Prerequisites

  • Laplace Transformation
  • Inverse Laplace Transformation
  • Kirchoff’s Loop Laws
  • Inductor Equation
  • Capacitor Equation

TL;DR

Note that this is grossly simplified as a cheatsheet.

We denote a Laplace Transform of f(t) as

\[\mathcal{L}\{f(t)\} = F(s)\] \[\mathcal{L}^{-1}\{F(s)\} = f(t)\]
Type t-domain s-domain
Resistor $x\Omega$ $x$
Ind. Voltage Source $xV$ $x/s$
Ind. Current Source $xA$ $x/s$
Inductor $xH$ $x*s$
Capacitor $xF$ $1/xs$

Introduction

Laplace Transform makes analysing a circuit much easier, here’s a taste of it.

Consider the following. We will use mesh-analysis to work this out:

Original Circuit

+--------{25H}--------+
|          →          |
|        ↑ 1 ↓        |
|          ←          |
+-------<6.25F>-------+
|          →          |
|        ↑ 2 ↓        |
|          ←          |
+--[800Ω]--[+ 30V -]--+
\[Loop 1\rightarrow-25*\frac{di_1(t)}{dt}-\frac{1}{6.25}*\int_{0^-}^{\inf}\frac{d(i_1(t)-i_2(t))}{dt}dt=0\] \[Loop 2\rightarrow-\frac{1}{6.25}*\int_{0^-}^{\inf}\frac{d(i_2(t)-i_1(t))}{dt}dt+30-800*i_2(t)=0\]

Well, forget Loop 2, Loop 1 is already daunting to calculate. This should be enough reason to want to learn Laplace.

The Laplace transformation transforms something from the t-domain to the s-domain.

In other words, we want to transform something in terms of time to frequency.

Laplace Circuit

+--------{25s}--------+
|          →          |
|        ↑*1*↓        |
|          ←          |
+-----<1/(6.25s)>-----+
|          →          |
|        ↑*2*↓        |
|          ←          |
+--[800]--[+ 30/s -]--+

Notes:

  • Currents are also Laplace Transformed
  • Assumption that Capacitor and Inductor aren’t charged before $0^-$
  • Detailed explanation of transformation later
\[Loop 1\rightarrow-25s*I_1(s)-\frac{I_1(s)-I_2(s)}{6.25s}=0\] \[Loop 2\rightarrow-\frac{I_2(s)-I_1(s)}{6.25s}+\frac{30}{s}-800*I_2(s)=0\]

Looks simpler, but now we have an extra variable s in there.

The main goal is the express everything that you need in terms of s, and decompose the fraction.

I’ll do $i_1$:

\[I_1(s)= \frac{0.0125}{s+160}-\frac{0.05}{s+40}+\frac{0.0375}{s}\] \[\mathcal{L}^{-1}\{I_1(s)\}= \mathcal{L}^{-1}\bigg\{\frac{0.0125}{s+160}\bigg\} - \mathcal{L}^{-1}\bigg\{\frac{0.05}{s+40}\bigg\} + \mathcal{L}^{-1}\bigg\{\frac{0.0375}{s}\bigg\}\] \[i_1(t) = \left[0.0125e^{-160t}-0.05e^{-40t}+0.0375\right]u(t)\]

We can find the voltage of the inductor

\[L\frac{di_1(t)}{dt} = v_{ind.} = 50e^{-40t}-50e^{-160t}\]

Proofs

To understand why it works, let’s look at the equations

Resistor

\[v(t) = r * i(t)\] \[\mathcal{L}\{v(t)\} = \mathcal{L}\{r*i(t)\}\] \[\mathcal{L}\{v(t)\} = r*\mathcal{L}\{i(t)\}\] \[V(s) = r*I(s)\]

Voltage Source

\[v(t) = v\] \[\mathcal{L}\{v(t)\} = \mathcal{L}\{v\}\] \[V(s) = \frac{v}{s}\]

