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