Engineering Solve RC Circuit Using Laplace Transforms

[photonsquared]

Find v(t) at t=800ms for the circuit in Figure 1.

Ans: 802mV



Writing a single node equation we have

\frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0.

Taking the Laplace transform we have

L\left\{ \frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0\right\} .

tu(t)\Rightarrow\frac{1}{s^{2}}

\frac{df}{dt}\Rightarrow sF(s)-f(0^{-})

\frac{V(s)}{5}-\frac{2}{5s^{2}}+0.1sV(s)-0.1v(0^{-})=0

Assume that v(0^{-})=0 we have

\frac{V(s)}{5}-\frac{2}{5s^{2}}+0.1sV(s)=0

multiplying through by 10 and combing like terms

V(s)\left(2+s\right)=\frac{4}{s^{2}}

Solving for V(s) we have

V(s)=\frac{4}{s^{2}(s+2)}

Applying the method of residues we have

\frac{4}{s^{2}(s+2)}=\frac{A}{s^{2}}+\frac{B}{s+2}

Multiplying throught by s^{2} we have

\frac{4}{s+2}=A+\frac{Bs^{2}}{s+2}

A=\frac{4-Bs^{2}}{s+2}\mid_{s=0}=2

Multiplying through by s+2 we have

\frac{4}{s^{2}}=\frac{A(s+2)}{s^{2}}+B

B=\frac{4-A(s+2)}{s^{2}}\mid_{s=-2}=1

Substituting A and B back into the equation we have

V(s)=\frac{2}{s^{2}}+\frac{1}{s+2}

Applying the known Laplace transform pairs we have

v(t)=2t+e^{-2t}

v(800ms)=2(0.8)+e^{-2(0.8s)}=1.802V

 

[user101]

I haven't checked all the details, but you seem to have the right idea.

What is the question?

 

[photonsquared]

Well my answer is different from that given in the book. My answer is too high by 1V.

 

[CEL]

Your partial fraction expansion is wrong. You should have:

\frac{4}{s^{2}(s+2)}=\frac{A}{s^{2}}+\frac{C}{s}+\frac{B}{s+2}
Using the residues you have A = 2 and B = 1 as you have found.
Rewrite the second member with the common denominator and equate the numerator to 4.
You will find C = -1. This is what you need to find the correct answer.

 

[photonsquared]

Thanks for not giving up on my question.

Kevin