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Convolution between 2 functions $f$ and $g$ is defined as $$(f\star g)(x) = \int_{-\pi}^\pi f(x') g(x-x') dx'.$$

Shift invariance of convolution is said to be the property that $$f(x-x_0) \star g(x) = (f \star g)(x-x_0).$$

Firstly, what does $f(x-x_0) \star g(x)$ represent and why $ f(x-x_0) \star g(x) = (f \star g)(x-x_0)$?

user10024395
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1 Answers1

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$$(f\star g)(x-x_0) = \int_{-\pi}^\pi f(x') g(x-x_0-x') dx'.$$ Aplly the transformation of variables $y=x'+x_0$, then $dy=dx'$ and the integral becomes $$\int_{-\pi+x_0}^{\pi+x_0} f(y-x_0) g(x-y) dy$$ But $$f(x-x_0) \star g(x)=\int_{-\pi}^{\pi} f(y-x_0) g(x-y) dy$$ Then if $f$ is periodic with period $2\pi$ then $f(x-x_0) \star g(x) = (f \star g)(x-x_0)$

QED
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  • How do you get your "but" statement? – user10024395 Nov 20 '17 at 03:18
  • This is how the convolution is defined. $g$ convoluted with shifted $f$. Think of $f(x-x_0)$ as a function $h(x)$ taking the value $f(a-x_0)$ at $x=a$ – QED Nov 20 '17 at 03:42
  • I am still confused. Should $f(x-x_0) \star g(x) = \int_{-\pi}^{\pi} f(x-x_0)g(x-x') dx'$ then? – user10024395 Nov 20 '17 at 07:30
  • $f(x-x_0) \star g(x) = \int_{-\pi}^{\pi} h(x')g(x-x') dx'=\int_{-\pi}^{\pi} f(x'-x_0)g(x-x') dx'$ – QED Nov 20 '17 at 08:25
  • yup which does not match what follows in "but". I meant shouldn't ... then? in my previous comment. – user10024395 Nov 20 '17 at 08:28
  • It follows exactly. Only the variable here is named $x'$ instead of $y$ used in the solution – QED Nov 20 '17 at 08:31
  • The problem is $y = x' + x_0$ as you define earlier so $y \neq x'$? How is that equal then? – user10024395 Nov 20 '17 at 08:34
  • It doesnt matter what the varibale of integration is. If $f:X\rightarrow Y$ is an integrable function then $\int f(y)dy$ and $\int f(x)dx$ bears the same meaning – QED Nov 20 '17 at 09:26