Johan’s theoretical math

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Evaluating continuous transfer functions

The transfer function can be evaluated using different inputs. We commonly use the impulse, step and sinusoidal input functions.

Let $$H$$ be a stable system with transfer function $$H(s)$$, input signal $$x(t)$$, and output $$y(t)$$. In this, “stable” implies that the poles are in left half of $$s$$-plane.

Impulse Response

Earlier, we derived the Laplace transform for the impulse function as $$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \delta(t)\laplace\Delta(s) = 1$$

Substituting this input function $$Y(s) = H(s)\,\Delta(s)$$

The response to an impulse input function, is the transfer function $$H(s)$$ itself $$\shaded{ Y(s)=H(s) }$$

Unit Step Response

The page Laplace Transforms gives the Laplace transform for the unit step function as $$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \gamma(t) \laplace \Gamma(s)\frac{1}{s}$$

Substitute the input function $$Y(s) = H(s)\,\Gamma(s)$$

The response to an unit step input function follows as $$\shaded{ Y(s) = \frac{H(s)}{s} }$$

Frequency Response

The frequency response is defined as the steady state response of system to a sinusoidal input.

Given sinusoidal input and transfer function \left\{\begin{align} x(t) &= \sin(\omega t)\gamma(t) &\mathrm{input\ signal} \nonumber \\ H(s) &= |H(s)|\,e^{j\angle H(s)} &\mathrm{transfer\ function} \nonumber \end{align}\right.

Find the output signal $$y(t)$$, using the Laplace transform of the sinusoidal input function \left.\begin{align} X(s) &= \frac{\omega}{s^2+\omega^2} \nonumber \\[8mu] Y(s) &= H(s)\,X(s)\nonumber \end{align}\right\}

Substitute $$X(s)$$ in $$Y(s)$$ \begin{align} Y(s) &= H(s)\,\frac{\omega}{s^2+\omega^2} \nonumber \\ &= H(s)\,\frac{\omega}{(s+j\omega)(s-j\omega)} \end{align}

According to Heaviside, this can be expressed as partial fractions [swarthmore, MIT-cu], where the term $$C_h(s)$$ represents the transient response resulting from $$H(s)$$. This term is independent of $$j\omega$$, and dies out for $$t\to\infty$$. $$Y(s) = \frac{\omega}{(s+j\omega)(s-j\omega)}H(s)=\underbrace{\frac{c_0}{s+j\omega}+\frac{c_1}{s-j\omega}}_{Y_{ss}(s)=\mathrm{steady\ state\ response}}+\underbrace{C_h(s)}_\mathrm{transient\ response} \label{eq:partialfractions}$$

Find $$c_0$$ \begin{align} H(s)\,\frac{\omega\cancel{(s+j\omega)}}{\cancel{(s+j\omega)}(s-j\omega)} &= \frac{c_0\cancel{(s+j\omega)}}{\cancel{s+j\omega}}+\frac{c_1(s+j\omega)}{s-j\omega}+C_h(s)(s+j\omega) \nonumber \\ \Rightarrow\, H(s)\,\frac{\omega}{s-j\omega} &= \left. c_0 + c_1\frac{s+j\omega}{s-j\omega} + C_h(s)(s+j\omega)\right|_{s=-j\omega} \nonumber \\ \Rightarrow\, H(-j\omega)\,\frac{\omega}{-j\omega-j\omega} &= c_0+c_1\frac{\cancelto{0}{-j\omega+j\omega}}{s-j\omega}+C_h(j\omega)(\cancelto{0}{-j\omega+j\omega}) \nonumber \\ \Rightarrow\, c_0 &= H(-j\omega)\,\frac{\cancel{\omega}}{-2j\cancel{\omega}} = \frac{H(-j\omega)}{-2j} \end{align}

