OK, I have a function, a transfer function, and it looks like this:
$$H(\omega) = \frac{1-\omega^2 LC}{1+\omega^2 LC – i \omega RC}$$
I want to separate the real and imaginary parts. This kin of algebra should be simple, and I feel like I am not getting something obvious.
Now, ordinarily if I wanted $a+bi$ form this would be easy, if the stuff in the denominator were in the numerator. But it isn't. So I thought of one way to do this was to separate the fraction out-
$$H(\omega) = \frac{1}{1+\omega^2 LC – i \omega RC}$$ into $$\frac{1}{1+\omega^2 LC – i \omega RC}$$ and $$\frac{-\omega^2 LC}{1+\omega^2 LC – i \omega RC}$$
but that doesn't seem to help much. Since the single i is in the denominator I can't seem to move it to the numerator.
Anyhow, this is probably simple algebra and maybe I am just tired. But if anyone can tell me what (probably obvious) thing I am missing here is that would help a lot. It's probably some property of complex numbers I forgot about. (I have no problem getting the absolute value of the function, for instance).
EDIT: somehow I know that when I see the answer I am going to say "D'OH!" 🙂
EDIT 2: Hey, wait a minute. If I multiply the numerator and denominator by the complex conjugate of the denominator that solves the problem I think. Correct?
Best Answer
if $\omega$ is real then write $\eta = \omega^2LC$ and $\nu = \omega RC$. your expression becomes $$ \frac{1-\eta}{1+\eta - i\nu} $$ multiplying by the conjugate of the denominator this becomes: $$ \left((1+\eta)^2 + \nu^2\right)^{-1}(1 - \eta^2+(1-\eta)\nu i) $$