Solution Notes -- Example Problem Set 8

1.

Member of this family of wavelets are self-Fourier, because they have exactly the same functional form as their Fourier Transforms. Just their parameters are interchanged or inverted in the functional form in the two domains.

2.

The modulation parameter $\mu_{0}$ in f(x) becomes the location parameter of the wavelet $F(\mu)$ which is its Fourier transform. The position parameter x₀ of wavelet f(x) becomes the modulation parameter of the wavelet $F(\mu)$ which is its Fourier transform. Hence modulation and shifting are dual. Finally, the scale parameter $\alpha$ sets the size of the wavelet f(x), and the reciprocal of $\alpha$ is the size of the wavelet $F(\mu)$ which is the Fourier transform of f(x); this expresses the duality of scaling (``similarity").

3.

The Fourier transform of the sum of any two Gabor wavelets is itself the sum of two Gabor wavelets.

4.

The Fourier transform of the product of any two Gabor wavelets is a single Gabor wavelet. This is because the product of any two Gabor wavelets is itself just a new Gabor wavelet (the arguments of the exponentials simply add within each exponential), and the conclusion then follows from (1).

5.

The Fourier transform of f⁽ⁿ⁾(x), the n^th-derivative of a Gabor wavelet, is:

$$F(\mu) = (i\mu)^{n}e^{-ix_{0}\mu} e^{-(\mu-\mu_{0})^{2}\alpha^{2}}$$

6.

The duality of convolution and multiplication implies that the convolution of any two Gabor wavelets has a Fourier transform which is the product of two Gabor wavelets. But we noted above in (4) that such a product is just a single Gabor wavelet. Combining this result with the self-Fourier property, we conclude that the convolution of any two Gabor wavelets is itself, yet again, just a single Gabor wavelet. Hence this family of functions is closed under convolution, just as it is under multiplication.