GATE 2026 EC Q.46: solution and answer

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Total energy (time domain): $E={\int }_0^{\infty }{\left(e^{-2t}\right)}^{2}dt={\int }_0^{\infty }{e}^{-4t}dt=\frac{1}{4}$.

2. Fourier transform in Hz: $x(t)=e^{-2t}u(t)\Rightarrow X(f)=\frac{1}{2+j2\pi f}$, so $|X(f){|}^2=\frac{1}{4+4{\pi }^2{f}^2}=\frac{1}{4\left(1+{\pi }^2{f}^2\right)}$.

3. Energy inside the band (Parseval / Rayleigh): for a real baseband signal, a bandwidth of $B$ Hz means the band $|f|\le B$, i.e. both sides of the spectrum. Since $|X(f){|}^2$ is even,
   $$E_B=2{\int }_0^B\frac{df}{4\left(1+{\pi }^2{f}^2\right)}=\frac{1}{2}\cdot \frac{1}{\pi }{\tan }^{-1}(\pi B)=\frac{1}{2\pi }{\tan }^{-1}(\pi B).$$

4. Energy fraction: $\frac{E_B}{E}=\frac{{\tan }^{-1}(\pi B)/(2\pi )}{1/4}=\frac{2}{\pi }{\tan }^{-1}(\pi B)$. As a check, this tends to $1$ as $B\to \infty$, as it must.

5. Set it equal to 0.99: $\frac{2}{\pi }{\tan }^{-1}(\pi B)=0.99\Rightarrow {\tan }^{-1}(\pi B)=0.495\pi$, so
   $$\pi B=\tan (0.495\pi )=\cot (0.005\pi )=\cot (0.015708)\approx \frac{1}{0.015708}-\frac{0.015708}{3}\approx \mathrm{63.66.}$$
   (For a small angle $\theta$, $\cot \theta \approx \frac{1}{\theta }-\frac{\theta }{3}$; the correction here is only about $0.005$, so $\pi B$ lies well inside $63<\pi B<64$.)

6. Bandwidth: $B=\frac{63.66}{\pi }\text{Hz}\approx 20.3\text{Hz}$.

7. Place it among the options: $63<63.66<64$, so $\frac{63}{\pi }\text{Hz}<B<\frac{64}{\pi }\text{Hz}$, which is option (B). Option (A) is far too large and option (D) far too small. Cross-check in rad/s: $|X(\omega ){|}^2=\frac{1}{4+{\omega }^2}$ gives $\frac{2}{\pi }{\tan }^{-1}\left(\frac{{\omega }_B}{2}\right)=0.99$, so ${\omega }_B=2\times 63.66=127.3\text{rad/s}$ and $B=\frac{{\omega }_B}{2\pi }=\frac{63.66}{\pi }\text{Hz}$, the same result.