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The figure given below shows the Fourier spectra of signal x(t) and y(t).
The Nyquist sampling rate for x(t) y(t) is_________.
The Nyquist sampling rate for x(t) y(t) is_________.
- a)250 kHz
- b)500 kHz
- c)50 kHz
- d)100 kHz
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The figure given below shows the Fourier spectra of signal x(t) and y(...
Concept:
Multiplication in the time domain is the convolution in the frequency domain and the maximum frequency of convolution of the signal is the sum of individual frequencies, i.e.
If z(t) = x(t) y(t)
The maximum frequency of Z(ω) = ωx + ωy
where ωx is the maximum frequency of signal x(t)
ωy is the maximum frequency of signal y(t)
Application:
The maximum frequency present in x(t) is:
Similarly, the maximum frequency present in y(t) is:
Now, the maximum frequency present in z(t) = x(t) y(t) will be:
fz = fx + fy
fz = 250 kHz
Nyquist sampling frequency will now be:
fN = 2 × fz
fN = 500 kHz
Multiplication in the time domain is the convolution in the frequency domain and the maximum frequency of convolution of the signal is the sum of individual frequencies, i.e.
If z(t) = x(t) y(t)
The maximum frequency of Z(ω) = ωx + ωy
where ωx is the maximum frequency of signal x(t)
ωy is the maximum frequency of signal y(t)
Application:
The maximum frequency present in x(t) is:
Similarly, the maximum frequency present in y(t) is:
Now, the maximum frequency present in z(t) = x(t) y(t) will be:
fz = fx + fy
fz = 250 kHz
Nyquist sampling frequency will now be:
fN = 2 × fz
fN = 500 kHz
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The figure given below shows the Fourier spectra of signal x(t) and y(t).The Nyquist sampling rate for x(t) y(t) is_________.a)250 kHzb)500 kHzc)50 kHzd)100 kHzCorrect answer is option 'B'. Can you explain this answer?
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