## Continuous transfer functions

Consider a black box with input signal $$x(t)$$ and output $$y(t)$$. This black box processes the input signal and it produces the output signal.

## The black box

When we model the transfer function of this black box as $$h(t)$$, the output signal $$y(t)$$ is a convolution ‘$$\ast$$’ of the input $$x(t)$$ and the transfer function $$h(t)$$. $$y(t) = h(t)*x(t)$$

Likewise, in the $$s$$-domain, the transfer function describes how the output signal $$Y(s)$$ responds to an arbitrary input signal $$X(s)$$. This convolution in the time-domain becomes a multiplication in the $$s$$-domain. Working in this $$s$$-domain makes the convolution, into a multiplication and is thereby easier to solve. $$Y(s) = H(s) X (s)$$

It allows one to determine the system response characteristics without having to solve the convolution.

Albeit for the discrete case, Discrete Transfer Functions describes why convolution is used in the time domain, and multiplication in the $$z$$ domain.

## Poles and zeroes

The generic form of the transfer function is $$H(s) = \frac{Y(s)}{X(s)}=\frac{b_ms^m+b_{m-1}s^{m-1}+\dots+b_1s+b_0}{a_ns^n+a_{n-1}s^{n-1}+\dots+a_1s+a_0} \label{eq:tf_polynominal}$$

where $$s=\sigma+j\omega$$. $$X(s)$$  and $$Y(s)$$ are the Laplace transform of the time representation of the input and output voltages $$x(t)$$ and $$y(t)$$. The highest power of the variable $$s$$ determines the order of the system, usually corresponding to total number of capacitors and inductors in the circuit.

It can be convenient to factor the polynomials in the numerator and denominator of the transfer function, and to write the function in terms of those factors [MIT] $$\begin{array}{cr} H(s)=K\frac{N(s)}{D(s)}=K\frac{(s-z_1)(s-z_2)\dots(s-z_m)}{(s-p_1)(s-p_2)\dots(s-p_n)},&K=\frac{b_m}{a_n} \end{array} \label{eq:tf_factors}$$

The $$z_i$$’s are the roots of the equation $$N(s)=0$$ and are defined as the system zeros.  The $$p_i$$’s are the roots of the equation $$D(s)=0$$ and are defined as the system poles.

The (complex) poles and zeros are properties of the transfer function, and therefore of the differential equation describing the input-output system dynamics. Together with the gain constant $$K$$ they completely characterize the differential equation, and provide a complete description of the system.

### Pole-Zero plot

The system dynamics may be represented graphically by plotting the pole and zero locations on the complex $$s$$-plane, whose axes represent the real and imaginary parts of the complex variable $$s$$. Such plots are known as pole-zero plots.

It is usual to mark a zero location by a circle ($$\circ$$) and a pole location a cross ($$\times$$). The location of the poles and zeros provide qualitative insights into the response characteristics of a system. [MIT]

## Transfer function

The transfer function may be evaluated for any value of $$s=\sigma+j\omega$$. It is common to express the complex value of the transfer function in polar form. $$H(s)=\left|H(s)\right|e^{j\angle H(s)}\label{eq:tf}$$

where magnitude $$|H(s)|$$ and phase $$\angle{H(s)}$$ are given by $$|H(s)| \equiv \sqrt{\Re\left\{H(s)\right\}^2 + \Im\left\{H(s)\right\}^2}$$ $$\angle{H(s)} \equiv \mathrm{atan2}\left( \Im\left\{H(s)\right\}, \Re\left\{H(s)\right\} \right)$$

where $$\Re$$ is the real operator, and $$\Im$$ is the imaginary operator, and $$\mathrm{atan2}$$ returns a value between $$-\pi$$ and $$\pi$$ [wiki], as defined in $$\mathrm{atan2}(y,x) = \begin{cases} \arctan\left(\frac{y}{x}\right) & x \gt 0 \\ \arctan\left(\frac{y}{x}\right)+\pi & x \lt 0 \land y \geq 0 \\ \arctan\left(\frac{y}{x}\right)-\pi & x \lt 0 \land y \lt 0 \\ \frac{\pi}{2} & x= 0 \land y \gt 0 \\ -\frac{\pi}{2} & x= 0 \land y \lt 0 \\ \text{undefined} & x= 0 \land y = 0 \end{cases}$$

## Visualization

The Laplace transform’s $$s$$-domain uses a rectangular coordinate system by defining $$s\triangleq\sigma+j\omega$$, where $$\sigma$$ on the horizontal axis represents the exponential decay, and $$\omega$$ on the vertical axis represents the frequency.

The factorized transfer function $$\eqref{eq:tf_factors}$$ can be written as $$H(z)=K \frac{\prod_{i=1}^m(z-q_i)}{\prod_{i=1}^n(z-p_i)}$$

In the complex plane, the difference between two number $$s_1$$ and $$s_2$$ can be visualized by an vector from $$s_2$$ to $$s_1$$ \begin{align} s_1-s_2 &= (\sigma_1+j\omega_1)-(\sigma_2+j\omega_2)\ nonumber \\ &= (\sigma_1-\sigma_2)+j(\omega_1-\omega_2) \end{align}

This can be visualized with an vector drawn from the tip of $$s_2$$ to the tip of $$s_1$$. Note that the length of the vector is unaffected by translation away from the origin. But the angle of the vector must be measured relative to a translated copy of the real axis.

Therefore, each of the factors in the numerator and denominator may be interpreted as a vector in the s-plane, originating from the zero $$z_i$$ or pole $$p_i$$ and directed to the point $$s$$ at which the function is to be evaluated.