Transforming a Current Source to a Voltage Source would make analysis easier

Inductor

\[\begin{align} v(t) &= L*\frac{di(t)}{dt}\\ \mathcal{L}\{v(t)\} &= \mathcal{L}\bigg\{L*\frac{di(t)}{dt}\bigg\} \\ V(s) &= L*\mathcal{L}\bigg\{\frac{di(t)}{dt}\bigg\} \\ &\stackrel{(1)}{=} L *\left[s*I(s)-i(0^-)\right] \\ V(s) &= LsI(s)-Li(0^-) \end{align}\]

Capacitor

\[\begin{align} i(t) &= C*\frac{dv(t)}{dt} \\ \mathcal{L}\{i(t)\} &= \mathcal{L}\bigg\{C*\frac{dv(t)}{dt}\bigg\} \\ I(s) &= C*\mathcal{L}\bigg\{\frac{dv(t)}{dt}\bigg\} \\ &\stackrel{(1)}{=} C *\left[s*V(s)-v(0^-)\right] \\ I(s) &= CsV(s)-Cv(0^-) \\ V(s) &= \frac{I(s)+Cv(0^-)}{Cs} \\ &= \frac{I(s)}{Cs} + \frac{v(0^-)}{s} \end{align}\]

Laplace of a Derivative (1)

Using the Identity:

\[s\mathcal{L}\{f(t)\}=\mathcal{L}\{f^{\prime}(t)\}+f(0^-)\]

We have

\[\mathcal{L}\{f^{\prime}(t)\} = s\mathcal{L}\{f(t)\}-f(0^-)\]

Putting it into a Circuit

Notice that we expressed everything in Voltage (excluding current source), this is so that we can insert them in easily.

  • $V_0, v_0$ are grounds

Resistor

\[v(t) = r * i(t)\] \[V(s) = r * I(s)\] 
 v        v       V       V
[0]      [a]     [0]     [a] 
 +--[xΩ]--+       +--[x]--+
    <---      =      <--   
    i(t)             I(t)

Note that if you measure $v_a$ and $V_a$, they tally with the transformation.

Voltage Source

\[v(t) = v\] \[V(s) = \frac{v}{s}\] 
 v            v       V         V
[0]          [a]     [0]       [a] 
 +--[- xV +]--+   =   +--[x/s]--+
    <-------             <----
      i(t)                I(t)

Transforming a Current Source to a Voltage Source would make analysis easier

Inductor

\[v(t) = L*\frac{di(t)}{dt}\] \[V(s) = LsI(s)-Li(0^-)\] 
 v        v       V                     V
[0]      [a]     [0]                   [a] 
 +--{xH}--+   =   +--{xs}--[+ xi(0) -]--+
    <---             <----------------
    i(t)                    I(t)

Notes:

  • The direction of the additional Voltage Source is important.
  • If the Inductor starts with 0 current, then we can omit it.

Capacitor

\[i(t) = C*\frac{dv(t)}{dt}\] \[V(s) = \frac{I(s)+Cv(0^-)}{Cs} = \frac{I(s)}{Cs} + \frac{v(0^-)}{s}\] 
 v        v       V                       V
[0]      [a]     [0]                     [a] 
 +--<xF>--+   =   +--<1/xs>--[- v(0)/s +]--+
    <---             <-------------------
    i(t)                     I(t)

Notes:

  • The direction of the additional Voltage Source is important.
  • If the Capacitor starts with 0 voltage, then we can omit it.

Summary

As mentioned in the TL;DR

At least you understand why now

\[\mathcal{L}\{f(t)\} = F(s)\] \[\mathcal{L}^{-1}\{F(s)\} = f(t)\]
Type t-domain s-domain
Resistor $x\Omega$ $x$
Ind. Voltage Source $xV$ $x/s$
Ind. Current Source $xA$ $x/s$
Inductor $xH$ $x*s$
Capacitor $xF$ $1/xs$