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When finding the overall system impulse,

how to do arithmetic with $\delta$ functions?

E.g. \begin{align} h[n]&=h_1[n]\star h_2[n]\\ &=\sum_{k=-\infty}^{\infty} h_1[k]h_2[n-k]\\ &=\sum_{k=-\infty}^{\infty} \left(2\delta[k-2]-3\delta[k+1])(\delta[n-k-1]+2 \delta[n-k+2]\right)\\ &=\sum_{k=-\infty}^{\infty} (2\delta[k-2]\delta[n-k-1]+2\delta[k-2]2 \delta[n-k+2]-3\delta[k+1]\delta[n-k-1]-3\delta[k+1]2 \delta[n-k+2]) \end{align}

The two middle terms are convolutions. But what about the two other? What to do with them?

Or am I supposed to leave the indices there, rather than evaluating the $\delta$ functions?

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  • $\begingroup$ you are sure they are multiplied and not convolved, right ? $\endgroup$ – Fat32 Mar 30 '17 at 14:50
  • $\begingroup$ Here it's multiplication, not convolution. But the above expression was found through convolution of the signals in $()()$. $\endgroup$ – mavavilj Mar 30 '17 at 14:54
  • $\begingroup$ You shall better formulate the convolution first, it seems there is an unconventional use of the variables n and k. $\endgroup$ – Fat32 Mar 30 '17 at 15:03
  • $\begingroup$ That seems to be the convolution between two sums of two deltas. What about doing the operation in the Fourier domain, and then coming back? $\endgroup$ – Tendero Mar 30 '17 at 15:05
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    $\begingroup$ @Tendero I haven't been taught Fourier transforms yet. $\endgroup$ – mavavilj Mar 30 '17 at 15:07
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You make it unnecessarily complicated.

Given $$\begin{align} h_1[n]&=2\delta[n-2]-3\delta[n+1]\\ h_2[n]&=\delta[n-1]+2 \delta[n+2] \end{align}$$ to calculate $$h[n]=h_1[n]\star h_2[n]$$

the easiest way is to use the sampling property of delta, which is

$$f[n]\star\delta(n\pm n_0)=f[n\pm n_0]$$

So you can write as follows

$$\begin{align} h_1[n]\star h_2[n] &= h_1[n]\star \left(\delta[n-1]+2 \delta[n+2]\right)\\ &=\color{green}{h_1[n-1]}+2\color{blue}{h_1[n+2]}\\ &=\color{green}{2\delta[(n-1)-2]-3\delta[(n-1)+1]}+2\color{blue}{\left(2\delta[(n+2)-2]-3\delta[(n+2)+1]\right)}\\ &=2\delta[n-3]-3\delta[n]+4\delta[n]-6\delta[n+3] \end{align}$$

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Using the sifting property, $$ x[n] \delta[n-m] = x[m]\delta[n-m] $$

specifically applied when $x[n]$ is $\delta[n-k]$ where $k$ is an integer, $$ \delta[n-k] \delta[n-m] = \delta[m-k]\delta[n-m] $$

which is equivalent to $\delta[n-m]$ when $k=m$ and otherwise identically $0$, when $m \neq k$ .

You can apply the result term by term. Give it a try and tell us your result.

Hint: (from your post) $$h_1[n] = 2\delta[n-2] - 3\delta[n+1]$$ $$h_2[n] = \delta[n-1] + 2\delta[n+2]$$ $$h[n]=(h_1[n]*h_2[n])$$

I suggest you at this point (for these very short impulse responses) to use the distribution property of convolution and the convolution property of the $\delta[n]$.

i.e. $$ (x_1[n] + x_2[n]) \star h[n] = x_1[n] \star h[n] + x_2[n] \star h[n] $$

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    $\begingroup$ I don't see how the distribution property applies here. Or does $\delta$ also have convolution property? $\endgroup$ – mavavilj Mar 30 '17 at 16:01
  • $\begingroup$ if you look at my last line, you can see the distrbution of convolution over addition there. Then you shall replace $x_1[n]$, $x_2[n]$ and $h[n]$ with appropriate signals. Then you will apply convolution property of impulse to obtain the final result. $\endgroup$ – Fat32 Mar 30 '17 at 21:12

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