# Circuit Analysis - Laplace notation

1. May 3, 2013

### anol1258

1. The problem statement, all variables and given/known data

The circuit below uses ideal components and disguises itself as a second order system when in fact it is really two first order systems. Prior to t=0 switch S is open. Then suddenly at t=0 switch S is closed. Find the peak output voltage $V_{0max}$, and the time taken for this peak to occur.
Hint: Use KVL loop equations formulated in Laplace notation.

circuit:

2. Relevant equations

V=IR

3. The attempt at a solution

Im stuck on most of it. I know how to take the inverse Laplace, but forming the right equations is getting me stuck.

Last edited: May 3, 2013
2. May 3, 2013

### Staff: Mentor

I can't see your image. By the URL, it looks like it's on a yahoo mail server. Can you upload a copy to the PF server and post it as an attachment? Use the paperclip icon in the edit frame.

By the way, Welcome to PF!

Last edited by a moderator: May 6, 2017
3. May 3, 2013

### Staff: Mentor

What do you know about expressing circuit elements in Laplace notation? What have you attempted so far?

4. May 3, 2013

### anol1258

I know when you differentiate you multiply by s, and when you integrate you divide by s in the frequency domain.

5. May 3, 2013

### anol1258

Here is what I have so far:

KVL on primary side and secondary side

I am doing this right?

6. May 3, 2013

### Staff: Mentor

You've got the right idea, but the mutual inductance terms should have an 's', and beware of the signs you assign to them; check the current directions with respect to the dots.

Current flowing into the dot in one loop will cause a current to flow out of the dot on the other. In terms of the controlled voltage source that is used to represent the mutual inductance, the voltage source should have a polarity that would want to drive the current in the same direction.

7. May 4, 2013

### anol1258

Is this correct now?

Would I now put these equations in matrix form and solve for either I1 or I2?

8. May 4, 2013

### Staff: Mentor

Sure. Looks like you'll need I2 in order to find Vout.

Since there are only two equations in two unknowns you could always resort to substitution and solving by hand. Whatever you're comfortable with.

9. May 4, 2013

### anol1258

Ok. Once I find I2 though how can I solve for Vout. This isn't a simple case where I can use Ohm's Law with the R2.

10. May 4, 2013

### Staff: Mentor

Sure it is. I2 flows through R2. Vout is across R2.

Of course, you'll need to take the inverse Laplace transform to I2 first...

11. May 4, 2013

### anol1258

But there isn't another node on the very bottom right part of the circuit. From looking at the diagram are you sure they are asking Vout across R2?

12. May 4, 2013

### Staff: Mentor

R2 is situated between the node labeled Vout and the reference node at the bottom (labeled with the ground symbol). So yes, I'm sure.

13. May 5, 2013

### anol1258

Thank you gneill

I'm having a hard time putting my current in laplace form so that I can do some inverse laplace on it. Here is the equation for the current:

14. May 6, 2013

### Staff: Mentor

Can you show how you arrived at this expression for I2 from your equations in post #7? The result doesn't look quite right to me.

15. May 6, 2013

### anol1258

I did Cramer's rule. I thought it would work for this problem, maybe not.
Here's cramers rule:

16. May 6, 2013

### anol1258

yeah I messed up should be:

17. May 6, 2013

### anol1258

How can i do this laplace. Its killin me

18. May 6, 2013

### Staff: Mentor

I don't think that your Laplace equation for I2 is quite right yet. Can you show me your two KVL loop equations? As I mentioned previously, pay careful attention to the signs of the mutual inductance terms. They are directly dependent upon your choice of loop current direction and the location of the coupling dots.

So, what directions have you chosen for the loop currents? Do they flow into or out of the dots on the coupled inductors? How does this affect the polarity of the induced-voltage source in the other loop?

As for dealing with the Laplace inversion, if you have a quadratic in the denominator you want to reduce the coefficient of the s2 term to unity (1), then factor the quadratic by finding its roots (say, $\alpha_1$ and $\alpha_2$. Write the quadratic as $(s - \alpha_1)(s - \alpha_2)$. That should get you going.