Laplace transforms: R series with RC parallel circuit

Problem Statement

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

Capacitor is uncharged at t(0-)

Fig (1)

$v(t0-)=0Volts\,$

$v(t)=10Volts\,$

$R_{1}=20{\boldsymbol {\Omega }}\,$

$R_{2}=30{\boldsymbol {\Omega }}\,$

$C=.1Farad\,$

Voltage equations:

equation 1

$v(t)=R1(i_{1}+i_{2})+R2(i_{1})\,$

$10=20(i_{1}+i_{2})+30(i_{1})\,$

$i_{1}=(10-20i_{2})/50\,$___________________________________equation (1)

equation 2

$v(t)=R1(i_{1}+i_{2})+{\dfrac {1}{C}}\int {i_{2}dt}\,$

$10=20(i_{1}+i_{2})+{\dfrac {1}{.1}}\int {i_{2}dt}\,$_______________________equation (2)

Solve equations (1) and (2) simultaneously

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

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

simplifies to...

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

Take the Laplace Transform to move to the S-domain

${\mathcal {L}}\left\{\int _{0-}^{\infty }{(i_{2})dt}\right\}=I_{2}/s\,$

${\mathcal {L}}\left\{(1)\right\}=1/s\,$

$12I_{2}+10I_{2}/S=6/S\,$

$I_{2}(12+10/S)=6/S\,$

$I_{2}=6/(12S+10)\,$

$I_{2}=(1/2)(1/(S+(5/6))\,$

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

$\lim _{s\rightarrow \infty }sF(s)=f(0)\,$

Final Value Theorem

$\lim _{s\rightarrow 0}sF(s)=f(\infty )\,$

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

Initial Value:

$\lim _{s\rightarrow \infty }sV(s)=6s/s-6s(1/(s+(5/6))\,$

$\lim _{s\rightarrow \infty }sV(s)=0\,$

Initial Value = 0 Volts

Final Value:

$\lim _{s\rightarrow 0}sV(s)=6s/s-6s(1/(s+(5/6))\,$

$\lim _{s\rightarrow 0}sV(s)=6\,$

Final Value = 6 Volts

$v(t0)=0\,$ Volts

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

Bode Plot

T-domain

$V_{in}(t)=10\,$

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

S-domain

$V_{in}(s)=10/s\,$

$V_{out}(s)=6/s-6(1/(s+5/6))\,$

Transfer Function

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

$H(S)=3/(6s+5)\,$

Bode Plot

Fig (1)

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How to use break points and asymptotes to obtain the magnitude frequency response of the system...

The break points are the values of s in H(s) that make the numerator and or the denominator 0.

The location of the break points determines the magnitude frequency response of the system at that frequency.

Zeros are where the numerator is equal to zero.

Poles are when the denominator is equal to zero.

Use Convolution to find the output of the system

$H(S)=3/(6s+5)\,$

$h(t)={\mathcal {L}}^{-1}\left\{(3/(6s+5))\right\},\,$

$h(t)=(1/2)*e^{-5t/6}\,$

$v(t)=v(t)*h(t)=\int _{0}^{t}{v(\tau )h(t-\tau )d\tau }\,$

$v(t)=\int _{0}^{t}{(10)((1/2)e^{(-5/6)(t-\tau )})d\tau }\,$

$v(t)=6-6e^{-5t/6}\,$

State Example

For the voltage on the capacitor...

${\begin{bmatrix}(dv/dt)v_{c}\end{bmatrix}}={\begin{bmatrix}-R1/C\end{bmatrix}}{\begin{bmatrix}(i_{2}+i_{3})\end{bmatrix}}+{\begin{bmatrix}1/C\end{bmatrix}}{\begin{bmatrix}V(t)\end{bmatrix}}$

${\begin{bmatrix}(dv/dt)v_{c}\end{bmatrix}}={\begin{bmatrix}-200\end{bmatrix}}{\begin{bmatrix}(i_{2}+i_{3})\end{bmatrix}}+{\begin{bmatrix}100\end{bmatrix}}{\begin{bmatrix}V(t)\end{bmatrix}}$

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Written by: Andrew Hellie

Checked by: Kendrick Mensink