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A discrete-time signal x[n] is obtained by sampling an analog signal at 10 kHz. The signal x[n] is filtered by a system with impulse response h[n] =  0.5{ δ[n] + δ[n - 1]}. The 3dB cutoff frequency of the filter is:
  • a)
    1.25 kHz
  • b)
    2.50 kHz
  • c)
    4 .00 kHz
  • d)
    5.00 kHz
Correct answer is option 'B'. Can you explain this answer?
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Assuming that the rest of the impulse response is not given, it is not possible to fully determine the system's characteristics. However, some information can be deduced based on the given portion of the impulse response.

Firstly, we can see that the system is a low-pass filter, since it attenuates higher frequencies more strongly than lower frequencies. This is because the cosine term in the expression for h[n] oscillates more rapidly at higher frequencies, causing cancellations.

Secondly, we can determine the cutoff frequency of the filter. The cutoff frequency is the frequency at which the filter's attenuation is -3 dB, or 50% in magnitude. From the given expression for h[n], we can see that the cosine term is equal to 1 when w = 0, and it becomes 0 at w = pi/2. Therefore, the cutoff frequency is pi/2, or approximately 1.57 radians/sample. In Hz, this corresponds to a frequency of 1.57/(2*pi) * 10000 = 250 Hz.

Finally, we can calculate the frequency response of the filter by taking the Fourier transform of h[n]. This gives:

H(w) = 0.5{1 + cos(w)} e^(-j*w*(N-1)/2)

where N is the length of the impulse response (which is not given). The frequency response is a complex function that describes how the filter modifies each frequency component of the input signal.
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A discrete-time signal x[n]is obtained by sampling an analog signal at 10 kHz. The signalx[n]is filtered by a system with impulse response h[n] = 0.5{ δ[n] + δ[n - 1]}. The 3dB cutofffrequency of the filter is:a)1.25 kHzb)2.50 kHzc)4 .00 kHzd)5.00 kHzCorrect answer is option 'B'. Can you explain this answer?
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