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[SOLVED] Circuit DE

dwsmith

Well-known member
Feb 1, 2012
1,673
Determine the differential equation relating \(v_i(t)\) and \(v_0(t)\) for the RLC circuit in the figure.
Screenshot from 2014-03-12 14:24:42.png
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 \(2\int i(t)dt\)?
 

chisigma

Well-known member
Feb 13, 2012
1,704
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 \(2\int i(t)dt\)?
Correct is...

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

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Correct is...

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

Kind regards

$\chi$ $\sigma$
Why do you have bounds of \((-\infty, t)\)?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Why do you have bounds of \((-\infty, t)\)?
... 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

$\chi$ $\sigma$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
... 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

$\chi$ $\sigma$
Is it possible to solve for \(v_0(t)\) using a Laplace transform?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Is it possible to solve for \(v_0(t)\) using a Laplace transform?
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
I don't understand. Can you demonstrate what you mean?

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

dwsmith

Well-known member
Feb 1, 2012
1,673
If I take the Laplace transform of
\[
e^{-3t}\mathcal{U}(t) = 3i + \frac{di}{dt} + 2\int_{-\infty}^ti(\tau)d\tau,
\]
I get
\[
i(t) = -\frac{3}{2}e^{-3t} + 2e^{-2t} - \frac{1}{2}e^{-t}.
\]
How do I get to \(v_0(t)\) for \(t > 0\)?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
I don't understand. Can you demonstrate what you mean?
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):
\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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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):
\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 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? 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:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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?
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 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*}
Shouldn't this be
$$\frac{1}{s + 3} = \frac{3}{2}(sV_0(s) - v_0(0)) + \frac{1}{2}(s^2V_0(s) -
sv_0(0) - v_0'(0)) \color{red}{+V_0(s)}?$$