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I am nearly at the end of finishing a problem in my textbook but I couldn't understand something in the answer; enter image description here

I did everything to the point I found the overall response of the system in terms of $h_{1}[n]$ which is $$h[n]=h_{1}[n]+2h_{1}[n-1]+h_{1}[n-2]$$ Now while comparing each $n$ value with the given graph for overall response, the answer is saying that for $n=0$; $$h[0]=h_{1}[0]$$ but I didn't get why it isn't like this: $$h[0]=h_{1}[0]+2h_{1}[-1]+h_{1}[-2]$$ How do we know $h_{1}[n]$ is zero for $n$ values smaller than zero?

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HINT:

Consider the cascade interconnection of three causal LTI systems, [...]"

What does that imply for the impulse response $h_1[n]$?

What is more, since the total impulse response is causal, and you know from its definition that $h_2[n]$ is causal, $h_1[n]$ must also be causal.

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  • $\begingroup$ I thought about that too actually but from the definition I remember for causality, the output should only be effected from present or past values, the $h_{1}[-1]$ and $h_{1}[-2]$ are past values and I thought they should be included. $\endgroup$
    – Tunguska
    Mar 24, 2021 at 13:58
  • $\begingroup$ @Tunguska: Present and past values of the input. This implies that the impulse response must be zero for negative times, because that part of the impulse response is the weight function for future input samples (remember the definition of convolution). $\endgroup$
    – Matt L.
    Mar 24, 2021 at 14:27
  • $\begingroup$ Oh, okay, I checked the convolution sum with the conditions of causality, the value of $h[n-k]$ should be zero for $n-k<0$ since for $y[n]$ and $x[k]$ we know that it should be $k>!n$ therefore $h[n]=0$ while $n<0$, thanks for pointing out! $\endgroup$
    – Tunguska
    Mar 24, 2021 at 14:55

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