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So my question looks like this:

Suppose that we have two LTI systems with impulse responses $$h_1(t) = \frac 12\delta(t-1)\quad\text{and}\quad h_2(t)=2\delta(t+1).$$ Determine whether these systems are inverses of one another.

Here's what I've done but I'm stuck. I'd appreciate some help with this

$$h_1(t)\star h_2(t)\overset{?}{=}\delta(t)$$ \begin{align} \int_{-\infty}^\infty h_1(\tau)h_2(t-\tau)d\tau &=\int_{-\infty}^\infty \frac12\delta(\tau-1)2\delta(t-\tau+1)d\tau\\ &=\int_{-\infty}^\infty \delta(\tau-1)\delta(t-\tau+1)d\tau\\ \end{align}

$$u = \tau - 1\implies \frac{du}{d\tau}=1\implies du = d\tau$$

$$\implies \int_{-\infty}^\infty h_1(\tau)h_2(t-\tau)d\tau = \int_{-\infty}^\infty \delta(u)\delta(t-u)d\tau$$

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    $\begingroup$ Welcome to SP.SE! Your question appears to be homework / self-study, however you've asked it as per our guidelines, so I think it's a good question. I'ved added tags of homework and self-study. Feel free to delete whichever one doesn't apply. $\endgroup$
    – Peter K.
    Oct 10, 2016 at 22:57

3 Answers 3

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So you have $$ \int_{-\infty}^{+\infty} \delta(u) \delta(t-u) du $$

Try using the sifting property of $\delta(t)$:

$$ \int_{-\infty}^{+\infty} f(u) \delta(T-u) du = f(T) $$

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  • $\begingroup$ Thanks, I'm not very strong in applying the sifting property yet. $\endgroup$
    – Drew U
    Oct 10, 2016 at 23:46
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You may also look at the transfer function ($s$-domain) of the two systems: $$H_1(s)=\frac{1}{2}e^{-s}$$ $$H_2(s)=2e^{s}$$ And the two systems in series will result in $$H_1(s)H_2(s)=1$$ which means they are inverse of each other.

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    $\begingroup$ (+1) but I would say, Z-domain for continuous time impulses ? $\endgroup$
    – Fat32
    Oct 11, 2016 at 0:56
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in addition to Peter K. who suggested you to use sifting property, I would also suggest you to use convolution property of delta function:

$$ x(t) \star \delta(t-a) = x(t-a) $$

You shall now replace $x(t)$ with the other impulse...

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  • $\begingroup$ I'm a little confused as to how I would apply this though. the x(t) function? you mean h1(t)? $\endgroup$
    – Drew U
    Oct 10, 2016 at 23:44
  • $\begingroup$ hmm $h_1(t) = \frac12 \delta (t-1)$ and $h_2(t) = 2 \delta (t+1)$ you should have applied convolution property as $h_1(t) \star h_2(t) = \frac12 \delta (t-1) \star 2 \delta (t+1) = \delta (t-1) \star \delta (t+1) = \delta ((t+1)-1) = \delta (t)$ ... And yes! the two systems are inverses. $\endgroup$
    – Fat32
    Oct 11, 2016 at 0:16
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    $\begingroup$ (+1) I think they are the same @Fat32 (the integral in the shifting property is convolution) $\endgroup$
    – msm
    Oct 11, 2016 at 0:31
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    $\begingroup$ @msm they are ? (sifting and convolution property) are not the same things. Multiplication with impulse is sifting: $x(t)\cdot \delta(t-a) = x(a) \delta (t-a)$ where as the convolution with impulse shifts the signal. I think when two impulses happen they seem to be the same. $\endgroup$
    – Fat32
    Oct 11, 2016 at 0:32
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    $\begingroup$ In that sense, you are right. I was referring to the integral form (sifting property). $\endgroup$
    – msm
    Oct 11, 2016 at 0:50

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