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I was reading on windowed fourier transform and wavelet transform, and i was thinking that the windowed fourier transform is a subset of wavelet transform. Is that true?

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  • $\begingroup$ alexandria.tue.nl/repository/books/612762.pdf $\endgroup$
    – user350
    Jan 14, 2014 at 15:16
  • $\begingroup$ there is a way to derive the continuous wavelet transform, at least for a class of mother wavelets in which the gabor wavelet is a subset, from the Short-Time Fourier Transform. $\endgroup$ Jan 14, 2014 at 17:32
  • $\begingroup$ Why is the Gabor wavelet a subset of the stft? @robertbristow-johnson $\endgroup$ Jan 14, 2014 at 17:35
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    $\begingroup$ it's not really, i am saying that you can derive the analysis and synthesis equations of the continuous wavelet transform from the STFT, if the mother wavelet is of the form $w(t) e^{j \omega t}$ . this derivation does not show it for more general wavelets. $\endgroup$ Jan 14, 2014 at 17:40

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Define the Fourier transform as $$ x(t) = \mathcal{F}^{-1}\big\{ X(\omega) \big\} \triangleq\frac{1}{2 \pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} \ d\omega $$

and $$ X(\omega) = \mathcal{F}\big\{ x(t) \big\} = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \ dt $$.

Define the real-valued and non-negative window function $w(t) \ge 0$: $$ \int_{-\infty}^{\infty} w^2(u) \ du = \int_{-\infty}^{\infty} \frac{1}{a} w^2\left( \frac{u}{a} \right) \ du = \int_{-\infty}^{\infty} \frac{1}{a} w^2\left( \frac{t-\tau}{a} \right) \ d\tau \triangleq \frac{c_g}{2 \pi} \quad \forall a \ne 0 \quad \forall t$$ .

Call this Eq (1): $$ \begin{align} x(t) & = x(t) \left[ \frac{2 \pi}{c_g} \int_{-\infty}^{\infty} \frac{1}{a} w^2\left( \frac{t-\tau}{a} \right) \ d\tau \right] \\ & = \frac{2 \pi}{c_g} \int_{-\infty}^{\infty} \left[ x(t) \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \right] \ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \ d\tau \\ \end{align} \quad \forall a>0 $$ .

Define the Short-Time Fourier Transform (STFT) as $$ X_{a,\tau}(\omega) \triangleq \mathcal{F}\left\{ x(t) \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \right\} = \int_{-\infty}^{\infty} x(t) \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) e^{-j\omega t} \ dt $$ .

That means that $$ x(t) \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) = \mathcal{F}^{-1}\big\{ X_{a,\tau}(\omega) \big\} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} X_{a,\tau}(\omega) e^{j\omega t} \ d\omega $$ .

From Eq (1), $$ \begin{align} x(t) & = \frac{2 \pi}{c_g} \int_{-\infty}^{\infty} \left[ x(t) \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \right] \ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \ d\tau \\ & = \frac{2 \pi}{c_g} \int_{-\infty}^{\infty} \left[ \frac{1}{2 \pi} \int_{-\infty}^{\infty} X_{a,\tau}(\omega) e^{j\omega t} \ d\omega \right] \ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \ d\tau \\ & = \frac{1}{c_g} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} X_{a,\tau}(\omega) e^{j\omega t} \ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \ d\omega \ d\tau \\ & = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \left[ \frac{2 \pi}{c_g} \int_{-\infty}^{\infty} X_{a,\tau}(\omega) \ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \ d\tau \right] \ e^{j\omega t} \ d\omega \\ & = \frac{1}{2 \pi} \int_{-\infty}^{\infty} X(\omega) \ e^{j\omega t} \ d\omega \\ \end{align} $$ .

This means that $$ X(\omega) = \frac{2 \pi}{c_g} \int_{-\infty}^{\infty} X_{a,\tau}(\omega) \ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \ d\tau \quad \forall a>0 $$

Since this is true for all $a>0$, we choose $a=\frac{1}{\omega}$ which means $$ X(\omega)=0 \quad \forall \omega \le 0 $$ and consequently $$ \Im\big[ x(t) \big] = \mathcal{H} \left\{ \Re\big[ x(t) \big] \right\} $$

where $\mathcal{H} \{ \cdot \} \ $ is the Hilbert Transform.

