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I have two functions in a convolution problem, and I'm struggling to wrap my head conceptually around what I'm supposed to do. The functions are: \begin{align*} h(t) &= \delta(t-1) + \delta(t-3)\\ x(t) &= \delta(t-3) - 2\delta(t-4), \end{align*}

and I want to find $y(t) = x(t)*h(t).$

paul garrett
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Samuel Tomp
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2 Answers2

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The spirit of the thing is that $\delta*f=f$ for almost any $f$. Similarly, letting $T_af(x)=f(x+a)$ be the translation operator, $(T_a\delta)*f=T_af$.

It does require something (there are several choices) to believe/prove/know that $\delta*\delta=\delta$... but, granting this, $(T_a\delta)*(T_b\delta)=T_{a+b}\delta$.

paul garrett
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The convolution is defined as \begin{align*} x(t) * h(t) &=\int_{-\infty}^{\infty}x(\tau)\,h(t-\tau)\,d\tau\\ &=\int_{-\infty}^{\infty}[\delta(\tau-3) - 2\delta(\tau-4)]\,[\delta(t-\tau-1) + \delta(t-\tau-3)]\,d\tau\\ &=\int_{-\infty}^{\infty}\delta(\tau-3)\,\delta(t-\tau-1)\,d\tau\\ &+\int_{-\infty}^{\infty}\delta(\tau-3)\,\delta(t-\tau-3)\,d\tau\\ &-2\int_{-\infty}^{\infty}\delta(\tau-4)\,\delta(t-\tau-1)\,d\tau\\ &-2\int_{-\infty}^{\infty}\delta(\tau-4)\,\delta(t-\tau-3)\,d\tau. \end{align*} Now the way the Dirac delta function works, it "picks out" values under the integral sign. The fact that the other function is a Dirac delta function doesn't change this. So we obtain: \begin{align*} x(t) * h(t) &=\delta(t-3-1)+\delta(t-3-3)-2\delta(t-4-1)-2\delta(t-7-3)\\ &=\delta(t-4)+\delta(t-6)-2\delta(t-5)-2\delta(t-10). \end{align*}

Adrian Keister
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  • Actually, the fact that the other function is a $\delta$ does change the nature of the game considerably. It matters a lot that it's a convolution, as opposed to just an integral. – paul garrett Oct 02 '19 at 17:45
  • @paulgarrett: I suppose if you had $\int_{-\infty}^{\infty}\delta(t-a),\delta(t-a),dt,$ then it would matter. Is that what you mean? – Adrian Keister Oct 02 '19 at 17:52
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    Yes, a purported pointwise square of $\delta$s is a problem, and some mildly serious care is required to see how "convolution" evidently side-steps this. – paul garrett Oct 02 '19 at 19:55
  • Ha! That's a great phrase: "mildly serious care". I'll have to remember that one! – Adrian Keister Oct 02 '19 at 20:22