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What is the nyquist rate of the signal x(t)=3cos(50*pi*t)+10sin(300*pi*t)-cos(100*pi*t)?(a) 50Hz(b) 100Hz(c) 200Hz(d) 300HzI got this question by my college professor while I was bunking the class.I'm obligated to ask this question of A2D and D2A Converters in portion DSP – Basic Signaling of Digital Signal Processing |
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Answer» RIGHT CHOICE is (d) 300Hz The best explanation: The FREQUENCIES present in the given signal are F1=25Hz, F2=150Hz, F3=50Hz Thus Fmax=150Hz and from the SAMPLING THEOREM, nyquist rate=2*Fmax Therefore, Fs=2*150=300Hz. |
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