So we have $$ x(t) = \frac{1}{c_g} \int_{-\infty}^{\infty} \int_{0}^{\infty} X_{1/\omega,\tau}(\omega) e^{j\omega t} \ \sqrt{\omega} w\left( \omega (t-\tau) \right) \ d\omega \ d\tau $$ .

Substituting in the integral: $\frac{1}{a} \rightarrow \omega$ and $\frac{-1}{a^2} da \rightarrow d\omega$,

$$ \begin{align} x(t) & = \frac{1}{c_g} \int_{-\infty}^{\infty} \int_{0}^{\infty} X_{a,\tau}\left(\frac{1}{a}\right) e^{j\frac{t}{a}} \ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) \frac{1}{a^2} \ da \ d\tau \\ & = \frac{1}{c_g} \int_{-\infty}^{\infty} \int_{0}^{\infty} \left[ X_{a,\tau} \left(\frac{1}{a}\right) e^{j\frac{\tau}{a}} \right] \ \left[ \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) e^{j\frac{t-\tau}{a}} \right] \frac{1}{a^2} \ da \ d\tau \\ \end{align} $$

where $$ g_{a,\tau}(t) \triangleq \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) e^{j\frac{t-\tau}{a}} $$ is the Wavelet,

$$ g(t) \triangleq g_{1,0}(t) = w(t) e^{jt} $$ is the Mother Wavelet, and

$$ g_{a,\tau}(t) = \frac{1}{\sqrt{a}} g\left( \frac{t-\tau}{a} \right) $$,

and

$$ \begin{align} X_{a,\tau} \left(\frac{1}{a}\right) e^{j\frac{\tau}{a}} & = \int_{-\infty}^{\infty} x(t) \frac{1}{\sqrt{a}} w\left( \frac{t-\tau}{a} \right) e^{-j\frac{t-\tau}{a}} \ dt \\ & = \int_{-\infty}^{\infty} x(t) \ g_{a,\tau}^*(t) \ dt \\ \end{align} $$

is the Continuous Wavelet Transform of $x(t)$. The scaler $c_g$ is

$$ c_g = 2 \pi \int_{-\infty}^{\infty} w^2(t) \ dt = 2 \pi \int_{-\infty}^{\infty} \big| g(t) \big|^2 \ dt = \int_{-\infty}^{\infty} \big| G(\omega) \big|^2 \ d\omega $$ .

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    $\begingroup$ Okay, to answer the original question, if $X_{a,\tau}(\omega)$ is the STFT of $x(t)$ as defined above given a particular window $w(t)$, then $ X_{a,\tau}(1/a) \ e^{j \tau/a} $ is the Continuous Wavelet Transform of $x(t)$. $\endgroup$ Jan 15, 2014 at 7:00
  • $\begingroup$ just for everyone's information, the notation using $g(t)$ for the mother wavelet and $c_g$ for that is from this 1988 Kronland & Martinet paper in Computer Music Journal which you can read online for free if you sign up with JSTOR. but they use $s(t)$ and $b$ instead of $x(t)$ and $\tau$ and they do not relate the CWT to the STFT and the Fourier Transform, which i try to do above. $\endgroup$ Jan 15, 2014 at 20:27
  • $\begingroup$ Robert: So the STFT and CWT are the same up to a complex exponential term? $\endgroup$ Feb 15, 2014 at 4:05
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Wavelets commonly factor the frequency of the basis vectors into their temporal widths, some with more than one temporal location. Windowed FFTs fix the temporal width of each basis vector to the length of the FFT, thus independent of any bin frequency, and with the basis location fixed to the center of the FFT.

I suppose one could consider the FFT to be similar to a wavelet transform with a bad set of wavelets (too many high frequency ones and with poor temporal/locality resolution, and too few low ones and with poor frequency resolution).

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