Fourier Transform¶

Complex form or exponential form of Fourier Integral

𝔉 [f (t)] = \int_{- \infty}^{\infty} e^{- i ω t} f (t) d t

assume that

\begin{matrix} \int_{- \infty}^{\infty} | f (x) | d x \end{matrix}

converges

Comparison
1. compare to Laplace transform
  
  $\begin{array}{r} s \to i ω \\ \int_{0}^{\infty} \to \int_{- \infty}^{\infty} \end{array}$
2. compare to Fourier Series Fourier Transform can be regarded as Fourier series whose $p \to \infty$ .

Amplitude Spectrum¶

The amplitude spectrum of $f (t)$ is a graph of $| 𝔉 [f (ω)] |$

Fourier & Inverse Fourier Transform¶

Fourier transform

$𝔉 [f (x)] = \int_{- \infty}^{\infty} e^{- i ω x} f (x) d x$
inverse Fourier transform

$𝔉^{- 1} [F (ω)] = \frac{1}{2 π} \int_{- \infty}^{\infty} F (ω) e^{i ω x} d ω = f (x)$

Fourier Sine Transform¶

Fourier sine transform

$𝔉_{s} [f (x)] = \int_{0}^{\infty} f (x) \sin (ω x) d x = F (ω)$
inverse Fourier sine transform

$𝔉_{s}^{- 1} [F (ω)] = \frac{2}{π} \int_{0}^{\infty} F (ω) \sin (ω x) d ω = f (x)$

Fourier Cosine Transform¶

Fourier cosine transform

𝔉_{c} [f (x)] = \int_{0}^{\infty} f (x) \cos (ω x) d x = F (ω)

inverse Fourier cosine transform

𝔉_{c}^{- 1} [F (ω)] = \frac{2}{π} \int_{0}^{\infty} F (ω) \cos (ω x) d ω = f (x)

Fourier Transform properties¶

Linearity¶

\begin{matrix} α, β \in ℝ \\ 𝔉 [α f (t) + β g (t)] = α 𝔉 [f (t)] + 𝔉 [β g (t)] \end{matrix}

Time Shifting¶

\begin{matrix} 𝔉 [f (t - t_{0})] = e^{- i ω t_{0}} 𝔉 [f (t)] \end{matrix}

Inverse Time Shifting

$\begin{matrix} 𝔉^{- 1} [e^{- i ω t_{0}} F (ω)] (t) = f (t - t_{0}) \end{matrix}$

Frequency Shifting¶

\begin{matrix} 𝔉 [e^{i ω_{0} t} f (t)] = F (ω - ω_{0}) \end{matrix}

Inverse Frequency Shifting

$\begin{array}{r} 𝔉^{- 1} [F (ω - ω_{0})] = e^{i ω_{0} t} f (t) \end{array}$

Scaling¶

\begin{matrix} 𝔉 [f (c t)] = \frac{1}{| c |} F (\frac{ω}{c}) \end{matrix}

Inverse Scaling

\begin{matrix} 𝔉^{- 1} [F (\frac{ω}{c})] = | c | f (c t) \end{matrix}

Time Reversal¶

\begin{matrix} 𝔉 [f (- t)] = F (- ω) \\ 𝔉^{- 1} [F (- ω)] = f (- t) \end{matrix}

Symmetry¶

\begin{matrix} 𝔉 [F (t)] = 2 π f (- ω) \end{matrix}

modulation¶

\begin{aligned} 𝔉 [f (t) \cos (ω_{0} t)] & = \frac{1}{2} [F (ω + ω_{0}) + F (ω - ω_{0})] \\ 𝔉 [f (t) \sin (ω_{0} t)] & = \frac{i}{2} [F (ω + ω_{0}) - F (ω - ω_{0})] \end{aligned}

particular example¶

prof. Laplace Transform

\begin{matrix} 𝔉 [H (t) e^{- a t}] = \frac{1}{a + i ω} \end{matrix}

Frequency Differentiation¶

\begin{matrix} 𝔉 [t^{n} f (t)] = (i)^{n} F^{(n)} \end{matrix}

Transforms of Derivatives¶

for Fourier transform

$\begin{array}{r} 𝔉 [f^{'} (x)] = i ω F (ω) \\ 𝔉 [f^{(n)}] = (i ω)^{n} F (ω) \end{array}$
for Fourier sine transform

