Generalized Fourier Transform: Delta Distribution¶
Sampling Property of Delta¶
Mathematical meaning of taking samples is multiplying by \(\delta\).
\[f \delta_a = f(a) \delta_a\]
Proof:
\[\begin{split}\langle f \delta_a, \phi \rangle & =
\langle \delta_a, f \phi \rangle
\\ & =
(f \phi) (a)
\\ & =
f(a) \phi (a)
\\ & =
\langle f(a) \delta_a, \phi \rangle\end{split}\]
\[\therefore
f \delta_a = f(a) \delta_a\]
\[\tag*{$\blacksquare$}\]
Convolution Property of Delta¶
\[f * \delta_a = f(x - a)\]
Proof:
\[\begin{split}\langle f * \delta_a, \phi \rangle & =
\langle f, \phi * \delta_a^- \rangle
\\ & =
\langle f, \phi * \delta_{-a} \rangle
\\ & =
\langle f(x), \int_{-\infty}^{+\infty} \delta_{-a} (x - y) \phi (y) dy \rangle
\\ & =
\langle f(x), \int_{-\infty}^{+\infty} \delta_{-a} (u) \phi (x - u) du \rangle
\\ & =
\langle f(x), \langle \delta_{-a} (u), \phi(x - u) \rangle \rangle
\\ & =
\langle f(x), \phi(x + a) \rangle
\\ & =
\int_{-\infty}^{+\infty} f(x) \phi(x + a) dx
\\ & =
\int_{-\infty}^{+\infty} f(u - a) \phi(u) du
\\ & =
\langle f(x - a), \phi(x) \rangle\end{split}\]
\[\therefore
f * \delta_a = f(x - a)\]
\[\tag*{$\blacksquare$}\]
Specifically:
\[\begin{split}f * \delta & = f
\\
\delta * \delta & = \delta
\\
\delta_a * \delta_b & = \delta_{a+b}\end{split}\]
Scaling Property of Delta¶
\[\delta (ax) = \frac{1}{|a|} \delta\]
Proof:
Suppose \(a > 0\):
\[\begin{split}\langle \delta (ax), \phi(x) \rangle & =
\int_{-\infty}^{+\infty} \delta (ax) \phi(x) dx
\\ & =
\frac{1}{a} \int_{-\infty}^{+\infty} \delta (u) \phi(\frac{u}{a}) du
\\ & =
\frac{1}{a} \phi(0)
\\ & =
\langle \frac{1}{a} \delta (x), \phi (x) \rangle\end{split}\]
\[\therefore
a \gt 0 \implies \delta (ax) = \frac{1}{a} \delta\]
Similarly:
\[ \begin{align}\begin{aligned}a \lt 0 \implies \delta (ax) = -\frac{1}{a} \delta\\\begin{split}\\\end{split}\\\therefore
\delta (ax) = \frac{1}{|a|} \delta\end{aligned}\end{align} \]
\[\tag*{$\blacksquare$}\]
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