GATE 2023 EC Q.27: solution and answer

Official answer: BSource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. Modulation property of the Fourier transform: with $m(t)\leftrightarrow M(\omega )$, $$x(t)=m(t)\cos {\omega }_{0}t\leftrightarrow X(\omega )=\frac{1}{2}\left[M(\omega -{\omega }_0)+M(\omega +{\omega }_0)\right]$$

2. $M(\omega )$ is zero outside $|\omega |\le W$, where $W$ is the bandwidth in rad/s ($W=B$ if $B$ is in rad/s, $W=2\pi B$ if $B$ is in Hz). The two shifted copies occupy ${\omega }_0-W\le \omega \le {\omega }_0+W$ and $-{\omega }_0-W\le \omega \le -{\omega }_0+W$.

3. The copies are disjoint when the lower edge of the right copy lies above the upper edge of the left copy: ${\omega }_0-W>-{\omega }_0+W$, i.e. ${\omega }_0>W$. With ${\omega }_0=10B$ this holds whichever unit $B$ is in, since $10B>B$ and $10B>2\pi B\approx 6.28B$. So the two copies do not overlap.

4. Parseval's theorem: $$E_x=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }|X(\omega ){|}^{2}d\omega =\frac{1}{4}\cdot \frac{1}{2\pi }{\int }_{-\infty }^{\infty }\left[|M(\omega -{\omega }_0){|}^2+|M(\omega +{\omega }_0){|}^2+2\text{Re}\{M(\omega -{\omega }_0)M^{\ast }(\omega +{\omega }_0)\}\right]d\omega$$ The cross term is zero at every $\omega$, because wherever one copy is non-zero the other is zero.

5. A frequency shift does not change the area under $|M{|}^2$, so each shifted copy carries energy $E$: $$E_x=\frac{1}{4}(E+E)=\frac{E}{2}$$

6. Time-domain check: ${\cos }^2{\omega }_{0}t=\frac{1}{2}(1+\cos 2{\omega }_{0}t)$, so $$E_x=\frac{1}{2}{\int }_{-\infty }^{\infty }{m}^2(t)dt+\frac{1}{2}{\int }_{-\infty }^{\infty }{m}^2(t)\cos 2{\omega }_{0}tdt$$ The second integral is the real part of the Fourier transform of $m^2(t)$ evaluated at $2{\omega }_0$. $m^2(t)$ is band-limited to $|\omega |\le 2W$ and $2{\omega }_0>2W$, so that integral is zero and $E_x=E/2$. Answer: (B).