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I have to prove that the expression

$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}}$$

is equal to

$$\frac{1}{3-( (\frac{\omega_r}{\omega})^2 + (\frac{\omega}{\omega_r})^2)}$$

where $\omega_r= \frac{1}{\sqrt{LC}}$.

My attempt:

What I started to do was to get rid of the denominators in the fraction and put everything together.

$$\frac{\omega^2C^2L-C}{\omega^2C^2L-C+\omega^2CL^2-L}$$

Then I divided the denominator by the numerator

$$\frac{1}{1+\frac{\omega^2CL^2-L}{\omega^2C^2L-C}}$$

And I'm kind of stuck now. Can someone give an hint on how should I proceed next? Or is there any easier way to start the proof? I'm just looking for a hint, thanks.

Robert Howard
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Granger Obliviate
  • 1,088

1 Answers1

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The expressions are NOT equal when $\omega = 2$ and $L=1$ and $C=1$:

$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}} \;\; = \;\; \frac{2\cdot1\; - \; \frac{1}{2\cdot 1}}{2\cdot 1 \; - \; \frac{1}{2\cdot1} \; + \; 2\cdot1 \; - \; \frac{1}{2\cdot 1}} \;\; = \;\; \frac{\frac{3}{2}}{\;4 - 1\;} \;\; = \;\; \frac{1}{2} $$

and

$$\frac{1}{3 \; - \; \left[ \left(\frac{\omega_r}{\omega}\right)^2 + \left(\frac{\omega}{\omega_r}\right)^2\right]} \; = \; \frac{1}{3 \; - \; \left[ \left(\frac{1}{2}\right)^2 + \left(\frac{2}{1}\right)^2\right]} \; = \; \frac{1}{\;3 \; - \; \frac{1}{4} \; - \; 4\;} \; = \; \frac{1}{\;-\frac{5}{4}\;} \; = \; -\frac{4}{5}, $$

where I've used the fact that $\;\omega_r= \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1\cdot 1\;}} = 1.$

Dave L. Renfro
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