Transfer Function for a || RLC Circuit


by TheAshlander
Tags: circuit, function, transfer
TheAshlander
TheAshlander is offline
#1
Jun26-08, 05:33 PM
P: 2
Hey all, I'm a new member, so nice to meet you all!

1. The problem statement, all variables and given/known data

I'm currently in the process for deriving the Transfer Function H(s) and H(jw) (w read as omega for angular frequency) for This Circuit:

I have taken several approaches but have not been able to land on anything completely. I will post my initial work, albeit is not much, but is the basis for my different techniques..

2. Relevant equations

Vo/Vi, s = jw


3. The attempt at a solution

The idea I see here is to take the Equivalent Impedance of the Inductor and Capacitor in parallel, along with the internal resistance of the inductor. Oh yeah, the output voltage is to be read across the capacitor (or the open circuit). Well, here's the math:

I'll call R the series addition of R3 and R4, and R5 will be the internal resistance of the inductor.

Equiv Imp. of parallel inductor & capacitor: ((Ls + R5)/(Cs))/(Ls+R5+1/cS)) can be simplified to ((Ls+R5)/(CLs^2 + R5Cs + 1))

into the formal equation: Vo/Vi => (1/Cs)/(R + ((Ls+R5)/(CLs^2 + R5Cs + 1))

i have been completely lost after this point.. Please help?

If you would like to see my failed attempts, I will surely post them.
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CEL
CEL is offline
#2
Jun27-08, 07:40 AM
P: 639
Call Z the impedance of the parallel inductor-capacitor. Using a voltage divider you get:
[tex]\frac{V_o}{V_i}=\frac{Z}{R+Z}[/tex]
TheAshlander
TheAshlander is offline
#3
Jun29-08, 01:32 PM
P: 2
If it's possible to please delete this post, please do so! I can't find the delete post button and I can't seem to edit my own post..


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