Fourier Transform

\(\require{AMSsymbols}\def\doubleunderline#1{\underline{\underline{#1}}}\)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 \(t<0\).

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

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.

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

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 \(\eqref{eq:fstar}\)

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

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 \(\frac{1}{2\pi}\)
$$
$$

@@to be finished@@

— some saved material from z transforms

\(X_s(j\Omega)\)

=============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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