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.


1 Answer 1


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.

  • $\begingroup$ 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). $\endgroup$ May 15, 2015 at 17:20
  • $\begingroup$ @minimalrisk: You would normally plot its magnitude. $\endgroup$
    – Matt L.
    May 15, 2015 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.