Spectrum of pulse

I am trying to find out if the pulse in the figure satisfies the Nyquist criterion for a modulation interval T. I'm used to the Nyqusit frequency, but I cannot grasp what the Nyquist criterion is about.

  • $\begingroup$ if you know what a Nyquist frequency is and know how to choose it, you know what you meed to know. The question is asking you to look at the spectrum and tell what sampling frequency should be chosen. $\endgroup$ – user28715 Dec 18 '19 at 14:17
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    $\begingroup$ @StanleyPawlukiewicz: I think it's not about sampling frequency here. The symbol rate $1/T$ is given, and the question is if a pulse with the given spectrum satisfies the Nyquist criterion for intersymbol interference (i.e., doesn't cause ISI). $\endgroup$ – Matt L. Dec 18 '19 at 15:56

The Nyquist criterion for avoiding intersymbol interference (ISI) requires the pulse to have zeros at multiples of the symbol interval $T$, such that symbols do not interfere with each other. The time domain requirement is


In the frequency domain, criterion $(1)$ can be formulated as


where $H(f)$ is the Fourier transform of $h(t)$. I.e., the shifted spectra (shifted by the symbol rate) must add up to a constant.

I leave it up to you to decide whether or not the given pulse satisfies $(1)$ and $(2)$.

I suggest you go through your lectures notes because it would be surprising if the Nyquist criterion has never been mentioned and you get an exercise like this.

  • $\begingroup$ Thanks for your response. If you don't mind me asking another question, if we use (2) , we will always get a sum of 1. How will that ever equal T? $\endgroup$ – jetaime341 Dec 20 '19 at 6:58
  • $\begingroup$ @jetaime341: What is important is that the shifted spectra add up to a constant. The exact value of that constant is irrelevant. Note that we only get $T$ on the right-hand side of $(2)$ because we required $h(0)=1$. You would get a different constant if we had generalized to $h(0)=c$. $\endgroup$ – Matt L. Dec 20 '19 at 9:37
  • $\begingroup$ @jetaime341: If the answer was helpful you can accept it by clicking on the green checkmark to its left, thanks! $\endgroup$ – Matt L. Dec 20 '19 at 9:37

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