GATE EC 2016 Set 1 — Question 20

Let the continuous-time sinusoid be $x(t) = \cos(2\pi f_s t)$ where $f_s = 33$ Hz.
Let the periodic Dirac impulse train be $p(t) = \sum_{n=-\infty}^{\infty} \delta(t - nT_p)$ where $T_p = 1/f_p$ and $f_p = 46$ Hz.The product signal is $y(t) = x(t)p(t)$. When a continuous-time signal is multiplied by an impulse train, the spectrum of the resulting signal consists of shifted versions of the original signal's spectrum, centered at multiples of the impulse train's frequency.The spectrum of $x(t)$ has impulses at $\pm f_s$.
The spectrum of $y(t)$ will have frequency components at $k f_p \pm f_s$ for integer values of $k$.Given $f_s = 33$ Hz and $f_p = 46$ Hz.
Let's list some of the absolute frequencies present in $y(t)$:
For $k=0$: $|0 \cdot f_p \pm f_s| = f_s = 33$ Hz
For $k=1$:
$|1 \cdot f_p - f_s| = |46 - 33| = 13$ Hz
$|1 \cdot f_p + f_s| = |46 + 33| = 79$ Hz
For $k=2$:
$|2 \cdot f_p - f_s| = |2(46) - 33| = |92 - 33| = 59$ Hz
$|2 \cdot f_p + f_s| = |2(46) + 33| = |92 + 33| = 125$ HzThe signal $y(t)$ is passed through an ideal analog low-pass filter with a cutoff frequency $f_c = 23$ Hz.
This filter will only allow frequency components less than or equal to $23$ Hz to pass.From the list of frequencies:

• $13$ Hz: This component will pass because $13 \le 23$ Hz.
• $33$ Hz: This component will not pass because $33 > 23$ Hz.
• $79$ Hz, $59$ Hz, $125$ Hz, etc.: These components will also not pass as they are all greater than $23$ Hz.

Therefore, the output signal from the low-pass filter will only contain the frequency component at $13$ Hz.The fundamental frequency of the output signal is the lowest non-zero frequency component present in the output. In this case, it is $13$ Hz.The final answer is $\boxed{13}$.