Resonance of LC

[nForce]

Oh, I get it now. Because we have reactive components LC connected in series and parallel we have two resonance frequencies.

Here, the Wolfram alpha solved the equation: https://www.wolframalpha.com/input/?i=(1%2F((1%2F(i*w*L))%2B(i*w*C))%2Bi*w*L+%3D+0++solve+for+w

I have some math questions tho. How do we know that we need to rationalize the fraction? Maybe some math wizz will answer now. First we need to learn solving by hand, then the software.

[Ratch]

Oh, I get it now. Because we have reactive components LC connected in series and parallel we have two resonance frequencies.

Here, the Wolfram alpha solved the equation: https://www.wolframalpha.com/input/?i=(1%2F((1%2F(i*w*L))%2B(i*w*C))%2Bi*w*L+%3D+0++solve+for+w

I have some math questions tho. How do we know that we need to rationalize the fraction? Maybe some math wizz will answer now. First we need to learn solving by hand, then the software.

You need to rationalize when the denominator contains a orthogonal term, also called a "j" term.  Don't forget the Schaum's reference I gave previously, where it is showed that a different L or C value can sometimes cause resonance at the same frequency.  Also Schaum showed that a circuit can sometimes be resonant at all frequencies.

Ratch

[kulky64]

You can easily find the impedance to be:
\$ Z(s)=\frac{L^2Cs^3+2Ls}{LCs^2+1}=\frac{Ls(LCs^2+2)}{LCs^2+1} \$
or
\$ Z(j\omega)=\frac{-jL^2C\omega^3+j2L\omega}{-LC\omega^2+1}=j\frac{L\omega(-LC\omega^2+2)}{-LC\omega^2+1} \$
From there you can see that impedance is zero for \$ \omega=0 \$ and \$ \omega_0=\frac{\sqrt{2}}{\sqrt{LC}} \$, and impedance goes to infinity for \$ \omega\rightarrow\infty \$ and \$ \omega_\infty=\frac{1}{\sqrt{LC}} \$.

[nForce]

Thank you kulky64, the best answer. 

I have a second question. How can I get the Q factor of this circuit?
This formula is from my book, but I don't understand it. I know that the energy stored in a capacitor is W= (CU^2)/2. If I understand it corectlly I need to find the energy of the whole circuit. But I don't know how.

Can someone explain to me, on my example of the circuit? Thank you.

[MrAl]

Hi,

Which circuit are you talking about here?

The Q for a bandpass filter is:
Q=F/BW

and so first calculate the center frequency F and then calculate the upper frequency FH and lower FL and then the bandwidth BW=FH-FL and then you can calculate the Q.

[nForce]

Hi,

Which circuit are you talking about here?

The Q for a bandpass filter is:
Q=F/BW

and so first calculate the center frequency F and then calculate the upper frequency FH and lower FL and then the bandwidth BW=FH-FL and then you can calculate the Q.

This circuit: http://shrani.si/f/k/oB/4jLK5yDq/circuit.jpg

It's so simplistic, so it could be solved by hand. Not software.

Yes that's other way to calculate the Q factor, but how do we get the FH and FL? I think that the formula in the attachment is better way, but I don't know how to the first or second way.

[Ratch]

nforce,

Because there is no resistance to consume power, I would think that the Q would be infinite in the theoretical circuit you are referencing.  Of course, in the real world, there is always resistance unless you are dealing with superconductors.

Ratch

[nForce]

nforce,

Because there is no resistance to consume power, I would think that the Q would be infinite in the theoretical circuit you are referencing.  Of course, in the real world, there is always resistance unless you are dealing with superconductors.

Ratch

Really? What if we put a resistor in series. Now what?

[The Electrician]

I've been wondering how rfeecs obtained the plots in reply #74 since the circuit has no resistance, and no component values.

[rfeecs]

I've been wondering how rfeecs obtained the plots in reply #74 since the circuit has no resistance, and no component values.

I just made up some values.  I set L=1000 nH, C=100 pF.  No resistance is needed to calculate impedance.
I agree this circuit by itself has infinite Q, since it has no resistance.

[Ratch]

nforce,

Because there is no resistance to consume power, I would think that the Q would be infinite in the theoretical circuit you are referencing.  Of course, in the real world, there is always resistance unless you are dealing with superconductors.

Ratch

Really? What if we put a resistor in series. Now what?

The Q will be omega*L/R.  As you can see, if R = 0, the Q will be infinite. Inserting a resistance in series will  not change the resonant frequency.

Ratch

[nForce]

nforce,

Because there is no resistance to consume power, I would think that the Q would be infinite in the theoretical circuit you are referencing.  Of course, in the real world, there is always resistance unless you are dealing with superconductors.

Ratch

Really? What if we put a resistor in series. Now what?

