Resonance of LC - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: February 24, 2017, 05:31:17 pm
content: I have a beginners question about resonance of LC circuits:I know that resonance frequency of just one capacitor and one inductor is sqrt(1/LC). But what if we have a network ofncapacitors andminductors? And they are all mixed connected in parallel and series. What about then?Somewhere I have read that if we have a mixed way RLC circuit or just LC circuit we find an impedance of the whole circuit and then take just imaginary part and equate with 0. Then we solve for w (omega). But this is just an approximation.Can someone tell more about this, and explain it? Thank you.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 24, 2017, 06:38:12 pm
content: As the circuit gets more complicated we start talking about "poles and zeros". Not really a beginner's concept. But you can google it and see what you find. Loosely speaking, poles are the frequencies where the impedance goes to infinity and zeros are the frequencies where the impedance goes to zero. These characteristic frequencies determine a lot about the circuit behavior.

author: Benta
author_rank: Super Contributor
timestamp: February 24, 2017, 06:52:04 pm
content: To get a feel for it, I'd encourage you to try it out yourself on some network with eg, two capacitors and two inductors.Calculate like if it was four resistors, but instead of using R as impedance, use 1/sC for capacitors and sL for inductors.When you've found the transfer function, subsistute jw for s and you can do the frequency analysis.Have fun.

author: MrAl
author_rank: Super Contributor
timestamp: February 24, 2017, 07:10:18 pm
content: Quote from: nForce on February 24, 2017, 05:31:17 pmHi,As others have pointed out, as you get into more components the analysis becomes more complicated and because of the way multiple parts work together the interpretation of 'resonance' becomes more application specific. In fact, even a simple RLC circuit can have up to three points that are deemed the 'resonant' points although the one that is usually studied the most is the physical resonant point.Add to that the many components may have several points that are locally a peak or dip, and so we find that we may have several points we might call 'resonant'. Depending on the application some or all may matter, or only one. In control theory for example often the lowest resonant point is the dominant so we may pay a lot of attention to that one while ignoring the rest.The best bet here is to follow along the footsteps of everyone else that came though your path already, and that is to first study in detail the series RLC circuit and the parallel RLC circuit. That will get you pretty far. Once you get comfortable with that you may want to move on to a dual RLC circuit and see what happens.if you know circuit analysis you can check out each circuit and try to find out the important points of each. In particular, AC circuit analysis, which is pretty simple if you know how to analyze DC circuits with voltage sources and resistors. A technique like Nodal Analysis is very general so you should try to learn that and then you can go a lot farther. Of course you can always try a simulator like the free LT Spice simulator and look for peaks and dips in the frequency response.Good luck to you.

author: nForce
author_rank: Frequent Contributor
timestamp: February 24, 2017, 11:02:43 pm
content: Quote from: rfeecs on February 24, 2017, 06:38:12 pmI have another question here tho. Do we find poles and zeros just at LC circuits in rational function? What about RLC circuits, is it different? Can we also find zeros and poles at RLC circuit?Thanks.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 24, 2017, 11:15:36 pm
content: Quote from: nForce on February 24, 2017, 11:02:43 pmYes. RLC circuits are treated just like LC circuits as far as poles and zeros.

author: Ratch
author_rank: Regular Contributor
timestamp: February 25, 2017, 04:39:44 am
content: Quote from: nForce on February 24, 2017, 05:31:17 pmYou should specify whether you are referring to resonance in series or parallel circuits. For both series and parallel circuits, resonance occurs when the orthogonal component of the series voltage sums to zero, or parallel current orthogonal components sum to zero. In series resonance, the lowest impedance occurs at resonance, and the voltage-current phase is zero. In parallel resonance, there are three frequencies of note. The frequency of resonance, the frequency of maximum impedance, and the frequency of zero voltage-current phase. If the Q of the circuit is high (greater than 10), the three frequencies will be very close together. It is also possible to make a parallel circuit resonant at all frequencies provided resistance is present in the circuit. Furthermore, parallel resonance can be made to occur at two different values of L or C if resistance is present.Ratch

author: MrAl
author_rank: Super Contributor
timestamp: February 25, 2017, 06:16:36 am
content: Quote from: nForce on February 24, 2017, 11:02:43 pmHi,Stationary poles and zeros can be found by doing some factoring of the expression of interest to get the denominator or numerator into a certain form. You can also do a root locus plot which tells how the poles and zeros migrate in some circuits.

author: nForce
author_rank: Frequent Contributor
timestamp: February 25, 2017, 05:29:45 pm
content: There is an important aspect that we forget.If I compute the impedance of the circuit containing L and C elements and equate that with zero. Will I get a current resonance or voltage?I still don't understand, are the resonant frequencies poles or zeros in rational function?

author: CraigHB
author_rank: Regular Contributor
timestamp: February 25, 2017, 05:58:59 pm
content: The poles are where the transfer function tends toward infinity. Those are the terms in the function's denominator that can go to zero. In science a zero denominator is never considered an infinite result to an equation, but rather a failure of the equation to be meaningful. In an actual circuit, the poles indicate where instabilities occur in the frequency response. For a circuit on paper with zero resistance, that results in oscillations of ever increasing magnitude. In an actual circuit there's always some resistance and a limit in voltage magnitude so it can never actually do that. Usually it will increase in magnitude until the circuit fails or it will get to some arbitrary limit and stay there. Can happen easily with control systems that have several poles in their functions.

author: Ratch
author_rank: Regular Contributor
timestamp: February 25, 2017, 07:09:49 pm
content: Quote from: nForce on February 25, 2017, 05:29:45 pmI don't think you read post #5 of this thread carefully. There, I explained that resonance occurs at the frequency where the reactance is zero. You determine resonance by by finding the frequency at which the reactance is zero. Also, you are getting side-tracked by poles, zeros, and the transfer functions. You seem to be confused about what resonance is. Why don't you post a schematic of a simple circuit and we can figure it out from there. Ask some more questions and analyze the answers carefully.Ratch

author: rstofer
author_rank: Super Contributor
timestamp: February 25, 2017, 07:30:31 pm
content: If there is a circuit, model it in LTSpice XVII and do an AC analysis over varying frequency. Then you will have something specific to talk about. Otherwise, it's Laplace Transforms, poles and zeros and other truly ugly math and we still don't have a circuit to model.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 25, 2017, 08:10:22 pm
content: Look a the original question:Quote from: nForce on February 24, 2017, 05:31:17 pmI'm thinking that sounds like a filter, for example:So what about then? What's the resonant frequency? How do we deal with it?This is why I said we don't think about resonant frequency any more, but poles and zeros.As I said, not really a beginner's concept, but if you want to go down that rabbit hole, you know what to search for.Maybe Ratch has another idea on how to find the resonant frequencies of the above circuit. A demonstration perhaps.

