# Fourier transform

$$\require{AMSsymbols}\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \require{cancel} \newcommand\ccancel[2][black] {\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black] {\color{#1}{\cancelto{#2}{\color{black}{#3}}}} \def\doubleunderline#1{\underline{\underline{#1}}}$$ Jean-Baptiste Joseph Fourier was a French mathematician and physicist well known for his Fourier Series. The Fourier transform is named in his honor.

## Laplace transform

The single-sided Laplace transform maps a time-domain function $$f(t)$$ to a $$s$$-domain function $$F(s)$$.

$$\mathfrak{L}\left\{\,f(t)\,\right\}=F(s)=\int_{0^-}^\infty e^{-st}f(t)\ \mathrm{d}t \label{eq:laplace1}$$

where $$s$$ represents a complex-frequency

$$s = \sigma + j\omega \label{eq:s}$$

Using the Laplace transform, a derivative such as $$f^\prime(t)=\frac{df(t)}{dt}$$ maps to the multiplication $$sF(s)$$. In other words, the Laplace transform maps a linear differential equation to a simple algebraic equation. This makes it useful for solving linear differential equations. For an example refer to RLC Filters.

## Causality

This single-sided (unilateral) Laplace transfer is suited for casual systems, where the output depends on past and current inputs but does not depend on future inputs. Any real physical system is a casual system. In a causal system, the impulse response, $$h(t)$$, is zero for time $$t\lt 0$$.

$$\forall_t\in\mathbb{R},t\lt 0:h(t) = 0$$

In non-causal systems, the output also depends on inputs. An example is a central moving average or image compression. For non-causal system, the two-sided (bilateral) Laplace transform must be used as defined by the integral

$$\mathfrak{B}\left\{\,f(t)\,\right\} = F(s) = \int_{-\infty}^\infty e^{-st}f(t)\ \mathrm{d}t \label{eq:laplace2}$$

## Fourier transform

If we set the real part, $$\sigma$$ of the complex variable $$s$$ to zero in equation $$\eqref{eq:s}$$

$$s = j\omega\label{eq:sjw}$$

Substituting $$\eqref{eq:sjw}$$ in $$\eqref{eq:laplace2}$$, results in $$F(j\omega)$$, which is essentially the frequency domain representation of $$f(t)$$. With $$j$$ a constant, we can write $$F(\omega)$$ instead of $$F(j\omega)$$ and define the double-sided Fourier Transform of a continuous-time signal $$f(t)\in \mathbb{C}$$ as

$$\shaded{ F(\omega)\ \triangleq \int_{-\infty}^{\infty}e^{-j\omega t}\ f(t)\ dt } \label{eq:fourier2}$$

provided that $$x(t)$$ is can be integrated, i.e. it does not go to infinity.

$$\int_{-\infty}^{\infty}|x(t)|\ dt\lt\infty$$

## Integrable

Many input functions, such as $$x(t)=t$$ and $$x(t)=e^t$$, are not integrable and their Fourier transform do not exist. Other signals have a Fourier transforms that contains non-conventional functions like $$\delta(\omega)$$, like $$x(t)=1$$ or $$x(t)=\cos(\omega_0 t)). To overcome this difficulty, we can multiply the given \(x(t)$$ by a exponential decaying factor

$$e^{-\sigma t},\ \sigma\in\mathbb{R}$$

so that $$x(t)e^{-\sigma t}$$ may be integrable for certain values $$\sigma$$. Applying this factor, the Fourier transform becomes

\begin{align}F(\omega)\ &= \int_{-\infty}^{\infty}e^{-j\omega t}\ e^{-\sigma t}\ f(t)\ dt \nonumber \\ &= \int_{-\infty}^{\infty}e^{-(\sigma+j\omega)t}\ f(t)\ dt\end{align}

The result is a function of a complex variable $$s=\sigma+j\omega$$, and is defined as the bilateral Laplace transform that we started with $$\eqref{eq:laplace2}$$.

$$F(s) = \int_{-\infty}^\infty e^{-st}f(t)\ \mathrm{d}t=\mathfrak{B}\left\{\,f(t)\,\right\}$$

### Inverse Fourier transform

For completeness the inverse Fourier transform:

$$f(t)\ =\ \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{j\omega t}\ F(\omega)\ d\omega \label{eq:fourier2inv}$$