$\begin{array}{r} 𝔉_{s} [f^{'} (x)] = - ω 𝔉_{c} [f (x)] \\ 𝔉_{s} [f^{″} (x)] = - ω^{2} 𝔉_{s} [f (x)] + ω f (0) \end{array}$
for Fourier cosine transform

$\begin{array}{r} 𝔉_{c} [f^{'} (x)] = ω 𝔉_{s} [f (x)] - f (0) \\ 𝔉_{c} [f^{″} (x)] = - ω^{2} 𝔉_{c} [f (x)] - f^{'} (0) \end{array}$

Transform of Integral¶

\begin{array}{r} 𝔉 [\int_{- \infty}^{t} f (τ) d τ] = \frac{1}{i ω} F (ω) \end{array}

Convolution¶

Definition

\begin{matrix} f (t) * g (t) = \int_{- \infty}^{\infty} f (τ) g (t - τ) d τ \end{matrix}

Property

\begin{aligned} f * g & = g * f \\ (α f + β g) * h & = α (f * h) + β (g * h) \end{aligned}

Theorem

\begin{matrix} \int_{- \infty}^{\infty} [f (t) * g (t)] d t = \int_{- \infty}^{\infty} f (t) d t \cdot \int_{- \infty}^{\infty} g (t) d t \\ 𝔉 [f (t) * g (t)] = F (ω) G (ω) \\ 𝔉 [f (t) g (t)] = \frac{1}{2 π} F (ω) * G (ω) \end{matrix}

Dirac Delta Function¶

Definition

$\begin{matrix} δ (t) = lim_{a \to 0} [\frac{1}{2 a} H (t + a) - H (t - a)] \end{matrix}$

Fourier Transform¶

\begin{matrix} 𝔉 [δ (t)] = 𝔉 [lim_{a \to 0} [\frac{1}{2 a} H (t + a) - H (t - a)]] = lim_{a \to 0} \frac{\sin (a ω)}{a ω} = 1 \end{matrix}

Filtering¶

\begin{matrix} \int_{- \infty}^{\infty} f (t) δ (t - t_{0}) d t = f (t_{0}) \end{matrix}

Discrete Fourier Transform¶

Let $u = {u_{j}}_{j = 0}^{N - 1}$

\begin{matrix} D [u] = U_{k} = \sum_{j = 0}^{N - 1} u_{j} \exp (- i ω_{0} j \frac{T}{N}) = \sum_{j = 0}^{N - 1} u_{j} \exp (- i 2 π j \frac{1}{N}) \end{matrix}

Linearity¶

\begin{matrix} D [a u + b v] = a U_{k} + b V_{k} \end{matrix}

Periodicity¶

\begin{matrix} U_{k + N} = U_{k} \end{matrix}

Inverse N-point DFT¶

\begin{matrix} u_{j} = \frac{1}{N} \sum_{j = 0}^{N - 1} U_{k} \exp (i ω_{0} j \frac{T}{N}) = \frac{1}{N} \sum_{j = 0}^{N - 1} U_{k} \exp (i 2 π j \frac{1}{N}) \end{matrix}

Complex Fourier Coefficients¶

\begin{matrix} d_{k} \approx \frac{1}{N} \cdot U_{k} \end{matrix}

note that the coefficient of complex Fourier $C_{ω}$

\begin{matrix} C_{ω} = \frac{1}{2 π} \cdot 𝔉 [f (t)] \end{matrix}

Sampled Fourier Series¶

\begin{matrix} S_{M} (\frac{j T}{n}) \approx \frac{1}{N} \sum_{k = 0}^{N - 1} V_{k} e^{i ω_{0} j k T / N} \end{matrix}

in which

V_{k} = {\begin{aligned} U_{k} & for k = 0, 1, \dots, M \\ 0 & for k = M + 1, \dots, N - M + 1 \\ U_{k} & for k = N - M, \dots, N - 1 \end{aligned}

Solving the BVP¶

BVP (Boundary Value Problem)

There are three possible condition below

$- \infty < x < \infty$

$\Rightarrow$ Fourier transform
$0 < x < \infty$ and $u (x, y) |_{x = 0} = 0$

$\Rightarrow$ Fourier sine transform
$0 < x < \infty$ and $\frac{\partial}{\partial x} u (x, y) |_{x = 0} = 0$

$\Rightarrow$ Fourier cosine transform

Method of Separation of Variable¶

kernel : assume that 1.

u (x, y) = X (x) Y (y)

2.

" f (x) " = " f (y) " = - λ