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

Fig (1)

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

$v(t)=10Volts\,$

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

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

$C=.1Farad\,$

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)

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

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

---

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

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

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

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

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

simplified to...

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

Fig (1)

---

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

---

Written by: Andrew Hellie

Checked by: