Question

A signal \(x(t) = 2\cos(180\pi t)\cos(60\pi t)\) is sampled at 200 Hz and then passed through an ideal low pass filter having cut-off frequency of 100 Hz. The maximum frequency present in the filtered signal in Hz is .......... (Round off to the nearest integer).

Hint:

For sampled signals, always reduce frequencies above Nyquist by aliasing: \(f_a = |f - n f_s|\). Then check low-pass filter cutoff.

Answer 1

Step 1: Simplify the signal.
\[ x(t) = 2\cos(180\pi t)\cos(60\pi t) \] Using product-to-sum identity: \[ 2\cos A \cos B = \cos(A+B) + \cos(A-B) \] Here \(A = 180\pi t, \, B = 60\pi t\). \[ x(t) = \cos(240\pi t) + \cos(120\pi t) \]

Step 2: Convert to Hz.
\(\cos(240\pi t)\) has frequency = \( \frac{240\pi}{2\pi} = 120 \, \text{Hz}\). \(\cos(120\pi t)\) has frequency = \( \frac{120\pi}{2\pi} = 60 \, \text{Hz}\). So original signal contains 60 Hz and 120 Hz.

Step 3: Sampling effect.
Sampling frequency = 200 Hz. Nyquist = 100 Hz. So frequencies above 100 Hz will alias. 120 Hz → alias = \(|120 - 200| = 80 \, \text{Hz}\). So sampled spectrum contains 60 Hz and 80 Hz.

Step 4: Filtering.
Low-pass filter with cut-off 100 Hz passes both 60 and 80 Hz. Maximum frequency = 80 Hz. Correction: 60 Hz and 90 Hz may arise if considering foldover from 110 Hz as well. Checking again: 120 Hz → alias = 200 - 120 = 80 Hz. So maximum = 80 Hz.

Final Answer:
\[ \boxed{80 \, \text{Hz}} \]