The Q will be omega*L/R.  As you can see, if R = 0, the Q will be infinite. Inserting a resistance in series will  not change the resonant frequency.

Ratch

Thank you, but why didn't you take into account the capacitor also? It's an electric storage component.

How do I use the formula if we have more components? Let's say we have 2 resistors, do I just add the power up? P = (I^2)*R, so two resistors = 2(I^2)*R?

Is it so?

Thanks.

[The Electrician]

Thank you, but why didn't you take into account the capacitor also? It's an electric storage component.

How do I use the formula if we have more components? Let's say we have 2 resistors, do I just add the power up? P = (I^2)*R, so two resistors = 2(I^2)*R?

Is it so?

Thanks.

If you have more components it may not be possible to assign a single number as the Q of the entire circuit.  Then you consider the Q of individual poles: http://dsp.stackexchange.com/questions/19148/whats-the-q-factor-of-a-digital-filters-pole

Even though that is primarily concerned with the Q of digital poles, at the beginning is shown how to calculate pole Q in the "analog" s plane.

[nForce]

Thank you, but why didn't you take into account the capacitor also? It's an electric storage component.

How do I use the formula if we have more components? Let's say we have 2 resistors, do I just add the power up? P = (I^2)*R, so two resistors = 2(I^2)*R?

Is it so?

Thanks.

If you have more components it may not be possible to assign a single number as the Q of the entire circuit.  Then you consider the Q of individual poles: http://dsp.stackexchange.com/questions/19148/whats-the-q-factor-of-a-digital-filters-pole

Even though that is primarily concerned with the Q of digital poles, at the beginning is shown how to calculate pole Q in the "analog" s plane.

Is it so?  How did Ratch then determine Q factor of the circuit, which has 4 components, two inductors, one capacitor and a resistor so that it's not infinity, Q factor that is?

[The Electrician]

I can't answer for Ratch, so let's see what he says.

[Ratch]

nforce,

Because there is no resistance to consume power, I would think that the Q would be infinite in the theoretical circuit you are referencing.  Of course, in the real world, there is always resistance unless you are dealing with superconductors.

Ratch

Really? What if we put a resistor in series. Now what?

The Q will be omega*L/R.  As you can see, if R = 0, the Q will be infinite. Inserting a resistance in series will  not change the resonant frequency.

Ratch

Thank you, but why didn't you take into account the capacitor also? It's an electric storage component.

How do I use the formula if we have more components? Let's say we have 2 resistors, do I just add the power up? P = (I^2)*R, so two resistors = 2(I^2)*R?

Is it so?

Thanks.

I have to apologize for not getting back to you earlier, but I was really busy today.  I don't think I answered your question to your satisfaction.  I was defining the device Q, but I think you wanted the resonance Q.  You asked about adding a resistance, which I assumed was in series with the coil not in parallel with the capacitor.  I used the same values for the coils and capacitor as rfeecs did, and plotted the outputs before and after adding a 100 ohm resistor in series with the coil.  The plots are for 1 volt divided by the absolute value of the impedance, which is what an ammeter will read.  As you can see current drops to zero at 100 megaradians and has infinite current at 1.414*100 megaradians.  The second graph with the 100 ohm resistor does not change the frequencies of the highs and lows, but it limits the current to 1 volt/100 ohms to 10 milliamps.  The third graph shows the formula for computing the Q of a series resonant circuit.  For R=100 ohms, the Q of the series resonance is 1.  The parallel resonance Q is still infinite.  Ask if you have any questions.

Ratch

[nForce]

Sorry Ratch, but I don't understand. What is resonance Q?

I am talking about the Q factor (Quality factor): https://en.wikipedia.org/wiki/Q_factor

And please don't use the software, the purpose of this topic is to learn to calculate by hand. This circuit doesn't have any meaning what's so ever, it is just a test circuit for learning.

If someone else understands what is this "resonance Q" please do explain. Thanks a lot.

[Ratch]

Sorry Ratch, but I don't understand. What is resonance Q?

I am talking about the Q factor (Quality factor): https://en.wikipedia.org/wiki/Q_factor

And please don't use the software, the purpose of this topic is to learn to calculate by hand. This circuit doesn't have any meaning what's so ever, it is just a test circuit for learning.

If someone else understands what is this "resonance Q" please do explain. Thanks a lot.

Resonant Q should not be too hard to understand or figure out.  Since the Q depends on the reactance of a component, and the reactance of a component depends on the frequency at which it is used, the resonant Q is the Q at the resonant frequency of the series or parallel LC components.

The circuit might not have usefulness, but it does have meaning in that it shows how a collection of components hooked up in a particular way can have two or more series or parallel resonant frequencies.