author: Ratch
author_rank: Regular Contributor
timestamp: February 25, 2017, 09:24:34 pm
content: Quote from: rfeecs on February 25, 2017, 08:10:22 pmThat circuit is not resonant, and does not depend on resonance to work. High frequency is blocked with inductance and shorted by capacitance. Just like a non-resonant RC filter does, but this circuit is more effective.Ratch

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 26, 2017, 12:54:42 am
content: Quote from: Ratch on February 25, 2017, 09:24:34 pmIt depends on the definition of resonant. That circuit is a fifth order Chebychev low pass filter. At two points in the pass band, the insertion loss goes to zero (dB) in the ideal case, close to zero in this case. At those two points, the input impedance is purely real. All of the inductive and capacitive reactances cancel. So you could say the circuit is resonant.This fits with nForce's original procedure of finding the frequencies where the imaginary part of the impedance goes to zero.What this has to do with poles and zeros, I don't know. My point was just that for complex RLC circuits, people tend to focus more on the frequencies of the poles and zeros as defining circuit behavior, rather than resonant frequencies.

author: Ratch
author_rank: Regular Contributor
timestamp: February 26, 2017, 03:10:56 am
content: Quote from: rfeecs on February 26, 2017, 12:54:42 amI calculated the reactance of the circuit from 0 to 2 GHz and plotted it. I used r=50 for the source and load impedance. It looks like resonance occurs at three points other than zero. The reactance appears to be negative most of the time. The calculation was too long to show here. Unfortunately, i don't seem to be able to post the plot, but it shows reactance at zero at 0.55 Ghz, 1.02 GHz, and 1.22 GHz.Ratch

author: nForce
author_rank: Frequent Contributor
timestamp: February 26, 2017, 03:07:58 pm
content: So if I understand correctly, we can find resonance frequencies with transfer function or by solving Im(Z) = 0?I don't know what I do with the "j" term? Even if I rationalize the denominator, I still have the "j" term somewhere.And Ratch what do you mean by series or parallel circuits, that there's a difference. We can have a mixed parallel and series elements.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 26, 2017, 07:09:58 pm
content: Quote from: Ratch on February 26, 2017, 03:10:56 amHmm. That's not what I get. I get zero reactance at about 81MHz, 95MHz, 162MHz, and 196MHz. I used LTSPICE. File is attached. Also attached is a plot of Zin in the real-imaginary plane.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 26, 2017, 07:11:52 pm
content: Here's the LTSPICE file.

author: orolo
author_rank: Frequent Contributor
timestamp: February 26, 2017, 08:02:12 pm
content: Quote from: rfeecs on February 26, 2017, 07:09:58 pmI agree. The impedance can be computed as a continued fraction:\$ z \quad = \quad \displaystyle 50 + \frac{1}{\displaystyle 18\cdot 10^{-12}s + \frac{1}{\displaystyle 65\cdot 10^{-9}s + \frac{1}{\displaystyle 33\cdot 10^{-12}s + \frac{1}{\displaystyle 65\cdot 10^{-9} + \frac{1}{\displaystyle 18\cdot 10^{-12}s + \frac{1}{\displaystyle 50}}}}}} \$Making s = 2*pi*I*f and drawing semilog plots of module and argument vs frequency to 2GHz, we get what one would expect from a Chebishev filter: 100 omhs in the passband, 50 ohms well beyond it, and a maximum near 300 ohms at almost 200MHz.Comparting this to a resonator, maybe it looks like a very lossy and low Q one, with series resonance near 160MHz, and parallel resonance at near 200MHz. That's far fetched, after all this is a low pass filter.

author: MrAl
author_rank: Super Contributor
timestamp: February 27, 2017, 01:14:03 pm
content: Quote from: Ratch on February 26, 2017, 03:10:56 amHi there,Yeah the definition of resonance is a little funny sometimes. The way i like to explain it is it is like where one view concentrates on the energy exchange and the other view concentrates on the electrical circuit behavior as observed in a given application circuit.The physicist guy says that resonance occurs with a certain exchange of energy, but the radio tuner guy says it occurs when his tank circuit peaks or dips. Still yet another guy unified these two views by relating the modes of resonance to Cassini Ovals.

author: nForce
author_rank: Frequent Contributor
timestamp: February 27, 2017, 05:58:04 pm
content: Sorry if this was already answered, but no one has answered these questions:Quote from: nForce on February 26, 2017, 03:07:58 pmTry answering in this manner: Yes this is true, because... (or) No, this is not true because... I don't understand these abstract answers. Thanks again.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 27, 2017, 06:24:45 pm
content: Quote from: nForce on February 27, 2017, 05:58:04 pmYes, this seems to be one definition of resonance, because at that point the capacitive reactance and inductive reactance cancel each other. I'm sure you saw the Wikipedia article:https://en.wikipedia.org/wiki/RLC_circuitIt discusses a lot of terms like natural frequency and driven resonance frequency.QuoteI don't understand what you are asking.

author: nForce
author_rank: Frequent Contributor
timestamp: February 27, 2017, 06:35:46 pm
content: Well with "j" I mean the imaginary unit. Because the impedance of inductor is "jwL" and impedance of capacitor is "1/(jwC).

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 27, 2017, 06:42:29 pm
content: What do you mean byQuote

author: nForce
author_rank: Frequent Contributor
timestamp: February 27, 2017, 07:03:59 pm
content: When I get the impedance of the whole circuit, probably as a rational function I rationalize the denominator and then equate the imaginary units with 0. (Im(Z) = 0). But I have the j terms in this equation. What do I have to do with them?

