I need to write the equations for the circuit in matrix form. What I have done is KVL equations for each loop and got
loop 1: I1(t)/C1 + R*I1'(t) - R*I2'(t) +L1*I1''(t) - V1(t) = 0 (after differentiating to get rid of the integral for C1)

What I need to do is reduce the second order ODE into a system of first order which I dont know how to do with the I2' in there so some help would be great.

Ignore the values in the picture the are irrelevant at this stage.

 

I need to write the equations for the circuit in matrix form. What I have done is KVL equations for each loop and got
loop 1: I1(t)/C1 + R*I1'(t) - R*I2'(t) +L1*I1''(t) - V1(t) = 0 (after differentiating to get rid of the integral for C1)

What I need to do is reduce the second order ODE into a system of first order which I dont know how to do with the I2' in there so some help would be great.

Ignore the values in the picture the are irrelevant at this stage.View attachment 92350

 

Hello there,

Here is a way to get from the single DE to the several ODE's.
[But now i see you are confused perhaps by the second current.]
[If that is true, then write *both* equations out and try again.]
[This is just a reduction method for a single DE.]

Start with:
ax''+b*x'+c*x=d

divide by the coefficient of highest derivative, here it is x'':

x''+b/a*x'+c/a*x=d/a <2>

Now the first n-1 derivatives are equal to the next up, so
starting with x1 list them:
x1'=x2

Now the last derivative (x2) is equal to the equation above <2>
with the sign of all of the terms on the left side changed:
x2=-b/a*x2-c/a*x1+d/a

Note the nth derivative turns out to be the (n+1)th x, so
x' changes to x2, while x changes to x1.


Now a third order example...

ax'''+b*x''+c*x'+d*x=e

x'''+b/a*x''+c/a*x'+d/a*x=e/a

and the last der x3 is this:

x3'=-b/a*x3-c/a*x2-d/a*x1+e/a

while the first two are simply:
x1'=x2
x2'=x3

so the set looks like this:
x1'=x2
x2'=x3
x3'=-b/a*x3-c/a*x2-d/a*x1+e/a

and those are your coupled ODE set.

 

Hello there,

Here is a way to get from the single DE to the several ODE's.
[But now i see you are confused perhaps by the second current.]
[If that is true, then write *both* equations out and try again.]
[This is just a reduction method for a single DE.]

Start with:
ax''+b*x'+c*x=d

divide by the coefficient of highest derivative, here it is x'':

x''+b/a*x'+c/a*x=d/a <2>

Now the first n-1 derivatives are equal to the next up, so
starting with x1 list them:
x1'=x2

Now the last derivative (x2) is equal to the equation above <2>
with the sign of all of the terms on the left side changed:
x2=-b/a*x2-c/a*x1+d/a

Note the nth derivative turns out to be the (n+1)th x, so
x' changes to x2, while x changes to x1.


Now a third order example...

ax'''+b*x''+c*x'+d*x=e

x'''+b/a*x''+c/a*x'+d/a*x=e/a

and the last der x3 is this:

x3'=-b/a*x3-c/a*x2-d/a*x1+e/a

while the first two are simply:
x1'=x2
x2'=x3

so the set looks like this:
x1'=x2
x2'=x3
x3'=-b/a*x3-c/a*x2-d/a*x1+e/a

and those are your coupled ODE set.

Cheers, that is a pretty clear explanation but I am still having trouble with the second current. I have the second equation down,
I2(t)/C2 + R*I2'(t) - R*I1'(t) +L2*I2''(t) - V2(t) = 0,
but just cant see what I am supposed to do. Something isn't clicking for me but it seems like something simple.

 

Hi,

I have a question first, is that inductor labeled "0 H" supposed to be a short? That is what it would be at zero Henries.

 

Hi,

I have a question first, is that inductor labeled "0 H" supposed to be a short? That is what it would be at zero Henries.

Hi,

It isnt supposed to be 0H, no values are given for the elements at this stage.

 

Hello,

Ok, no problem. We assume no values were given.

One more question then, what made you decide that this was a second order problem and not a fourth order?

 

Hello,

Ok, no problem. We assume no values were given.

One more question then, what made you decide that this was a second order problem and not a fourth order?

The problem was given a while ago but I recall the tutor explaining second order and that stuck with me. I didn't even think of it being fourth from the 2 capacitors and 2 inductors since this is for my math class when we haven't really covered it in my circuits class.

 

Hello again,

Hey wait a minute, isnt your first equation supposed to have -R*I2 not -R*I2' ?
The loops common resistor has I1 and I2 flowing right?

Also, what do you consider the output to be in this circuit?

BTW you can also try superposition.

 

Hi,

Originally that was the current and this 1/C1 * integral ( i(t) ) + R(I1(t) - I2(t)) +L1*I1'(t) - V1(t) = 0 was the KVL equation but looking at some examples of single loop RCL they took the derivative with respect to t to get rid of the integral. Also I just noticed a typo, it should be V1'(t) and V2'(t) for both equations.

As for the output, the only other information given is that the currents are the variables.

I might just attempt the question for mechanical or civil engineers instead since a lot of students are having trouble with the latter parts of this one. Thanks for your help with the single DE to several ODE's, that will probably come in handy with the other questions.