Laplace transforms: R series with RC parallel circuit

Problem Statement

Find the Voltage across the capacitor for t>=0:

Voltage across capacitor at t({0-})=0

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Fig (1)

$v(t0-)=0Volts$

$v(t)=10Volts$

$R_1=20\mathrm{\Omega }$

$R_2=30\mathrm{\Omega }$

$C=.1Farad$

Use Loop Equations to solve for the currents in $i_1$ and $i_2$

Loop 1 (Resistor Branch)

$v(t)=R1(i_1+i_2)+R2(i_1)$

$10=20(i_1+i_2)+30(i_1)$

$i_1=(10-20i_2)/50$___________________________________equation (1)

Loop 2 (Capacitor Branch)

$v(t)=R1(i_1+i_2)+\frac{1}{C}\int i_{2}dt$

$10=20(i_1+i_2)+\frac{1}{.1}\int i_{2}dt$_______________________equation (2)

Solve equations (1) and (2) simultaneously

Substituting equation (1) into equation (2) gives...

$20((10-20i_2)/50+i_2)+\frac{1}{.1}\int i_{2}dt=10$

simplifies to...

$12i_2+10\int i_{2}dt=6$

Take the Laplace Transform to move to the S-domain

${ℒ}\left\{\underset{0-}{\overset{\mathrm{\infty }}{\int }}(i_2)dt\right\}=I_2/s$

${ℒ}\{(1)\}=1/s$

$12I_2+10I_2/S=6/S$

$I_2(12+10/S)=6/S$

$I_2=6/(12S+10)$

$I_2=(1/2)(1/(S+(5/6))$

Take the inverse Laplace transform to move back into the t-domain

$i_2=(1/2)(e^{-(5/6)t})$

substitute this equation back into equation (1)

$i_1=(10-20(.5e^{-(5t/6)}))/50$

$i_1=(1/5)(1-e^{-(5t/6)})$

Voltage on Capacitor

$v_{capacitor}=10/20(i_1+i_2)$

$v_{capacitor}=10-20((1/5)(1-e^{-(5t/6)})+(1/2)(e^{-(5t/6)}))$

$v_{capacitor}=10-4+4e^{-(5t/6)}-10e-(5t/6)$

Answer

$v_{capacitor}=6-6e^{-(5t/6)}$ Volts

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Apply the Initial and Final Value Theorems to find the initial and final values

Initial Value Theorem

$\underset{s\to \mathrm{\infty }}{\lim }sF(s)=f(0)$

Final Value Theorem

$\underset{s\to 0}{\lim }sF(s)=f(\mathrm{\infty })$

$V(S)=6/s-6(1/(s+(5/6))$

Initial Value:

$\underset{s\to \mathrm{\infty }}{\lim }sV(s)=6s/s-6s(1/(s+(5/6))$

$\underset{s\to \mathrm{\infty }}{\lim }sV(s)=0$

Initial Value = 0 Volts

Final Value:

$\underset{s\to 0}{\lim }sV(s)=6s/s-6s(1/(s+(5/6))$

$\underset{s\to 0}{\lim }sV(s)=6$

Final Value = 6 Volts

$v(t0)=0$ Volts

$v(t\mathrm{\infty })=6$ Volts

Bode Plot

$V_{in}(t)=10$

$V_{out}(t)=6-6*e^{-}((5/6)t)$

$H(S)=V(s{)}_{in}/V(s{)}_{out}$

$H(S)=1/(6s)-1/(6(s-5))$

simplified to...

$H(S)=-30/(36s(s-5))$

Fig (1)

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Use Convolution to find the output of the system

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Written by: Andrew Hellie

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