—-

### Discrete Time Fourier Transform

As part of the Z-transform, we defined:

$$z\triangleq\mathrm{e}^{sT} \nonumber$$ where $$s=\sigma+j\omega$$

To find the frequency response, we follow the same methodology as for the Discrete Time Fourier Response and evaluate the $$z$$-domain expression $$F(z)$$ along $$s=j\omega$$ where $$z = \left.\mathrm{e}^{sT}\right|_{s=\omega T} = \mathrm{e}^{j\omega T} \label{eq:zunitcircle}$$

In the $$z$$-plane, angular frequency are shown in normalized form, where the normalized angular frequency $$\omega T$$ is the angle with the positive horizontal axis, what places $$\omega T=0$$ at $$1$$ on the positive horizontal axis. Positive frequencies go in a counter-clockwise pattern from there, occupying the upper semicircle. Negative frequencies form the lower semicircle. The positive and negative frequencies meet at the common point of $$\omega T=\pi$$ and $$\omega T=-\pi$$. This implies that the expression $$\mathrm{e}^{j\omega T}$$ corresponds to the unit circle. As we will see later, this circular geometry corresponds to the periodicity of the frequency spectrum of the discrete signal.

This evaluation along $$\mathrm{e}^{j\omega T}$$ is called the Discrete Time Fourier Transform (DTFT). \begin{align} f[n] \fourier & \left.F(z)\right|_{z=\mathrm{e}^{j\omega T}}=\left.\sum_{n=0}^\infty z^{-n}\ f[n]\right|_{z=\mathrm{e}^{j\omega T}} \nonumber \\ \fourier & F(\mathrm{e}^{j\omega T})=\sum_{n=0}^\infty (\mathrm{e}^{j\omega T})^{-n}\ f[n] \end{align}

The Discrete Time Fourier Transform follows as $$\shaded{ f[n]\,\fourier\, F(\mathrm{e}^{j\omega T})=\sum_{n=0}^\infty \mathrm{e}^{-jn\omega T}\ f[n] }$$

Since $$\omega$$ is a continuous variable, there are an infinite number of possible values for $$\omega T$$ from $$0$$ to $$2\pi$$ or from $$-\pi$$ to $$\pi$$. [src: MIT-ocw]

Let there be $$N$$ samples equally spaced around the unit circle $$w_k=\frac{2\pi k}{N},\quad k=0,1,\ldots,N-1$$ and define the N samples of $$F(\mathrm{e}^{j\omega T})$$ \begin{align} F[k] &\triangleq \left.F(\mathrm{e}^{j\omega T})\right|_{\omega=\frac{2\pi k}{N}}\nonumber\\ &= \sum_{n=0}^\infty \mathrm{e}^{-jn\frac{2\pi k}{N} T}\ f[n] \end{align}

Define a shorthand notation $$W_n\,\triangleq\,\mathrm{e}^{-j\frac{2\pi}{N}}$$ so that \begin{align} F[k] &= \sum_{n=0}^\infty (W_{\small N})^{nkT}\ f[n] \end{align}

Since $$f[n]$$ is can be infinity long and the summation is to infinity, we can still not compute $$F[k]$$. Even if $$f[n]$$ were finite, we $$F[k]$$ is sampling the DTFT, and might not be able to recover $$f[n]$$. There are some required condition for which we willl be able to recover $$f[n]$$ from the $$F[k]$$.

If you restrict the Z-transform to the unit circle in the $$z$$-plane, you then get the Discrete Time Fourier Transform.

see http://www.ece.rutgers.edu/~psannuti/ece345/FT-DTFT-DFT.pdf see https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-341-discrete-time-signal-processing-fall-2005/lecture-notes/lec15.pdf see https://ccrma.stanford.edu/~jos/st/DFT_Definition.html

To derive the frequency-domain relation of ideal sampling, consider the Fourier transform of $$f^{\star}$$ from equation $$\eqref{eq:fstar}$$

$$f^{\star}(t) \triangleq f(t)\Delta(t) = f(t)\sum_{n=0}^{\infty}{\delta(t-nT)} \nonumber$$

Since $$f^{\star}(t)$$ is the product of $$f^(t)$$ and $$\Delta(t)$$, the Fourier transform is the convolution of the Fourier transforms $$f(t)$$ and $$\Delta(t)$$. When expressed in Hz, the transform is scaled by $$\frac{1}{2\pi}$$ 

