Fourier Series
We first introduce fourier series, where we decompose a periodic signal into a sum of sines and cosines.
We use the complex form of fourier series.
Some book uses the triangular form of fourier series, that is,
S ( t ) = ∑ k = 0 k = + ∞ a k cos ( 2 π k t ) + b k sin ( 2 π k t ) S(t) = \sum_{k=0}^{k=+\infty} a_k \cos(2 \pi k t) + b_k \sin(2 \pi k t) S ( t ) = k = 0 ∑ k = + ∞ a k cos ( 2 πk t ) + b k sin ( 2 πk t ) Dirichlet tells us that, for a periodic signal, if, for any given t 0 t_0 t 0 f ( t 0 − 0 ) f(t_0-0) f ( t 0 − 0 ) f ( t 0 + 0 ) f(t_0+0) f ( t 0 + 0 ) α > 0 \alpha > 0 α > 0
∫ 0 α ∣ ∣ f ( t 0 + τ ) − f ( t 0 + 0 ) ∣ ∣ τ d τ \int_{0}^{\alpha} \frac{||f(t_0 + \tau) - f(t_0 + 0)||}{\tau} \mathrm{d} \tau ∫ 0 α τ ∣∣ f ( t 0 + τ ) − f ( t 0 + 0 ) ∣∣ d τ ∫ 0 α ∣ ∣ f ( t 0 + τ ) − f ( t 0 − 0 ) ∣ ∣ τ d τ \int_{0}^{\alpha} \frac{||f(t_0 + \tau) - f(t_0 - 0)||}{\tau} \mathrm{d} \tau ∫ 0 α τ ∣∣ f ( t 0 + τ ) − f ( t 0 − 0 ) ∣∣ d τ Then there exists a way to use,
S ( t ) = ∑ k = − ∞ k = + ∞ a i e j k ω t S(t) = \sum_{k=-\infty}^{k=+\infty} a_i e^{j k \omega t} S ( t ) = k = − ∞ ∑ k = + ∞ a i e jkω t Where,
ω = 2 π T \omega = \frac{2 \pi}{T} ω = T 2 π To express the periodic signal f ( t ) f(t) f ( t )
Where,
S ( t 0 ) = 1 2 ( f ( t 0 + 0 ) + f ( t 0 − 0 ) ) S(t_0) = \frac{1}{2} (f(t_0 + 0) + f(t_0 - 0)) S ( t 0 ) = 2 1 ( f ( t 0 + 0 ) + f ( t 0 − 0 )) As for the coefficients, we can have
∫ 0 T e j p ω t e j q ω t d t = ∫ 0 T e j ( p + q ) ω t d t = ∫ 0 T cos ( ( p + q ) ω t ) d t + j ∫ 0 T sin ( ( p + q ) ω t ) d t = { 0 if p + q ≠ 0 T if p + q = 0 \int_{0}^{T} e^{jp\omega t} e^{jq\omega t} \mathrm{d}t \\
= \int_{0}^{T} e^{j(p+q)\omega t} \mathrm{d}t \\
= \int_{0}^{T} \cos((p+q)\omega t) \mathrm{d}t + j \int_{0}^{T} \sin((p+q)\omega t) \mathrm{d}t \\
= \begin{cases}
0 & \text{if } p + q \neq 0 \\
T & \text{if } p + q = 0
\end{cases} ∫ 0 T e j p ω t e j q ω t d t = ∫ 0 T e j ( p + q ) ω t d t = ∫ 0 T cos (( p + q ) ω t ) d t + j ∫ 0 T sin (( p + q ) ω t ) d t = { 0 T if p + q = 0 if p + q = 0 So we can conclude that, the coefficients are,
a k = 1 T ∫ 0 T f ( t ) e − j k ω t d t a_k = \frac{1}{T} \int_{0}^{T} f(t) e^{-j k \omega t} \mathrm{d}t a k = T 1 ∫ 0 T f ( t ) e − jkω t d t This is the definition of fourier series.
