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I don't understand a part of the solution of an exercise and would be grateful if you could help.

Note: $\psi(t)$ and $x(t)$ are time continuous signals (german: "zeitkontinuierliche Signale"). The signal $\psi(t)$ is the same as in the Wavelet Transform https://en.wikipedia.org/wiki/Continuous_wavelet_transform

The follwing equality needs to hold:

$\int_{-\infty}^{\infty} x(\tau)\frac{1}{a}\psi (\frac{\tau -t}{a}) d\tau \overset{!}{=} \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau$

So $h(t-\tau) \overset{!}{=} \frac{1}{a}\psi (\frac{\tau -t}{a})$, i.e. $h(t) \overset{!}{=} \frac{1}{a}\psi (\frac{-t}{a})$

My question is about the last equality, $h(t) \overset{!}{=} \frac{1}{a}\psi (\frac{-t}{a})$. Why does shifting $h(t-\tau)$ by $+ \tau$ implies shifting $\psi (\frac{\tau -t}{a})$ by $-\tau$ and not $+\tau \ $?

Thanks for you help !

Martin Argerami
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2 Answers2

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Because your emphasis is on "shifting", I'm not entirely sure what you are asking. But if $u=t-\tau $, then $\tau-t=-u $. It's just that substitution.

Martin Argerami
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  • So the $t$ in the last equation is not the same $t$ as before, is that so ? Like $t_{new}=t_{old}- \tau$ thus $-t_{new}= \tau -t_{old}$ Is that correct ? –  Jan 12 '20 at 14:46
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    Yes exactly. When you think about it, variable names are irrelevant, it's just a symbol. – Martin Argerami Jan 12 '20 at 14:49
  • Ah okay Thanks a lot for your help. I just find it really confusing that they don't tell us that they use the same variable names for different purposes in the same exercise. –  Jan 12 '20 at 14:50
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You don't shift the function, but the variable $t$: if you shift $t$ by $+\tau$, you get

$$t-\tau\rightarrow (t+\tau)-\tau=t$$ And $$\tau-t\rightarrow\tau-(t+\tau)=-t$$

Francesco Bernardini
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