Ordinary Diferential Equations in electric circuit

[ffp]

Homework Statement

An electric circuit has its model represented by the following ordinary diferential equation:

$V_0''(t)+4V_0'(t)+3V(t)=V_i'(t)+2V_i(t)$

Where

-$V_i(t)$ is the input of the voltage source of the circuit given by $V_i(t)=3e^{-2t}$, for t≥0 and null for t>0

-$V_0(t)$ is the output measured signal of the circuit.

Based on the data provided e considering, in the mathematical model, the initial conditions null, the expression of the output for t≥0 is

Relevant Equations

$V_0''(t)+4V_0'(t)+3V_0(t)=V_i'(t)+2V_i(t)$

$V_i(t)= 3e^{-2t}$

It's a multiple choice exercise and I have managed to find the characteristic equation V0(t) which is $V_0(t)= C_1e^{-t}+C_2e^{-3t}$

Initially I thought that it was a non homogeneous ODE, but doing the math for the right part of the equation, I found out that it equals to 0.

So, I need help finding the constants $C_1$ and $C_2$ and I don't know how to handle the initial conditions.

If I make $V_0(0)=0$ and $V_0'(0)=0$ both $C_1$ and $C_2$ would be equal to 0, which is wrong.

So, I tried calculating $V_i(t)$ and $V_i'(t)$ and doing $V_i(0)=0$ and $V_i'(0)=0$, but still got the wrong answer $(C_1=C_2=3/2)$, while the answer is $C_1=3/2$ and $C_2=-3/2$.

So, how should I handle the initial conditions in this exercise?

 

[erobz]

Homework Statement: An electric circuit has its model represented by the following ordinary diferential equation:

V0''(t)+4V0'(t)+3V(t)=Vi'(t)+2Vi(t)

Where

-Vi(t) is the input of the voltage source of the circuit given by Vi(t)=3e^(-2t), for t≥0 and null for t>0

- V0(t) ia the output measured signal of the circuit.

Based on the data provided e considering, in the mathematical model, the initial conditions null, the expression of the output for t≥0 is
Relevant Equations: V0''(t)+4V0'(t)+3V(t)=Vi'(t)+2Vi(t)

Vi(t)= 3e^(-2t)

It's a multiple choice exercise and I have managed to find the characteristic equation V0(t) which is V0(t)= C1*e^(-t)+C2*e^(-3t)

Initially I thought that it was a non homogeneous ODE, but doing the math for the right part of the equation, I found out that it equals to 0.

So, I need help finding the constants C1 and C2 and I don't know how to handle the initial conditions.

If I make V0(0)=0 and V0'(0)=0 both C1 and C2 would be equal to 0, which is wrong.

So, I tried calculating Vi(t) and Vi'(t) and doing Vi(0)=0 and Vi'(0)=0, but still got the wrong answer (C1=C2=3/2), while the answer is C1=3/2 and C2=-3/2.

So, how should I handle the initial conditions in this exercise?

Please format your mathematics using LaTeX Guide

What is $V(t)$ because you have it in there too?

 

[berkeman]

Please format your mathematics using LaTeX Guide

Agreed, I have sent the OP some extra tips on how to post math using LaTeX.

 

[ffp]

Please format your mathematics using LaTeX Guide

What is $V(t)$ because you have it in there too?

Ok, I think I got it right now. And no, there is no $V(t)$, only $V_0(t)$ and $V_i(t)$, that are the output and input voltage respectively.

 

[erobz]

Ok, I think I got it right now. And no, there is no $V(t)$, only $V_0(t)$ and $V_i(t)$, that are the output and input voltage respectively.

Well if you don’t mind how did you do it? Because I get a system for the coefficients whose only solution is $C_1=C_2=0$ like you said before. It’s somewhat confusing because as you say the RHS is identically 0 for all times $t$, making the general solution that of a homogeneous ODE?

 

[ffp]

Well if you don’t mind how did you do it? Because I get a system for the coefficients whose only solution is $C_1=C_2=0$ like you said before. It’s somewhat confusing because as you say the RHS is identically 0 for all times $t$, making the general solution that of a homogeneous ODE?

If you can understand my handwriting, this is how I did it.

Until up to the characteristic equation, I believe I got it right. My issue is with the constants calculation.

As I posted in OP, I did $V_i(0)=0$ and $V_i'(0)=0$, but I don't know if this logic is correct.

PS: Are $V_i(0)=V_0(0)$ and $V_i'(0)=V_0'(0)$ ?

PPS: Oh, in my last post, the "I think I got it right now", was meant to using Latex, lol. Not the solution of the question

 

[erobz]

The general solution is $$V_O(t) = C_1e^{-3t} + C_2 e^{-t} $$

Its not dependent on the inputs is it? The inputs add to $0$ for all time $t$. I think we are going to need a bit more of mathematician here...Its not making sense to me.

If the RHS of the ODE was not zero you would add a particular solution $y_p$ assuming the form of the generalized input ( or forcing term ), but we have literally nothing on the RHS. It's just a homogenous ODE!

Also, initial conditions are on $V_O$ and its derivatives. Not $V_i$.

