How is hd(n) infinite duration when it is from -π to +π. Book says as it is infinite duration, we in the next step take-:

h[n]=hd[n] from n=-(N-1)/2 to (N-1)/2 and 0 otherwise.

I can't see how this hd[n] is infinite. What am I missing?

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1 Answer 1


The integral (by the way, an inverse Fourier transform of sorts) doesn't "remove" the $n$ from the exponent, so this remains something that still contains the complex exponential $e^{j\omega n}$ – and that has infinite extension along the $n$ axis.

  • $\begingroup$ I see. the integral we get is j*sin(nπ)/π. So you mean n value ranges from -∞ to +∞. I forgot what this n is about? What is the meaning of n here? Is it number of samples? $\endgroup$
    – achhainsan
    Commented Aug 2, 2021 at 12:44
  • $\begingroup$ as you can see in your formula, it's the "running" variable. You can call it "time", if that fits in your case, makes sense since you're asking about "duration". You seem to be a bit confused in general what is which your material's framework – I can't help you with that, as I don't have the same material. The only way to help yourself here is go back and read the notation very carefully as it's introduced. $\endgroup$ Commented Aug 2, 2021 at 12:45
  • $\begingroup$ Thanks you cleared my confusion. I will understand n meaning later. $\endgroup$
    – achhainsan
    Commented Aug 2, 2021 at 12:47
  • $\begingroup$ I'll urge you to understand it now. You're basing your learning on looking at things without understanding otherwise. $\endgroup$ Commented Aug 2, 2021 at 12:49

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