Calculating the Resonant Frequency in RLC circuits

Thread Starter

WilkinsMicawber

Joined Jun 5, 2017
29
When finding resonant frequency, you find the equivalent admittance or impedance and set the imaginary part equal to 0 or infinity. I choose to use admittance because the resulting calculations are usually easier. The problem I'm having, however, is in knowing when the admittance is infinite and when it is zero. In traditional parallel circuits admittance is infinite and in series circuits it is zero. What I don't understand is, what is it when the circuit is a mixture of both?. Can anyone clarify? The following examples are some problems where it isn't clear. In the first and second circuits, the admittance is equal to zero. In the third, it's equal to infinity. I can't figure out why, however.
 

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MrAl

Joined Jun 17, 2014
7,949
When finding resonant frequency, you find the equivalent admittance or impedance and set the imaginary part equal to 0 or infinity. I choose to use admittance because the resulting calculations are usually easier. The problem I'm having, however, is in knowing when the admittance is infinite and when it is zero. In traditional parallel circuits admittance is infinite and in series circuits it is zero. What I don't understand is, what is it when the circuit is a mixture of both?. Can anyone clarify? The following examples are some problems where it isn't clear. In the first and second circuits, the admittance is equal to zero. In the third, it's equal to infinity. I can't figure out why, however.

Hi,

This is why you should learn a more general method rather than rely on fixed reference circuits with solutions that are memorized.

If you can work with complex numbers then you should have no problem with any of these circuits. Just replace L with s*L and replace C with 1/(s*C) and go from there. After you simplify, you can replace 's' with j*w and you should be able to solve just about anything from there.
 

The Electrician

Joined Oct 9, 2007
2,815
Have a look at this page: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/serres.html#c2

There's also a page about parallel resonance: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html

If the definition you want to use for resonance is the phase definition, then you want to find the frequency where the phase of current with respect to voltage is zero. That occurs when the imaginary part is zero for both the impedance and admittance formulation. If the imaginary part is infinite then the phase angle would be 90 degrees, not zero degrees.

For the admittance to be infinite is not the same as the imaginary part of the admittance being infinite. It's entirely possible for the admittance to be infinite (due to the real part being infinite) with the imaginary part of the admittance being zero.

Again, resonance according to the phase definition occurs when the imaginary part of impedance or admittance is zero.

You might have a look at these pages:

https://forum.allaboutcircuits.com/threads/series-and-parallel-natural-frequencies.11647/#post-71029

https://forum.allaboutcircuits.com/threads/resonance-in-rlc-circuits.120109/#post-956980
 

MrAl

Joined Jun 17, 2014
7,949
Right, but when do i know when the imaginary part should be 0 and when it should be infinity?
Hello again,

Show an example of where you set the imaginary part of the admittance to infinity and get the right result.
Use either a regular RLC series or parallel circuit, or both.
 

Thread Starter

WilkinsMicawber

Joined Jun 5, 2017
29
For the file attached in the first post of this thread, circuit b can be solved by setting the imaginary part of admittance to infinity. Perhaps this was just a fluke. Setting it to zero makes it impossible to solve. From what I gather, for the phase definition, setting either the imaginary part of the impedence or admittance to zero works, as long as doing so gives a solveable equation.

In certain cases setting the impedance to 0 is much more difficult than setting the admittance to 0.
 

The Electrician

Joined Oct 9, 2007
2,815
You don't set the impedance to zero; you set the imaginary part of the impedance to zero.

Since this is homework, I'm not going to give the complete solution, but it's no more difficult to set the imaginary part of the impedance to zero than it is to set the imaginary part of the admittance to zero.

For us to help you with this problem, it would be good to start with deriving an expression for the impedance (or the admittance) of circuit a.

Can you do this, and post your result?
 

MrAl

Joined Jun 17, 2014
7,949
For the file attached in the first post of this thread, circuit b can be solved by setting the imaginary part of admittance to infinity. Perhaps this was just a fluke. Setting it to zero makes it impossible to solve. From what I gather, for the phase definition, setting either the imaginary part of the impedence or admittance to zero works, as long as doing so gives a solveable equation.

In certain cases setting the impedance to 0 is much more difficult than setting the admittance to 0.
Hi,

I agree with Electrician in that something does not sound right here, that's why i was hoping to see your actual work so we can go over it.
For example, i would have to ask where you found a reference that you should set the imaginary part of anything to infinity, and i would like to see that work for how you actually did that.

If we look at a general impedance:
Z=a+b*j

and set the imaginary part to zero we get:
b=0

although 'b' here would really be a function of w in most cases so we'd have f(w)=0..

Now if we compute the admittance:
A=a/(a^2+b^2)-b/(a^2+b^2)*j

when we set the imaginary part of that to zero the denominator gets factored out as well as the minus sign so we are again left with:
b=0

or as before f(w)=0.