I think something is not right about all this, but I'm not anywhere near an expert in solving differential equations...so take what I say lightly.

 

[erobz]

Maybe @berkeman would move this to the math forums. It's just an exercise in solving ODE's as far as I can tell.

 

[Mark44]

$V_0''(t)+4V_0'(t)+3V(t)=V_i'(t)+2V_i(t)$

It's a multiple choice exercise and I have managed to find the characteristic equation V0(t) which is $V_0(t)= C_1e^{-t}+C_2e^{-3t}$

I've taught college-level differential equations a number of times but have never seen a problem quite like this. A typical problem will have the unknown function and one or more of its derivatives on one side and possibly a forcing function on the other.

The unknown function and derivatives would typically be what you have on the right side, with your input voltage and its derivative. The forcing function would typically be the output stuff.

Minor nit -- you have $V_i$ as the input voltage. For output, use 'o', not 0 (zero). IOW, $V_o$, not $V_0$.

So, I need help finding the constants $C_1$ and $C_2$ and I don't know how to handle the initial conditions.

It doesn't seem to me that you are given initial conditions.

If I make $V_0(0)=0$ and $V_0'(0)=0$ both $C_1$ and $C_2$ would be equal to 0, which is wrong.

You can't just make up your own initial conditions. Have you provided us with the complete problem statement?

 

[ffp]

@Mark44 , yes I provided the complete statement. Had to translate it, because it's not in english, but there is no aditional info...

 

[erobz]

When you went with, and after:

$$ C_1 + C_2 =3$$

I see no justification for that. How did you reason that out?

Initial conditions apply to $V_o(t)$, and $V’_o(t)$. If they are both null, and with no forcing function they just stay put at $0$

 

[berkeman]

-$V_i(t)$ is the input of the voltage source of the circuit given by $V_i(t)=3e^{-2t}$, for t≥0 and null for t>0

That last inequality is reversed, no? Shouldn't it be $V_i(t) = 0$ for t<0?

And with the initial condition that $V_i(t)$ starts off at zero and has been zero for all time before that, it would seem that would also imply that $V_o$ is zero at t=0- as well (unless there are independent voltage sources involved or something, which is not implied in the problem).

 

[ffp]

When you went with, and after:

$$ C_1 + C_2 =3$$

I see no justification for that. How did you reason that out?

Initial conditions apply to $V_o(t)$, and $V’_o(t)$. If they are both null, and with no forcing function they just stay put at $0$

That was a desperate atempt to find the answer. I did it, not knowing it was right (tbh I thought it was wrong). Now I know is definitely wrong.

This question is from a public test for a government company in my country. So, it should have been disregarded, right (it wasn't, though).

 

[ffp]

That last inequality is reversed, no? Shouldn't it be $V_i(t) = 0$ for t<0?

And with the initial condition that $V_i(t)$ starts off at zero and has been zero for all time before that, it would seem that would also imply that $V_o$ is zero at t=0- as well (unless there are independent voltage sources involved or something, which is not implied in the problem).

Yes, you are right, that was my mistake when typing the statement. Thank you for pointing out. The correct inequity is t<0.

I really think the question is missing something now, reading all you guys answers. This is all all we can get, right?

 

[willem2]

I really think the question is missing something now, reading all you guys answers. This is all all we can get, right?

This is a calculation of the impulse response of a second order linear system (mass - spring - damper).
Are you allowed to use the Dirac delta function? in that case the right side of the differential equation is just δ(t), and you can use the laplace transform.
A an equation with f(V'', V', v) = δ(t) and initial values V = V'= 0 will have the same solution as an equation with f(V'', V', v) and initial values V(0) = 0 and V'(0) = 1.
Since we have two exponentials in the solution, this is an overdamped system.
Impulse Response of Second-Order Systems

 

[ffp]

I

This is a calculation of the impulse response of a second order linear system (mass - spring - damper).
Are you allowed to use the Dirac delta function? in that case the right side of the differential equation is just δ(t), and you can use the laplace transform.
A an equation with f(V'', V', v) = δ(t) and initial values V = V'= 0 will have the same solution as an equation with f(V'', V', v) and initial values V(0) = 0 and V'(0) = 1.
Since we have two exponentials in the solution, this is an overdamped system.
Impulse Response of Second-Order Systems

I'm not sure I understand what you said. But I still have to calculate the C1 and C2 constants...

 

[alan123hk]

$$V_0''(t)+4V_0'(t)+3V(t)=V_i'(t)+2V_i(t)~~~~~~~\text V_i(t)= 3e^{-2t}$$
$$V_0''(t)+4V_0'(t)+3V(t)= -6V_ie^(-2t)+6e^{-2t}=0$$
Then we have the general solution of this differential equation (forcing function=0)
$$V_O(t) = C_1e^{-3t} + C_2 e^{-t}=0~~~~,~~~~V'_O(t)= 3C_1e^{-3t}+C_2e^{-t}=0$$
At time zero we have
$$C_1+C2=0~~~~\text{and}~~~~~3C_1+C_2=0$$
Therefore C1 = C2 =0, what am I doing wrong?

Can you describe how you arrived at answer $~C_1=C_2=3/2$?
Thanks