Fourier transform

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

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

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). $$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathscr{F}}} \begin{align} f[n] \fourier & \left.F(z)\right|_{z=\mathrm{e}^{j\omega T}}=\left.\sum_{n=0}^\infty z^{-n}\ f[n]\right|_{z=\mathrm{e}^{j\omega T}} \nonumber \\ \fourier & F(\mathrm{e}^{j\omega T})=\sum_{n=0}^\infty (\mathrm{e}^{j\omega T})^{-n}\ f[n] \end{align} $$

The Discrete Time Fourier Transform follows as $$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathscr{F}}} \shaded{ f[n]\,\fourier\, F(\mathrm{e}^{j\omega T})=\sum_{n=0}^\infty \mathrm{e}^{-jn\omega T}\ f[n] } $$

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 will 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

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathscr{F}}} F(t)\fourier F(j\Omega)=\int_{-\infty}^{\infty}f(t)\,\mathrm{e}^{-j\Omega t}\mathrm{d}t\nonumber $$

Applying the CTFT to $\eqref{eq:xst2}$ $$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathscr{F}}} \begin{align} F(t)\fourier F(j\Omega) &=\int_{-\infty}^{\infty}\overbrace{\sum_{n=-\infty}^{\infty}{x_c(nT)\ \delta(t-nT)}}^{x_s(t)}\,\mathrm{e}^{-j\Omega t}\mathrm{d}t \end{align} $$ Impulse 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 $$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathscr{F}}} \begin{align} F(t)\fourier F(j\Omega) &=\int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}{x_c(nT)\ \delta(t-nT)}\,\mathrm{e}^{-j\Omega nT}\mathrm{d}t\nonumber\\ &=\sum_{n=-\infty}^{\infty}\left({x_c(nT)\,\mathrm{e}^{-j\Omega nT}\underbrace{\int_{-\infty}^{\infty}\delta(t-nT)}_{=1}\,\mathrm{d}t}\right)\nonumber\\ &=\sum_{n=-\infty}^{\infty}{x_c(nT)\,\mathrm{e}^{-j\Omega nT}\,\mathrm{d}t}\nonumber\\ \end{align} $$ Since $x[n]\triangleq x_c(nT)$ $$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathscr{F}}} \begin{align} x(t)\fourier X(j\Omega) &=\sum_{n=-\infty}^{\infty}{x[n]\,\mathrm{e}^{-j\Omega nT}\,\mathrm{d}t}\label{eq:ft4} \end{align} $$ With the Z-transform

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathscr{Z}}} \begin{align} f[n] \ztransform F(z)&=\sum_{n=0}^\infty z^{-n}\ f[n]\nonumber\\ \text{where }z&=\mathrm{e}^{sT}\nonumber\\ \text{and }f[n]&=f(nT)\nonumber \end{align}\nonumber $$

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 frequency 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

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathscr{F}}} \mathrm{e}^{jat}\fourier 2\pi \delta (\omega-a)\nonumber $$

The Fourier transform of the impulse train follows as $$ \begin{align} s(t)\fourier S(j\omega) &=\frac{1}{T}\sum_{n=-\infty}^{\infty}2\pi\delta(\omega-n\omega_0)\nonumber\\ &=\frac{2\pi}{T}\sum_{n=-\infty}^{\infty}\delta(\omega-n\omega_0) \end{align} $$

2 Replies to “Fourier transform”

John Ansah-Antwi says:

2022-03-25 at 08:31

I enjoy your explanations and proof of theory in the laplace section but it appears this page is still in code form
Coert Vonk says:

2022-03-25 at 19:54

Thanks for letting me know. Should be fixed.

Comments are closed.

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Laplace transform

Causality