A sinusoid of 10 kHz is sampled at 15 k samples/s. The resulting signal is passed through an ideal low pass filter (LPF) with cut-off frequency of 25 kHz. The maximum frequency component at the output of the LPF (in kHz) is

Frequency of the signal (f${}_m$) = 10 kHz

Sampling rate is 15 k samples/s i.e. fₛ=15 k samples/s

The cut-off frequency of low pass filter = 25 kHz

The Impulse frequency response of the given signal is shown below.



$X_S\left(f\right)=f_s\sum _{n=-\infty }^{\infty }X\left(f-n{f}_s\right)$

Redrawing the frequency spectrum after passing through the low pass filter (LPF) heaving the cut-off frequency of 25 kHz.



Frequencies present at filter output are = 5k, 10k, 20k, 25 kHz.

Maximum frequency present at filter output is 25 kHz

Alternate method:

After sampling, frequencies present will be: ± f${}_m$ ± nfₛ

Where, n = 0, 1, 2, 3, 4, ...

For n = 0, frequencies present = ±10 kHz

For n = 1, frequencies present = ± 10 k ± 15 k = ±25 kHz, ±5 kHz

After passing through ideal LPF having a cut-off frequency of 25 kHz, the highest frequency component present at filter output = 25 kHz.

Therefore, the maximum frequency present at filter output is 25 kHz