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My background: until very recently in my studies I was dealing with analog systems and signals and now we are being taught discrete signals.

I am stuck at this question:

Suppose the impulse response of a discrete linear and time invariant system is $h(n) = u(n)$ Find the output signal if the input signal is $x(n) = u(n-1)-u(n-5)$

When $n<1$ the input signal doesn't overlap with the impulse response so the convolution is 0.

When $1<n<5$ part of the input signal overlaps with the impulse response (from $0$ to $n-1$) so the result of the convolution is $n$?

But what if $n>5$?. Isn't it correct that the signal overlap from $n-1$ until $n-5$ so the convolution must be equal to $4$?

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Almost!

When n<1 the input signal doesn't overlap with the impulse response so the convolution is 0.

That's correct.
Note: strictly speaking, that's correct if by "overlap" you mean "overlap when both signals have a value of $1$". The unit step function is defined everywhere, so the signals overlap everywhere. For example, at $n=0$, the impulse response has value $1$ and the input has value $0$. That doesn't mean they don't "overlap".

You need to be careful with your limits. You have an error with your expression for $x[n-k$]. Change $\leq$ to $<$ to get:

$$ x[n-k] = \begin{cases} 1 &n-5 < k\leq n-1\\ 0 &\text{otherwise} \end{cases} $$

When 1<n<5 part of the input signal overlaps with the impulse response (from 0 to n−1) so the result of the convolution is n?

is then correct with $1\leq n < 5$, and the caveat on overlap mentioned earlier.

But what if $n>5$? Isn't it correct that the signal overlap from $n−1$ until $n−5$ so the convolution must be equal to $4$?

Again, assuming you mean "overlap only at values where both signals are $1$", this should read: but what if $n\geq5$? Isn't it correct that the signals overlap from $n−4$ to $n−1$ so the convolution must be equal to $4$?


To summarize: $$ y[n] = h[n]*x[n] = \begin{cases} 0 & n < 1\\ n &1 \leq n < 5\\ 4 &\text{otherwise} \end{cases} $$

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