Each of these vectors may be written in polar form, for example for a pole $$p_i=\sigma_i+j\omega_i$$, the magnitude and angle of the vector to the point $$s=\sigma+j\omega$$ are \begin{aligned} |s-p_i| &= \sqrt{(\sigma-\sigma_i)^2+(\omega-\omega_i)^2} \\ \angle(s-p_i) &= \mathrm{atan2}\left(\omega-\omega_i,\sigma-\sigma_i\right) \end{aligned}

## Multiplication and division

While we’re on the subject, a quick note: Multiplication and division of complex numbers is most easily done in polar form $$\begin{eqnarray} Z_1Z_2 & =|Z_1|e^{j\angle{Z_1}}\ |Z_2|e^{j\angle{Z_2}} & =|Z_1||Z_2|e^{j(\angle{Z_1}+\angle{Z_2{)}}}\\ \frac{Z_1}{Z_2} & =\frac{|Z_1|e^{j\angle{Z_1}}}{|Z_2|e^{j\angle{Z_2}}} & =\frac{|Z_1|}{|Z_2|}e^{j(\angle{Z_1}-\angle{Z_2{)}}} \end{eqnarray}$$

Applying $$|K|=K$$ and $$\angle{K}=\mathrm{atan2}(0,K)=0$$, the magnitude and angle of the complete transfer function $$H(s)$$ may be written as \left\{ \begin{align} H(s) &= |H(s)|\ e^{j\angle{H(s)}}\nonumber\\ |H(s)| &= K \frac{\prod_{i=1}^m\left|(s-z_i)\right|}{\prod_{i=1}^n\left|(s-p_i)\right|} \nonumber \\ \angle{H(s)}&=\sum_{i=1}^m\angle(s-z_i)-\sum_{i=1}^n\angle(s-p_i) \nonumber \end{align} \right.

Notes

• A time-continuous system is unstable when poles are in the right half of the $$s$$-plane.
• In the $$s$$-plane, the values along the vertical axis are equal to the frequency response of the system. That is, the Fourier transform is the Laplace transform evaluated at $$\sigma=0$$.

Suggested next reading is Evaluating Transfer Functions.

## Laplace transform and proofs

Around 1785, Pierre-Simon marquis de Laplace, a French mathematician and physicist, pioneered a method for solving differential equations using an integral transform. This Laplace transform turns differential equations in time, into algebraic equations in the Laplace domain thereby making them easier to solve.

## Definition

Pierre-Simon Laplace introduced a more general form of the Fourier Analysis that became known as the Laplace transform.  It transforms a time-domain function, $$f(t)$$, into the $$s$$-plane by taking the integral of the function multiplied by $$e^{-st}$$ from $$0^-$$ to $$\infty$$, where $$s$$ is a complex number with the form $$s=\sigma +j\omega$$. Coordinates in the $$s$$-plane use ‘$$j$$’ to designate the imaginary component, in order to distinguish it from the ‘$$i$$’ used in the normal complex plane. [wiki]

The one-sided Laplace transform is defined as

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

In this equation

• $$\mathfrak{L}$$ symbolizes the Laplace transform.  $$F(s)$$ is the Laplace domain equivalent of the time domain function $$f(t)$$.
• The lower limit of $$0^-$$ emphasizes that the value at $$t=0$$ is entirely captured by the transform.
• Since the upper limit of the integral is $$\infty$$, we must ask ourselves if the Laplace Transform, $$F(s)$$, even exists.  That is the function $$f(t)$$ doesn’t grow faster than an exponential function.

## Overview

The sections below introduce commonly used properties, common input functions and initial/final value theorems, referred to from my various Electronics articles.  Many are based on the excellent notes from the linear physics group at Swarthmore College, and reproduced here mainly for my own understanding and reference.

### Properties

Properties in time and Laplace domains
Time domain Laplace domain
Linearity $$a\cdot f(t)+b\cdot g(t)\nonumber$$ $$a\cdot F(s) + b\cdot G(s)\nonumber$$ proof
First Derivative $$\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\nonumber$$ $$s\,F(s)-f(0^-)\nonumber$$ proof
Second Derivative $$\tfrac{\mathrm{d}^2}{\mathrm{d}t^2}f(t)\nonumber$$ $$s^2F(s)-sf(0^-) – f'(0^-)\nonumber$$ proof
Integration $$\int_{0^-}^t f(\tau)\mathrm{\tau}\nonumber$$ $$\frac{1}{s}F(s)\nonumber$$ proof
Convolution $$f(t)\ast g(t)\nonumber$$ $$F(s)\,G(s)\nonumber$$ proof

### Functions

Functions in time and Laplace domains
Time domain Laplace domain
Impulse $$\delta(t)\nonumber$$ $$1\nonumber$$ proof
Unit Step $$\gamma(t)\nonumber$$ $$\frac{1}{s}\nonumber$$ proof
Ramp $$t\,\gamma(t)\nonumber$$ $$\frac{1}{s^2}\nonumber$$ proof
Exponential $$e^{-at}\gamma(t)\nonumber$$ $$\frac{1}{s+a},\ \forall_{a>0}\nonumber$$ proof
Sine $$\sin(\omega t)\,\gamma(t)\nonumber$$ $$\frac{\omega}{s^2+\omega^2}\nonumber$$ proof
Cosine $$\cos(\omega t)\,\gamma(t)\nonumber$$ $$\frac{s}{s^2+\omega^2}\nonumber$$ proof
Decaying Sine $$e^{-\alpha t}\sin(\omega t)\,\gamma(t)\nonumber$$ $$\frac{\omega}{(s+\alpha)^2+\omega^2}\nonumber$$ proof
Decaying Cosine $$e^{-\alpha t}\cos(\omega t)\,\gamma(t)\nonumber$$ $$\frac{s+\alpha}{(s+\alpha)^2+\omega^2}\nonumber$$ proof
Time Delayed $$f(t-a)\,\gamma(t-a)\nonumber$$ $$e^{-su}F(s)\nonumber$$ proof

### Initial and Final Value Theorem

Initial and final value theorem
Time domain Laplace domain
Initial Value $$f(0^+)\nonumber$$ $$\nonumber$$ proof
Final Value $$f(\infty)\nonumber$$ $$\nonumber$$ proof

The proof for each of these transforms can be found below.

