How to find the resonant frequency of this circuit?

[Nat3]

Homework Statement

Given this circuit:

[PLAIN]http://img546.imageshack.us/img546/2921/resonance.png

Homework Equations

How does one find the resonant frequency? (Since there are no values for the components, I assume this means find the formula).

The Attempt at a Solution

I read that the resonant frequency occurs when the imaginary part of the impedance is equal to zero, so I tried solving it that way but was not able to isolate the imaginary part of the impedance from the real part.

Any help would be very much appreciated!

 

[skeptic2]

Can you show us your calculations?

 

[Nat3]

Hello skeptic,

Thank you for your reply.

I tried this:

[PLAIN]http://img546.imageshack.us/img546/2921/resonance.png

Z = L || (C+R) = \frac{(L)(C+R)}{L+(C+R)} = \frac{(j\omega L)(\frac{1}{j\omega C}+R)}{j\omega L+\frac{1}{j\omega C}+R}=\frac{\frac{L}{C} + j\omega LR}{j\omega L+\frac{1}{j\omega C}+R}

But I don't know how to separate the real part from the imaginary part.

 

[Nat3]

If it was something simple, like:

Z = R + j\omega L + \frac{1}{j\omega C}

then I could just factor out the j:

Z = R + j(\omega L - \frac{1}{\omega C})

and set the imaginary/complex part to zero:

\omega L - \frac{1}{\omega C} = 0

and solve for \omega to get the formula for the resonant frequency, right?

But I can't figure out what to do for the circuit with L in parallel with R and C.

 

[Nat3]

I would really appreciate any help anyone might be able to give on this

 

[gneill]

You're doing alright with your impedance math. In order to separate into real and imaginary parts you first want to clear the imaginary part of the denominator. To do that, multiply the numerator and denominator both by the complex conjugate of the denominator. In other words:

\frac{a + jb}{c + jd} \cdot \frac{c - jd}{c - jd} = \frac{ac + bd + j(bc - ad)}{c^2 + d^2}

You can then pick out the imaginary part quite easily.

 

[Nat3]

Thanks a lot! I didn't think of multiplying by the conjugate After doing that it was kind of messy, but worked out.

Thanks again.