Important Examples¶
\[\DeclareMathOperator{\sinc}{sinc}\]
General Tips¶
Integration techniques
integration by substitution
integration by parts
Rectangular Function¶
\[\begin{split}\pi (t) =
\begin{cases}
1 & -0.5 \le t \le 0.5
\\
0 & t < -0.5 \lor t > 0.5
\end{cases}
\implies
\mathcal{F} \pi (s) = \sinc (s)\end{split}\]
Proof:
\[\begin{split}\mathcal{F} \pi (s) & =
\int_{-\infty}^{+\infty} e^{-2 \pi i s t} \pi(t) dt
\\ & =
\int_{-0.5}^{0.5} e^{-2 \pi i s t} dt
\\ & =
-\frac{1}{2 \pi i s} \cdot e^{-2 \pi i s t} |_{-0.5}^{0.5}
\\ & =
-\frac{1}{2 \pi i s} (e^{- \pi i s} - e^{\pi i s})
\\ & =
-\frac{1}{2 \pi i s}
(\cos (\pi s) - i \cdot \sin (\pi s) - \cos (\pi s) - i \cdot \sin (\pi s))
\\ & =
-\frac{- 2 i \cdot \sin (\pi s)}{2 \pi i s}
\\ & =
\frac{\sin (\pi s)}{\pi s}
\\ & =
\sinc(x)\end{split}\]
\[\tag*{$\blacksquare$}\]
Triangle Function¶
\[\begin{split}\Lambda (t) =
\begin{cases}
t + 1 & -1 \le t \lt 0
\\
-t + 1 & 0 \le t \le 1
\\
0 & t < -1 \lor t > 1
\end{cases}
\implies
\mathcal{F} \Lambda (s) = \sinc^2 (s)\end{split}\]
Proof:
\[ \begin{align}\begin{aligned}\begin{split}\mathcal{F} \Lambda (s) & =
\int_{-\infty}^{\infty} e^{-2 \pi i s t} \Lambda (t) dt
\\ & =
\int_{-1}^0 e^{-2 \pi i s t} (t + 1) dt +
\int_0^1 e^{-2 \pi i s t} (-t + 1) dt
\\ & =
\int_{-1}^0 e^{-2 \pi i s t} t dt +
\int_{-1}^0 e^{-2 \pi i s t} dt -
\int_0^1 e^{-2 \pi i s t} t dt +
\int_0^1 e^{-2 \pi i s t} dt
\\ & =
\int_{-1}^0 e^{-2 \pi i s t} t dt -
\int_0^1 e^{-2 \pi i s t} t dt +
\int_{-1}^1 e^{-2 \pi i s t} dt\end{split}\\\begin{split}\\\end{split}\\\begin{split}\because
\int_{-1}^0 e^{-2 \pi i s t} t dt -
\int_0^1 e^{-2 \pi i s t} t dt & =
\int_{-1}^0 e^{-2 \pi i s t} t dt +
\int_1^0 e^{-2 \pi i s t} t dt
\\ & =
\int_{-1}^0 e^{-2 \pi i s t} t dt +
\int_{-1}^0 e^{-2 \pi i s (-u)} (-u) d(-u)
\\ & =
\int_{-1}^0 e^{-2 \pi i s t} t dt +
\int_{-1}^0 e^{2 \pi i s u} u du
\\ & =
\int_{-1}^0 (e^{-2 \pi i s t} + e^{2 \pi i s t}) t dt
\\ & =
\int_{-1}^0 (
\cos (2 \pi s t) -
i \cdot \sin (2 \pi s t) +
\cos (2 \pi s t) +
i \cdot \sin (2 \pi s t)) t dt
\\ & =
2 \int_{-1}^0 \cos (2 \pi s t) t dt
\\ & =
2 \frac{1}{2 \pi s} \int_{-1}^0 t d \sin (2 \pi s t)
\\ & =
\frac{1}{\pi s} (t \cdot \sin (2 \pi s t) |_{-1}^0 -
\int_{-1}^0 \sin (2 \pi s t) dt)
\\ & =
- \frac{-1 \cdot \sin (2 \pi s \cdot (-1))}{\pi s} -
\frac{1}{-2 \pi^2 s^2} \cos (2 \pi s t) |_{-1}^0
\\ & =
- \frac{\sin (2 \pi s)}{\pi s} +
\frac{1}{2 \pi^2 s^2} (1 - \cos (2 \pi s))
\\ & =
- \frac{\sin (2 \pi s)}{\pi s} +
\frac{\sin^2 (\pi s)}{\pi^2 s^2}
,\end{split}\\\begin{split}\\
\int_{-1}^1 e^{-2 \pi i s t} dt & =
-\frac{1}{2 \pi i s} \cdot e^{-2 \pi i s t} |_{-1}^{1}
\\ & =
-\frac{1}{2 \pi i s} (e^{-2 \pi i s} - e^{2 \pi i s})
\\ & =
-\frac{1}{2 \pi i s}
(\cos (2 \pi s) - i \cdot \sin (2 \pi s) - \cos (2 \pi s) - i \cdot \sin (2 \pi s))
