Circuit with 2 resistors, inductor, and capacitor

[Vapen]

Homework Statement

[PLAIN]http://img534.imageshack.us/img534/228/helps.png

I have trouble completing (b), as I do not know how to solve for the damping factor in this circuit, and hence cannot characterize this system. I have the poles of the system, where s ~= -3048.059 and s~= -8201.941. I also have the transfer function, which I used to find the poles.

Homework Equations

I found the transfer function to be:

G(s) = \frac{1}{CLs^2 + \frac{L+R^2C}{R}s + 2}

The Attempt at a Solution

I just don't know how to solve for zeta.

 

[vela]

I suggest you use Laplace transforms to solve the differential equation for the damped harmonic oscillator and compare what you get there with G(s). Hopefully, you can then see how to solve for \zeta.

 

[Vapen]

I don't understand how I would apply that.

\ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega_0^2 x = 0

After Laplace transform (assuming initial conditions are zero):

X(s)s^2 + 2\zeta\omega_0X(s)s + \omega_0^2X(s)

Are you suggesting to equate this to the denominator? Your reply is ambiguous, I can't relate this to the original transfer function unless I use the reciprocal of it or some other method.

 

[vela]

You actually want to look at the forced damped harmonic oscillator.

<br /> \ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega_0^2 x = f(t)<br />

Depending on what f(t) is, you get a different response x(t) from the system. In the s-domain, you'll find X(s)=H(s)F(s) for some function H(s), which is what I'm suggesting you find. With the circuit, e_i plays the role of the forcing function, and the response of the system is e_o. In the s-domain, you have E_o(s) = G(s)E_i(s).

 

[Vapen]

I understand that part, but I still don't know what you mean by applying that second order differential equation to this. Are you implying that I need to know the actual values for the forcing function to characterize this circuit?

 

[vela]

No. I'm saying compare H(s) to G(s). The transfer function G(s) describes a forced damped harmonic oscillator, but it's not written in terms of the quantities \omega_0 and \zeta that characterize the oscillator. By comparing H(s) and G(s), you can identify how the coefficients in G(s) relate to those quantities.

 

[Vapen]

Hmm ok, you edited the previous post.

H(s) = G(s)
Then:
\frac{X(s)}{F(s)} = \frac{1}{CLs^2 + \frac{L+R^2C}{R}s + 2}
\frac{s^2 + 2\zeta\omega_0s + \omega_0^2}{F(s)} = \frac{1}{CLs^2 + \frac{L+R^2C}{R}s + 2}

But I'm left with the F(s) which I cannot do anything with?

I found the poles to both be unique, real roots, and thus characterized the system as overdamped. I would still like to find zeta regardless, though.

 

[vela]

I didn't say set them equal to each other, and X(s) isn't equal to that polynomial.

First, take the Laplace transform of

\ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega_0^2 x = f(t)

to find the oscillator's transfer function, H(s)=X(s)/F(s).

 

[Vapen]

Sorry, I typed that wrong.

X(s)s^2 + 2\zeta\omega_0X(s)s + \omega_0^2X(s)=F(s)
H(S) = \frac{X(S)}{F(S)} = \frac{1}{s^2 + 2\zeta\omega_0s + \omega_0^2}

 

[vela]

Now, if you pull a factor of 1/LC out front, you have

G(s) = \frac{1}{LC} \left[\frac{1}{s^2+(L/R+RC)s+2/LC}\right]

The factor of 1/LC in front is just a constant; it only affects the overall gain of the circuit. Comparing what's in the square bracket and what you found for H(s), you should be able to identify what \omega_0^2 and 2\zeta\omega_0 are equal to and then solve for \zeta.

 

[Vapen]

Okay, that's initially what I thought I had to do when I read this question. The constant in the denominator (the 2) was the part that I was confused about. I realize now that the numerator, when constant, will have no effect on the denominator when finding a value for \omega_0. Thanks for your help.

 

[Vapen]

I believe your s term is wrong, it seems that it should be the reciprocal of that.

\frac{1}{RC} + \frac{R}{L}

Just in case anybody else sees this post.

 

[vela]

I just started from what you had posted for G(s) earlier, assuming you had gotten it right, but yes, you're right otherwise the units don't work out correctly.