Laplace Transform
4.8 Application: Electrical Circuits
A. Introduction
This section briefly shows the practical use of the Laplace Transform in electrical engineering for solving differential equations and systems of such equations associated with electric circuits. The Laplace Transform is particularly beneficial for converting these differential equations into more manageable algebraic forms.
We start by looking at a single initial value problem (IVP) from a basic RLC circuit. We demonstrate how the Laplace transform can simplify finding the circuit’s current as a function of time by translating a differential equation into an algebraic equation.
Consider an RLC series circuit with a resistor of [asciimath]0.06\ Omega[/asciimath] and an inductor of [asciimath]0.01\ "H"[/asciimath], and a capacitor of [asciimath]50/89\ "F"[/asciimath] powered by a voltage [asciimath]E(t)=0.1sin(10t)\ "V"[/asciimath]voltage source. Initially, the current and charge on the capacitor are zero. Determine the current in the circuit as a function of time.
Show/Hide Solution
Given information:
- Resistor: [asciimath]R=0.06\ Omega[/asciimath]
- Inductor: [asciimath]L=0.01 \ "H"[/asciimath]
- Capacitor: [asciimath]C=50/89\ "F"[/asciimath]
- Voltage source: [asciimath]E(t)=0.1sin(10t)\ "V"[/asciimath]
- Initial current on capacitor: [asciimath]I(0)=0[/asciimath]
- Initial charge on capacitor: [asciimath]q(0)=I'(0)=0[/asciimath]
In Example 3.9.1, we developed the initial value problem governing this RLC circuit.
[asciimath](d^2I)/(dt^2)+6 (dI)/dt+178I=100cos(10t),[/asciimath] [asciimath]I(0)=0, \ I'(0)=0[/asciimath]
Applying the Laplace Transform to the differential equation results in
[asciimath]\mathcal{L}{I''}+6\mathcal{L}{I'}+178\mathcal{L}{I}=(100s)/(s^2+10^2)[/asciimath]
Letting [asciimath]\mathcal{L}{I(t)}=J(s)[/asciimath], we have
[asciimath]\mathcal{L}{I''}=s^2J(s)-sI(0)-I'(0)=s^2J(s)[/asciimath]
[asciimath]\mathcal{L}{I'}=sJ(s)-sI(0)=sJ(s)[/asciimath]
Since [asciimath]I(0)[/asciimath] and [asciimath]I'(0)[/asciimath] are both zero, the equation simplifies to
[asciimath]s^2J(s)+6sJ(s)+178J(s)=(100s)/(s^2+10^2)[/asciimath]
Solving for[asciimath]J(s)[/asciimath], we find
[asciimath]J(s)=(100s)/((s^2+10^2)(s^2+6s+178))[/asciimath]
Breaking [asciimath]J(s)[/asciimath] down by partial fraction expansion, we obtain
[asciimath]J(s)=1/807((650s+5000)/(s^2+10^2))-[/asciimath] [asciimath]1/807((650s+8900)/(s^2+6s+178))[/asciimath]
To simplify the second fraction, we complete the square.
[asciimath]J(s)=650/807(s/(s^2+10^2))+5000/807(1/(s^2+10^2)) -[/asciimath] [asciimath]650/807(s/((s+3)^2+13^2)) -[/asciimath] [asciimath]8900/807(1/((s+3)^2+13^2))[/asciimath]
Applying the inverse Laplace Transforms to [asciimath]J(s)[/asciimath] yields the current [asciimath]I(t)[/asciimath].
[asciimath]I(t)=650/807cos(10t)+500/807sin(10t)+[/asciimath] [asciimath]e^(-3t)(-650/807cos(13t)-6950/10491sin(13t))[/asciimath]
This result is consistent with what we obtained in Example 3.9.1 by solving the initial value problem using the method of undetermined coefficients.
B. Solving Systems of Linear Equations with the Laplace Transform
The Laplace Transform can be applied to turn certain systems of differential equations with initial values into systems of algebraic equations in the s-domain. Solving these algebraic equations allows us to find functions of [asciimath]s[/asciimath], which we can then convert back into time-domain solutions using the inverse Laplace Transform. Next, we address a more complex example involving a series-parallel RL circuit, which results in a system of differential equations.
a) For the given electrical circuit diagram, derive the system of differential equations that describes the currents in various branches of the circuit. Assume that all initial currents are zero. b) Once the system of differential equations and initial conditions are established, solve the system for the currents in each branch of the circuit.
Diagram Description
Show/Hide Solution
We denote the current passing through the main branch by [asciimath]I_1[/asciimath], the current through the 2ohm-resistor by [asciimath]I_2[/asciimath] and the current passing through the 0.1H-inductor by [asciimath]I_3[/asciimath].
Given the voltage drop across a resistor is [asciimath]RI[/asciimath] and across an inductor is [asciimath]L(dI)/dt[/asciimath], we apply Kirchhoff’s voltage law to the electrical network.
In the main loop including 0.1 H-inductor, we find
[asciimath]4I_1+0.1(dI_3)/dt+0.2(dI_1)/dt=12[/asciimath]
In the sub-branch including the 2 Ω-resistor and 0.1 H-inductor, we find
[asciimath]0.1(dI_3)/dt-2I_2=0[/asciimath]
Also, since current [asciimath]I_1[/asciimath] is split into [asciimath]I_2[/asciimath] and [asciimath]I_3[/asciimath], we have
[asciimath]I_1=I_2+I_3[/asciimath]
Thus the system of equations describing the currents in the circuit is
[asciimath]{(4I_1+0.1I'_3+0.2I'_1=12),(0.1I'_3-2I_2=0), (I_1-I_2-I_3=0) :}[/asciimath] [asciimath];[/asciimath] [asciimath]I_1(0)=I_2(0)=I_3(0)=0[/asciimath](4.8.1)
b)
To solve the system, we apply the Laplace Transform to each equation in the system.
