Sampling of Signal x(t)

- The signal x(t) = 100 cos (24π × 10^3t) is sampled with a sampling period of 50 μs.
- The sampling frequency is given by fs = 1/Ts, where Ts is the sampling period.
- Therefore, fs = 1/50 μs = 20 kHz.
- According to the Nyquist sampling theorem, the signal can be reconstructed perfectly if the sampling frequency is at least twice the highest frequency component of the signal.
- The highest frequency component of x(t) is 24π × 10^3 Hz, which is less than half of the sampling frequency of 20 kHz.
- Hence, the sampling is ideal and there is no aliasing.

Filtering of Sampled Signal

- The sampled signal can be expressed as x(nTs) = 100 cos (24π × 10^3nTs), where n is an integer.
- The sampled signal is passed through an ideal low-pass filter with a cutoff frequency of 15 kHz.
- The frequency response of the ideal low-pass filter is given by H(f) = 1 for |f| ≤ 15 kHz and H(f) = 0 for |f| > 15 kHz.
- The output of the filter can be obtained by convolving the input signal with the impulse response of the filter, which is a sinc function.
- The Fourier transform of the sinc function is a rectangular function, which is equal to 1 for frequencies between -fc and fc, and zero otherwise.
- Therefore, the output of the filter will be the product of the Fourier transforms of the input signal and the filter impulse response.

Calculation of Filter Output Frequencies

- The Fourier transform of the sampled signal x(nTs) is given by X(f) = 0.5[δ(f - fs) + δ(f + fs)] * Xc(f), where Xc(f) is the Fourier transform of the continuous-time signal x(t).
- The Fourier transform of the ideal low-pass filter impulse response is given by H(f) = rect(f/2fc), where fc is the cutoff frequency.
- Therefore, the Fourier transform of the filter output can be obtained as Y(f) = X(f) * H(f), where * denotes convolution.
- Substituting the expressions for X(f) and H(f), we get Y(f) = 0.5[rect((f - fs)/2fc) + rect((f + fs)/2fc)] * Xc(f).
- The rectangular functions can be simplified using the identity rect(a)rect(b) = rect(min(a, b)).
- Therefore, Y(f) = rect((f - fs)/2fc) * Xc(f) for |f - fs| ≤ fc, and Y(f) = rect((f + fs)/2fc) * Xc(f) for |f + fs| ≤ fc.
- Substituting the values of fs and fc, we get Y(f) = rect((f - 20 kHz)/30 kHz) * Xc(f) for 5 kHz ≤ f ≤ 35 kHz, and Y(f) = rect((f + 20 kHz)/30 kHz) * Xc(f) for -35 kHz ≤ f ≤ -5 kHz.
- The Fourier transform of the filter output can be inverse-transformed to obtain the time-domain signal.
- The time