.

Laplace Transforms

Pierre-Simon,_marquis_de_Laplace_(1745-1827)_-_Guérin

source wikipedia.org

In 1785, Pierre-Simon, marquis de Laplace, a French mathematician and physicist, discovered an integral transform later named after him.  This Laplace transform turns differential equations in time, into algebraic equations in the Laplace domain thereby making them easier to solve.\(\)

Definition

In 1809, Piere-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

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
\mathfrak{L}\left\{\,f(t)\,\right\}&=F(s)=\int_{0^-}^\infty e^{-st}f(t)\ \mathrm{d}t
\end{aligned}
}\label{eq:laplace}\tag{1}$$

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

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

Functions

Time domain Laplace domain
Impulse $$\delta(t)1$$ $$1$$ proof
Unit Step $$\gamma(t)$$ $$\frac{1}{s}$$ proof
Ramp $$t\,\gamma(t)$$ $$\frac{1}{s^2}$$ proof
Exponential $$e^{-at}\gamma(t)$$ $$\frac{1}{s+a},\ \forall_{a>0}$$ proof
Sine $$\sin(\omega t)\,\gamma(t)$$ $$\frac{\omega}{s^2+\omega^2}$$ proof
Cosine $$\cos(\omega t)\,\gamma(t)$$ $$\frac{s}{s^2+\omega^2}$$ proof
Decaying Sine $$e^{-\alpha t}\sin(\omega t)\,\gamma(t)$$ $$\frac{\omega}{(s+\alpha)^2+\omega^2}$$ proof
Decaying Cosine $$e^{-\alpha t}\cos(\omega t)\,\gamma(t)$$ $$\frac{s+\alpha}{(s+\alpha)^2+\omega^2}$$ proof
Time Delayed $$f(t-a)\,\gamma(t-a)$$ $$e^{-su}F(s)$$ proof
Decaying Oscillation $$e^{-at}\left(
B\cos(\omega_d t)+
\frac{C-aB}{\omega_d}\sin(\omega_d t)
\right)\,\gamma(t)$$
$$\frac{\omega_d}{(s+a)^2+\omega_d}$$

 Initial and Final Value Theorem

Time domain Laplace domain
Initial Value $$f(0^+)$$ $$\lim_{s\to\infty}\left(s\,F(s)\right)$$ proof
Final Value $$f(\infty)$$ $$\lim_{s\to0}\left( s\,F(s) \right)$$ proof

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

Linearity Property

The linearity property in the time domain

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

Transformed to the Laplace domain

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

From which follows

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
a\cdot f(t)+b\cdot g(t)
\overset{\mathfrak{L}}\longleftrightarrow
a\cdot F(s) + b\cdot G(s)
}\label{eq:linearity}\tag{2}$$

own workFirst Derivative Property

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

$$u(t)=\frac{\mathrm{d}}{\mathrm{d}t}f(t)\tag{3}$$

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

$$\begin{aligned}
\mathfrak{L}\left\{\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)\right\} &= \int_{0^-}^{\infty}
e^{-st}\tfrac{\mathrm{d}f(t)}{\mathrm{d}t}
\mathrm{d}t
=\int_{0^-}^{\infty}
\underbrace{e^{-st}}_{u(t)}
\underbrace{\tfrac{\mathrm{d}f(t)}{\mathrm{d}t}}_{v'(t)} \mathrm{d}t\Rightarrow
\end{aligned}\label{eq:derivative1}\tag{4}$$

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

$$\left\{
\begin{aligned}
\int_a^b u(t)\ v'(t)\ \mathrm{d}t&=\left[ u(t)\ v(t)\right]_a^b
-\int_a^b u'(t)\ v(t)\ \mathrm{d}t\\
u(t)&=\int_{0^-}^t f(\tau)\mathrm{d}\tau
\Rightarrow u'(t)=f(t)\\
v'(t)&=e^{-st}
\Rightarrow v(t)=-\tfrac{1}{s}e^{-st}
\end{aligned}
\right.\label{eq:intbyparts}\tag{5}$$

Solve \(\eqref{eq:derivative1}\) using integration by parts \(\eqref{eq:intbyparts}\)

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

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 \(s\).

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\require{cancel}
\begin{aligned}
\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)
\overset{\mathfrak{L}}\longleftrightarrow
s\,F(s)-f(0^-)
\end{aligned}}\label{eq:derivative}\tag{6}$$

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

$$u(t)=\frac{\mathrm{d}^2}{\mathrm{d}t^2}f(t)\tag{7}$$

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

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

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 \(\frac{\mathrm{d}}{\mathrm{d}t}f(t)\)

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

Substitute \(G(s)\) in \(U(s)\)

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

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

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\require{cancel}
\begin{aligned}
\tfrac{\mathrm{d}^2}{\mathrm{d}t^2}f(t)
\overset{\mathfrak{L}}\longleftrightarrow
s^2F(s)-sf(0^-) – f'(0^-)
\end{aligned}}
\label{eq:secondderivative}\tag{8}$$

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

Integral PropertyIntegration Property

Determine the Laplace transform of the integral

$$u(t)=\int_{0^-}^t f(\tau)\mathrm{d}\tau\tag{9}$$

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

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

Again, solve using integration by parts \(\eqref{eq:intbyparts}\)

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

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 \(s\).