[asciimath]{(4\mathcal{L}{I_1}+0.1\mathcal{L}{I'_3}+0.2\mathcal{L}{I'_1}=12/s),(0.1\mathcal{L}{I'_3}-2\mathcal{L}{I_2}=0), (\mathcal{L}{I_1}-\mathcal{L}{I_2}-\mathcal{L}{I_3}=0) :}[/asciimath](4.8.2)
Letting [asciimath]\mathcal{L}{I_1}=J_1(s)[/asciimath], [asciimath]\mathcal{L}{I_2}=J_2(s)[/asciimath], and [asciimath]\mathcal{L}{I_3}=J_3(s)[/asciimath], we have
[asciimath]\mathcal{L}{I'_1}=sJ_1(s)-I_1(0)[/asciimath] [asciimath]=sJ_1(s)[/asciimath]
[asciimath]\mathcal{L}{I'_3}=sJ_3(s)-I_3(0)[/asciimath] [asciimath]=sJ_3(s)[/asciimath]
Since initial currents are zero, system 4.8.2 simplifies to
[asciimath]{(4J_1(s)+0.1sJ_3(s)+0.2sJ_1(s)=12/s),(0.1sJ_3(s)-2J_2(s)=0), (J_1(s)-J_2(s)-J_3(s)=0) :}[/asciimath]
In the third equation, we express [asciimath]J_2(s)[/asciimath] in terms of the other two variables.
[asciimath]J_2(s)=J_1(s)-J_3(s)[/asciimath] (4.8.3)
Next, we substitute this expression for [asciimath]J_2(s)[/asciimath] into the second equation, which reduces the system to two equations with two unknowns.
[asciimath]{(4J_1(s)+0.1sJ_3(s)+0.2sJ_1(s)=12/s),(0.1sJ_3(s)-2(J_1(s)-J_3(s))=0) :}[/asciimath]
[asciimath]={((4+0.2s)J_1(s)+0.1sJ_3(s)=12/s),(-2J_1(s)+(0.1s+2)J_3(s)=0) :}[/asciimath] (4.8.4)
To eliminate [asciimath]J_3(s)[/asciimath], we multiply the first equation by [asciimath](0.1s+2)[/asciimath] and the second equation by [asciimath]-0.1s[/asciimath] and then add both equations. This results in
[asciimath](0.1s+2)(4+0.2s)J_1(s)+0.2sJ_1(s)[/asciimath] [asciimath]=(0.1s+2)12/s[/asciimath]
Rearranging for [asciimath]J_1(s)[/asciimath] gives
[asciimath]J_1(s)=(1.2+24/s)/(0.02s^2+s+8)[/asciimath]
To eliminate decimal and rational terms, we multiply the numerator and the denominator by [asciimath]50s[/asciimath].
[asciimath]J_1(s)=(60s+1200)/(s^3+50s^2+400s)[/asciimath] [asciimath]=(60s+1200)/(s(s^2+50s+400))[/asciimath] [asciimath]=(60s+1200)/(s(s+10)(s+40))[/asciimath]
Breaking [asciimath]J_1(s)[/asciimath] down by partial fraction expansion, we get
[asciimath]J_1(s)=3/s-2/(s+10)-1/(s+40)[/asciimath]
By substituting [asciimath]J_1(s)[/asciimath] in the second equation in system 4.8.4, we find [asciimath]J_3(s)[/asciimath].
[asciimath]J_3(s)=(2J_1(s))/(0.1s+2)[/asciimath]
[asciimath]=(120s+2400)/(s(s+10)(s+40)(0.1s+2))[/asciimath] [asciimath]=(1200(0.1s+2))/(s(s+10)(s+40)(0.1s+2))[/asciimath]
This simplifies to
[asciimath]J_3(s)=1200/(s(s+10)(s+40))[/asciimath]
Breaking [asciimath]J_3(s)[/asciimath] down by partial fraction expansion yields
[asciimath]J_3(s)=3/s-4/(s+10)+1/(s+40)[/asciimath]
By substituting the expressions for [asciimath]J_1(s)[/asciimath] and [asciimath]J_3(s)[/asciimath] in Equation 4.8.3, we find [asciimath]J_2(s)[/asciimath].
[asciimath]J_2(s)=J_1(s)-J_3(s)=2/(s+10)-2/(s+40)[/asciimath]
Finally, applying the inverse Laplace Transforms to [asciimath]J_1[/asciimath], [asciimath]J_2[/asciimath], and [asciimath]J_3[/asciimath], we determine the current in the branches of the circuit.
[asciimath]\mathcal{L}^-1{J_1(s)} =I_1(t)[/asciimath][asciimath]=3-2e^(-10t)-e^(-40t)[/asciimath]
[asciimath]\mathcal{L}^-1{J_2(s)} =I_2(t)[/asciimath][asciimath]=2e^(-10t)-2e^(-40t)[/asciimath]
[asciimath]\mathcal{L}^-1{J_3(s)} =I_3(t)[/asciimath][asciimath]=3-4e^(-10t)+e^(-40t)[/asciimath]