You can eschew computers and either plot by hand or use a calculator to plot 50 points for drawing a graph. But not me or most other folks.  It is too slow or tedious.  Besides, "plug 'n chug" does not teach you what is happening.

Keep asking questions.

Ratch

[nForce]

No, I would like to know how to determine Q factor, not resonant Q. That is other story.

Here is the wikipedia, quote:

In electrical systems, the stored energy is the sum of energies stored in lossless inductors and capacitors; the lost energy is the sum of the energies dissipated in resistors per cycle.

Here is the example from my work book:



It's not hard, it is barely one equation, but why don't we here give into account also the inductor. Wikipedia qoute: sum of energies stored in lossless inductors and capacitors.

[Ratch]

No, I would like to know how to determine Q factor, not resonant Q. That is other story.

Here is the wikipedia, quote:

In electrical systems, the stored energy is the sum of energies stored in lossless inductors and capacitors; the lost energy is the sum of the energies dissipated in resistors per cycle.

Here is the example from my work book:



It's not hard, it is barely one equation, but why don't we here give into account also the inductor. Wikipedia qoute: sum of energies stored in lossless inductors and capacitors.

I can see that you can factor 1/2 and |U|, whatever that is, from your equation.  OK, lets go through 3 examples.  Assume R=L=C=V=w=1.  As an aside, this circuit is resonant at all frequencies at RLC = 1 values.  For w=1, the reactive power of 1/2 watt is stored in L and 1 watt is stored in C.  There is more reactive power in C, so we take the reactive power of C which is 1, and use the formula Q= reactive power/resistive power for a Q of 2. Similarly, we get a Q of 5/8 when w=1/2, and a Q of 10 for w=2.  Different component values will product different results.  If you have any question about how to calculate the reactive or resistive power, just ask.

Edit:  I goofed in describing how to calculate the Q of the above circuit.  Hopefully I corrected my mistake.

Ratch

[nForce]

It's not a reactive power, but energy of reactive components. This is in the numerator. And it should be the sum.

[Ratch]

It's not a reactive power, but energy of reactive components. This is in the numerator. And it should be the sum.

My textbook definitely says Q = reactive power/real power for both series and parallel circuits.  I know there are other definitions for Q involving bandwidth and other factors, but the one I gave is one of the correct definitions.

The sum of the reactive power will not work for resonant circuits because both L and C store and release the same energy at the same rate.  The total energy is constant at resonance and is equal to the maximum energy stored in either L or C.  Therefore, the best method I can think of is to use the maximum power of whatever L or C component stores more energy.  For off resonant series circuits, C energy storage will predominate at the lower than resonant frequency, and L at the above resonant frequency.  Vice-versa for parallel resonant circuits.

All the texts and literature I have come across want to pontificate about the Q at resonance, and don't cover any method of determining the Q of off resonant frequencies.  When it comes down to it, what is Q good for other than determining the characteristics of resonant circuits?  I will keep alert for any information about Q at off resonant frequencies.

Ratch 

[MrAl]

It's not a reactive power, but energy of reactive components. This is in the numerator. And it should be the sum.

My textbook definitely says Q = reactive power/real power for both series and parallel circuits.  I know there are other definitions for Q involving bandwidth and other factors, but the one I gave is one of the correct definitions.

The sum of the reactive power will not work for resonant circuits because both L and C store and release the same energy at the same rate.  The total energy is constant at resonance and is equal to the maximum energy stored in either L or C.  Therefore, the best method I can think of is to use the maximum power of whatever L or C component stores more energy.  For off resonant series circuits, C energy storage will predominate at the lower than resonant frequency, and L at the above resonant frequency.  Vice-versa for parallel resonant circuits.

All the texts and literature I have come across want to pontificate about the Q at resonance, and don't cover any method of determining the Q of off resonant frequencies.  When it comes down to it, what is Q good for other than determining the characteristics of resonant circuits?  I will keep alert for any information about Q at off resonant frequencies.

Ratch

Hello,

Q is a factor that can be applied to a lone inductor.  That's because of the always present ESR of the inductor.  So the circuit is really a resistor in series with an inductor, and together it has a Q factor.

Q=w*L/R

[Ratch]

Mr. Al,

Q is a factor that can be applied to a lone inductor.  That's because of the always present ESR of the inductor.  So the circuit is really a resistor in series with an inductor, and together it has a Q factor.

Q=w*L/R

Of course.  However, nforce and I were discussing a more complicated LCR circuit involving an off resonant frequency.

Ratch

[nForce]

Ok, Ratch it is indeed true for your formula of Q factor is correct. So there are three formulas, one is from your textbook, one is from mine and one provided MrAI.

But I would like to know my way which is closer to me, because they are examples in this book.

MrAI, to which circuit are you reffering to. First or the second?