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
\int_{0^-}^t f(\tau)\mathrm{\tau}
\overset{\mathfrak{L}}\longleftrightarrow
\frac{1}{s}F(s)
\end{aligned}}\label{eq:integration}\tag{10}$$

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

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

Substitute \(u=t-\lambda\)

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

Gives us the Laplace transfer for the convolution property

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
f(t)\ast g(t)
\overset{\mathfrak{L}}\longleftrightarrow
F(s)\,G(s)
}\label{eq:convolution}\tag{11}$$

own workImpulse Function

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

$$u(t)=\delta(t)=\begin{cases}
0 & t<0 \\
\mathrm{undefined} & t=0 \\
0 & t>0
\end{cases}\label{eq:impuls_def1}\tag{12a}$$

and satisfies the condition

$$\int_{-\infty}^{\infty}\delta(t)=1\label{eq:impuls_def2}\tag{12b}$$

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 \(t=0\)

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

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

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
\delta(t)
\overset{\mathfrak{L}}\longleftrightarrow
1
\end{aligned}
}\label{eq:impulse}\tag{13}$$

own workUnit Step Function

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

$$u(t)=\gamma(t)=\begin{cases}
0 & t<0 \\
1 & t\geq 0 \\
\end{cases}\label{eq:unitstep_def_a}\tag{14}$$

The unit step function is related to the impulse function as

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

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

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

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

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

Gives us the Laplace transfer of the unit step function

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
\gamma(t)
\overset{\mathfrak{L}}\longleftrightarrow
\frac{1}{s}
\end{aligned}
}\label{eq:unitstep}\tag{15}$$

own workRamp 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 Unterschied, which means “difference”.  The capital letter of \(\gamma\) is \(\Gamma\) what looks a bit like the step function.

$$u(t)=t\,\gamma(t)\label{eq:ramp_def_a}\tag{16}$$

The ramp function is related to the unit step  function as

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

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

$$\begin{aligned}
U(s)=\mathcal{L}\left\{\,t\,\right\}\,&=\int_{0^-}^\infty \underbrace{e^{-st}}_{v'(t)}\,\underbrace{t}_{u(t)}\,\mathrm{d}t
\end{aligned}\label{eq:ramp1}\tag{17}$$

Use integration by parts

$$\left\{
\begin{aligned}
\int_a^b u(t)\ v'(t)\ \mathrm{d}t&=\left[ u(t)\ v(t)\right]_a^b
-\int_a^b u'(t)\ v(t)\ \mathrm{d}t\\
u(t)&=t
\Rightarrow u'(t)=1\\
v'(t)&=e^{-st}
\Rightarrow v(t)=-\tfrac{1}{s}e^{-st}
\end{aligned}
\right.\label{eq:intbyparts2}\tag{18}$$

Solve \(\eqref{eq:ramp1}\) using integration by parts \(\eqref{eq:intbyparts2}\)

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

Gives us the Laplace transfer of the ramp function

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
t\,\gamma(t)
\overset{\mathfrak{L}}\longleftrightarrow
\frac{1}{s^2}
\end{aligned}
}\label{eq:ramp}\tag{19}$$

own workExponential 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

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

Gives us the Laplace transform of the exponential time function

$$\require{cancel}
\def\doubleunderline#1{\underline{\underline{#1}}}
\begin{aligned}
\doubleunderline{
e^{-at}\gamma(t)
\overset{\mathfrak{L}}\longleftrightarrow
\frac{1}{s+a}
}
&&\forall_{a>0}
\end{aligned}\label{eq:exponential}\tag{20}$$

own workSine Function

Another popular input signal is the sine wave, starting at \(t=0\)

$$u(t)=f(t)=\sin(\omega t)\,\gamma(t)\label{eq:sin_def}\tag{21}$$

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

$$\begin{aligned}
\mathcal{L}\left\{f(t)\right\}=F(s)
&=\int_{0^-}^{\infty}e^{-st}\sin(\omega t)\gamma(t)
\,\mathrm{d}t\\
&=\int_{0^-}^{\infty}e^{-st}\sin(\omega t)
\,\mathrm{d}t
\end{aligned}\label{eq:sinlaplace}\tag{22}$$