Similarly, find $$c_1$$ \begin{align} H(s)\,\frac{\omega\cancel{(s-j\omega)}}{(s+j\omega)\cancel{(s-j\omega)}} &= \frac{c_0(s-j\omega)}{s+j\omega}+\frac{c_1\cancel{(s-j\omega)}}{\cancel{s-j\omega}}+C_h(s)(s-j\omega) \nonumber \\ \Rightarrow\, H(s)\,\frac{\omega}{s+j\omega} &= \left. c_0\frac{s-j\omega}{s-j\omega}+c_1+C_h(s)(s-j\omega)\right|_{s=j\omega} \nonumber \\ \Rightarrow\, H(j\omega)\,\frac{\omega}{j\omega+j\omega} &= c_0\frac{\cancelto{0}{j\omega-j\omega}}{s-j\omega}+c_1+C_h(j\omega)(\cancelto{0}{j\omega-j\omega}) \nonumber \\ \Rightarrow\, c_1 &= H(j\omega)\,\frac{\cancel{\omega}}{2j\cancel{\omega}} = \frac{H(j\omega)}{2j} \end{align}

Inverse Laplace transform of $$\eqref{eq:partialfractions}$$ back to the time domain $$y(t) = \underbrace{c_0e^{-j\omega t}+c_1e^{j\omega t}}_{y_{ss}(t)=\mathrm{steady\ state\ response}}+\underbrace{\cancelto{0\mathrm{\ as\ }t \to\infty}{\mathfrak{L}^{-1}\left\{C_h(s)\right\}}}_{\mathrm{transient\ response}}$$

Substitute $$c_0$$ and $$c_1$$ in $$y_{ss}(t)$$ \begin{align} y_{ss}(t) &= \frac{H(-j\omega)}{-2j}e^{-j\omega t}+\frac{H(j\omega)}{2j}e^{j\omega t} \nonumber \\ &= \frac{H(j\omega)e^{j\omega t}-H(-j\omega)e^{-j\omega t}}{2j}\label{eq:yss} \end{align}

Based on Euler’s formula we can express $$H(s)$$ in polar coordinates

\left\{ \begin{align} H(s) &= |H(s)|\,e^{j\angle H(s)} \nonumber \\ |H(j\omega)| &= K \frac{\prod_{i=1}^m \sqrt{\left(\Re\{{z_i}\}\right)^2+\left(\omega+\Im\{z_i\}\right)^2}}{\prod_{i=1}^n \sqrt{\left(\Re\{{p_i}\}\right)^2+\left(\omega+\Im\{p_i\}\right)^2}} \nonumber \\ \angle{H(s)} &= \sum_{i=1}^m\mathrm{atan2}\left(\omega+\Im\{z_i\}, \Re\{{z_i}\}\right) -\sum_{i=1}^n\mathrm{atan2}\left(\omega+\Im\{p_i\}, \Re\{{p_i}\}\right) \nonumber \\ \end{align} \right. \label{eq:euler}

Substitute the polar representation of $$H(s)$$, in $$y_{ss}$$ \begin{align} y_{ss}(t) &= |H(j\omega)|\left( \frac{e^{j\angle H(j\omega)}e^{j\omega t}-e^{j\angle H(-j\omega)}e^{-j\omega t}}{2j} \right) \nonumber \\ &= |H(j\omega)|\left( \frac{e^{j\left(\angle H(j\omega)+\omega t\right)}-e^{j\left(\angle H(-j\omega)-\omega t\right)}}{2j} \right) \nonumber \\ &= |H(j\omega)|\, \underbrace{\frac{e^{j(\omega t + \angle{H(j\omega)})}-e^{-j(\omega t + \angle H(j\omega))}}{2j}}_{\sin(\omega t+\angle H(j\omega))} \end{align}

In this, we recognize the Laplace transfer for a sinusoidal function

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

The frequency response follows as $$\shaded{ y_{ss}(t) = |H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t) }$$

In other words, for a linear system a sinusoidal input generates a sinusoidal output with the same frequency, but different amplitude and phase.

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.

Mechanical systems

Using Laplace transforms to solve mechanical ordinary differential equations.

Elements

Before we look at examples of mechanical systems, let’s recall the equations of mechanical element.

Spring

According to Hooke’s law, the reactive force is linear proportional to the displacement and opposite the direction of the force

$$f_s(t)=S\cdot x(t) \label{eq:spring}$$ where $$S$$ is the spring stiffness [N/m].

