I am given the frequency response for a continuous time signal $X(j\omega)$ = 2 at $\omega=0$ and 0 at $\omega = -10000\pi$ and $10000 \pi$. Looks like a triangle. I am told to sketch $X(e^{jw})$ the frequency response of a sampled signal given T=0.0005 sec.

How could I/should I sample the CTFT to arrive at the DTFT. Do I need to go back and find $x(t)$ which is the original continuous time signal and sample that first or what?

  • $\begingroup$ Do you know what the sampling theorem is ? $\endgroup$
    – Hilmar
    Apr 29, 2022 at 19:15
  • 1
    $\begingroup$ Hm... signals don't have a frequency response, only systems do. $\endgroup$
    – MBaz
    Apr 29, 2022 at 19:56

1 Answer 1


It is important to understand the property that sampling in one domain is periodicity in the other domain. If that isn't clear please review these other posts that help detail why this is the case:

Aliasing after downsampling

Nyquist Theorem - Why unique frequencies upto Fs/2 and not Fs? f+Fs is start of Aliasing

Where should I set my anti-aliasing filter corner frequency for this signal?

Applying Nyquist's sampling theorem to a real signal

Zero padding affects the DTFT?

Once understood and adding to the above posts that go deeper into the mathematical details, we see how this applies to the CTFT (Continuous-Time Fourier Transform), DTFT (Discrete-Time Fourier Transform) and DFT (Discrete Fourier Transform) as bottom lined in the graphic below. What we see is that the DTFT will be a periodically repeating (in frequency) version of the CTFT. Further and importantly, if the CTFT has spectral content that extends beyond the Nyquist boundary given by $\pm f_s/2$ where $f_s$ is the sampling rate, then in the process of being periodic we will also see the results of aliasing. Finally for completeness to the graphic shown, the DFT is the discrete frequency (sampled) version of the DTFT.

high level view of CTFT, DTFT and DFT


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