## Property proofs

### Linearity Property

The linearity property in the time domain

$$u(t) = a\cdot f(t)+b\cdot g(t)$$

Transformed to the Laplace domain

\begin{align} \mathfrak{L}\left\{\,a\cdot f(t)+b\cdot g(t)\,\right\} &=\int_{0^-}^{\infty}\left( a\cdot f(t)+ b\cdot g(t) \right) * e^{-st}\mathrm{d}t \nonumber\\ &= a\underbrace{\int_{0^-}^{\infty} f(t) * e^{-st}\mathrm{d}t}_{F(s)} + b\underbrace{\int_{0^-}^{\infty} g(t) * e^{-st}\mathrm{d}t}_{G(s)} \end{align}

From which follows

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ a\cdot f(t)+b\cdot g(t) \laplace a\cdot F(s) + b\cdot G(s) } \label{eq:linearity}$$

### First Derivative Property

The first derivative in time is used in deriving the Laplace transform for capacitor and inductor impedance. The general formula

$$u(t) = \frac{\mathrm{d}}{\mathrm{d}t}f(t)$$

Transformed to the Laplace domain using $$\eqref{eq:laplace}$$

$$\mathfrak{L}\left\{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\right\} = \int_{0^-}^{\infty} e^{-st}\tfrac{\mathrm{d}f(t)}{\mathrm{d}t} \mathrm{d}t = \int_{0^-}^{\infty} \underbrace{e^{-st}}_{u(t)} \underbrace{\tfrac{\mathrm{d}f(t)}{\mathrm{d}t}}_{v'(t)} \mathrm{d}t\Rightarrow \label{eq:derivative_}$$

Recall integration by parts, based on the product rule, from your favorite calculus class

\left\{ \begin{align} \int_a^b u(t)\ v'(t)\ \mathrm{d}t&=\left[ u(t)\ v(t)\right]_a^b -\int_a^b u'(t)\ v(t)\ \mathrm{d}t \nonumber \\ u(t)&=\int_{0^-}^t f(\tau)\mathrm{d}\tau \Rightarrow u'(t)=f(t) \nonumber \\ v'(t)&=e^{-st} \Rightarrow v(t)=-\tfrac{1}{s}e^{-st} \nonumber \end{align} \right. \label{eq:intbyparts}

Solve $$\eqref{eq:derivative_}$$ using integration by parts

\begin{align} \mathfrak{L}\left\{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\right\}&= \left[ e^{-st}f(t)\right]_{0^-}^{\infty} – \int_{0^-}^\infty (-s)e^{-st}f(t)\mathrm{d}t\nonumber\\ &= \cancel{e^{-s\infty}f(\infty)} – \bcancel{e^{-s0^-}}f(0^-)+ s \underbrace{\int_{0^-}^\infty e^{-st}f(t)\mathrm{d}t}_{\mathfrak{L}f(t) = F(s)} \end{align}

The first term goes to zero because $$f(\infty)$$ is finite which is a condition for existence of the transform. The last term is simply the definition of the Laplace Transform multiplied by $$s$$.

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \laplace s\,F(s)-f(0^-) } \label{eq:derivative}$$
The initial condition is taken at $$t=0^-$$. This means that we only need to know this initial conditions before the input signal started.

### Second Derivative Property

The second derivative in time is found using the Laplace transform for the first derivative $$\eqref{eq:derivative}$$. The general formula

$$u(t) = \frac{\mathrm{d}^2}{\mathrm{d}t^2}f(t)$$

Introduce $$g(t)=\frac{\mathrm{d}}{\mathrm{d}t}f(t)$$

\left\{ \begin{align} u(t) &= \frac{\mathrm{d}}{\mathrm{d}t}g(t) \nonumber \\ g(t) &= \frac{\mathrm{d}}{\mathrm{d}t}f(t) \nonumber \end{align} \right.

From the transform of the first derivative $$\eqref{eq:derivative}$$, we find the Laplace transforms of $$\frac{\mathrm{d}}{\mathrm{d}t}g(t)$$ and $$\frac{\mathrm{d}}{\mathrm{d}t}f(t)$$

\left. \begin{align} U(s) &= \mathfrak{L}\left\{\,\frac{\mathrm{d}}{\mathrm{d}t}g(t)\,\right\} = s\,G(s)-g(0^-) \nonumber \\ G(s)&=\mathfrak{L}\left\{\,\frac{\mathrm{d}}{\mathrm{d}t}f(t)\,\right\} = s\,F(s)-f(0^-) \nonumber \end{align} \right\} \overset{subst} \Rightarrow

Substitute $$G(s)$$ in $$U(s)$$

\begin{align} U(s)&=s\left(sF(s)-f(0^-) \right) – g(0^-)\nonumber\\ &=s^2F(s)-sf(0^-) – \left.\frac{\mathrm{d}}{\mathrm{d}t}f(t)\right|_{0^-} \end{align}

This brings us to the Laplace transform of the second derivative of $$f(t)$$

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \tfrac{\mathrm{d}^2}{\mathrm{d}t^2}f(t) \laplace s^2F(s)-sf(0^-) – f'(0^-) } \label{eq:secondderivative}$$

The initial conditions are taken at $$t=0^-$$. This means that we only need to know these initial conditions before the input signal started.

### Integration Property

Determine the Laplace transform of the integral

$$u(t) = \int_{0^-}^t f(\tau)\mathrm{d}\tau$$

Apply the Laplace transform definition $$\eqref{eq:laplace}$$

$$\mathfrak{L}\left\{ \int_{0^-}^t f(\tau)\mathrm{\tau} \right\} = \int_{0^-}^{\infty} \underbrace{\left( \int_{0^-}^t f(\tau)\mathrm{d}\tau \right)}_{u(t)} \underbrace{e^{-st}}_{v'(t)} \mathrm{d}t\Rightarrow$$

\left. \begin{align} \mathfrak{L}\left\{ \int_{0^-}^t f(\tau)\mathrm{\tau} \right\}&=\int_{0^-}^{\infty}u(t)\ v'(t)\ \mathrm{d}t \nonumber \\ \int_a^b u(t)\ v'(t)\ \mathrm{d}t&=\left[ u(t)\ v(t)\right]_a^b – \int_a^b u'(t)\ v(t)\ \mathrm{d}t \nonumber \\ u(t)&=\int_{0^-}^t f(\tau)\mathrm{d}\tau \Rightarrow u'(t)=f(t) \nonumber \\ v'(t)&=e^{-st} \Rightarrow v(t)=-\tfrac{1}{s}e^{-st} \nonumber \end{align} \right\}

