Engineering Transfer Function for a || RLC Circuit

AI Thread Summary
The discussion focuses on deriving the transfer function H(s) and H(jω) for a parallel RLC circuit. The user has attempted to calculate the equivalent impedance of the inductor and capacitor while considering the internal resistance of the inductor, but is struggling to simplify the equations. They have expressed their confusion after reaching a certain point in the calculations and are seeking assistance. The user also mentions the output voltage is measured across the capacitor and offers to share their previous attempts for further guidance. Overall, the thread highlights the challenges faced in circuit analysis and the need for collaborative problem-solving.
TheAshlander
Messages
2
Reaction score
0
Hey all, I'm a new member, so nice to meet you all!

Homework Statement



I'm currently in the process for deriving the Transfer Function H(s) and H(jw) (w read as omega for angular frequency) for http://www.ebfreedom.com/fwajx/cDiag.jpg"

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..

Homework Equations



Vo/Vi, s = jw


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.
 
Last edited by a moderator:
Physics news on Phys.org
Call Z the impedance of the parallel inductor-capacitor. Using a voltage divider you get:
\frac{V_o}{V_i}=\frac{Z}{R+Z}
 
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..
 

Similar threads

Replies
17
Views
6K
Replies
1
Views
2K
Replies
7
Views
5K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
40
Views
5K
Back
Top