I have a question regarding the Continuous Wavelet Transform (CWT) of non finite energy functions, such as $g(t) = a\exp(i\omega_0t)$. We know that the CWT is defined for functions in the Hilbert space $L^2(\mathbb{R})$, but I've found in Chapter 4 of "A Wavelet Tour of Signal Processing" that the CWT of $g(t)$ is $(W_\psi g)(u,s) = a\sqrt{s}\hat{\psi^*}(s\omega_0)\exp(i\omega_0u)$, where $\psi$ is a wavelet. How is this possible ? (since $g \not\in L^2(\mathbb{R})$)
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- $\begingroup$ @username $f$ is a function in $L^2(\mathbb{R})$, $s$ and $u$ are the scale and time localization parameters of the CWT. $W_\psi f$ is the CWT operator applied on $f$. $\endgroup$Humberto Gimenes Macedo– Humberto Gimenes Macedo2021-03-29 19:03:15 +00:00Commented Mar 29, 2021 at 19:03
- $\begingroup$ @username I'm sorry, it is the CWT of $g$. $\endgroup$Humberto Gimenes Macedo– Humberto Gimenes Macedo2021-03-29 19:16:31 +00:00Commented Mar 29, 2021 at 19:16
- $\begingroup$ @username You are right, the correct expression is $(W_\psi g)(u,s) = a\sqrt{s}\hat{\psi^*}(s\omega_0)\exp(i\omega_0u)$. $\endgroup$Humberto Gimenes Macedo– Humberto Gimenes Macedo2021-03-29 19:26:42 +00:00Commented Mar 29, 2021 at 19:26
- 1$\begingroup$ If $\psi$ is well behaved, that is, $\psi$ is compactly supported, or exponentially decreasing at infinity, then the transform is well defined as well, isn't it? $\endgroup$username– username2021-03-29 20:02:45 +00:00Commented Mar 29, 2021 at 20:02
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1 Answer
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For a Gabor wavelet, I guess that, up to a different normalization, you have
\begin{align} &(W_\psi g)(y,\eta)=\langle g(x),e^{-π(x-y)^2} e^{2iπ (x-y/2)\eta}\rangle_{L^2(\mathbb R)} =\int g(x)e^{-π(x-y)^2} e^{-2iπ (x-y/2)\eta} dx \\&=\int g(x+y)e^{-πx^2} e^{-2iπ (x+y/2)\eta} dx =a\int e^{2iπ\omega_0(x+y)}e^{-πx^2} e^{-2iπ (x+y/2)\eta} dx \\&=ae^{2iπ\omega_0 y-iπ y\eta} \int e^{-πx^2}e^{-2iπ x(\eta-\omega_0)} dx= ae^{iπy(2\omega_0 -\eta)}e^{-π(\eta-\omega_0)^2}, \end{align} a formula rather similar to yours.
