How Does Resonance Frequency Calculation Work in LC Circuits?

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Homework Help Overview

The discussion revolves around the calculation of resonance frequency in LC circuits, particularly focusing on the relationship between the circuit components and the gain measured across a load resistor. Participants are exploring the mathematical derivation of the resonance frequency and questioning the implications of complex frequency values.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivation of the transfer function and the process of taking derivatives with respect to complex frequency. There is a question about the validity of obtaining a complex resonant frequency and its interpretation in the context of known formulas.

Discussion Status

Some participants are providing guidance on the differentiation process and suggesting alternative approaches to simplify the calculations. There is an ongoing exploration of the mathematical steps involved, with no clear consensus reached yet.

Contextual Notes

Participants note that the topic has not been covered in their coursework, indicating a potential gap in foundational knowledge regarding resonance in circuits. There is also mention of using software tools for calculations, which may influence the understanding of the manual derivation process.

Little Dump
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This isn't really a homework question. I'm just trying to figure things out for myself about resonance in a circuit that they haven't taught us yet.

Say you have a circuit

Rp || L || C and you have a sine source and some load connected to it call it Re and you want to mesaure the gain. Vi being across the sine source, Vo being across the load resistor.

So you get somehing like this:

[tex] H = \frac{V_o}{V_i} = \frac{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p}{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p + j*w*L*Rp}[/tex]

So you take the derivative with respect to j*w and get something even uglier then that.

Set it equal 0 and solve for j*w and you get

[tex] j*w = \frac{1}{\sqrt{L*C}}[/tex]

But don't you actually want to solve for w? Not j*w? In which case if you solve for w you get a complex resonant frequency. Which to me makes no sense. And is wrong because I know the formula for the circuit in question is:

[tex] w_o = \frac{1}{\sqrt{L*C}}[/tex]

What did I over look?
 
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Little Dump said:
This isn't really a homework question. I'm just trying to figure things out for myself about resonance in a circuit that they haven't taught us yet.

Say you have a circuit

Rp || L || C and you have a sine source and some load connected to it call it Re and you want to mesaure the gain. Vi being across the sine source, Vo being across the load resistor.

So you get somehing like this:

[tex] H = \frac{V_o}{V_i} = \frac{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p}{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p + j*w*L*Rp}[/tex]

So you take the derivative with respect to j*w and get something even uglier then that.

Set it equal 0 and solve for j*w and you get

[tex] j*w = \frac{1}{\sqrt{L*C}}[/tex]

But don't you actually want to solve for w? Not j*w? In which case if you solve for w you get a complex resonant frequency. Which to me makes no sense. And is wrong because I know the formula for the circuit in question is:

[tex] w_o = \frac{1}{\sqrt{L*C}}[/tex]

What did I over look?

Did you take the derivative with respect to jw properly... wherever
you have w^2, rewrite it as -(jw)^2. Also, I'd then replace all the jw, with a variable x, and take the derivative with respect to x. This just makes things a little easier to look at...
 
That's exactly how I did it. I got maple to do it for me, and then I did it by hand. Same thing.
 
Try again with
[tex]H=\frac{|V_0|}{|V_i|}[/tex]
 

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