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Real Part of Gain of LRC Circuit

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  1. Feb 1, 2016 #1
    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

    I start with

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

    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.
     

    Attached Files:

  2. jcsd
  3. Feb 1, 2016 #2

    gneill

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    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.
     
  4. Feb 1, 2016 #3
    I thought that ##\frac{v_{out}}{v_{in}} was a complex value function, do I not need to take the real part to get the magnitude? Also we were told we are looking for the gain of the function, is that different?
     
  5. Feb 1, 2016 #4

    gneill

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    Staff: Mentor

    The transfer function is complex. The magnitude of a complex value is determined by both of the components: ##|a + jb| = \sqrt{a^2 + b^2}## .

    The ratio of the output to the input is the gain function, also known as the transfer function. It's magnitude at a given frequency is called the gain at that frequency.
     
  6. Feb 1, 2016 #5
    Ok, so I still need the function to be of the form ##a+jb## to be able to evaluate the magnitude? For the ##\frac{v_{out}}{v_{in}}## I should start by working with

    $$\frac{v_{out}}{v_{in}}=\frac{\frac{Z_C Z_L}{Z_C+Z_L}}{\frac{Z_C Z_L}{Z_C+Z_L}+R}$$

    I believe?
     
  7. Feb 1, 2016 #6

    gneill

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    Staff: Mentor

    Sure. You may find it expedient to divide the through the top and bottom by ##\frac{Z_C Z_L}{Z_C+Z_L}## to begin with in order to concentrate all the required simplification in one place :wink:

    Also, note that ##\left| \frac{a + jb}{c + jd} \right| = \frac{|a + jb|}{|c + jd|}## That may help.
     
  8. Feb 1, 2016 #7
    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.
     
  9. Feb 1, 2016 #8

    gneill

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    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.
     
  10. Feb 1, 2016 #9
    Ah now I see, I need that eye for detail! Thanks again!
     
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