Again, solve using integration by parts

\begin{align} \mathfrak{L}\left\{ \int_{0^-}^t f(\tau)\mathrm{\tau} \right\} &= \int_{0^-}^{\infty} \underbrace{\left( \int_{0^-}^t f(\tau)\mathrm{d}\tau \right)}_{u(t)} \underbrace{e^{-st}}_{v'(t)} \mathrm{d}t \nonumber\\ &= \left[ \left(\int_{0^-}^t f(\tau)\mathrm{d}\tau\right) \left(-\frac{1}{s}e^{-st }\right) \right]_{0^-}^{\infty} -\int_{0^-}^\infty f(t) \left( -\frac{1}{s}e^{-st} \right) \mathrm{d}t \nonumber \\ &= -\frac{1}{s} \left[ e^{-st } \int_{0^-}^t f(\tau)\mathrm{d}\tau \right]_{0^-}^{\infty} + \frac{1}{s} \underbrace{ \int_{0^-}^\infty f(t) e^{-st} \mathrm{d}t }_{\mathfrak{L}f(t)=F(s)} \nonumber \\ &= -\frac{1}{s} \left( \cancel{e^{-s\infty }\int_{0^-}^\infty f(\tau)\mathrm{d}\tau} \ -\ e^{-s0^- }\cancel{\int_{0^-}^{0^-} f(\tau)\mathrm{d}\tau} \right) + \frac{1}{s}F(s) \end{align}

The first term goes to zero because $$f(\infty)$$ is finite which is a condition for existence of the transform. In the second term, the exponential goes to one and the integral is $$0$$ because the limits are equal. The last term is simply the definition of the Laplace Transform over $$s$$.

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \int_{0^-}^t f(\tau)\mathrm{\tau} \laplace \frac{1}{s}F(s) } \label{eq:integration}$$

### Convolution Property

Just to show the strength of the Laplace transfer, we show the convolution property in the time domain of two causal functions

$$u(t)=f(t) \ast g(t) = \int_{-\infty}^{\infty}f(\lambda)\,g(t-\lambda)\,\mathrm{d}\lambda$$
where $$\ast$$ is the convolution operator.

Transformed to the Laplace domain

\begin{align} \mathfrak{L}\left\{\,f(t) \ast g(t)\,\right\} &=\int_{0^-}^{\infty} \left( \int_{-\infty}^{\infty}f(\lambda)\,g(t-\lambda)\,\mathrm{d}\lambda \right) e^{-st}\mathrm{d}t \nonumber\\ &= \int_{-\infty}^{\infty} \int_{0^-}^{\infty}f(\lambda)\,g(t-\lambda)\, e^{-st}\mathrm{d}t\,\mathrm{d}\lambda &\mathrm{change\ order\ of\ integration}\nonumber\\ &= \int_{-\infty}^{\infty} f(\lambda) \int_{0^-}^{\infty}g\underbrace{(t-\lambda)}_{u}\, e^{-st}\mathrm{d}t\,\mathrm{d}\lambda &\mathrm{f(\lambda) \mathrm{\ independent\ of\ }t} \end{align}

Substitute $$u=t-\lambda$$

\begin{align} \mathfrak{L}\left\{\,f(t) \ast g(t)\,\right\} &= \int_{-\infty}^{\infty} f(\lambda) \int_{\underline{(-\lambda)^-}}^{\infty}g(u)\, e^{-s(u+\lambda)}\mathrm{d}u\,\mathrm{d}\lambda &g(u)=0,\ \forall u\lt 0 \nonumber\\ &=\int_{-\infty}^{\infty} f(\lambda) \int_{0}^{\infty}g(u)\, e^{-su}\underline{e^{-s\lambda}}\mathrm{d}u\,\mathrm{d}\lambda &e^{-s\lambda}\mathrm{\ independent\ of\ }u \nonumber\\ &=\int_{-\infty}^{\infty} f(\lambda)e^{-s\lambda} \underline{\int_{0}^{\infty}g(u) e^{-su}\mathrm{d}u}\,\mathrm{d}\lambda &\mathrm{inner\ intergral\ independent\ on\ }\lambda \nonumber\\ &=\int_{\underline{-\infty}}^{\infty} f(\lambda)e^{-s\lambda} \mathrm{d}\lambda\ \int_{0}^{\infty}g(u) e^{-su}\mathrm{d}u &f(\lambda)=0,\ \forall \lambda\lt 0 \nonumber\\ &=\underbrace{\int_{0^-}^{\infty} f(\lambda)e^{-s\lambda} \mathrm{d}\lambda}_{F(s)}\forall \lambda\lt 0 \nonumber \\ &= \underbrace{\int_{0}^{\infty}g(u) e^{-su}\mathrm{d}u}_{G(s)} &\mathrm{these\ are\ Laplace\ transforms} \end{align}

Gives us the Laplace transfer for the convolution property

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ f(t)\ast g(t) \laplace F(s)\,G(s) } \label{eq:convolution}$$

## Function proofs

### Impulse Function

The impulse function $$\delta(t)$$ is often used as an theoretical input signal to study system behavior.  The definition is

$$u(t)=\delta(t) = \begin{cases} \mathrm{undefined}, & t=0 \\ 0, & \neq 0 \end{cases} \label{eq:impuls_def1}$$
and satisfies the condition
$$\int_{-\infty}^{\infty}\delta(t) = 1 \label{eq:impuls_def2}$$

in other words, the area is 1 so that $$\delta(t)$$ is as high, as $$\mathrm{d}t$$ is narrow.