Non-periodic signals are just signals with infinite period. Previously, a i a_i a i F ( ω ) F(\omega) F ( ω )
F ( ω ) = ∫ − ∞ + ∞ f ( t ) e − j ω t d t F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{-j \omega t} \mathrm{d}t F ( ω ) = ∫ − ∞ + ∞ f ( t ) e − jω t d t And inversely,
f ( t ) = 1 2 π ∫ − ∞ + ∞ F ( ω ) e j ω t d ω f(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(\omega) e^{j \omega t} \mathrm{d}\omega f ( t ) = 2 π 1 ∫ − ∞ + ∞ F ( ω ) e jω t d ω This is the fourier transform.
We can note,
F ( f ( t ) ) = F ( ω ) \mathcal{F}(f(t)) = F(\omega) F ( f ( t )) = F ( ω ) Or,
f ( t ) → F F ( ω ) f(t) \xrightarrow{\mathcal{F}} F(\omega) f ( t ) F F ( ω ) Some integration doesn't converge, and we force them to be by using the cauchy principle, which is,
P . V . ∫ − ∞ + ∞ f ( t ) e − j ω t d t = lim T → + ∞ ∫ − T + T f ( t ) e − j ω t d t P.V. \int_{-\infty}^{+\infty} f(t) e^{-j \omega t} \mathrm{d}t = \lim_{T \to +\infty} \int_{-T}^{+T} f(t) e^{-j \omega t} \mathrm{d}t P . V . ∫ − ∞ + ∞ f ( t ) e − jω t d t = T → + ∞ lim ∫ − T + T f ( t ) e − jω t d t However, this is sometimes also hard to calculate. We sometime also utilize the property that such functions are limits of other converge functions to get a value.
Delta Function
∫ − ∞ + ∞ δ ( t ) e − j ω t d t = 1 \int_{-\infty}^{+\infty} \delta(t) e^{-j \omega t} \mathrm{d}t = 1 ∫ − ∞ + ∞ δ ( t ) e − jω t d t = 1 δ ( t ) → F 1 \delta(t) \xrightarrow{\mathcal{F}} 1 δ ( t ) F 1 Single Side Exponential Function
∫ − ∞ + ∞ e − α t e − j ω t u ( t ) d t = 1 j ω + α \int_{-\infty}^{+\infty} e^{-\alpha t} e^{-j \omega t} u(t) \mathrm{d}t = \frac{1}{j \omega + \alpha} ∫ − ∞ + ∞ e − α t e − jω t u ( t ) d t = jω + α 1 e − α t u ( t ) → F 1 j ω + α e^{-\alpha t} u(t) \xrightarrow{\mathcal{F}} \frac{1}{j \omega + \alpha} e − α t u ( t ) F jω + α 1 Even Double Side Exponential Function
∫ − ∞ + ∞ ( u ( t ) e − α t + u ( − t ) e α t ) e − j ω t d t = 1 α + j ω + 1 α − j ω = 2 α α 2 + ω 2 \int_{-\infty}^{+\infty} (u(t)e^{-\alpha t} + u(-t)e^{\alpha t}) e^{-j \omega t} \mathrm{d}t \\= \frac{1}{\alpha + j\omega} + \frac{1}{\alpha - j\omega} = \frac{2 \alpha }{\alpha^2 + \omega^2} ∫ − ∞ + ∞ ( u ( t ) e − α t + u ( − t ) e α t ) e − jω t d t = α + jω 1 + α − jω 1 = α 2 + ω 2 2 α Constant
c = lim α → 0 c ( u ( t ) e − α t + u ( − t ) e α t ) c = \lim_{\alpha \to 0} c(u(t)e^{-\alpha t} + u(-t)e^{\alpha t}) c = α → 0 lim c ( u ( t ) e − α t + u ( − t ) e α t ) We can prove that,
α π ( α 2 + ω 2 ) \frac{\alpha }{\pi(\alpha^2 + \omega^2)} π ( α 2 + ω 2 ) α is a valid probability density.
