite and the inequality is trivially true.

3. (B) is ruled out because it is the exact negation of the bound just proved, so it is false for every signal at every $\omega$. For example, for any $f(t)\ge 0$, at $\omega =0$ we get $|F(0)|=\int |f(t)|dt$ (equality, not $>$).

4. (C) fails for a signal that integrates to zero. With $\text{rect}(t)=1$ for $|t|<\frac{1}{2}$ and $0$ otherwise, take $f(t)=\text{rect}(t-\frac{1}{2})-\text{rect}(t+\frac{1}{2})$: a $+1$ pulse on $(0,1)$ and a $-1$ pulse on $(-1,0)$. Then $\int fdt=F(0)=0$, while $$F(\omega )=\frac{1-e^{-j\omega }}{j\omega }-\frac{e^{j\omega }-1}{j\omega }=\frac{2-2\cos \omega }{j\omega }=\frac{4{\sin }^2(\omega /2)}{j\omega }$$ At $\omega =\pi$, $|F(\pi )|=\frac{4}{\pi }>0$, so $|F(\omega )|\le 0$ is false.

5. (D) fails at a spectral null. Take $f(t)=\text{rect}(t)$: $\int fdt=1$, while $F(\omega )=\frac{\sin (\omega /2)}{\omega /2}$ is $0$ at $\omega =2\pi$. Then $|F(2\pi )|=0<1$.

6. Only (A) is always true.