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 27, 2017, 07:13:32 pm
content: That doesn't make sense.If Z is of the form a+jb. If Im(Z)=0, then b=0.Maybe share an example where you have the problem.

author: nForce
author_rank: Frequent Contributor
timestamp: February 27, 2017, 09:36:51 pm
content: Oh, sorry my bad you are right.In this way I get a result, but how do I know if the computed value is a current resonance or voltage resonance? How to differ them?

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 27, 2017, 09:56:42 pm
content: Quote from: nForce on February 27, 2017, 09:36:51 pmI have no idea. I'm not sure that it is possible just by looking at impedance.

author: orolo
author_rank: Frequent Contributor
timestamp: February 27, 2017, 11:38:30 pm
content: To make sure there is resonance, you need to look a bit around the points of real impedance.First, imagine a load made of lots of passives, Rs, Ls and Cs. Combining them in series and parallel, you wind up with an equivalent load ZL whose value is a formula dependent on the frequency, some rational function p(s)/q(s), where s is the complex frequency s = I·w.Imagine that, for some (angular) frequency w0, that load impedance is a real number. Why can you consider that there is a resonance? Start with the fact that the load impedance at that frequency is \$Z_L(\omega_0) = R_L\$, some real number. That isn't saying too much: we only know that at that very precise frequency you have something like a resistor load. So let us see what happens at nearby frequencies. Since rational functions are smooth enough, near the resonat frequency the impedance must be, to first order, a small variation of that real load:\$Z_L(\omega_0 + \mathrm{d}\omega) \ = \ R_L + (a + bi)\mathrm{d}\omega \$where a and b are some real numbers. Further, imagine that b is not zero (that would be a rather singular case). If we aggregate things a bit, near the resonant frequency we have:\$Z_L(\omega_0 + \mathrm{d}\omega) \ = \left(R_L + a\mathrm{d}\omega\right) \, + \, ib\mathrm{d}\omega \$What this formula is saying us is that, near the resonant frequency, as frequency rises impedance goes from capacitive to inductive (if b>0), or as inductive to capacitve (if b<0). If b is a big number, that transition will be very fast: this is somehow related to the Q of the resonance.So a resonant frequency can be considered a point where load impedance goes from capacitive to inductive. How is this related to oscillation? Imagine that load is connected to some non-inverting gain element, for example, the virtual ground of an operational amplifier. If the gain is big enough, at resonance the opamp will feed back a signal in phase and cause oscillation at frequency w0. However, as you move away from w0, the phase shift caused by the increasing reactance (the i·b·dw term) will make oscillation impossible. You get a sinusoidal oscillation only at frequency w0.The low pass filter of the previous posts is not a great resonator because the transitions were really slow: impedance was real (about 100 + 0j) at 162MHz, and reached about 290 + 79j ohms at 200MHz. That's a slope b = 79/(200-162) = 2 ohms/MHz! The 200MHz resonance is a bit better, but compare that to a crystal resonator, who can have slopes of hundreds of ohms perHertz. Now that is resonance.To summarize, just considering an LC network for its resonances is a restricted point of view. You get a convoluted variation of impedance with increasing frequency, and you can achieve a lot more than selecting a very specific frequency with that behavior. Besides, as MrAl said, there are other points of view about resonance, for example, tracking the stored energy in the system when you inject a sinusoid into it. A good part of the problem is defining resonance of the LC network in uninequivocal terms.

author: nForce
author_rank: Frequent Contributor
timestamp: February 28, 2017, 09:54:05 am
content: QuoteRatch what do you mean by series and parallel circuit? So series = voltage, and parallel = current?

author: MrAl
author_rank: Super Contributor
timestamp: February 28, 2017, 01:38:08 pm
content: Quote from: nForce on February 26, 2017, 03:07:58 pmHello again,One way to investigate is to find the transfer function Hs and then find |Hs| and then solve this:d|Hs|/dw=0and look for all local or absolute min and max's.The 'j' is the imaginary operator. To calculate the amplitude you can convert to the general case:Hs=a+b*jand then use:|Hs|=sqrt(a^2+b^2)wherea is the real part,b is the imaginary part (with the 'j').The phase shift if you need it is:Ph=atan2(b,a)"atan2" is the two argument inverse tangent function which covers the full 0 to 360 degree range.

author: nForce
author_rank: Frequent Contributor
timestamp: February 28, 2017, 02:22:19 pm
content: MrAI,do you understand this: please answer if do. Thank you.Quote from: nForce on February 28, 2017, 09:54:05 am

author: MrAl
author_rank: Super Contributor
timestamp: February 28, 2017, 03:01:34 pm
content: Quote from: nForce on February 28, 2017, 02:22:19 pmHello again,It appears that you may be after something that may not matter to the circuit at hand.For the circuit at hand, which is the one with the multiple components, i tried to show what the important thing is for the analysis. It is certainly up to you though if you want to per something else that may or may not be related to the circuit, and i realize that other things come up sometimes too.With that in mind, the critical frequencies i found are close to:5.0653345640775*10^7,1.371919896056*10^8,9.536195775181*10^7,1.629471368511*10^8,2.0*10^8except for that last one which is the supposed 3db down point.The relative amplitudes at those frequencies are:[0.9962,0.994,1.0,1.0,0.73618]and as we can see that means the circuit functions mostly as a low pass filter because the other amplitudes do not vary that much from nearly 0db. That last one is the 200MHz amplitude, and we see that it is not exactly 3db down, but for this circuit there would be other factors that set in anyway like the parasitic elements.So what i am really saying here is that for the electrical guy physical resonance may not be what we want to look at anyway.I would also suggest that we look at some simpler examples.

author: Ratch
author_rank: Regular Contributor
timestamp: February 28, 2017, 03:48:08 pm
content: Quote from: nForce on February 26, 2017, 03:07:58 pmYou would do well to Google for some of your questions.http://pediaa.com/difference-between-series-and-parallel-resonance/Setting the j term of the impedance to zero will give you the frequency of resonance. There might be several frequencies at which a circuit resonates, or it might resonate at all frequencies. I can give you an example if you wish.Ratch

author: Ratch
author_rank: Regular Contributor
timestamp: February 28, 2017, 04:07:12 pm
content: Quote from: rfeecs on February 26, 2017, 07:09:58 pmI did not use LTSpice. I calculated the impedance of the circuit using 50 ohms resistance for both the source and load impedance. Then I plotted the reactance and found the zeros. If you really give a damn, I can post my calculations, if I can get this forum to allow me to include jpg files.Ratch

author: nForce
author_rank: Frequent Contributor
timestamp: February 28, 2017, 06:14:00 pm
content: I think, it's better if I give a simple example of a circuit, and my attempt to solve:http://shrani.si/f/k/oB/4jLK5yDq/circuit.jpgI don't know how to insert Latex code here, so I have photograph the "solution". How to continue here?

author: MrAl
author_rank: Super Contributor
timestamp: February 28, 2017, 07:00:13 pm
content: Quote from: nForce on February 28, 2017, 06:14:00 pmHi,What are you trying to calculate there, when the impedance is zero?

