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 \(F(s)\).
where \(s\) represents a complex-frequency
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 \(t\lt 0\).
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
Fourier transform
If we set the real part, \(\sigma\) of the complex variable \(s\) to zero in equation \(\eqref{eq:s}\)
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
provided that \(x(t)\) is can be integrated, i.e. it does not go to infinity.
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
so that \(x(t)e^{-\sigma t}\) may be integrable for certain values \(\sigma\). Applying this factor, the Fourier transform becomes
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 \(\eqref{eq:laplace2}\).
Inverse Fourier transform
For completeness the inverse Fourier transform:
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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 frequencyApply 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 $$ \begin{align} s(t)\fourier S(j\omega) &=\frac{1}{T}\sum_{n=-\infty}^{\infty}2\pi\delta(\omega-n\omega_0)\nonumber\\ &=\frac{2\pi}{T}\sum_{n=-\infty}^{\infty}\delta(\omega-n\omega_0) \end{align} $$
I enjoy your explanations and proof of theory in the laplace section but it appears this page is still in code form
Thanks for letting me know. Should be fixed.