\\ & =
-\frac{-2 i \cdot \sin(2 \pi s)}{2 \pi i s}
\\ & =
\frac{\sin (2 \pi s)}{\pi s}\end{split}\\\begin{split}\\
\therefore
\mathcal{F} \Lambda (s) & =
\int_{-1}^0 e^{-2 \pi i s t} t dt -
\int_0^1 e^{-2 \pi i s t} t dt +
\int_{-1}^1 e^{-2 \pi i s t} dt
=
- \frac{\sin (2 \pi s)}{\pi s} +
\frac{\sin^2 (\pi s)}{\pi^2 s^2} +
\frac{\sin (2 \pi s)}{\pi s}
\\ & =
\frac{\sin^2 (\pi s)}{\pi^2 s^2}
\\ & =
\sinc^2 (s)\end{split}\end{aligned}\end{align} \]
\[\tag*{$\blacksquare$}\]
Gaussian function¶
\[f (t) = e^{- \pi t^2}
\implies
\mathcal{F} f (s) = f (s)\]
Proof:
\[ \begin{align}\begin{aligned}F(s) = \mathcal{F} f (s)
=
\int_{-\infty}^{\infty} e^{-2 \pi i s t} f (t) dt
=
\int_{-\infty}^{\infty} e^{-2 \pi i s t} e^{- \pi t^2} dt\\\begin{split}\\\end{split}\\\begin{split}F'(s) & = \int_{-\infty}^{\infty}
\frac{d}{ds} e^{-2 \pi i s t} e^{- \pi t^2} dt
\\ & =
\int_{-\infty}^{\infty}
e^{- \pi t^2} \frac{d}{ds} e^{-2 \pi i s t} dt
\\ & =
\int_{-\infty}^{\infty}
e^{- \pi t^2} \cdot (-2 \pi i t) \cdot e^{-2 \pi i s t} dt
\\ & =
i \cdot \int_{-\infty}^{\infty}
e^{-2 \pi i s t} \cdot (-2 \pi t) \cdot e^{- \pi t^2} dt
\\ & =
i \cdot \int_{-\infty}^{\infty}
e^{-2 \pi i s t} d e^{- \pi t^2}
\\ & =
i \cdot e^{-2 \pi i s t} \cdot e^{- \pi t^2} |_{-\infty}^{+\infty} -
i \cdot \int_{-\infty}^{+\infty}
e^{- \pi t^2} d e^{-2 \pi i s t}\end{split}\\\begin{split}\\\end{split}\\\begin{split}\because
\lim_{t \to -\infty} e^{-2 \pi i s t} \cdot e^{- \pi t^2} & =
\lim_{t \to +\infty} e^{2 \pi i s t} \cdot e^{- \pi t^2}
\\ & =
\lim_{t \to +\infty} \frac{e^{2 \pi i s t}}{e^{\pi t^2}}
\\ & =
\lim_{t \to +\infty}
\frac{\cos (2 \pi s t) + i \cdot \sin(2 \pi s t)}{e^{\pi t^2}}
\\ & =
0,\end{split}\\\begin{split}\\\end{split}\\\begin{split}\lim_{t \to +\infty} e^{-2 \pi i s t} \cdot e^{- \pi t^2} & =
\lim_{t \to +\infty} e^{-2 \pi i s t} \cdot e^{- \pi t^2}
\\ & =
\lim_{t \to +\infty} \frac{e^{-2 \pi i s t}}{e^{\pi t^2}}
\\ & =
\lim_{t \to +\infty}
\frac{\cos (2 \pi s t) - i \cdot \sin(2 \pi s t)}{e^{\pi t^2}}
\\ & =
0\end{split}\\\begin{split}\\\end{split}\\\begin{split}\therefore
F'(s) & =
i \cdot e^{-2 \pi i s t} \cdot e^{- \pi t^2} |_{-\infty}^{+\infty} -
i \cdot \int_{-\infty}^{+\infty}
e^{- \pi t^2} d e^{-2 \pi i s t}
\\ & =
-i \cdot \int_{-\infty}^{+\infty}
e^{- \pi t^2} d e^{-2 \pi i s t}
\\ & =
-i \cdot \int_{-\infty}^{+\infty}
(-2 \pi i s) \cdot e^{- \pi t^2} \cdot e^{-2 \pi i s t} dt
\\ & =
-2 \pi s \int_{-\infty}^{+\infty}
e^{-2 \pi i s t} \cdot e^{- \pi t^2} dt
\\ & =
-2 \pi s \cdot F(s)\end{split}\\\begin{split}\\\end{split}\\\begin{split}\therefore
\mathcal{F} f(s) & =
F(s)
=
e^{- \pi s^2}
\\ & =
f(s)\end{split}\end{aligned}\end{align} \]
\[\tag*{$\blacksquare$}\]
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