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 28, 2017, 07:06:00 pm
content: Quote from: nForce on February 28, 2017, 06:14:00 pmI think you have a sign error. An additional problem with that circuit is that the resonance occurs where the impedance goes to infinity.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 28, 2017, 07:12:01 pm
content: Just to add more confusion, the attached circuit clearly has a resonance, but the impedance is purely real, equal to 50 ohms.So finding resonances by looking at the impedance doesn't always work.

author: Ratch
author_rank: Regular Contributor
timestamp: February 28, 2017, 08:19:52 pm
content: Quote from: nForce on February 28, 2017, 06:14:00 pmYour calculations are wrong. The impedance of inductance is j*L*omega and the impedance of C is -j/(C*omega). Unfortunately, I am unable to post jpg files on this forum, so I cannot show you.Ratch

author: MrAl
author_rank: Super Contributor
timestamp: February 28, 2017, 08:43:00 pm
content: Quote from: rfeecs on February 28, 2017, 07:12:01 pmHi,That looks like it could be a very good example.I dont know why anyone would have so much difficulty accepting the critical points calculation outlined earlier in this thread. Maybe because it leads to complicated complex algebraic expressions?

author: Ratch
author_rank: Regular Contributor
timestamp: February 28, 2017, 08:58:09 pm
content: Quote from: rfeecs on February 28, 2017, 07:12:01 pmThat is an example of a circuit which is resonant at all frequencies. I mentioned such a circuit in post #34 of this thread.Ratch

author: nForce
author_rank: Frequent Contributor
timestamp: February 28, 2017, 09:06:48 pm
content: QuoteI took the reactance, but because there is a "j" term in the fraction so I have to take the whole.QuoteHow do you know?QuoteYes it is. If you look at the fraction it's 1/1/(jwC) which is equal to jwC.

author: Ratch
author_rank: Regular Contributor
timestamp: February 28, 2017, 10:07:11 pm
content: Quote from: nForce on February 28, 2017, 09:06:48 pmYou are correct. My mistake, sorry. But you made a mistake in your algebra. You divided by j to eliminate it from the equation. Unfortunately, one of the terms you divided into had a 1/j literal, so the term became minus, and instead of canceling, the two terms add negatively.Ratch

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 28, 2017, 10:12:28 pm
content: Quote from: Ratch on February 28, 2017, 08:58:09 pmHow is it resonant at all frequencies? Explain that one.Oh, because the reactance is zero at all frequencies, maybe? Then a resistor is resonant at all frequencies.I don't get it. I think it has a very definite resonance at one frequency, around 159MHz.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 28, 2017, 10:46:42 pm
content: Quote from: rfeecs on February 28, 2017, 10:12:28 pmOK, I guess I see it now. If you look at the two main arms of the circuit, the capacitive and inductive susceptance are equal and opposite at all frequencies and they cancel out.I guess you could say that is resonating at all frequencies. Very strange way to look at it. But OK.

author: Ratch
author_rank: Regular Contributor
timestamp: March 01, 2017, 12:50:36 am
content: rfeecs,Yes, you have the right idea. A single resistor only dissipates energy. The circuit you submitted transfers energy between its energy storage components. If only I could attach jpg files to my posts, I could show you some more interesting aspects of those kind of circuits. Every time I choose a file, it shows up in the file attachment box, but i cannot move it into my message.Ratch

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 01, 2017, 01:12:23 am
content: Quote from: Ratch on March 01, 2017, 12:50:36 amThe attachments don't show up in the preview, but they will show up at the end of the post as a thumbnail.Here's a thread that talks about how to put an image inline with your message:https://www.eevblog.com/forum/chat/how-to-include-images-in-a-post/

author: Ratch
author_rank: Regular Contributor
timestamp: March 01, 2017, 02:00:53 am
content: Quote from: rfeecs on March 01, 2017, 01:12:23 amI will try again.OK, I think I got the idea. It's clunky and clumsy compared to the other forums, but i think I got it now. I will post more about resonant circuits shortly.lRatch

author: Ratch
author_rank: Regular Contributor
timestamp: March 01, 2017, 02:39:03 am
content: rfeecs,Here is the plot I made of the first circuit you posted earlier. I suppressed the output of the calculations after the first two components because the output contained toomany terms to display. After the calculation was complete, there were over 100 terms for the j-component alone. That circuit must be very sensitive because just rounding off the terms caused the plot to vary significantly. The "%" sign in the calculations is the symbol for the previous result. I made two plots of the reactance, 0-1GHz and 0-2Ghz. As you can see, I started from the right alternating between parallel and series. I hope I did it right.Ratch

author: Ratch
author_rank: Regular Contributor
timestamp: March 01, 2017, 03:15:23 am
content: To the Ineffable All,The two attached files contain two pages from the Schaum's Outline Series,Electric Circuits. It contains a concise description of parallel resonance. It also shows how two different values of a L or C component can have the same resonant frequency, and how the circuit can be resonant at all frequencies or not resonant at any frequency. Enjoy.Ratch

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 01, 2017, 08:28:48 am
content: Quote from: Ratch on March 01, 2017, 02:39:03 amIt looks to me like your results are in radians/sec, not Hz.

author: MrAl
author_rank: Super Contributor
timestamp: March 01, 2017, 10:44:40 am
content: Quote from: rfeecs on February 28, 2017, 10:46:42 pmHi,Which circuit are you talking about here?

author: Ratch
author_rank: Regular Contributor
timestamp: March 01, 2017, 02:03:30 pm
content: Quote from: The Electrician on March 01, 2017, 08:28:48 amCorrect, I probably should have plotted the curve in hertz.Ratch

author: Ratch
author_rank: Regular Contributor
timestamp: March 01, 2017, 02:10:44 pm
content: Quote from: MrAl on March 01, 2017, 10:44:40 amThe one depicted in posts #12 and #13.Ratch

author: MrAl
author_rank: Super Contributor
timestamp: March 01, 2017, 03:01:26 pm
content: Quote from: Ratch on March 01, 2017, 02:10:44 pmHi,No the circuit that you said was "resonant at all frequencies".Where is that one?

