Laplace transform, proof of properties and functions (2024)

\(\)Around 1785, Pierre-Simon marquis de Laplace, a French mathematician and physicist, pioneered a method for solving differential equations using anintegral transform. ThisLaplace 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 theLaplace 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 introducecommonly usedproperties, common input functions and initial/final valuetheorems, referred to from my variousElectronics articles. Many are based on the excellent notes from the linear physics group atSwarthmore College, and reproduced here mainly for my own understanding and reference.

Properties

Functions

Initial and Final Value Theorem

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 Laplacedomain

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

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 DerivativeProperty

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\(\eqref{eq:laplace}\)

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

Recallintegration 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 \(\eqref{eq:derivative_}\) usingintegration 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 isa 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 istaken at\(t=0^-\). This means that we only need to know thisinitial conditions before theinput signal started.

SecondDerivativeProperty

The secondderivative 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 aretaken at\(t=0^-\). This means that we only need to know these initial conditions before theinput 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 usingintegration 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 isa condition for existence of the transform. In the secondterm, 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 showtheconvolution 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 Laplacedomain

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

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 convolutionproperty

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

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 maybereplaced 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 orHeaviside stepfunction, 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

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

RampFunction

The unit orHeaviside stepfunction, denoted with \(\gamma(t)\) is defined as below [smathmore]. We use \(\gamma(t)\), to avoid confusion with theEuropean symbol for voltage source \(u(t)\), where \(u\)stands for “Potentialunterschied”,which means potential difference. The capital letter of \(\gamma\) is \(\Gamma\) whatlooks a bit like thestep function.

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

The rampfunction is related to the unit step function as

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

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

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

Useintegration 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 \(\eqref{eq:ramp1}\) usingintegration 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 rampfunction

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

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

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

CosineFunction

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 SineFunction

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 CosineFunction

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 simplifiesLaplace transformbecause \(\gamma(t-a)\) 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 theLaplace transform. Together it gives us the Laplace transform ofa 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 dependon \(s\), 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

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

FinalValueTheorem

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’ta 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, westart with the First Derivative \(\eqref{eq:derivative}\) and apply thedefinition 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 \(s\to0\)

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

Take the terms out of the limit that don’t dependon \(s\), 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 finalvalue theorem follows as

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