We can integrate it. You can either use the triangular function,
∫ − ∞ + ∞ α π ( α 2 + ω 2 ) d ω = ∫ − ∞ + ∞ 1 π ( 1 + ( ω α ) 2 ) d ω α = tan θ = ω α ∫ − π 2 π 2 1 π d θ = 1 \int_{-\infty}^{+\infty} \frac{\alpha }{\pi(\alpha^2 + \omega^2)} \mathrm{d} \omega \\
= \int_{-\infty}^{+\infty} \frac{1}{\pi(1 + (\frac{\omega}{\alpha})^2)} d\frac{\omega}{\alpha} \\
\overset{\tan \theta = \frac{\omega}{\alpha}}{=} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\pi} d\theta = 1 ∫ − ∞ + ∞ π ( α 2 + ω 2 ) α d ω = ∫ − ∞ + ∞ π ( 1 + ( α ω ) 2 ) 1 d α ω = t a n θ = α ω ∫ − 2 π 2 π π 1 d θ = 1 Or use the residual theorem,
∫ − ∞ + ∞ α π ( α 2 + ω 2 ) d ω = ∫ − ∞ + ∞ 1 2 π ( 1 α − j ω + 1 α + j ω ) d ω = 2 π 2 π = 1 \int_{-\infty}^{+\infty} \frac{\alpha }{\pi(\alpha^2 + \omega^2)} \mathrm{d} \omega \\
= \int_{-\infty}^{+\infty} \frac{1}{2\pi} (\frac{1}{\alpha - j\omega} + \frac{1}{\alpha + j \omega}) \mathrm{d} \omega \\
= \frac{2\pi}{2\pi} = 1 ∫ − ∞ + ∞ π ( α 2 + ω 2 ) α d ω = ∫ − ∞ + ∞ 2 π 1 ( α − jω 1 + α + jω 1 ) d ω = 2 π 2 π = 1 This is actually the Lorentzian Distribution.
As α → 0 \alpha \to 0 α → 0
lim α → 0 α π ( α 2 + ω 2 ) = { 0 ω ≠ 0 + ∞ ω = 0 \lim_{\alpha \to 0} \frac{\alpha }{\pi(\alpha^2 + \omega^2)} = \begin{cases}
0 & \omega \neq 0 \\
+\infty & \omega = 0
\end{cases} α → 0 lim π ( α 2 + ω 2 ) α = { 0 + ∞ ω = 0 ω = 0 That is to say,
lim α → 0 α π ( α 2 + ω 2 ) = δ ( ω ) \lim_{\alpha \to 0} \frac{\alpha }{\pi(\alpha^2 + \omega^2)} = \delta(\omega) α → 0 lim π ( α 2 + ω 2 ) α = δ ( ω ) So,
c → F 2 π c δ ( ω ) c \xrightarrow{\mathcal{F}} 2\pi c \delta(\omega) c F 2 π cδ ( ω ) ∫ − ∞ + ∞ ( u ( t ) e − α t − u ( − t ) e α t ) e − j ω t d t = 1 α + j ω − 1 α − j ω = 2 j ω α 2 + ω 2 \int_{-\infty}^{+\infty} (u(t)e^{-\alpha t} - u(-t)e^{\alpha t}) e^{-j \omega t} \mathrm{d}t \\= \frac{1}{\alpha + j\omega} - \frac{1}{\alpha - j\omega} = \frac{2 j\omega }{\alpha^2 + \omega^2} ∫ − ∞ + ∞ ( u ( t ) e − α t − u ( − t ) e α t ) e − jω t d t = α + jω 1 − α − jω 1 = α 2 + ω 2 2 jω Sign Function
s g n ( t ) = lim α → 0 ( u ( t ) e − α t − u ( − t ) e α t ) sgn(t) = \lim_{\alpha \to 0} (u(t)e^{-\alpha t} - u(-t)e^{\alpha t}) s g n ( t ) = α → 0 lim ( u ( t ) e − α t − u ( − t ) e α t ) Thus,
s g n ( t ) → F 2 j ω sgn(t) \xrightarrow{\mathcal{F}} \frac{2 j }{\omega} s g n ( t ) F ω 2 j Step Function
You may think,
u ( t ) = lim α → 0 e − α t u ( t ) u(t) = \lim_{\alpha \to 0} e^{-\alpha t} u(t) u ( t ) = α → 0 lim e − α t u ( t ) So,
u ( t ) → F 1 j ω u(t) \xrightarrow{\mathcal{F}} \frac{1}{j \omega} u ( t ) F jω 1 This is wrong because, for integration that doesn't converge, we use the Cauchy Principal Value.