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 01, 2017, 06:11:31 pm
content: Quote from: MrAl on March 01, 2017, 03:01:26 pmReply #39

author: MrAl
author_rank: Super Contributor
timestamp: March 02, 2017, 03:16:56 pm
content: Hi,Thanks, and i wont quote all that text again here :-)Yeah that circuit is interesting from a theoretical standpoint, but we should probably look at the practical significance also. I have a feeling the practical circuit when in use for a real application will have a different overall effect. I have to say though that of course i cant think of every possible use under the sun. For example, if someone wants to create a very complicated resistor then i cant stop them from doing so :-) Of course to the contrary is if there is an output that would mean constant loading to the previous stage, which would of course be more than very practical.

author: snarkysparky
author_rank: Frequent Contributor
timestamp: March 02, 2017, 03:36:12 pm
content: resonance is where there are complex conjugate pole pairs in the denominator of the transfer function. The resonant points are I think the real part of the pole.

author: MrAl
author_rank: Super Contributor
timestamp: March 02, 2017, 05:57:47 pm
content: Quote from: snarkysparky on March 02, 2017, 03:36:12 pmHi,Well i think what some people in this thread have been doing is just looking at the imaginary part of the transfer function and finding when that goes to zero, because that indicates physical resonance.What i was doing was trying to point out that the resonance found that way may not be the only resonance and other methods may have to be employed. In fact, that view may end up being the most insignificant relative to the operation of the circuit when used in an application. Also, we run into special theoretical examples where the outcome is far from a practical one. This happens more than we might like because theory does not transfer immediately to the practical with even the tiniest imperfection sometimes. There's no doubt though that theory may uncover a good way to do things when we take measures to reduce the imperfections or use it in a way that takes advantage of that theoretical aspect.

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 02, 2017, 06:22:33 pm
content: Quote from: MrAl on March 02, 2017, 03:16:56 pmYes, one application for that type of circuit is a reflectionless filter:https://www.microwaves101.com/15-homepage/1317-reflection-less-filtersWhile most filters reflect power in the stopband, this type absorbs the stopband power and presents a constant impedance.

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 02, 2017, 06:44:21 pm
content: Quote from: Ratch on March 01, 2017, 03:15:23 amThe admittance of LC circuit presented there (attached) in general has two poles and two zeros. The poles are at -1/RC and -L/R. To make it "resonant at all frequencies", the R,L,C are chosen to set the frequencies of the zeros to be the same as the poles. So all frequency dependent terms cancel. But that also means that all the poles and zeros are real (no complex conjugate pairs). In the time domain response, you just have decaying exponential responses, no sinusoids with energy bouncing back and forth.It seems to me this another example of satisfying a mathematical definition of resonance, but not really what most people think of a resonant circuit.It is also an example where looking at the poles and zeros provides a better insight into what the circuit is really doing.

author: Ratch
author_rank: Regular Contributor
timestamp: March 02, 2017, 07:13:01 pm
content: Quote from: MrAl on March 02, 2017, 05:57:47 pmAs I mentioned in post #6 of this thread, some folks think that besides zero reactance, an alt right definition of parallel resonance is the frequency where the phase is zero or the impedance is maximum. At high Q levels, these three frequencies are quite close to each other. I believe that zero reactance is the proper definition and the other two are spurious. There is no ambiguity in series resonance because all definitions have the same frequency.Ratch

author: kulky64
author_rank: Regular Contributor
timestamp: March 02, 2017, 11:28:07 pm
content: I manually derived transfer function for circuit in Reply #39:\$ Z(s)=\frac{V(s)}{I(s)}=\frac{R_1L_1C_1L_2C_2s^4+(R_1R_2L_1C_1C_2+L_1L_2C_2)s^3+(R_1L_1C_1+R_1L_2C_2+R_2L_1C_2)s^2+(R_1R_2C_2+L_1)s+R_1}{L_1C_1L_2C_2s^4+(R_1+R_2)L_1C_1C_2s^3+(L_1C_1+L_1C_2+L_2C_2)s^2+(R_1+R_2)C_2s+1} \$ ,and then in Matlab calculated poles, zeros and gain for component values as given in Reply #39. From there it is clear why this circuit behaves as pure resistance at all frequencies. It has two equal pairs of complex conjugate poles and exactly the same zeros. So zeros in numerator cancel out with poles in denominator, remains only real 50 ohms.

author: MrAl
author_rank: Super Contributor
timestamp: March 03, 2017, 08:50:49 am
content: Hi,Well one striking characteristic of the circuit as discussed so far is the criterion:1/sqrt(L1*C1)=1/sqrt(L2*C2) [which of course simplifies]Change that relationship a little and see what happens :-)

author: kulky64
author_rank: Regular Contributor
timestamp: March 03, 2017, 11:33:39 am
content: Only L1*C1=L2*C2 is not sufficient condition for this circuit to work, there are total 4 conditions that must be met:L2=4*L1C2=(1/4)*C1R1=2*sqrt(L1/C1)R2=R1So you can choose arbitrarily only value of L1 and C1, other component values must be calculated according to these conditions. Break any one of them and this circuit will no longer be resistive for all frequencies.

author: MrAl
author_rank: Super Contributor
timestamp: March 03, 2017, 03:43:48 pm
content: Hi,Perhaps so, but what jumps right out at you is L1*C1=L2*C2. Right off we see someone was up to something here :-)There could be other interesting things too.

author: nForce
author_rank: Frequent Contributor
timestamp: March 13, 2017, 02:33:33 pm
content: Quote from: rfeecs on February 28, 2017, 07:06:00 pmSorry, but I have to get to the bottom of this. How do you know for my circuit that the resonance is when the impedance goes to infinity? Please do explain.And which case is in my example? Because we have parallel and series connected components. It can't be just series resonance or parallel resonance which I read on the net.Thank you.

