If I have a delta function of the following type

$$ n\delta(5n-20) * 3^{5n}u[n] $$ How can I calculate this convolution?

I have thought of the property $$ \delta[n-n_0] *x[n] = x[n-n_0] $$

Then if I say $$ k=5n $$ and $$ t[n]= 3^{5n}u[n] $$

It should be that the convolution is equal to :

$$ nt[\frac{n}{5}-20] $$ and replacing t $$ n 3^{5(\frac{n}{5}-20)}u[\frac{n}{5}-20]$$

Is this right? Or am I wrong somewhere?


You can use the following argumentation to find the result. The discrete time unit-sample function $\delta[n]$ has the following property for integer $M$: $$ \delta[Mn] = \delta[n] $$ and more generally you can conlcude that for integer $M$ and $d$ we have $$ \delta[M(n-d)] = \delta[n-d] $$

Therefore you can replace $\delta[5n-20] = \delta[5(n-4)]$ with $\delta[n-4]$ and proceed as usual to find the result of the convolution as:

$$y[n] = n\delta(5n-20) \star 3^{5n}u[n] = n\delta(n-4) \star 3^{5n}u[n] = 4 \cdot 3^{5(n-4)}u[n-4] $$

where in this last line we have used the sifting and shifting properties of the impulse function.

  • $\begingroup$ why in the end it is 4 * .. , shouldn't it be n*.. $\endgroup$ – Maverick98 Jan 13 '18 at 15:48
  • 2
    $\begingroup$ sifting property: $n \cdot \delta[n-d] = d \cdot \delta[n-d] $ ... $\endgroup$ – Fat32 Jan 13 '18 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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