GATE 2020 EC – Question 52

Communications · Analog Communications · 2 marks · Multiple choice

For the modulated signal $x(t)=m(t)\cos(2\pi f_ct)$, the message signal $m(t)=4\cos(1000\pi t)$ and the carrier frequency $f_c$ is 1 MHz. The signal $x(t)$ is passed through a demodulator, as shown in the figure below. The output $y(t)$ of the demodulator is

Diagram for GATE 2020 EC question 52

$\cos(460\pi t)$.
$\cos(920\pi t)$.
$\cos(1000\pi t)$.
$\cos(540\pi t)$.

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Show answer and explanation

Correct answer: (B) $\cos(920\pi t)$.

Explanation

Multiplying by $\cos(2\pi(f_c+40)t)$ gives $4\cos(1000\pi t)\cdot\frac12\left[\cos(80\pi t)+\cos(2\pi(2f_c+40)t)\right]$. The low-pass filter keeps only the low-frequency part: $2\cos(1000\pi t)\cos(80\pi t)=\cos(1080\pi t)+\cos(920\pi t)$. These are at 540 Hz and 460 Hz, and the filter cut-off of 510 Hz passes only 460 Hz. So $y(t)=\cos(920\pi t)$.