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 13, 2017, 04:35:39 pm
content: You circuit has two resonances. It has a parallel resonance where the impedance goes to infinity, and a series resonance where the impedance goes to zero. Plots of the impedance are attached showing the two resonance points.

author: nForce
author_rank: Frequent Contributor
timestamp: March 13, 2017, 04:53:16 pm
content: Hey, thanks for these graphs.How did you make them? Did you calculate the transfer function by hand, and then plot in Octave/Matlab or I don't know what is this.Can you show me how to continue calculating the impedance...

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 13, 2017, 05:01:37 pm
content: I used a circuit simulator, Keysight ADS, I had handy at work. You could also use LTSPICE, which is free. LTSPICE has a .net directive that lets you calculate Z parameters.Octave or Matlab would also work if you work out the impedance equation by hand. Any plotting program, even Excel would work in that case.

author: Richard Crowley
author_rank: Super Contributor
timestamp: March 13, 2017, 05:19:05 pm
content: Not being a great wiz at math, I always appreciated this Frequency-Reactance Nomograph.I first saw it on the OpAmp Labs website, but now it is available at: RF Cafe...http://www.rfcafe.com/references/electrical/frequency-reactance-nomograph.htm

author: kulky64
author_rank: Regular Contributor
timestamp: March 13, 2017, 07:50:04 pm
content: Quote from: rfeecs on March 13, 2017, 04:35:39 pmHow did you get these plots? Phase in this circuit can be only +90 deg or -90 deg. Nothing in between.

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 13, 2017, 09:28:37 pm
content: Quote from: kulky64 on March 13, 2017, 07:50:04 pmI plotted phase of S11 instead of Z11. Attached I have added phase of Z11. You are right. It only takes values of +90 or -90.

author: nForce
author_rank: Frequent Contributor
timestamp: March 14, 2017, 04:14:03 pm
content: Oh, I get it now. Because we have reactive components LC connected in seriesandparallel 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+wI 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.

author: Ratch
author_rank: Regular Contributor
timestamp: March 14, 2017, 04:40:09 pm
content: Quote from: nForce on March 14, 2017, 04:14:03 pmYou 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

author: kulky64
author_rank: Regular Contributor
timestamp: March 14, 2017, 04:50:41 pm
content: 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}} \$.

author: nForce
author_rank: Frequent Contributor
timestamp: March 15, 2017, 10:52:51 am
content: 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.

author: MrAl
author_rank: Super Contributor
timestamp: March 15, 2017, 07:00:40 pm
content: Hi,Which circuit are you talking about here?The Q for a bandpass filter is:Q=F/BWand 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.

author: nForce
author_rank: Frequent Contributor
timestamp: March 15, 2017, 08:33:41 pm
content: Quote from: MrAl on March 15, 2017, 07:00:40 pmThis circuit:http://shrani.si/f/k/oB/4jLK5yDq/circuit.jpgIt'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.

author: Ratch
author_rank: Regular Contributor
timestamp: March 15, 2017, 08:45:39 pm
content: 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

author: nForce
author_rank: Frequent Contributor
timestamp: March 15, 2017, 09:22:51 pm
content: Quote from: Ratch on March 15, 2017, 08:45:39 pmReally? What if we put a resistor in series. Now what?

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 15, 2017, 10:48:05 pm
content: I've been wondering how rfeecs obtained the plots in reply #74 since the circuit has no resistance, and no component values.

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 16, 2017, 12:06:58 am
content: Quote from: The Electrician on March 15, 2017, 10:48:05 pmI 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.

author: Ratch
author_rank: Regular Contributor
timestamp: March 16, 2017, 04:59:01 am
content: Quote from: nForce on March 15, 2017, 09:22:51 pmThe 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

author: nForce
author_rank: Frequent Contributor
timestamp: March 16, 2017, 09:10:03 am
content: Quote from: Ratch on March 16, 2017, 04:59:01 amThank 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.

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 16, 2017, 01:42:17 pm
content: Quote from: nForce on March 16, 2017, 09:10:03 amIf 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-poleEven 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.

author: nForce
author_rank: Frequent Contributor
timestamp: March 16, 2017, 06:29:32 pm
content: Quote from: The Electrician on March 16, 2017, 01:42:17 pmIs 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?

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 16, 2017, 07:54:24 pm
content: I can't answer for Ratch, so let's see what he says.

author: Ratch
author_rank: Regular Contributor
timestamp: March 17, 2017, 05:07:53 am
content: Quote from: nForce on March 16, 2017, 09:10:03 amI 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

author: nForce
author_rank: Frequent Contributor
timestamp: March 17, 2017, 08:47:28 am
content: 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_factorAnd 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.

author: Ratch
author_rank: Regular Contributor
timestamp: March 17, 2017, 01:50:44 pm
content: Quote from: nForce on March 17, 2017, 08:47:28 amResonant 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

author: nForce
author_rank: Frequent Contributor
timestamp: March 17, 2017, 04:49:26 pm
content: No, I would like to know how to determine Q factor, not resonant Q. That is other story.Here is the wikipedia, quote:QuoteHere 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 losslessinductorsand capacitors.

author: Ratch
author_rank: Regular Contributor
timestamp: March 17, 2017, 11:28:13 pm
content: Quote from: nForce on March 17, 2017, 04:49:26 pmI 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

author: nForce
author_rank: Frequent Contributor
timestamp: March 18, 2017, 10:03:58 am
content: It's not a reactive power, but energy of reactive components. This is in the numerator. And it should be the sum.

author: Ratch
author_rank: Regular Contributor
timestamp: March 18, 2017, 01:57:08 pm
content: Quote from: nForce on March 18, 2017, 10:03:58 amMy 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

author: MrAl
author_rank: Super Contributor
timestamp: March 18, 2017, 04:13:41 pm
content: Quote from: Ratch on March 18, 2017, 01:57:08 pmHello,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

author: Ratch
author_rank: Regular Contributor
timestamp: March 18, 2017, 04:49:33 pm
content: Mr. Al,QuoteOf course. However, nforce and I were discussing a more complicated LCR circuit involving an off resonant frequency.Ratch

author: nForce
author_rank: Frequent Contributor
timestamp: March 18, 2017, 08:46:09 pm
content: 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?

