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How to decide circuit resonance frequency?

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triquent

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In the circuit theory course, how to decide the circuit resonance frequency? for example, for a circuit one inductance L1 is in serial with one capacitance C and then in parallel with another inductance L2 which is inseries with a resistance R. What is the circuit resonance frequency? how many resonance frequency does it have?
the circuit is as following:

|--L1---C---|
|--L2---R---|

which book is talking about this?
 

They superimpose. The L1-R path provides a lowpass function, and the L2-C path provides a bandpass function. They will only interact a lot if the lowpass corner frequency is near the bandpass center frequency.

To a first order, the resonance frequency will simply be f=1/2Π (L2*C)^0.5
 

some people told me, there are two resonance frequency.
one is f1=1/2pi(L1*C)^0.5, another is f2=1/2pi[(L2//L2)*C]^0.5,(L1//L2 means: L1 parallel with L2). Is this right?
which book talks about this?

biff44 said:
They superimpose. The L1-R path provides a lowpass function, and the L2-C path provides a bandpass function. They will only interact a lot if the lowpass corner frequency is near the bandpass center frequency.

To a first order, the resonance frequency will simply be f=1/2Π (L2*C)^0.5

Added after 7 minutes:

i read some books. said for resonance frequency calculation: first find the impedance for the circuit, Z=R+jX. then let X=0 to get the resonance frequency. But by this method, i only get one resonance frequency. but it should be two. what is the right way to calculate the resonance frequency?
 

Convert the L2-R pair into parallel, it will give you a resistor and inductor connected parallel and the calculate the transfer function.This circuit will give you 2 resonance frequency, it's sure.One of them is parallel , the other is serial at certain frequencies.
 

i still didn't get it, how to convert the L2-R pair into parallel and get serial and parallel resonance frequency respectively. Anyone can give detailed calculation? here L1=10uH, C=50pF, L2=50uH, R=50 Ohm.
When I use Im(Z)=0, I got w1=0(f=0), w2=18.2Mrad/s(f2=2.9MHz). then resonance frequency could be 0Hz?
BigBoss said:
Convert the L2-R pair into parallel, it will give you a resistor and inductor connected parallel and the calculate the transfer function.This circuit will give you 2 resonance frequency, it's sure.One of them is parallel , the other is serial at certain frequencies.
 

Serial transfer function of that circuit is as follows..

(1+LCs²)*(sL2+R)
-----------------------
(LC+L2C)s²+RCs+1

The poles will give you resonance frequency, e.g. roots of denominator( which force denominator equal is zero) .But however zero's ( which make numerator equal is zero) will give you the other resonance points in series connection.The zeros will give you the frequencies that makes circuit is shorted.

In this case, you will find ( if available) two zero ( maybe coincided) and two poles ( probably discrete and independent ).

Note that: s=jω

This circuit looks like a crystal's equivalent circuit.At 0Hz, this circuit will behave a single resistor.(Put s=0 an look at )
Check this function and verify yourself...
 

Find the total impedence of the circuit and then let X=0 (Z=R+jX or Z=R-jX)
you'll find the resonant freq.
 

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