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simple question, but I cant seem to understand how they got the answer. I have $x(t)$ as a signal, and I'm told that its Nyquist frequency is $\omega_0$

I'm asked - what is the Nyquist frequency of this next signal: $$ x(t)\cos^2(\omega_0t)$$

after playing with it I got the Fourier transform of it $$0.5(j\omega)+0.25X(j(\omega+2\omega_0))+0.25X((j(\omega-2\omega_0))$$

So I got the same signal (different amplitude) and two more of it shifted in frequency. I draw it and concluded that the Nyquist frequency is $3\omega_0$ Buy I'm wrong and don't know how to get to the correct answer. The formal answer is $$2(\omega_0/2 +2\omega_0)=5\omega_0$$ Thanks.

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Normally the Nyquist frequency is a property of a discrete-time signal, and not of a continuous-time signal, but I guess that what is meant is the minimum rate at which the signal $x(t)$ must be sampled in order to satisfy the sampling theorem. That would be twice its bandwidth.

So the signal $x(t)$ has bandwidth $\omega_0/2$ and it is multiplied by $\cos^2(\omega_0t)=\frac12 (1+\cos(2\omega_0t))$. This creates a shifted spectrum centered at $2\omega_0$. So the maximum frequency of the modulated signal is $2\omega_0+\omega_0/2$, i.e. the new center frequency plus the original signal's bandwidth. The corresponding Nyquist rate is twice this maximum frequency, which gives the result that you stated in your question.

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  • $\begingroup$ a follow up question, for aמ odd real signal x(t), we get from fourier transform X(jw) which is odd and pure imaginary. What I dont quite understand is, how can I differ a Pure imaginary fourier transform from a pure real fourier transform graphically (with frequency axis). $\endgroup$
    – minimal risk
    May 15, 2015 at 17:20
  • $\begingroup$ @minimalrisk: You would normally plot its magnitude. $\endgroup$
    – Matt L.
    May 15, 2015 at 18:03

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