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I am studying wavelets and it has been given that

$$ \psi_{a,b} = \frac{1}{\sqrt{|a|}} \psi \left(\frac{t-b}{a}\right) $$

now the function

$$ \psi(t)= \begin{cases} 1,& \text{if } 0\leq t<\frac 12\\ -1, & \text{if } \frac 12\leq t<1\\ 0& \text{otherwise} \end{cases} $$

is given as in terms of previous equation

$$ \psi_{a,b}= \frac {1} {\sqrt{a}}\left[u(t-a)-2u\left(t-b-\frac a2\right)+u(t-b-a)\right] $$

when $a>0$ and

$$ \psi_{a,b}=- \frac {1} {\sqrt{-a}}\left[u(t-a)-2u\left(t-b-\frac a2\right)+u(t-b-a)\right] $$

when $a<0$.

My issue is, how can $a$ which is a dilation parameter lead to something like a negative function when $a<0$?

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Simply put, $\frac{t}{-a} =\frac{-t}{a}$.

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    $\begingroup$ Thanks!!! I got it. I was not able to do it graphically, but now I can. $\endgroup$ – Userhanu Nov 29 '16 at 15:31
  • $\begingroup$ Perfect, I was almost confused of the short response. It is quite interested to look at the property in the Fourier domain: $\phi(at)\to \frac{1}{|a|}\Phi(f/a)$. You can see how the sign change on $f$ due to $a=-1$ can transfer to $t$ in the $e^{-i2\pi ft}$, and you have to intervert the two $-\infty$ and $\infty$ bounds on the integral, hence the sign change. $\endgroup$ – Laurent Duval Nov 29 '16 at 15:37

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