RLC Circuit Differential Equation problem

[MathsDude69]

Homework Statement

For a RLC circuit with RC = 1/2 and LC = 1/16 determine the differential equation that describes the relationship between the input and output voltages. An image of the circuit is shown with RLC all in series with the input voltage Vi(t) across all 3 components. The voltage drop across the capacitor is labelled Vo(t)

Homework Equations

Kirchoff's Voltage Law

The Attempt at a Solution

From Kirchoff's voltage law:

Vi(t) = Vr(t) + Vc(t) + Vl(t)

Vr(t) = Rir(t) = RC(Vc(t))' = RC(Vo(t))' Using the prime to indicate differentiation

The voltage drop across the resistor can now be descirbed as above henceforth we now have:

Vi(t) = RC(Vo(t))' + Vc(t) + Vl(t)

Given that Vc(t) = Vo(t) we can also write:

Vi(t) = RC(Vo(t))' + Vo(t) + Vl(t)

The voltage drop across the inductor can be expressed as:

Vl(t) = L(il(t))' and as ir = ic = il Vl(t) = LC(Vo(t))'

Henceforth the differential equation is:

Vi(t) = RC(Vo(t))' + Vo(t) + LC(Vo(t))'Is this solution correct or have I flamingo'd up somewhere?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[madmike159]

Can you upload the image of the circuit?

 

[MathsDude69]

Sure. Here you go.

 

[madmike159]

I hate these questions. Have you considered that V = L(di/dt) and i = C(dv/dt)?

 

[vela]

The voltage drop across the inductor can be expressed as:

Vl(t) = L(il(t))' and as ir = ic = il Vl(t) = LC(Vo(t))

You made a mistake here when you substituted for i_l.

Henceforth the differential equation is:

Vi(t) = RC(Vo(t))' + Vo(t) + LC(Vo(t))'

This isn't correct either. You will get a second-order equation for an RLC circuit.

 

[MathsDude69]

Im guessing then that it should be along the lines of:

L*d²i/dt² + R*di/dt + 1\C*i

what is confusing is that all the information I have pertains to RL/RC circuits which are first order systems.

Is it even possible to rearrange a second order system such as this to obtain a second order differential equation using only the relationship between voltage and time?

The whole purpose of this exercize is to find the laplace transform and then the laplace frequency response H(s). :-s

 

[vela]

You almost had it with your original attempt. Just go back and do the math a little more carefully.

How did you go from V_L(t)=L(i_L(t))', which is correct, to V_L(t)=LC(V_o(t)), which is incorrect?

 

[MathsDude69]

You almost had it with your original attempt. Just go back and do the math a little more carefully.

How did you go from V_L(t)=L(i_L(t)), which is correct, to V_L(t)=LC(V_o(t)), which is incorrect?

I assumed that in a series connection that the current was equal between all parts, thus the magnitude of the current given by the voltage drop across the capacitor would also indicate the level of current in the inductor. But then again you know what they say about assumption

The only other example I have is an LR circuit which stipulates that:

i_L(t) = i_L(0^-) + 1/L * integral between 0^- and t of V_L(tau)d(tau) .. (sorry I am not too sure how to do the mathematical signs)

Should I sustitue this into the original equation??

 

[vela]

I assumed that in a series connection that the current was equal between all parts

This is right. It's a consequence of Kirchoff's current law.

thus the magnitude of the current given by the voltage drop across the capacitor would also indicate the level of current in the inductor.

What you mean by this part isn't so clear to me. Could you provide an equation for what you mean?

 

[MathsDude69]

Sure. I figured i_c(t) = C*(V_C(t))'

In my case V_C(t) and V_O(t) are the same quantity.

 

[vela]

OK, so V_L(t)=L(i_L(t))'=L(i_C(t))'=...?

 

[MathsDude69]

OK, so V_L(t)=L(i_L(t))'=L(i_C(t))'=...?

...erm... L(C*(Vc(t))'') ??

 

[vela]

Right, so you end up with

V_i = LC V''_o + RC V'_o + V_o

which is the second-order differential equation you want.

 

[MathsDude69]

Awsome. Thanks a lot!