- Thread starter
- #1

- Thread starter dwsmith
- Start date

- Thread starter
- #1

- Feb 13, 2012

- 1,704

Correct is...Determine the differential equation relating \(v_i(t)\) and \(v_0(t)\) for the RLC circuit in the figure.

View attachment 2100

Would this just be

\[

v_i(t) = 3i + \frac{di}{dt} + 2\int i(t)dt

\]

but \(v_0 = 2\int i(t)dt\). Do I need write it as \(v_0\) or as \(2\int i(t)dt\)?

$\displaystyle v_{i} (t) = 3\ i + \frac{d i}{d t} + 2\ \int_{- \infty}^{t} i(\tau)\ d \tau$

Kind regards

$\chi$ $\sigma$

- Thread starter
- #3

Why do you have bounds of \((-\infty, t)\)?Correct is...

$\displaystyle v_{i} (t) = 3\ i + \frac{d i}{d t} + 2\ \int_{- \infty}^{t} i(\tau)\ d \tau$

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

... the voltage across a capacitor is proportional to the amount of charge You have pumped on it. If You don't like the term $\infty$ it is possible to write $\displaystyle v_{C} (t) = v_{c} (0) + \frac{1}{c}\ \int_{0}^{t} i(\tau)\ d \tau$...Why do you have bounds of \((-\infty, t)\)?

Kind regards

$\chi$ $\sigma$

Last edited:

- Thread starter
- #5

Is it possible to solve for \(v_0(t)\) using a Laplace transform?... the voltage across a capacitor is proportional to the amount of charge You have pumped on it. If You don't like the term $\infty$ it is possible to write $\displaystyle v_{C} (t) = v_{c} (0) + \frac{1}{c}\ \int_{0}^{t} i(\tau)\ d \tau$...

Kind regards

$\chi$ $\sigma$

- Admin
- #6

- Jan 26, 2012

- 4,203

Yes. In fact, electrical engineers will typically write down circuit elements on the diagram with their Laplace-transformed expressions, so that the very first equation they write down is the LT'ed equation.Is it possible to solve for \(v_0(t)\) using a Laplace transform?

- Thread starter
- #7

I don't understand. Can you demonstrate what you mean?Yes. In fact, electrical engineers will typically write down circuit elements on the diagram with their Laplace-transformed expressions, so that the very first equation they write down is the LT'ed equation.

Suppose \(v_i(t) = e^{-3t}\mathcal{U}(t)\). Then how can we get back \(v_0\) from a Laplace transform?

Last edited:

- Thread starter
- #8

- Admin
- #9

- Jan 26, 2012

- 4,203

Sorry for the delay. This is called $s$-domain circuit analysis. You have the $s$-domain impedances (kinda like what you have in steady-state sinusoidal analysis with phasors):I don't understand. Can you demonstrate what you mean?

\begin{align*}

Z_{R}&=R \\

Z_{L}&=Ls \\

Z_{C}&= \frac{1}{Cs}.

\end{align*}

These come straight from the V-I characteristics of resistors, capacitors, and inductors: just LT those relationships, and compare with $V=IZ$, Ohm's Law, to get the $s$-domain impedances (these are defined with the initial conditions zeroed out).

There are tricks for dealing with initial conditions as well - you usually insert a voltage or current source to handle that. Here is a good summary of the method.

- Thread starter
- #10

I had the solutions in my office which I retrieved yesterday. It has thatSorry for the delay. This is called $s$-domain circuit analysis. You have the $s$-domain impedances (kinda like what you have in steady-state sinusoidal analysis with phasors):

\begin{align*}

Z_{R}&=R \\

Z_{L}&=Ls \\

Z_{C}&= \frac{1}{Cs}.

\end{align*}

These come straight from the V-I characteristics of resistors, capacitors, and inductors: just LT those relationships, and compare with $V=IZ$, Ohm's Law, to get the $s$-domain impedances (these are defined with the initial conditions zeroed out).

There are tricks for dealing with initial conditions as well - you usually insert a voltage or current source to handle that. Here is a good summary of the method.

\[

i(t) = \frac{1}{2}\frac{dv_0(t)}{dt}

\]

where does this identity come from? I know we have that \(\dot{v}_0(t) = 2\) but how does that equate to \(2i(t)\)?

Edit 1:

I also worked out \(V_0(s)\) and obtained:

If \(v_i(t) = e^{-3t}\mathcal{U}(t)\), then

\begin{align*}

e^{-3t}\mathcal{U}(t) &= \frac{3}{2}\frac{dv_0(t)}{dt} +

\frac{1}{2}\frac{d^2v_0(t)}{dt^2} + v_0(t)\\

\frac{1}{s + 3} &= \frac{3}{2}(sV_0(s) - v_0(0)) + \frac{1}{2}(s^2V_0(s) -

sv_0(0) - v_0'(0))\\

\frac{2}{s + 3} &= 3sV_0(s) - 1 + s^2V_0(s) - s - 2\\

V_0(s)(s^2 + 3s) &= \frac{2}{s + 3} + s + 3\\

V_0(s) &= \frac{2}{(s + 3)(s^2 + 3s)} + \frac{s + 3}{s^2 + 3s}\\

&= \frac{2 + (s + 3)^2}{s(s + 3)^2}\\

&= \frac{2 + s^2 + 6s + 9}{s(s + 3)^2}\\

&= \frac{s^2 + 6s + 11}{s(s + 3)^2}

\end{align*}

but the solution says that answer is

\[

\frac{2(s^2 + 5s + 7)}{(s + 1)(s + s)(s + 3)}.

\]

Is the solution wrong? I don't see how we can get that.

Last edited:

- Admin
- #11

- Jan 26, 2012

- 4,203

This comes from the $VI$ characteristic of a capacitor. For any capacitor of capacitance $C$, it is true that $i=C dV/dt$. Since you have a series $RLC$ circuit, all the current goes through the capacitor. Since $C=1/2$ in your case, that's how you get $i=(1/2) dv_0/dt$.I had the solutions in my office which I retrieved yesterday. It has that

\[

i(t) = \frac{1}{2}\frac{dv_0(t)}{dt}

\]

where does this identity come from?

Shouldn't this beI know we have that \(\dot{v}_0(t) = 2\) but how does that equate to \(2i(t)\)?

Edit 1:

I also worked out \(V_0(s)\) and obtained:

If \(v_i(t) = e^{-3t}\mathcal{U}(t)\), then

\begin{align*}

e^{-3t}\mathcal{U}(t) &= \frac{3}{2}\frac{dv_0(t)}{dt} +

\frac{1}{2}\frac{d^2v_0(t)}{dt^2} + v_0(t)\\

\frac{1}{s + 3} &= \frac{3}{2}(sV_0(s) - v_0(0)) + \frac{1}{2}(s^2V_0(s) -

sv_0(0) - v_0'(0))\\ \end{align*}

$$\frac{1}{s + 3} = \frac{3}{2}(sV_0(s) - v_0(0)) + \frac{1}{2}(s^2V_0(s) -

sv_0(0) - v_0'(0)) \color{red}{+V_0(s)}?$$