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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?

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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.

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

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