# Circuit Problem:Find transfer function and value of C2 for poles at specific location

1. Dec 5, 2007

### VinnyCee

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

http://img252.imageshack.us/img252/410/prelab4problem1tz5.jpg [Broken]

Find transfer function of the circuit above (i.e. - $\frac{V_o(s)}{V_i(s)}$)

$$\frac{V_o(s)}{V_i(s)}\,=\,\frac{a_1}{s^2\,+\,a_2\,s\,+\,a_3}$$

1) Find a1, a2, a3 in terms of R, C1 and C2

2) Given that $C_1\,=\,100\,\mu\,F$ and $R\,=\,10\,K\Omega$, find $C_2$ such that the system has a pair of complex conjugate poles located at $-1\,\pm\,j\,\sqrt{399}$.

2. Relevant equations

KCL, OP Amp rules, complex numbers.

3. The attempt at a solution

Ok, I went through a nodal analysis, I'm not going to post the steps here, but here are the results...

$$\frac{V_o}{V_i}\,=\,\frac{1}{C_1\,C_2\,R\,s^2\,+\,2\,C_2\,R\,s\,-\,1}$$

$$\frac{V_o}{V_i}\,=\,\frac{\frac{1}{C_1\,C_2\,R}}{s^2\,+\,\frac{2}{C_1}\,s\,-\,\frac{1}{C_1\,C_2\,R}}$$

So that means that...

$$a_1\,=\,\frac{1}{C_1\,C_2\,R}$$

$$a_2\,=\,\frac{2}{C_1}$$

$$a_3\,=\,\frac{1}{C_1\,C_2\,R}$$

That's for part one, does that seem right?

For part two, we want to MAKE the roots of the following equation (denominator):

$$s^2\,+\,2000\,s\,+\,\frac{1}{C_2}\,=\,0$$

EQUAL TO...

$$-1\,\pm\,j\,\sqrt{399}$$

How do I make that happen?

Last edited by a moderator: May 3, 2017
2. Dec 5, 2007

### The Electrician

The first problem is that you've got the transfer function wrong. See the first attachment for the correct transfer function.

And as to your last question, don't you remember the quadratic formula? See the second attachment.

Substitute the coefficients a, b and c, and then select C2 to get the roots you want.

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3. Dec 5, 2007

### VinnyCee

So the C2 value for part 2 is ... calculating ... brb!

Last edited: Dec 5, 2007
4. Dec 5, 2007

### The Electrician

Well, what did you get?

And, now, you can answer a question for me. How do you paste those mathematical expressions into your post? When I right click on one and select properties, it appears that it is a Latex image.