Apply the Laplace transform definition $$\eqref{eq:laplace}$$

$$\mathcal{L}\left\{\delta(t)\right\} = \Delta(s) = \int_{0^-}^{\infty}e^{-st}\delta(t)\,\mathrm{d}t$$

Since the impulse is $$0$$ everywhere but at $$t=0$$, the upper limit of the integral can be changed to $$0^+$$.

$$\Delta(s) = \int_{0^-}^{0^+}e^{-st}\delta(t)\,\mathrm{d}t$$

The function $$e^{-st}$$ is continuous at $$t=0$$, and may be replaced by its value at $$t=0$$

$$\Delta(s)=\left.e^{-st}\right|_{t=0}\int_{0^-}^{0^+}\delta(t)\,\mathrm{d}t = \int_{0^-}^{0^+}\delta(t)\,\mathrm{d}t$$

Substituting the condition $$\int_{-\infty}^{\infty}\delta(t)=1$$ from $$\eqref{eq:impuls_def2}$$ gives us the Laplace transform of the impulse function

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \delta(t) \laplace 1 } \label{eq:impulse}$$

### Unit Step Function

The unit or Heaviside step function, denoted with $$\gamma(t)$$ is defined as a function of $$\gamma(t)$$.

$$u(t) = \gamma(t) = \begin{cases} 0 & t<0 \\ 1 & t\geq 0 \\ \end{cases} \label{eq:unitstep_def_a}$$

The unit step function is related to the impulse function as

$$\gamma(t) = \int\delta(t)\,\mathrm{d}t$$

Apply the Laplace transform definition $$\eqref{eq:laplace}$$

\begin{align} \Gamma(s)\,&=\int_{0^-}^\infty e^{-st}\,\gamma(t)\,\mathrm{d}t \nonumber \\ &= \int_{0^-}^\infty\,e^{-st}\,1\,\mathrm{d}t \nonumber \\ &= -\frac{1}{s}\left[e^{-st}\right]_{0^-}^\infty \end{align}

The upper limit of the integral only goes to zero if the real part of the complex variable $$s$$ is positive, so that $$\left.e^{-st}\right|_{s\to\infty}$$

\begin{align} \Gamma(s)\,&=-\frac{1}{s}\left(e^{-s\infty}-e^{-s0}\right) = -\frac{1}{s}\left(0-1\right) \end{align}

Gives us the Laplace transfer of the unit step function

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \gamma(t) \laplace \frac{1}{s} } \label{eq:unitstep}$$

### Ramp Function

The unit or Heaviside step function, denoted with $$\gamma(t)$$ is defined as below [smathmore]. We use $$\gamma(t)$$, to avoid confusion with the European symbol for voltage source $$u(t)$$, where $$u$$ stands for “Potentialunterschied”, which means potential difference.  The capital letter of $$\gamma$$ is $$\Gamma$$ what looks a bit like the step function.

$$u(t) = t\,\gamma(t) \label{eq:ramp_def_a}$$

The ramp function is related to the unit step  function as

$$u(t) = \int\gamma(t)\,\mathrm{d}t$$

Apply the Laplace transform definition $$\eqref{eq:laplace}$$

$$U(s) = \mathcal{L}\left\{\,t\,\right\}\,=\int_{0^-}^\infty \underbrace{e^{-st}}_{v'(t)}\,\underbrace{t}_{u(t)}\,\mathrm{d}t \label{eq:ramp1}$$

Use integration by parts

\left\{ \begin{align} \int_a^b u(t)\ v'(t)\ \mathrm{d}t&=\left[ u(t)\ v(t)\right]_a^b – \int_a^b u'(t)\ v(t)\ \mathrm{d}t\nonumber\\ u(t) &= t \Rightarrow u'(t)=1 \nonumber \\ v'(t) &= e^{-st} \Rightarrow v(t)=-\tfrac{1}{s}e^{-st} \nonumber \end{align} \right. \label{eq:intbyparts2}

Solve $$\eqref{eq:ramp1}$$ using integration by parts

\begin{align} \mathfrak{L}\left\{\,t\,\right\}&= \left[ (t) \cdot (-\frac{1}{s}e^{-st})\right]_{0^-}^{\infty} -\int_{0^-}^\infty 1\cdot (-\frac{1}{s}e^{-st})\mathrm{d}t\nonumber\\ &= -\left[ \frac{t}{s}e^{-st}\right]_{0^-}^{\infty} +\frac{1}{s}\underbrace{\int_{0^-}^\infty e^{-st}\mathrm{d}t}_{\Gamma(t)=\frac{1}{s}}\nonumber\\ &= -\left(\frac{\infty}{s}e^{-s\infty} -\cancel{\frac{0}{s}e^{-s0}} \right) +\frac{1}{s^2}\nonumber\\ &= -\left(\cancel{\frac{\infty}{se^{s\infty}}} -0 \right) +\frac{1}{s^2} \end{align}

Gives us the Laplace transfer of the ramp function

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ t\,\gamma(t) \laplace \frac{1}{s^2} } \label{eq:ramp}$$

### Exponential Function

An exponential function time domain, starting at $$t=0$$

$$u(t)= e^{-ax}\cdot \gamma(t)$$

The step function becomes 1 at the lower limit of the integral, and is $$0$$ before that

\begin{align} \mathfrak{L}\left\{\, e^{-at}\gamma(t) \right\} &= \int_{0^-}^{\infty} e^{-at}\gamma(t)\, e^{-st}\mathrm{d}t \nonumber\\ &= \int_{0^-}^{\infty} e^{-(s+a)t}\mathrm{d}t \nonumber\\ &= \left[ \frac{1}{-(s+a)}e^{-(s+a) t} \right]_{0^-}^{\infty} \nonumber\\ &= -\frac{1}{s+a}\left( \bcancel{e^{-(s+a) \infty}} – \cancelto{1}{e^{-(s+a) 0^-}} \right) \nonumber\\ &= \frac{1}{s+a} & a<0 \end{align}

Gives us the Laplace transform of the exponential time function

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ e^{-at}\gamma(t) \laplace \frac{1}{s+a} },\ \forall_{a>0 } \label{eq:exponential}$$

### Sine Function

Another popular input signal is the sine wave, starting at $$t=0$$

$$u(t) = f(t) = \sin(\omega t)\,\gamma(t) \label{eq:sin_def}$$

Apply the definition of the Laplace transform $$\eqref{eq:laplace}$$

\begin{align} \mathcal{L}\left\{f(t)\right\}=F(s) &=\int_{0^-}^{\infty}e^{-st}\sin(\omega t)\gamma(t) \,\mathrm{d}t \nonumber \\ &=\int_{0^-}^{\infty}e^{-st}\sin(\omega t) \,\mathrm{d}t \end{align}\label{eq:sinlaplace}

