Correct Answer - Option 3 : 30 Hz
Period of sampling train, Ts = 20 ms
\(\therefore {f_s} = \frac{1}{{20 \times {{10}^{ - 3}}}} = 50{\rm{Hz}}\)
If frequency of x(t) is fx, then after sampling the signal, the sampled signal has the frequency,
fs - fx = 50 - fx and fs + fx = 50 + fs
Now, the sampled signal is applied to and ideal low pass filter with cut off frequency.
fc = 25 Hz
Now, the o/p of filter carried a single frequency component of 20 Hz
∴, only \(\left( {{f_s} - {f_x}} \right)\) component passes through the filter, ie.
fs - fx < 25
and fs - fx = 20
50 - fx = 20
fx = 50 - 20 = 30 Hz