# Convolution of delta function in discrete time with parameter

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:
$$y[n] = n\delta(5n-20) \star 3^{5n}u[n] = n\delta(n-4) \star 3^{5n}u[n] = 4 \cdot 3^{5(n-4)}u[n-4]$$
• sifting property: $n \cdot \delta[n-d] = d \cdot \delta[n-d]$ ... – Fat32 Jan 13 '18 at 16:04