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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?

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1 Answer 1

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Your first solution is correct, your second isn't. Using correct notation, you have

$$u[n-5]=u[n]-\sum_{k=0}^4\delta[n-k]\tag{1}$$

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 $u[n-5]$.

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