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Finding half-power frequency.

by perplexabot
Tags: frequency, halfpower
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perplexabot
#1
May31-12, 07:37 PM
P: 210
Hi all. I'm trying to do a circuit problem in my book. The circuit consists of a resistor, capacitor, inductor and a voltage source all in series. The first part of the question says: Find the resonance frequency and half-power frequencies. My book does go on and solve this problem with two different methods using given formulas for series RLC circuits, however, I would like to solve this circuit without formulas (if possible). My problem is calculating the half-power frequency not the resonance frequency.

Given:
R = 2 Ohms
L = 1 mF
C = .4 microH

My approach:
Transfer function: H(s) = R + sL + 1/sC => H(ω) = j(ω*R*C) + (1 - [ω^2 * L * C])

Magnitude of Transfer function: |H(ω)| = √( (ω*R*C)^2 + (1 - [ω^2 * L * C])^2 )
Set Magnitude of Transfer function equal to 1/√2 or set Magnitude of Transfer function squared equal to 1/2:
|H(ω)| = 1/√2 or |H(ω)|^2 = 1/2
Solving for this I get ω1 = -27k or ω2 = -65k

Obviously this is wrong due to the negative ω's. Not only is the sign wrong but also the magnitude. The book achieved the answers: ω1 = 49k or ω2 = 51k

Can anyone please tell me what I'm doing wrong (don't tell me to use formulas please)?
Thank you for your time.

PS: This is NOT a homework problem.
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vk6kro
#2
May31-12, 10:04 PM
Sci Advisor
P: 4,016
Resonance happens when the reactances of the capacitor and inductor are equal.

The Q (or Q factor) of a circuit is equal to the reactance of the inductor at resonance divided by the resistance of the resistor.

So, you may like to look up Q factor on Wikipedia or a text book to see how to derive the bandwidth.
perplexabot
#3
May31-12, 11:32 PM
P: 210
Can you get the halfpower frequency from the transfer function?

the_emi_guy
#4
Jun1-12, 12:34 AM
P: 585
Finding half-power frequency.

At DC your power dissipation is 0 (cap is blocking current). At very high freq, power dissipation is 0 (inductor is blocking current). At resonance your power dissipation is
(V^2)/R since the LC is series resonant and behaves like a short, all of the input voltage is across the resistor.

There will be two half power frequencies, one above and one below resonance, where the circuit power dissipation is (V^2)/2R, or the voltage across the resistor is 0.707Vin.

Your transfer function needs to be in the form of output/input.
In this case, since we are finding power consumption, which only occurs in the resistor, our output is the voltage across the resistor. Vr/Vin = R/(R + sL + 1/sc).
Next, set Vr/Vin = 0.707 and solve for freq.
NascentOxygen
#5
Jun1-12, 09:00 PM
HW Helper
Thanks
P: 5,142
Quote Quote by perplexabot View Post
My approach:
Transfer function: H(s) = R + sL + 1/sC => H(ω) = j(ω*R*C) + (1 - [ω^2 * L * C])
=> H(ω) = 1/(jωC) ( j(ω*R*C) + (1 - [ω^2 * L * C]))
Magnitude of Transfer function: |H(ω)| = √( (ω*R*C)^2 + (1 - [ω^2 * L * C])^2 )
Obviously can't be right.
perplexabot
#6
Jun27-12, 03:00 AM
P: 210
Quote Quote by the_emi_guy View Post
At DC your power dissipation is 0 (cap is blocking current). At very high freq, power dissipation is 0 (inductor is blocking current). At resonance your power dissipation is
(V^2)/R since the LC is series resonant and behaves like a short, all of the input voltage is across the resistor.

There will be two half power frequencies, one above and one below resonance, where the circuit power dissipation is (V^2)/2R, or the voltage across the resistor is 0.707Vin.

Your transfer function needs to be in the form of output/input.
In this case, since we are finding power consumption, which only occurs in the resistor, our output is the voltage across the resistor. Vr/Vin = R/(R + sL + 1/sc).
Next, set Vr/Vin = 0.707 and solve for freq.
Thank you. Now I know that in order to solve for the cutoff frequencies my transfer function has to be in the form of Vo/Vi and not in the form of impedence as I had tried. Using Vo/Vi works. Thanks again and sorry for the super late reply.


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