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Why is $\delta[an]=\frac 1 a \delta[n]$ in discrete time? Prove.

Hi, I want to prove it but I don’t know on what facts to rely. In continuous time we have integral and dt in that integral that turns to be dt/a for continuous time delta function. But in discrete time, we don’t have dt. So how to prove it then?

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  • $\begingroup$ Where did you find that "identity"? $\endgroup$ – Matt L. Apr 26 '20 at 19:39
  • $\begingroup$ Here: pilot.cnxproject.org/content/collection/col10064/latest/module/… $\endgroup$ – Vitali Pom Apr 26 '20 at 19:39
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    $\begingroup$ This is a property of the Dirac distribution and follows from a substitution of the integration variable. It does not have an equivalent for the Kronecker unit impulse. $\endgroup$ – Jazzmaniac Apr 26 '20 at 19:50
  • $\begingroup$ Thanks! Marking as correct. $\endgroup$ – Vitali Pom Apr 26 '20 at 19:52
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It's actually good that you couldn't prove it because the claim is wrong. Note that in discrete time we have $\delta[n]=1$ for $n=0$ and $\delta[n]=0$ for $n\neq 0$. So if $a\neq 0$, you simply get $\delta[an]=\delta[n]$ (note that $a$ must be an integer for this to make sense). If $a=0$ then $\delta[an]=1$ (for all $n$).

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  • $\begingroup$ Thanks Matt. Let’s wait for one more opinion before I mark your answer as the right answer, since I already saw a few things I considered to be wrong and meanwhile my teacher proves I’m the only wrong time after time. Thanks for your answer, it raises my level of confident (besides seemingly being true). $\endgroup$ – Vitali Pom Apr 26 '20 at 19:46

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