However,
P . V . ∫ − ∞ + ∞ e − j ω t u ( t ) d t ≠ lim α → 0 ∫ + 0 + ∞ e − α t e − j ω t d t P.V. \int_{-\infty}^{+\infty} e^{-j \omega t} u(t) \mathrm{d}t \neq \lim_{\alpha \to 0} \int_{+0}^{+\infty} e^{-\alpha t} e^{-j \omega t} \mathrm{d}t P . V . ∫ − ∞ + ∞ e − jω t u ( t ) d t = α → 0 lim ∫ + 0 + ∞ e − α t e − jω t d t Since the integration doesn't approach two ends as the same speed.
Because,
u ( t ) = 1 2 + 1 2 s g n ( t ) u(t) = \frac{1}{2} + \frac{1}{2} sgn(t) u ( t ) = 2 1 + 2 1 s g n ( t ) And since integration is linear, the fourier transform should also be linear, and thus,
u ( t ) → F π δ ( t ) + 1 j ω u(t) \xrightarrow{\mathcal{F}} \pi \delta(t) + \frac{1}{j\omega} u ( t ) F π δ ( t ) + jω 1 Gate Function
∫ − ∞ + ∞ rect ϵ ( t ) e − j ω t d t = ∫ − ϵ + ϵ 1 2 ϵ e − j ω t d t = − 1 2 ϵ ∫ − j ω ϵ + j ω ϵ e − j ω t d ( − j ω t ) = 1 2 ϵ j ω ( e j ω ϵ − e − j ω ϵ ) = 2 ϵ sin ( ω ϵ ) ω ϵ \int_{-\infty}^{+\infty} \text{rect}_{\epsilon}(t) e^{-j \omega t} \mathrm{d}t \\
= \int_{-\epsilon}^{+\epsilon} \frac{1}{2\epsilon} e^{-j \omega t} \mathrm{d}t \\
= -\frac{1}{2\epsilon} \int_{-j\omega\epsilon}^{+j\omega\epsilon} e^{-j \omega t} \mathrm{d}(-j\omega t) \\
= \frac{1}{2\epsilon j \omega} (e^{j \omega \epsilon} - e^{-j \omega \epsilon}) \\
= 2\epsilon \frac{\sin(\omega \epsilon)}{\omega \epsilon} ∫ − ∞ + ∞ rect ϵ ( t ) e − jω t d t = ∫ − ϵ + ϵ 2 ϵ 1 e − jω t d t = − 2 ϵ 1 ∫ − jω ϵ + jω ϵ e − jω t d ( − jω t ) = 2 ϵ jω 1 ( e jω ϵ − e − jω ϵ ) = 2 ϵ ω ϵ sin ( ω ϵ ) This function is important. Thus we define sinc \text{sinc} sinc
We define,
sinc ( t ) = sin ( π t ) π t \text{sinc}(t) = \frac{\sin(\pi t)}{\pi t} sinc ( t ) = π t sin ( π t ) Thus,
Rect ϵ ( t ) = 2 ϵ sinc ( ω ϵ π ) \text{Rect}_\epsilon(t) = 2\epsilon \text{sinc}(\frac{\omega \epsilon}{\pi}) Rect ϵ ( t ) = 2 ϵ sinc ( π ω ϵ )