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I have a question about the expression of the output of a discrete time filter described by its impulse response $h(k)$. Looking at the defintion of a discrete filter with input $u(k)$ and output $y(k)$, if holds:

$$y(k) = h(k) * u(k) = \sum_{n\in\mathbb{Z}}h(n)u(k-n) = \sum_{n\in\mathbb{Z}}h(k-n)u(n) $$

I agree with these definitions but in several examples I found on different forums and in lab exercices, when $h(k)$ is a causal filter, it is possible to go from the infinite sum above to the following definition :

$$y(k) = \sum_{n = 0} ^kh(n)u(k-n) $$

I get it that $h(k)=0$ for $k<0$, so we can start the sum at $n=0$, but why not until $n=+\infty$ ?

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    $\begingroup$ this formula only applies if BOTH the signal and the impulse response are causal. $\endgroup$
    – Hilmar
    Oct 25 at 10:08
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When the system is causal, i.e., $h[n]=0$ for $n<0$, the sum starts at $0$ and goes to infinity, as you said. The upper limit of the sum becomes the current time index only if also the input signal vanishes for negative time indices, because then $x[k-n]=0$ for $n>k$.

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