2BD: finish this

## my scrap book

#### $$X_s(j\Omega)$$

=============begin (don’t think we need this) Once more, let’s start with equation $$\eqref{eq:fstarnT}$$ and $$\eqref{eq:fstar0}$$, again bilateral
$$x_s(t)=x_c(t)\,s(t)=\sum_{n=-\infty}^{\infty}{x_c(nT)\ \delta(t-nT)}\label{eq:xst2}$$
The Continuous Time Fourier Transform (CTFT) is defined as
$$F(t)\fourier F(j\Omega)=\int_{-\infty}^{\infty}f(t)\,\mathrm{e}^{-j\Omega t}\mathrm{d}t\nonumber$$
Applying the CTFT to $$\eqref{eq:xst2}$$ \begin{align} F(t)\fourier F(j\Omega) &=\int_{-\infty}^{\infty}\overbrace{\sum_{n=-\infty}^{\infty}{x_c(nT)\ \delta(t-nT)}}^{x_s(t)}\,\mathrm{e}^{-j\Omega t}\mathrm{d}t \end{align} Tmpulse function $$\delta(t-nT)$$ is $$0$$ everywhere but at $$t=nT$$, the “sifting property”, so we can replace $$\mathrm{e}^{-j\Omega t}$$ with $$\mathrm{e}^{-j\Omega nt}$$, and bring all that is independent of $$t$$ out of the integration \begin{align} F(t)\fourier F(j\Omega) &=\int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{x_c(nT)\ \delta(t-nT)}\,\mathrm{e}^{-j\Omega nT}\mathrm{d}t\nonumber\\ &=\sum_{n=-\infty}^{\infty}\left({x_c(nT)\,\mathrm{e}^{-j\Omega nT}\underbrace{\int_{-\infty}^{\infty}\delta(t-nT)}_{=1}\,\mathrm{d}t}\right)\nonumber\\ &=\sum_{n=-\infty}^{\infty}{x_c(nT)\,\mathrm{e}^{-j\Omega nT}\,\mathrm{d}t}\nonumber\\ \end{align} Since $$x[n]\triangleq x_c(nT)$$ \begin{align} x(t)\fourier X(j\Omega) &=\sum_{n=-\infty}^{\infty}{x[n]\,\mathrm{e}^{-j\Omega nT}\,\mathrm{d}t}\label{eq:ft4} \end{align} With the Z-transform
\begin{align} f[n] \ztransform F(z)&=\sum_{n=0}^\infty z^{-n}\ f[n]\nonumber\\ \text{where }z&=\mathrm{e}^{sT}\nonumber\\ \text{and }f[n]&=f(nT)\nonumber \end{align}\nonumber
Evaluate this $$F(z)$$ at $$z=j\omega$$, so that $$z=\mathrm{e}^{j\omega T}$$ \begin{align} \left.F(z)\right|_{s=j\omega} &=\left.\sum_{n=0}^\infty z^{-n}\,f[n]\right|_{s=j\omega}&\Rightarrow\nonumber\\ F(\mathrm{e}^{j\omega}) &=\sum_{n=0}^\infty (\mathrm{e}^{j\omega T})^{-n}\,f[n]\nonumber\\ &=\sum_{n=0}^\infty \mathrm{e}^{-j\omega nT}\,f[n]\label{eq:fjomega} \end{align} Combining equation $$\eqref{eq:fjomega}$$ to $$\eqref{eq:ft4}$$ $$X(j\Omega) =\left.X(\mathrm{e}^{j\omega})\right|_{\omega=\Omega T} =X(\mathrm{e}^{j\Omega T})\label{eq:Xjo}$$ The term $$X(\mathrm{e}^{j\omega})$$ is simply a frequency-scaled version of $$X(j\Omega)$$, with the frequency scaling specified by $$\omega=\Omega T$$. This scaling can be thought of as a normalization of the frequency axis so that the frequency $$\Omega=\Omega_s$$ in (X(j\Omega)\) is normalized to $$\omega=2\pi$$ for $$X(\mathrm{e}^{j\omega})$$. The fact that there is a frequency scaling or normalization in the transformation from $$X(j\Omega)$$ to $$X(\mathrm{e}^{j\omega})$$ is directly associated with the fact that there is a time normalization in the transformation from $$x_s(t)$$ to $$x[n]$$. Specifically, $$x_s(t)$$ remains a spacing between