Operations
Linear Combination
\[\mathcal{F}(a \cdot f + b \cdot g) =
a \cdot \mathcal{F}f + b \cdot \mathcal{F}g\]
Can be proved by applying definition directly.
Shift
\[(\mathcal{F} f(t - b))(s) = e^{-2 \pi i s b} \cdot \mathcal{F} f (s)\]
Proof:
\[ \begin{align}\begin{aligned}(\mathcal{F} f(t - b))(s) =
\int_{-\infty}^{+\infty}
e^{-2 \pi i s t} f(t - b) dt\\\begin{split}\\\end{split}\\Let: u = t - b \implies t = u + b\\\begin{split}\\\end{split}\\\begin{split}(\mathcal{F} f(t - b))(s) & =
\int_{-\infty}^{+\infty}
e^{-2 \pi i s \cdot (u + b)} f(u) du
\\ & =
e^{-2 \pi i s b}
\int_{-\infty}^{+\infty}
e^{-2 \pi i s u} f(u) du
\\ & =
e^{-2 \pi i s b} \cdot \mathcal{F} f (s)\end{split}\end{aligned}\end{align} \]
\[\tag*{$\blacksquare$}\]
Stretch
squashed in time (signal) corresponds to stretched in phase and squashed
in magnitude
stretched in time (signal) corresponds to squeezed in phase and stretched
in magnitude
a signal cannot be both localized in time and frequency
\[(\mathcal{F} f(a t))(s) = |\frac{1}{a}| F(\frac{s}{a})\]
Proof:
\[ \begin{align}\begin{aligned}(\mathcal{F} f(a t))(s) =
\int_{-\infty}^{+\infty} e^{-2 \pi i s t} f(at) dt\\\begin{split}\\\end{split}\\Let: u = a t \implies t = \frac{1}{a} u\\\begin{split}\\\end{split}\\\begin{split}(\mathcal{F} f(a t))(s) & =
\int_{-\infty}^{+\infty}
e^{-2 \pi i s \cdot \frac{u}{a}} f(u) dt
\\ & =
\int_{-\infty}^{+\infty}
e^{-2 \pi i \cdot \frac{s}{a} \cdot u} f(u) dt
\\ & =
|\frac{1}{a}| \int_{-\infty}^{+\infty}
e^{-2 \pi i \cdot \frac{s}{a} \cdot u} f(u) du
\\ & =
|\frac{1}{a}| F(\frac{s}{a})\end{split}\end{aligned}\end{align} \]
\[\tag*{$\blacksquare$}\]
Differentiation
Fourier transform turns differentiation into multiplication.
The Derivitive Theorem of Fourier Transform:
\[ \begin{align}\begin{aligned}\mathcal{F} f^{(n)} & =
(2 \pi i s)^n \cdot \mathcal{F} f\\\begin{split}\\\end{split}\\\frac{\partial^n \mathcal{F} f}{\partial s^n} & =
\mathcal{F} ((-2 \pi i x)^n f(x))\end{aligned}\end{align} \]
Rigorous proof can be found in Differentiation Formulas.
Warning
These formulas hold only if function \(f\) satisfies some specific conditions
e.g. vanish at infinity
Convolution
Convolution Theorem of Fourier Transform:
First statement:
\[ \begin{align}\begin{aligned}\mathcal{F} f \cdot \mathcal{F} g = \mathcal{F}(f * g)\\\begin{split}\\\end{split}\\\mathcal{F}^{-1} F \cdot \mathcal{F}^{-1} G = \mathcal{F}^{-1} (F * G)\end{aligned}\end{align} \]
Second statement:
\[ \begin{align}\begin{aligned}\mathcal{F} (f \cdot g) = \mathcal{F} f * \mathcal{F} g\\\begin{split}\\\end{split}\\\mathcal{F}^{-1} (F \cdot G) = \mathcal{F}^{-1} F * \mathcal{F}^{-1} G\end{aligned}\end{align} \]
Proof:
\[ \begin{align}\begin{aligned}\begin{split}\mathcal{F} g(s) \cdot \mathcal{F} f(s) & =
\int_{-\infty}^{+\infty} e^{-2 \pi i s t} g(t) dt \cdot
\int_{-\infty}^{+\infty} e^{-2 \pi i s t} f(t) dt
\\ & =
\int_{-\infty}^{+\infty} e^{-2 \pi i s t} g(t) dt \cdot
\int_{-\infty}^{+\infty} e^{-2 \pi i s x} f(x) dx
\\ & =
\int_{-\infty}^{+\infty}
\int_{-\infty}^{+\infty}
e^{-2 \pi i s t} \cdot e^{-2 \pi i s x} \cdot g(t) \cdot f(x) dt dx
\\ & =
\int_{-\infty}^{+\infty}
\int_{-\infty}^{+\infty}
e^{-2 \pi i s (t + x)} \cdot g(t) \cdot f(x) dt dx
\\ & =
\int_{-\infty}^{+\infty} (
\int_{-\infty}^{+\infty}
e^{-2 \pi i s (t + x)} \cdot g(t) dt
) f(x) dx\end{split}\\\begin{split}\\\end{split}\\Let: u = t + x \implies t = u - x\\\begin{split}\\\end{split}\\\begin{split}\mathcal{F} g(s) \cdot \mathcal{F} f(s) & =
\int_{-\infty}^{+\infty} (
\int_{-\infty}^{+\infty}
e^{-2 \pi i s (t + x)} \cdot g(t) dt
) f(x) dx
\\ & =
\int_{-\infty}^{+\infty} (
\int_{-\infty}^{+\infty}
e^{-2 \pi i s u} \cdot g(u - x) du
) f(x) dx
\\ & =
\int_{-\infty}^{+\infty} e^{-2 \pi i s u} (
\int_{-\infty}^{+\infty}
g(u - x) \cdot f(x) dx
) du\end{split}\\\begin{split}\\\end{split}\\Let: (f * g)(u) = \int_{-\infty}^{+\infty} g(u - x) \cdot f(x) dx\\\begin{split}\\\end{split}\\\begin{split}\mathcal{F} g(s) \cdot \mathcal{F} f(s) & =
\int_{-\infty}^{+\infty} e^{-2 \pi i s u} (
\int_{-\infty}^{+\infty}
g(u - x) \cdot f(x) dx
) du
\\ & =
\mathcal{F}(f * g)(s)\end{split}\end{aligned}\end{align} \]
similarly:
\[\mathcal{F}^{-1} F \cdot \mathcal{F}^{-1} G = \mathcal{F}^{-1} (F * G)\]
To prove the second statement we start with the first statement of the inverse
Fourier transform:
\[ \begin{align}\begin{aligned}\mathcal{F}^{-1} F \cdot \mathcal{F}^{-1} G = \mathcal{F}^{-1} (F * G)\\\begin{split}\\\end{split}\\\therefore
f \cdot g =
\mathcal{F}^{-1} \left( \mathcal{F} f * \mathcal{F} g \right)\\\begin{split}\\\end{split}\\\therefore
\mathcal{F} \left( f \cdot g \right) =
\mathcal{F} f * \mathcal{F} g\end{aligned}\end{align} \]
Similarly:
\[\mathcal{F}^{-1} (F \cdot G) = \mathcal{F}^{-1} F * \mathcal{F}^{-1} G\]
\[\tag*{$\blacksquare$}\]
Back to Fourier Transform.