Apply the Euler identity for sine

\begin{align} F(s)&=\int_{0^-}^{\infty}e^{-st}\,\frac{e^{j\omega t}-e^{-j\omega t}}{2j} \,\mathrm{d}t \nonumber \\ &=\frac{1}{2j}\int_{0^-}^{\infty}e^{-st}\,\left(e^{j\omega t}-e^{-j\omega t}\right) \,\mathrm{d}t \nonumber \\ &=\frac{1}{2j}\int_{0^-}^{\infty}e^{-st}\,e^{j\omega t}\,\mathrm{d}t -\frac{1}{2j}\int_{0^-}^{\infty}e^{-st}\,e^{-j\omega t} \,\mathrm{d}t \nonumber \\ &=\frac{1}{2j}\int_{0^-}^{\infty}e^{(-s+j\omega) t}\,\mathrm{d}t -\frac{1}{2j}\int_{0^-}^{\infty}e^{(-s-j\omega)t} \,\mathrm{d}t \end{align} \label{eq:sin2}

The simple definite integral $$\int_{0^-}^{\infty}e^{-(s+a) t}\,\mathrm{d}t$$, was already solved as part of $$\eqref{eq:exponential}$$

$$\int_{0^-}^{\infty}\ e^{-(s+a) t}\,\mathrm{d}t = \frac{1}{s+a} ,\ a \label{eq:sin3}$$

Substitute $$\eqref{eq:sin3}$$

\begin{align} F(s) &= \frac{1}{2j}\left( \frac{1}{s-j\omega} \right) – \frac{1}{2j}\left( \frac{1}{s+j\omega} \right) \nonumber \\ &= \frac{1}{2j}\left( \frac{1}{s-j\omega} – \frac{1}{s+j\omega} \right) \end{align}

Bring it under a common denominator

\begin{align} F(s)&= \frac{1}{2j}\left( \frac{1}{(s-j\omega)} \frac{(s+j\omega)}{(s+j\omega)} – \frac{1}{(s+j\omega)} \frac{(s-j\omega)}{(s-j\omega)} \right) \nonumber \\ &= \frac{1}{2j} \frac{(s+j\omega)-(s-j\omega)}{s^2-2j\omega-j^2\omega^2} \nonumber\\ &= \frac{1}{\bcancel{2j}} \frac{\bcancel{2j}\omega}{s^2\cancel{-js\omega+js\omega}+\omega^2} \end{align}

Et voilà, the Laplace transform of sine function

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \sin(\omega t)\,\gamma(t) \laplace \frac{\omega}{s^2+\omega^2} }\ label{eq:sine}$$

### Cosine Function

Yet another popular input signal is the cosine wave, starting at $$t=0$$

$$u(t) = f(t) = \cos(\omega t)\,\gamma(t) \label{eq:cos_def}$$

The Laplace transforms of the cosine is similar to that of the sine function, except that it uses Euler’s identity for cosine

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \cos(\omega t)\,\gamma(t) \laplace \frac{s}{s^2+\omega^2} } \label{eq:cosine}$$

### Decaying Sine Function

Consider a decaying sine wave, starting at $$t=0$$

$$u(t) = f(t) = e^{-\alpha t}\sin(\omega t)\,\gamma(t) \label{eq:decayingsine_def}$$

Apply the Euler identity for sine

\begin{align} f(t) &= e^{-\alpha t}\sin(\omega t)\,\gamma(t) \nonumber \\ &= e^{-\alpha t}\frac{e^{j\omega t}-e^{-j\omega t}}{2j}\,\gamma(t)\nonumber \\ &= \frac{e^{(j\omega-\alpha)t}-e^{-(j\omega+\alpha) t}}{2j}\,\gamma(t) \nonumber \\ &= \frac{1}{2j}\left(e^{(j\omega-\alpha)t}-e^{-(j\omega+\alpha) t}\right)\gamma(t) \end{align}

We recognize the exponential functions, and apply their Laplace transforms $$\eqref{eq:exponential}$$

\begin{align} F(s) &= \frac{1}{2j}\left( \frac{1}{s-(j\omega-\alpha)}- \frac{1}{s+(j\omega+\alpha)} \right)\nonumber\\ &= \frac{1}{2j}\left( \frac{1}{s+\alpha-j\omega}- \frac{1}{s+\alpha+j\omega} \right) \end{align}

Put over a common denominator

\begin{align} F(s) &= \frac{1}{2j}\left( \frac{1}{s+\alpha-j\omega}- \frac{1}{s+\alpha+j\omega} \right) \nonumber \\ &= \frac{1}{2j}\left( \frac{1}{(s+\alpha-j\omega)}\frac{(s+\alpha+j\omega)}{(s+\alpha+j\omega)} – \frac{1}{(s+\alpha+j\omega)}\frac{(s+\alpha-j\omega)}{(s+\alpha-j\omega)} \right) \nonumber \\ &= \frac{1}{2j}\left( \frac{(s+\alpha+j\omega) -(s+\alpha-j\omega)}{(s+\alpha)^2-(j\omega)^2} \right) =\frac{1}{2j}\left( \frac{\cancel{s}\cancel{+\alpha}+j\omega\cancel{-s}\cancel{-\alpha}+j\omega}{(s+\alpha)^2+\omega^2} \right) \nonumber \\ &= \frac{1}{\cancel{2j}}\left( \frac{\cancel{2j}\omega}{(s+\alpha)^2+\omega^2} \right) \end{align}

The Laplace transforms of the decaying sine

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ e^{-\alpha t}\sin(\omega t)\,\gamma(t) \laplace \frac{\omega}{(s+\alpha)^2+\omega^2} } \label{eq:decayingsine}$$

### Decaying Cosine Function

Consider a decaying cosine wave, starting at $$t=0$$

$$u(t) = f(t) = e^{-\alpha t}\cos(\omega t)\,\gamma(t) \label{eq:decayingcosine_def}$$

The Laplace transforms of the decaying cosine is similar to that of the decaying sine function, except that it uses Euler’s identity for cosine.