author: MrAl
author_rank: Super Contributor
timestamp: March 18, 2017, 09:22:34 pm
content: Quote from: nForce on March 18, 2017, 08:46:09 pmHi,I thought i already stated that it was for a lone inductor, which of course has ESR, or for an ideal inductor in series with a resistor. In both cases we have just one resistor and one inductor.There will be formulas for the Q and which one you use will depend on the application. The ones we have been discussing i think fall into the category of impedances, while at least one other one:Q=F/BWfalls into the category of filters.The center frequency F is found and then the BW by using something like d|H(jw)| /dw=0.We could talk about that more too.

author: MrAl
author_rank: Super Contributor
timestamp: March 18, 2017, 09:31:44 pm
content: Quote from: Ratch on March 18, 2017, 04:49:33 pmHi again,Yes i was replying to your statement/question in reply 96 of:"When it comes down to it, what is Q good for other than determining the characteristics of resonant circuits? "and wanted to broaden the field where Q might be reasonable, even in the absence of a resonance of any kind.And then we have the Q of a low pass filter, which may not be thought of as having any resonance or may be thought of that way too. d=1/2/Q, d the damping factor.I think my favorite is the power and energy statement of Q, but easier to calculate i think is the reactive power over real power, for impedance elements.

author: kulky64
author_rank: Regular Contributor
timestamp: March 19, 2017, 12:48:48 am
content: Quote from: nForce on March 17, 2017, 04:49:26 pmI think you don't understand equation for calculating Q from your work book very well. Because this equation IS for calculation of Q at resonant frequency. It will not work for arbitrary frequency. And the reason why there is not explicitly calculated energy stored in inductor in numerator of your equation is because at resonance, energy stored in inductor is numerically equal to energy stored in capacitor. So you if you know for example energy stored in capacitor, you take double of that and you have sum of energy stored in capacitor and inductor.

author: nForce
author_rank: Frequent Contributor
timestamp: March 19, 2017, 12:28:40 pm
content: kulky64, the best answer. You should come here more frequentlyOther please continue from here what kulky64 said. It is important.kulky64, you said that it's double of energy stored in the capacitor. But in the equation in the numerator is just for one capacitor. (CU^2)/2, so it should be double of that, like this: (CU^2).

author: kulky64
author_rank: Regular Contributor
timestamp: March 19, 2017, 02:06:00 pm
content: In equation from your textbook, numerator and denominator is expanded by 1/2 for some unknown to me reason. If you delete this 1/2 from numerator and denominator, it will start making sense. In numerator should be only \$ C\left|U\right|^2 \$, power dissipated in resistor is \$ \frac{\left|U_R\right|^2}{R} \$, so this should be in denominator, not \$ \frac{1}{2}\frac{\left|U_R\right|^2}{R} \$.

author: Ratch
author_rank: Regular Contributor
timestamp: March 19, 2017, 03:38:47 pm
content: Quote from: nForce on March 18, 2017, 08:46:09 pmI have been trying to wrap my mind around what the Q with respect to a circuit driven by a non-resonant frequency means. I don't think it has any significance. All I can find in the literature relates the circuit Q to the resonant frequency. At resonance, the maximum energy stored in the L and C components is the same, and the totals of the energy in the L and C components are equal at any time. Therefore, the Q for a series resonant circuit is the reactance of either L or C divided by the resistance. The Q for a parallel circuit is R divided by the the reactance of either L or C. The Q in terms of the LCR components is derived and presented in the attachment.Ratch

author: kulky64
author_rank: Regular Contributor
timestamp: March 19, 2017, 04:06:20 pm
content: I think it's best to derive the expression for Q for particular circuit in hand and not fix yourself to some textbook formulas. For example if we further manipulate expression for Q from nForce example, we get:\$ Q=\frac{1}{R}\sqrt{\frac{L}{C}-R^2} \$

author: MrAl
author_rank: Super Contributor
timestamp: March 19, 2017, 04:10:59 pm
content: Quote from: Ratch on March 19, 2017, 03:38:47 pmHi again Ratch,in particular i am replying to the statements:"I have been trying to wrap my mind around what the Q with respect to a circuit driven by a non-resonant frequency means. I don't think it has any significance. "We already talked a little about the Q of an inductor, and we could talk about the Q of a capacitor, but it woudl be a similar situation, so we know at least there is a calculation that reveals the Q of either element, taken alone by itself.The way to proceed from here is to find a relationship between the individual Q's to other uses such as in a resonant circuit, which answers the question:"I have two inductors one with Q=5 and one with Q=10, if i use it in a circuit with a capacitor with Q=10 which one will give me an overall higher circuit Q?"So if you can find a way to combine the Q's of the two you could know right away which component is the better one to use with your total circuit (which also includes a cap). That in itself without any other reasoning, gives significance to the Q of a lone element without it being in resonance.For example, prove (or disprove) the following:Q=1/(1/QL+1/QC)where Q is the total circuit Q and QL and QC are the Q's of the inductor and cap respectively.But you may not want to stop there. You may find other relationships, but it will take some work on your part to do some calculations and see what you can come up with. I know you do a lot of math too so you might find this interesting to do. See what you can find :-)The Q of an LC circuit off resonance may be interesting when the frequency is not set to the desired resonant frequency. This information may be used to combine with other circuits as well. You could try designing a simple resonant circuit, calculate the Q at arbitrary frequency, then see what happens when it combines with another circuit with some other arbitrary Q (at the same frequency).