Apply the Euler identity for sine to \(\eqref{eq:sinlaplace}\)

$$\begin{aligned}
F(s)&=\int_{0^-}^{\infty}e^{-st}\,\frac{e^{j\omega t}-e^{-j\omega t}}{2j}
\,\mathrm{d}t
\\
&=\frac{1}{2j}\int_{0^-}^{\infty}e^{-st}\,\left(e^{j\omega t}-e^{-j\omega t}\right)
\,\mathrm{d}t
\\
&=\frac{1}{2j}\int_{0^-}^{\infty}e^{-st}\,e^{j\omega t}\,\mathrm{d}t
-\frac{1}{2j}\int_{0^-}^{\infty}e^{-st}\,e^{-j\omega t}
\,\mathrm{d}t
\\
&=\frac{1}{2j}\int_{0^-}^{\infty}e^{(-s+j\omega) t}\,\mathrm{d}t
-\frac{1}{2j}\int_{0^-}^{\infty}e^{(-s-j\omega)t}
\,\mathrm{d}t
\end{aligned}\label{eq:sin2}\tag{23}$$

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

$$\require{cancel}
\begin{aligned}
\int_{0^-}^{\infty}\ e^{-(s+a) t}\,\mathrm{d}t
&= \frac{1}{s+a} & a>0
\end{aligned}\label{eq:sin3}\tag{24}$$

Substitute \(\eqref{eq:sin3}\) in \(\eqref{eq:sin2}\)

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

Bring it under a common denominator

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

Et voilà, the Laplace transform of sine function

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
\sin(\omega t)\,\gamma(t)
\overset{\mathfrak{L}}\longleftrightarrow
\frac{\omega}{s^2+\omega^2}
\end{aligned}
}\label{eq:sine}\tag{25}$$

Cosine Function

Yet another popular input signal is the cosine wave, starting at \(t=0\)

$$u(t)=f(t)=\cos(\omega t)\,\gamma(t)\label{eq:cos_def}\tag{26}$$

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

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
\cos(\omega t)\,\gamma(t)
\overset{\mathfrak{L}}\longleftrightarrow
\frac{s}{s^2+\omega^2}
\end{aligned}
}\label{eq:cosine}\tag{27}$$

Decaying Sine Function

Consider a decaying sine wave, starting at \(t=0\)

$$u(t)=f(t)=e^{-\alpha t}\sin(\omega t)\,\gamma(t)\label{eq:decayingsine_def}\tag{28}$$

Apply the Euler identity for sine

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

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

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

Put over common denominator

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

The Laplace transforms of the decaying sine

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
e^{-\alpha t}\sin(\omega t)\,\gamma(t)
\overset{\mathfrak{L}}\longleftrightarrow
\frac{\omega}{(s+\alpha)^2+\omega^2}
\end{aligned}
}\label{eq:decayingsine}\tag{29}$$

Decaying Cosine Function

Consider a decaying cosine wave, starting at \(t=0\)

$$u(t)=f(t)=e^{-\alpha t}\cos(\omega t)\,\gamma(t)\label{eq:decayingcosine_def}\tag{30}$$

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.

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
e^{-\alpha t}\cos(\omega t)\,\gamma(t)
\overset{\mathfrak{L}}\longleftrightarrow
\frac{s+\alpha}{(s+\alpha)^2+\omega^2}
\end{aligned}
}\label{eq:decayingcosine}\tag{31}$$

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

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

Substitute \(u=t-a\)

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

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

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
f(t-a)\,\gamma(t-a)
\overset{\mathfrak{L}}\longleftrightarrow
e^{-su}F(s)
}\label{eq:timedelay}\tag{32}$$

Initial Value Theorem

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

$$\begin{aligned}
\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)
\overset{\mathfrak{L}}\longleftrightarrow
s\,F(s)-f(0^-)
\end{aligned}$$

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

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

Take the limit as \(s\to\infty\)

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

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 \(0\)

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

The initial value theorem follows as

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
f(0^+)
=
\lim_{s\to\infty}\left(s\,F(s)\right)
\end{aligned}}\label{eq:initialvalue}\tag{33}$$

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

$$\begin{aligned}
\tfrac{\mathrm{d}}{\mathrm{d}t}f(t)
\overset{\mathfrak{L}}\longleftrightarrow
s\,F(s)-f(0^-)
\end{aligned}$$

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

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

Take the limit as \(s\to 0\)

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

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

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

The integral doesn’t depend on \(s\)

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

The final value theorem follows as

$$\def\doubleunderline#1{\underline{\underline{#1}}}
\doubleunderline{
\begin{aligned}
f(\infty)
=
\lim_{s\to0}\left( s\,F(s) \right)
\end{aligned}}\label{eq:finalvalue}\tag{34}$$

Next

Suggested next reading is Transfer Functions.

Coert Vonk

Coert Vonk

Independent Firmware Engineer at Los Altos, CA
Welcome to the things that I couldn’t find.This blog shares some of the notes that I took while deep diving into various fields.Many such endeavors were triggered by curious inquiries from students. Even though the notes often cover a broader area, the key goal is to help the them adopt, flourish and inspire them to invent new technology.
Coert Vonk

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