# Discrete convolution of a causal signal with a delayed unit step

I am computing the discrete convolution of a signal $$x[n] = a^nu[n]$$ with a delayed unit step, say $$h[n] = u[n-5]$$.

If I place both directly into the discrete convolution formula I end up with: $$\frac{1 - a^{(n-4)}}{1 - a}u[n-5]$$, which I think is correct.

However, if I make $$u[n-5] = u[n] - \delta[4] - \delta[3] \cdots -\delta[0]$$ it becomes:

$$\sum_{m = 0}^{n} a^m - \sum_{n = 0}^{4} a^n$$ $$\frac{1-a^{n+1}}{1-a} - \frac{1-a^5}{1-a} = \frac{a^5 - a^{(n+1)}}{1-a}u[n]$$

Which is obviously non zero for n = 0. What I am getting wrong here? Is the decomposition of $$u[n - 5]$$ not correct or is it something else?

$$u[n-5]=u[n]-\sum_{k=0}^4\delta[n-k]\tag{1}$$
Using $$(1)$$, we get for the result of the convolution of $$x[n]$$ with $$u[n-5]$$
$$y[n]=u[n]\sum_{k=0}^nx[k]-\sum_{k=0}^4x[n-k]\tag{2}$$
From $$(2)$$ it can be seen that the two terms on the right-hand side cancel for all $$n<5$$. For $$n\ge 5$$ the result is the same as your first result, even though the approach using $$(1)$$ is more tedious than computing the convolution directly using $$x[n]$$ and $$u[n-5]$$.