\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ \begin{align*} e^{-\alpha t}\cos(\omega t)\,\gamma(t) \laplace \frac{s+\alpha}{(s+\alpha)^2+\omega^2} \end{align*} } \label{eq:decayingcosine}

### Time Delayed Function

A delay in the time domain, starting at $$t-a=0$$

$$u(t) = f(t-a)\cdot \gamma(t-a)$$

The delayed step function $$\gamma(t)$$

$$\gamma(t-a) = \begin{cases} 0 & t&a \\ 1 & t\geq a \\ \end{cases}$$

The delayed step function simplifies Laplace transform because $$\gamma(t-a)$$ is $$1$$ starting at $$t=-a$$, and is $$0$$ before

\begin{align} \mathfrak{L}\left\{\,f(t-a)\cdot \gamma(t-a)\,\right\} &=\int_{o^-}^{\infty}\left(\, f(t-a)\cdot \gamma(t-a)\, \right) e^{-st}\mathrm{d}t \nonumber \\ &= \int_{a^-}^{\infty} f(t-a)\cdot e^{-st}\mathrm{d}t \end{align}

Substitute $$u=t-a$$

\begin{align} \mathfrak{L}\left\{\,f(t-a)\cdot \gamma(t-a)\,\right\} &= \int_{a^-}^{\infty} f(u) e^{-s(u+a)}\mathrm{d}u,&u=t-a \nonumber \\ &= e^{-sa} \underbrace{\int_{0^-}^{\infty} f(u) e^{-su}\mathrm{d}u}_{F(s)} \end{align}

The last integral is simply the definition of the Laplace transform.  Together it gives us the Laplace transform of a time delayed function.

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \shaded{ f(t-a)\,\gamma(t-a) \laplace e^{-su}F(s) } \label{eq:timedelay}$$

### Initial Value Theorem

The right sided initial value of a function $$f(0^+)$$ follows from its Laplace transform of the derivative $$\eqref{eq:derivative}$$

\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \begin{align} \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \laplace s\,F(s)-f(0^-) \end{align}

Invoke the definition of the Laplace transform for the First Derivative theorem $$\eqref{eq:derivative}$$, and split the integral

\begin{align} s\,F(s)-f(0^-)&=\mathfrak{L}\left\{\, \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \,\right\} \nonumber\\ &= \int_{0^-}^{\infty} \underbrace{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)}_{f'(t)}\,e^{-st}\mathrm{d}t &f’=\tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \nonumber\\ &= \int_{0^-}^{\infty} f'(t)\,e^{-st}\mathrm{d}t &\mathrm{split\ integral} \nonumber\\ &= \int_{0^-}^{0^+} f'(t)\,e^{-st}\mathrm{d}t + \int_{0^+}^{\infty} f'(t)\,e^{-st}\mathrm{d}t \end{align}

Take the limit as $$s\to\infty$$

\begin{align} \lim_{s\to\infty}\left(s\,F(s)-f(0^-)\right) &= \lim_{s\to\infty}\left( \int_{0^-}^{0^+} f'(t)\,e^{-st}\mathrm{d}t + \int_{0^+}^{\infty} f'(t)\,e^{-st}\mathrm{d}t \right) \end{align}

Take the terms out of the limit that don’t depend on $$s$$, and when substituting $$s=\infty$$ in the second integral, that goes to $$0$$

\begin{align} \lim_{s\to\infty}\left(s\,F(s)\right) – f(0^-) &= \int_{0^-}^{0^+} f'(t)\,e^{-st}\mathrm{d}t + \lim_{s\to\infty}\left( \int_{0^+}^{\infty} f'(t)\,\cancelto{0}{e^{-st}}\mathrm{d}t \right) \nonumber\\ &= \int_{0^-}^{0^+} f'(t)\,e^{-st}\mathrm{d}t, & \mathrm{where\ }\int f'(t)=f(t) \nonumber\\ &= \left[f(t)\right]_{0^-}^{0^+} \nonumber\\ \lim_{s\to\infty}\left(s\,F(s)\right) – \cancel{f(0^-)} &= f(0^+)-\cancel{f(0^-)} \end{align}

The initial value theorem follows as

$$\shaded{ f(0^+) = \lim_{s\to\infty}\left(s\,F(s)\right) } \label{eq:initialvalue}$$

### Final Value Theorem

The final value of a function $$f(\infty)$$ follows from its Laplace transform of the derivative $$\eqref{eq:derivative}$$.  Note that functions such as sine, and cosine don’t a final value

\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \begin{align} \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \laplace s\,F(s)-f(0^-) \end{align}

Similarly to the initial value theorem, we start with the First Derivative $$\eqref{eq:derivative}$$ and apply the definition of the Laplace transform $$\eqref{eq:laplace}$$, but this time with the left and right of the equal sign swapped, and split the integral

\begin{align} s\,F(s)-f(0^-) &= \mathfrak{L}\left\{\, \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \,\right\} \nonumber \\ &= \int_{0^-}^{\infty} \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \,e^{-st}\mathrm{d}t \end{align}

Take the limit as $$s\to 0$$

$$\lim_{s\to0}\left( s\,F(s)-f(0^-) \right) = \lim_{s\to0}\left( \int_{0^-}^{\infty} \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \,e^{-st}\mathrm{d}t \right)$$

Take the terms out of the limit that don’t depend on $$s$$, and $$\lim_{s\to0}e^{-st}=1$$ inside the integral

$$\lim_{s\to0}\left( s\,F(s)\underline{-f(0^-)} \right) = \lim_{s\to0}\left( \int_{0^-}^{\infty} \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \,\cancelto{1}{e^{-st}}\mathrm{d}t \right)$$

The integral doesn’t depend on $$s$$

\begin{align} \lim_{s\to0}\left( s\,F(s) \right)-f(0^-) &= \int_{0^-}^{\infty} \tfrac{\mathrm{d}}{\mathrm{d}t}f(t) \,\mathrm{d}t ,&\int\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\mathrm{d}t=f(t) \nonumber\\ &= [f(t)]_{O^-}^{\infty} \nonumber\\ \lim_{s\to0}\left( s\,F(s) \right)\cancel{-f(0^-)} &= f(\infty)\cancel{-f(0^-)} \end{align}

The final value theorem follows as

$$\shaded{ f(\infty) = \lim_{s\to0}\left( s\,F(s) \right) } \label{eq:finalvalue}$$

Suggested next reading is Transfer Functions.