Resistance

A dashpot provides friction force linear proportional with the velocity and opposite the direction of the force.

$$f_r(t)=R\cdot \frac{\mathrm{d}x(t)}{\mathrm{d}t} \label{eq:damper}$$ where $$R$$ is the resistance, or damping coefficient [Ns/m]

Mass

According to Newton’s second law of motion the reactive force is linear proportional to the acceleration and opposite the direction of the force

$$f_m(t)=M\cdot \frac{\mathrm{d}^2x(t)}{\mathrm{d}t^2} \label{eq:mass}$$ where $$M$$ is the mass [kg]

First Order Example

Constant force on horizontal parallel damper and spring, starting at $$t=0$$

$$f_{e}(t) = F_0\gamma(t) \label{eq:applied}$$

Sum of the forces must be zero. This combine the equations for the external force $$\eqref{eq:applied}$$ with the equations for spring $$\eqref{eq:spring}$$ and damper $$\eqref{eq:damper}$$. \begin{align} f_r(t) + f_s(t) &= f_e(t) \nonumber \\ R\frac{\mathrm{d}x(t)}{\mathrm{d}t} + S x(t) &= F_0\gamma(t) \end{align}

Transform this to the ordinary differential equation, knowing that the system starts from rest, therefor $$\frac{\mathrm{d}x(0)}{\mathrm{d}t}=0$$ and $$x(0)=0$$.

$$R\,s\,X(s)+ S X(s) = F_0\frac{1}{s}$$

Solve for $$X(s)$$ \begin{align} X(s) (s R +S)&=F_0\frac{1}{s} \nonumber \\ \Rightarrow\ X(s)&=\frac{F_0}{s(s R +S)} \end{align}

To return back to the time domain $$x(t)$$, we need to find a reverse Laplace transform. There is none. According to Heaviside, this can be expressed as partial fractions. [swarthmore] $$X(s)=\frac{F_0}{s(sR+S)}\equiv\frac{c_0}{s}+\frac{c_1}{sR+S} \label{eq:heaviside}$$

The constants $$c_{0,1}$$ are found using Heaviside’s Cover-up Method [swarthmore, MIT-cu]: multiply $$\eqref{eq:heaviside}$$ with respectively $$s$$ and $$(sR-S)$$. \left\{ \begin{align} \frac{\cancel{s}F_0}{\cancel{s}(sR+S)} &\equiv\frac{\cancel{s}c_0}{\cancel{s}}+\frac{sc_1}{sR+S} \nonumber\\ \frac{\cancel{(sR+S)}F_0}{\cancel{(sR+S)}s}&\equiv\frac{(sR+S)c_0}{s}+\frac{\cancel{(sR+S)}c_1}{\cancel{sR+S}}\nonumber \end{align} \right.

Given that these equations are true for any value of $$s$$, choose two convenient values to find $$c_0$$ and $$c_1$$ $$\left\{ \begin{eqnarray} c_0=\left.\frac{F_0}{sR+S}-\frac{s\,c_1}{sR+S}\right|_{s=0}&=\frac{F_0}{S}-\frac{0}{0+S}&=\frac{1}{S}F_0\label{eq:constants1}\\ c_1=\left.\frac{F_0}{s}-\frac{(sR+S)c_0}{s}\right|_{s=-\frac{S}{R}}&=-\frac{R}{S}F_0-\frac{0}{-\frac{S}{R}}&=-\frac{R}{S}F_0\nonumber \end{eqnarray} \right.$$

The unit step response $$x(t)$$ follows from the inverse Laplace transform of $$\eqref{eq:heaviside}$$ \begin{align} x(t)&= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{c_1}{sR+S}\right\} & t\geq0 \nonumber \\ &= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\frac{1}{R}\mathcal{L}^{-1}\left\{\frac{c_1}{s+\frac{S}{R}}\right\} & t\geq0 \nonumber \\ &= c_0+\frac{1}{R}c_1e^{-\frac{S}{R}t} & t\geq0 \end{align}

Substituting the constants $$\eqref{eq:constants1}$$ gives the unit step response \begin{align} x(t)&=\frac{1}{S}F_0+\frac{1}{\cancel{R}}(-\frac{\cancel{R}}{S}F_0)e^{-\frac{S}{R}t} & t\geq0 \nonumber \\ &=\frac{1}{S}F_0\left(1-e^{-\frac{S}{R}t}\right) & t\geq0\\ \end{align}

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.