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If: $y(t) = x(t)*h(t)$ and $g(t) = x(9t)*h(9t)$

(Where * is convolution)

How can I use properties of the Fourier transform to show:

$g(t) = Ay(Bt)$

and find constants?

I think A should be $1/81$ but I can only see that from first principles and not from a property.

user2290362
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  • The abuse of notation makes your question unclear but I think I get it. Hint: how does Fourier transform behaves with respect to convolution? – Siméon Aug 22 '14 at 10:08
  • The convolution is the inverse transform of the product of the transforms. I know that Y(jw) would therefore be X(jw)H(jw) but how can I express G(jw) in the same way? – user2290362 Aug 22 '14 at 10:20

1 Answers1

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For $a>0$,

$\mathcal{F}\{g(a t)\} = \int_{-\infty}^\infty g(a t) e^{- j \omega t} d t = \int_{-\infty}^\infty g(u) e^{-j \omega {u/a}} \frac{d u}{a} = \int_{-\infty}^\infty g(u) e^{-j (\omega/a) u} \frac{d u}{a} = G(\omega/a)/a$

via the $u$-substitution $u=at$, where $\mathcal{F}\{g(t)\} = G(\omega)$. The case for $a<0$ can be treated similarly.

Now use the convolution turning into multiplication by Fourier transform.

Batman
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