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.

z-plane frequency response

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 $$ \begin{align} s(t)\fourier S(j\omega) &=\frac{1}{T}\sum_{n=-\infty}^{\infty}2\pi\delta(\omega-n\omega_0)\nonumber\\ &=\frac{2\pi}{T}\sum_{n=-\infty}^{\infty}\delta(\omega-n\omega_0) \end{align} $$
Embedded software developer
Passionately curious and stubbornly persistent. Enjoys to inspire and consult with others to exchange the poetry of logical ideas.

Leave a Reply

Your email address will not be published. Required fields are marked *

 

This site uses Akismet to reduce spam. Learn how your comment data is processed.