If we instead set it equal to infinity we get:
-b/(a^2+b^2)=inf

and now are we using plus infinity or minus infinity, but in any case it looks like we get nonsense.

I'll wait for your reply though before making any judgment about what is going on here, but it sounds like you may be confusing the impedance tendency with the tendency of the imaginary part of something, which are two different things entirely. The entire impedance may go to infinity (with no resistor) but that's different, and that is a different form of resonance which i am not sure you are looking for. It sounds more like you are looking for what is sometimes referred to as "physical resonance" which involves energy and not just electrical resonance which can involve either voltage or current alone with little consideration for the energy in the circuit. We can look at this more if you like too.
 

Thread Starter

WilkinsMicawber

Joined Jun 5, 2017
29
I understand you must set the imaginary part of either admittance or impedance to zero, not infinity, but for those interested, here is how I calculated the resonant frequency by setting the imaginary part of admittance to infinity:

(I am in class, but when I get home I can do a hand-written analysis if the following is too hard to follow)
For circuit B in the picture in the first post:

Y = ( 1 / ( (1/ jwc) + (jwl) ) ) + ( 1/R )

= ( 1 / (1 - (w^2)CL) / (jwc) ) + (1/R)

= jwc / (1 - (w^2)CL) + (1/R)

Setting the imag. part to infinity:

wC/ (1 - (w^2)CL) = inf.

(1- (w^2)CL) = 0
w = sqr(1/CL)

The electrician:

For the circuit in A of the first post, if you set the imaginary part of the admittance to zero, you end up having to multiply by a much smaller conjugate than in the case where you set the imaginary part of the impedance to zero. I can post the worked-out solutions to both methods when I get home.
 
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The phase criterion for resonance only applies to parallel resonance. For parallel resonance there are 3 criteria for resonance, one of which is the phase criterion.

Circuit b is a case of series resonance for which there is only one definition, so resonance is defined as the admittance reaching a maximum, which in this case is the same as the imaginary part reaching a maximum. Since the inductor and capacitor are taken to be lossless, maximum means infinite.
 

MrAl

Joined Jun 17, 2014
7,949
I understand you must set the imaginary part of either admittance or impedance to zero, not infinity, but for those interested, here is how I calculated the resonant frequency by setting the imaginary part of admittance to infinity:

(I am in class, but when I get home I can do a hand-written analysis if the following is too hard to follow)
For circuit B in the picture in the first post:

Y = ( 1 / ( (1/ jwc) + (jwl) ) ) + ( 1/R )

= ( 1 / (1 - (w^2)CL) / (jwc) ) + (1/R)

= jwc / (1 - (w^2)CL) + (1/R)

Setting the imag. part to infinity:

wC/ (1 - (w^2)CL) = inf.

(1- (w^2)CL) = 0
w = sqr(1/CL)

The electrician:

For the circuit in A of the first post, if you set the imaginary part of the admittance to zero, you end up having to multiply by a much smaller conjugate than in the case where you set the imaginary part of the impedance to zero. I can post the worked-out solutions to both methods when I get home.

Hello again,

It is apparent that these tiny circuits are not representative of what constitutes a general idea of how these things work. In other words it looks like we have a special case here in which an unusual technique happens to work, and what is even more interesting is it may be the only way it works in the manner in which the solution is sought after. That tells me that it's not a good idea to dwell on it except as an electronic novelty item, although sometimes these novelties are interesting in there own right anyway.

If you review the method you used, you'll find that the reason infinity worked in this one case was because there was nothing else to affect the resonance calculation. Just as in the case i was describing in my last post, there is no R to affect the calculation even though it is in the circuit. That's because R does not affect the resonance in that circuit so it goes out of the calculation. In a more typical circuit we would also have a small resistance in series with L which could affect the resonance or at least make it harder to calculate using the idea of setting the imaginary part of the admittance to infinity, which is really setting the reciprocal of the imaginary part of the admittance to zero which is not in general the same as setting the imaginary part of the impedance to zero because the reciprocal of the imaginary part of the admittance is not the imaginary part of the impedance, so we should expect to get results that are not equal in the more general case.

The conclusion that seems to come from this is that setting the imaginary part of the impedance to zero is the more general method with the other method using the imaginary part of the admittance only working in some cases where resistances factor out. This means it may be difficult to predict when the admittance method works and when it does not in the general case, because it could be a much more complicated circuit.

To see the difference, try placing a second resistance RL in series with the inductor and see what comes of it. I dont think you will get the right result with the admittance/infinity method, but then do the impedance/zero method and see the difference.
 
The electrician:

For the circuit in A of the first post, if you set the imaginary part of the admittance to zero, you end up having to multiply by a much smaller conjugate than in the case where you set the imaginary part of the impedance to zero. I can post the worked-out solutions to both methods when I get home.
Don't go back to an older post and add a note like the red part above. This belongs in a later post. Normally, I wouldn't have gone back to see this; it was just accidental that I noticed it.

By all means, post your work.
 
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