## 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

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

## 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). \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}

## Impedance

Electrical impedance is a complex-valued measure of the opposition that a circuit presents to a time-varying electric current when a voltage is applied.  Combined with the Laplace transform it allows us to use algebra to calculate electrical networks.

$$u$$Instead of $$\Delta v$$, we use the European symbol for voltage difference: $$u$$. The letter ‘u’ stands for “Potentialunterschied”.

## Capacitor impedance

Capacitors store energy in an electric field.  The amount of charge $$q$$ stored in a capacitor is linearly proportional to the voltage $$u$$ over the capacitor. [MIT] $$q(t) = Cu(t)\label{eq:c_equiv}$$ where $$C$$ is a constant called capacitance.  The SI unit for capacitance is Farad with values typically range from from 2.2 pF to 470 μF.

The electrical charge $$q(t)$$ is a function of the current accumulated over time, assuming that there is no initial charge. $$q(t)=\int\limits_{0}^{t} \! i(\tau) \, \mathrm{d}\tau \label{eq:c_integral}$$

Combining equation $$\eqref{eq:c_equiv}$$ and $$\eqref{eq:c_integral}$$ and solving for the current and taking the derivative on both sides, gives the current as a function of the voltage \begin{align} \int\limits_{0}^{t} \! i(\tau) \, \mathrm{d}\tau = C\,u(t) \\ \Rightarrow i(t) &= C\frac{\mathrm{d}u(t) }{\mathrm{d}t} \end{align}

The Laplace transform allows us to use algebra in complex frequency domain, instead of working with differential equations.  Using the Laplace transform of the first derivative \left. \begin{align} I(s) &= \mathfrak{L}i(t) = \mathfrak{L}\left\{ C\frac{\mathrm{d}u(t)}{\mathrm{d}t} \right\} = C\mathfrak{L}\left\{ \frac{\mathrm{d}}{\mathrm{d}t}u(t) \right\} \nonumber \\ \mathfrak{L}\left\{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\right\}&= -f(0^-)+s\mathfrak{L}\left\{f(t)\right\} \nonumber \\ f(0^-) &= 0 \nonumber \end{align} \right\} \Rightarrow I(s) = s\,C\,U(s) \label{eq:c_laplace} where capital letters are used to indicate complex domain variables, such as $$I(s)=\mathfrak{L}i(t)$$ and $$U(s)=\mathfrak{L}u(t)$$.

Solving $$\eqref{eq:c_laplace}$$ for the complex impedance $$Z_{C}(s) \equiv\frac{U(s)}{I(s)}$$ we get $$\shaded{ Z_{C}(s) = \frac{1}{sC} } \label{eq:c_impedance}$$

The plot below shows the magnitude of the capacitor impedance $$Z_C$$ as a function of the frequency where $$s=\sigma+j\omega$$.

## Inductor impedance

An inductor stores energy in a magnetic field.  The magnetic flux $$\phi$$ in the inductor is linearly proportional to the current $$i$$ through the inductor.  The role played by the inductor in this magnetic case is analogous to that of a capacitor in the electric case. $$\phi(t) = L i(t) \label{eq:l_equiv}$$ where $$L$$ is a constant called inductance.  The SI unit for inductance is Henry with values typically range from from 0.1 µH to 1 mH.

According to Faraday’s law of induction [MIT], an inductor opposes changes in current by developing a voltage $$\varepsilon=-u$$ proportional to the negative of the rate of change of magnetic flux $$\phi$$. $$\varepsilon(t) = -u(t) = -\frac{\mathrm{d}\phi(t) }{\mathrm{d}t} \label{eq:l_emf}$$

Combining equation $$\eqref{eq:l_equiv}$$ and $$\eqref{eq:l_emf}$$ yields \begin{align} u(t) = -\varepsilon(t) = +\frac{\mathrm{d}\phi(t) }{\mathrm{d}t} = L \frac{\mathrm{d}i(t) }{\mathrm{d}t} \end{align}

Once more, the Laplace transform allows us to use algebra in complex frequency domain, instead of working with differential equations.  Using the Laplace transform of the first derivative \left. \begin{align} U(s) &=\mathfrak{L}u(t) =\mathfrak{L}\left\{ L\tfrac{\mathrm{d}i(t)}{\mathrm{d}t} \right\} =L\mathfrak{L}\left\{ \tfrac{\mathrm{d}}{\mathrm{d}t}i(t) \right\} \nonumber \\ \mathfrak{L}\left\{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\right\} &= -f(0^-)+s\mathfrak{L}\left\{f(t)\right\} \nonumber \\ f(0^-)&=0 \nonumber \end{align} \right\} \Rightarrow U(s) = s\,L\,I(s) \label{eq:l_laplace} where capital letters are used to indicate complex domain variables, such as $$I(s)=\mathfrak{L}i(t)$$ and $$U(s)=\mathfrak{L}u(t)$$.

Solving $$\eqref{eq:l_laplace}$$ for the complex impedance $$Z_L(s) \equiv\frac{U(s)}{I(s)}$$ we get $$\shaded{ Z_L(s) = sL } \label{eq:c_inductance}$$

The plot below shows the magnitude of the inductor impedance $$Z_L$$ as a function of the frequency

## Resistor impedance

For completeness we show the resistor impedance $$Z_r$$ as a function of the frequency $$\shaded{ Z_r(s) = R } \label{eq:r_inductance}$$

The plot below shows the magnitude of the resistor impedance $$Z_r$$ as a function of the frequency

## Overview

Electric quantities
name symbol unit impedance
Voltage $$u(t)$$ Volt
Current $$i(t)$$ Ampere
Resistance $$R$$ $$u(t)=R\cdot i(t)$$ $$Z_r=R$$
Capacitance $$C$$ $$u(t)=\frac{1}{C}\int_0^t i(\tau)\,\mathrm{d}\tau$$ $$Z_c=\frac{1}{j\omega C}$$
Inductance $$L$$ $$u(t)=L\frac{\mathrm{d}i(t)}{\mathrm{d}t}$$ $$Z_l=j\omega L$$
$$\sum u(t)=0$$ $$\sum i(t)=0$$

My write-ups that use the impedance formulas include