author: Ratch
author_rank: Regular Contributor
timestamp: March 19, 2017, 05:41:26 pm
content: Hi MrAl,I don't see any problem figuring out the Q of a resonant circuit from the Q's of the individual components. If the resistance does not affect the frequency of circuit resonance, then the resistance of each component can be found from the Q of the component together with the reactance of the component, and combined into one resistance. If the resistance affects the frequency, then solving the problem with nonlinear methods is in order. I don't think trying to figure out how to combine the Q's of each component into a circuit is practical except for the simplest of circuit configurations.I still don't think that the concept of circuit Q at an off resonant frequency means anythingRatch

author: Ratch
author_rank: Regular Contributor
timestamp: March 19, 2017, 05:47:18 pm
content: Quote from: kulky64 on March 19, 2017, 04:06:20 pmWould you be so kind to provide a post number or a circuit schematic designating which nforce example you are referencing? I don't want to chase my tail perusing the wrong circuit.Ratch

author: kulky64
author_rank: Regular Contributor
timestamp: March 19, 2017, 06:04:36 pm
content: Reply #93

author: Ratch
author_rank: Regular Contributor
timestamp: March 19, 2017, 07:11:26 pm
content: Quote from: kulky64 on March 19, 2017, 06:04:36 pmThank you. Usually textbooks are more accurate and knowledgeable than the folks who read them. As you can see in the attachment, the Q formula you presented is the "inductance" Q, not the circuit Q. If I needed to calculate the circuit Q, I would first transform the series R & L into a parallel R & L at the resonant frequency. Then I would use R*Sqrt(C/L) formula to find the circuit Q. I am also including an interesting link.http://www.pronine.ca/qbw.htmRatch

author: MrAl
author_rank: Super Contributor
timestamp: March 20, 2017, 02:59:25 pm
content: Quote from: Ratch on March 19, 2017, 05:41:26 pmHi again Ratch,Well there is a difference between thinking and knowing. In one case you have proof, in the other case you still dont know. What i dont understand here though is how you could quote a passage that actually uses this concept and provides a useful result, then state that you are still thinking about if it means anything.Q=1/(1/QL+1/QC)is not resonant specific, and yet it provides a useful result. Yes it could be in a resonant circuit, but each individual Q is based on the frequency of operation which may or may not be at a resonant frequency. But even if we dont consider anything other than the resonant frequency we still are facing the concept of calculating the individual Q's at several frequencies, which we dont have to take as being any resonant point.To state this a little more clearly:Q(w)=1/(1/QL(w)+1/QC(w))however we may want to know say QL(w) for several frequencies.If we design a filter with two parts, each part may operate at a different 'resonant' frequency. Since they are both in the same circuit however, the overall Q during either resonant point will be of concern because that will be the Q of the whole circuit, and there will always be one part that is not operating at any resonant point even when one point is. This is partly why i suggested that you do a few numerical experiments and see what you can find out. If you arent sure about something, go ahead and try to find out for sure. Discovery can be as interesting as book learning, and often more.

author: Ratch
author_rank: Regular Contributor
timestamp: March 20, 2017, 05:44:06 pm
content: Quote from: MrAl on March 20, 2017, 02:59:25 pm

author: MrAl
author_rank: Super Contributor
timestamp: March 20, 2017, 07:12:24 pm
content: Quote from: Ratch on March 20, 2017, 05:44:06 pmHi,I think i understand now why you are not seeing this for it's simplicity.A quick analogy would be adding series resistors to get the total resistance:RT=R1+R2+R3Does it make sense to do this? Sure it does. But what if each resistor could somehow change in value with frequency? Still makes sense to do it:RT(f)=R1(f)+R2(f)+R3(f)but what stands out is that each resistance is independent of each other resistance, so we dont really have to keep adding all three if we know that two of them add up to say 5. If we know even one of them, we dont have to keep calculating the whole bunch. Say we only know R3 at 100Hz equals 10 ohms. Then we reduce it to:RT(100)=R1(100)+R2(100)+10If we knew that R2(100) was equal to 20, we could reduce this to:RT(100)=R1(100)+30Maybe this doesnt work out, so we try anther resistor to replace R1:RT(100)=R4(100)+30The whole point here is that we did not have to keep calculating the values of R2 and R3 over and over because they are independent of one anotherAs for the circuit with two parts and two resonant points, if we had the SEPARATE data for each part of the circuit, maybe we could combine them without having to calculate the Q for the entire circuit twice by means of doing the entire calculation over again. So if we knew that circuit 1 was 10 and the calculated circuit 2 as 20, we might be able to calculate the total Q for the two frequencies just knowing the Q at those two frequencies for both sections. This means we could combine sections without constantly calculating both parts, only the part that we changed.For a single inductor and capacitor, we have:QT=1/(1/QL+1/QC)and so we could try different C's for example, all the while knowing QL, and so we dont have to keep calculating the Q of the entire circuit with both elements.What could be done is simply calculate some Q's of some circuits, then combine them, calculate the total Q, then try to see if you can find a way to combine the individual Q's in order to arrive at the total Q.If we did this with the L and C we would end up with the formula above. It may be more complicated with more complicated sub circuits, but you could find that out. I realize it takes some work here, but that's what we do :-)We could also do a search :-)

author: Ratch
author_rank: Regular Contributor
timestamp: March 20, 2017, 08:38:59 pm
content: Hi MrAl,All right, let's try that idea out. Suppose we have two coils each with a Q of 10 at a given frequency. That means each coil has x amount of resistance and 10x amount of inductive reactance. We put them in series. The circuit then has 2x amount of resistance and 20x amount of reactance. That makes a total Q of 10 for the two coils in series. Now, you want me to figure out a universal formula that will give a Q value just by knowing only the Q of the coils? How about if they are in parallel? What if they are different inductance and Q values. Isn't it just easier to compute the Q values by resistance and reactance?Ratch

author: MrAl
author_rank: Super Contributor
timestamp: March 21, 2017, 07:29:14 pm
content: Quote from: Ratch on March 20, 2017, 08:38:59 pmHi again,That's true if you only have to do circuits that are constantly so different that you never can do anything else. But as i am sure you know, we dont always calculate things the same way.Power=I^2*RPower=I*ETwo different ways for the same thing.If you are this against it, dont bother. You wanted to know why we might consider this, so i told you. That doesnt mean that you MUST do it, only if you want to, that's all.

author: Ratch
author_rank: Regular Contributor
timestamp: March 22, 2017, 12:55:27 am
content: Quote from: MrAl on March 21, 2017, 07:29:14 pmI don't understand the point of your answer. I go with what is the simplest, easies,t and works. Are you suggesting I figure out the total Q for every circuit configuration I use a lot? OK, maybe. However, I don't use the same circuits many times. Something is always different.Ratch

author: MrAl
author_rank: Super Contributor
timestamp: March 22, 2017, 11:27:07 pm
content: QuoteThat is the problem and i'm tired of trying to explain this simple concept to you.