# Homework Help: Real Part of Gain of LRC Circuit

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1. Feb 1, 2016

### PatsyTy

1. The problem statement, all variables and given/known data
Given the circuit below (image uploaded) driven at a frequency $ω=2πƒ$ show that

$$|\frac{v_{out}}{v_{in}}|=\frac{1}{\sqrt{1+(\frac{1}{ωτ_{l}}-ωτ_{c})^2}}$$

where $τ_{c}=RC$ and $τ_{l}+L/R$

2. Relevant equations
$Z_R=R$, $Z_L=jωL$ and $Z_C=-j/ωC$ where $j=\sqrt-1$

3. The attempt at a solution

$$\frac{v_{out}}{v_{in}}=\frac{Z_R}{Z_C Z_L / (Z_C+Z_L)+Z_R}=\frac{Z_R(Z_C+Z_L)}{Z_C Z_L+Z_R(Z_C+Z_L)}$$

where the impedance of the resistor, inductor and capacitor are $Z_R$, $Z_L$ and $Z_C$ respectively. I sub in values for the impedances seeing as I don't think there is any more simplification I can do here

$$\frac{v_{out}}{v_{in}}=\frac{R(-j/ωC+jωL)}{-j/ωC jωL+R(-j/ωC+jωL)}$$

I know I have to get this into a form where I can take the real part of $$\frac{v_{out}}{v_{in}}$$, in other words I want this in the form

$$\frac{v_{in}}{v_{out}}=\frac{A+jB}{C+jD}$$

however I have no idea how to do this, I'm not sure if I have set up the problem wrong. I've tried multiplying through by $\frac{j}{j}$ and $\frac{-j}{-j}$ but it doesn't seem to get me anywhere. I feel I know how to solve most of this problem, I am just unsure how to rearrange this expression to get a usable result.

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2. Feb 1, 2016

### Staff: Mentor

You've used a voltage divider approach, which is fine, but it looks like you're taking the output as being across the resistor rather than the LC section. Note that the reference node is at the bottom.

What you're looking for is the magnitude of the transfer function, $|\frac{v_{out}}{v_{in}}|$. That's not the same thing as the real part of the transfer function.

3. Feb 1, 2016

7. Feb 1, 2016

### PatsyTy

Thanks, I'll do so!

Just one quick question, your last line I'm not sure if I'm missing something but it looks like you repeated the same thing on both sides of your equation.

8. Feb 1, 2016

### Staff: Mentor

Not quite. Note the extent of the "|" operators in each case. For the magnitude of a fraction you can take the magnitudes of the numerator and denominator separately.

9. Feb 1, 2016

### PatsyTy

Ah now I see, I need that eye for detail! Thanks again!