samples equal to the sampling period $$T$$. In contrast, the spacing of sequence values $$x[n]$$ is always unity: i.e. the time axis is normalize by a factor $$T$$. Correspondingly, in the trequency domain, the frequency axis is normalized by a factor of $$f_s=\frac{1}{T}$$. Combining equation $$\eqref{eq:XsjOmega}$$ to $$\eqref{eq:Xjo}$$ \begin{align} X_s\left(\mathrm{e}^{j\Omega T}\right) &=\frac{1}{T}\sum_{k=-\infty}^{\infty}X_c{\large(j\Omega}-jk\Omega_s{\large)} \end{align} Substitute $$\omega=\Omega T$$ and $$\Omega_s=\frac{2\pi}{T}$$ \begin{align} X_s\left(\mathrm{e}^{j\omega}\right) &=\frac{1}{T}\sum_{k=-\infty}^{\infty}X_c \left(j \left[\tfrac{\omega}{T}-\tfrac{2\pi k}{T}\right]\right) \end{align} =============end Fourier transform of Dirac Comb Let’s start with the Fourier transform of the periodic impulse train $$s(t)$$ [stackexchange] $$s(t)=\sum_{n=-\infty}^{\infty}\delta(t-nT)\label{eq:st5}$$ The Fourier transform is defined as
\begin{align} x(t)&=\sum_{n=-\infty}^{\infty}c_n\,e^{jn\omega_0t}\nonumber\\ \text{where }c_n&=\frac{1}{T}\int_{-T/2}^{T/2}x(t)\,\mathrm{e}^{-jn\omega_0t}\mathrm{d}t\nonumber \end{align}\nonumber where $$c_n$$ are exponential Fourier series coefficients and $$\omega_o$$ is the fundamental frequency
Apply the exponential Fourier series to equation $$\eqref{eq:st5}$$ \begin{align} s(t)&=\sum_{n=-\infty}^{\infty}c_n\,e^{jn\omega_0t}\label{eq:xt7}\\ c_n &=\frac{1}{T}\int_{-T/2}^{T/2}\sum_{n=-\infty}^{\infty}\delta(t-nT)\,\mathrm{e}^{-jn\omega_0t}\mathrm{d}t\nonumber\\ &=\frac{1}{T}\sum_{n=-\infty}^{\infty}\underbrace{\int_{-T/2}^{T/2}\delta(t-nT)}\,\mathrm{e}^{-jn\omega_0t}\mathrm{d}t \end{align} Observe the integral period from $$-\frac{T}{2}$$ to $$-\frac{T}{2}$$. During this period, only a single impulse $$\delta(t)$$ exists. All the other impulses occur before or after the integration period. Consequently, we can rewrite $$c_n$$ as $$c_n=\frac{1}{T}\underbrace{\sum_{n=-\infty}^{\infty}\delta(t)}\,\mathrm{e}^{-jn\omega_0t}\mathrm{d}t$$ Applying the sifting property, $$\delta(t)$$ only has a value at $$t=0$$ $$c_n=\frac{1}{T}\,\mathrm{e}^{-jn\omega_0\color{blue}{0}}=\frac{1}{T}$$ Substitute the value of $$c_n$$ in $$\eqref{eq:xt7}$$ $$s(t)=\sum_{n=-\infty}^{\infty}\frac{1}{T}\,e^{jn\omega_0t} =\frac{1}{T}\sum_{n=-\infty}^{\infty}e^{jn\omega_0t}\nonumber\\$$ Recall the Fourier transform
$$\mathrm{e}^{jat}\fourier 2\pi \delta (\omega-a)\nonumber$$
The Fourier transform of the impulse train follows as \begin{align} s(t)\fourier S(j\omega) &=\frac{1}{T}\sum_{n=-\infty}^{\infty}2\pi\delta(\omega-n\omega_0)\nonumber\\ &=\frac{2\pi}{T}\sum_{n=-\infty}^{\infty}\delta(\omega-n\omega_0) \end{align}
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Passionately curious and stubbornly persistent. Enjoys to inspire and consult with others to exchange the poetry of logical ideas.

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