Anyone understand resonant frequency in a LC network??

[fuzzy645]

Just trying to understand in "laymans terms" what the resonant frequency means in a LC network.  I guess I'm looking both in musical terms as well as technically.  Is it that the frequency that "resonates" the most, or rather is it the frequency that will be dipped?

The frequency should be:


...here L is the inductance in henries, and C is the capacitance in farads.

So if we had a .047 uf cap in series with a 1.5 h inductor, the math would be

1 / ( 2 * PI * sqrt( 1.5  *  0.000000047) )

...yiedling 599 hz.  

Like I said, what does this mean?  It is somewhat similar to the limit frequency concept (for example such as a high pass filter which allows frequencies higher than the limit frequency to pass and lower frequencies are rolled off).

Thanks!

[amptramp]

It will depend on whether the L and C are connected in series or parallel.  In series, the impedance is low only for the resonant frequency and high for everything else.  In parallel, the impedance is high for the resonant frequency and low for everything else.  The next question is how you connect it.  If you have an input going to the L and an output from the junction of L and C with the other side of C going to ground, you get a lowpass filter.  If the signal goes in on the C and comes out at the junction of C and L with the other side of the L grounded, you have a highpass filter.  If you have a series LC going to ground with a signal across it, it will pass all frequencies except the tuned frequency.  If you have a parallel tuned circuit to ground, you get a filter that only passes one frequency, and this is commonly used in radio / TV and communication channels where a signal must be tuned.  The tuning refers to changing the L or C to get the circuit to accept the correct frequency and reject all others.

[fuzzy645]

It will depend on whether the L and C are connected in series or parallel.  In series, the impedance is low only for the resonant frequency and high for everything else.  In parallel, the impedance is high for the resonant frequency and low for everything else.  The next question is how you connect it.  If you have an input going to the L and an output from the junction of L and C with the other side of C going to ground, you get a lowpass filter.  If the signal goes in on the C and comes out at the junction of C and L with the other side of the L grounded, you have a highpass filter.  If you have a series LC going to ground with a signal across it, it will pass all frequencies except the tuned frequency.  If you have a parallel tuned circuit to ground, you get a filter that only passes one frequency, and this is commonly used in radio / TV and communication channels where a signal must be tuned.  The tuning refers to changing the L or C to get the circuit to accept the correct frequency and reject all others.

Thanks for your reply.

I will then clarify my question.  Lets say its hooked up in series as follows follows then:

input from guitar ----> .047 cap --->  1.5 H  (connected in series) ---> ground

In this case you mentioned the impedance is low only for the resonant frequency and high for everything else, so the resonant frequency should (I hate to say it) "resonates" more than the other frequencies....correct?  You also mentioned if you have a series LC going to ground with a signal across it, it will pass all frequencies except the tuned frequency, which is consistent with the last statement.

So in summary, my initial statement that is "cuts" (or dips) the resonant frequency is incorrect.  Rather, it is quite the opposite. It "cuts" everything else above/below the resonant frequency (bleeds to ground) and leaves the resonant frequency in tact.  Sound right?

Thanks!!!

[R.G.]

1. Go read about impedances, on the web, and in the various discussions here, using the search function.
2. Capacitive impedance decreases with increasing frequency.
3. Inductive impedance increases with increasing frequency.

With that as background, a "perfect" low impedance source driving the capacitor -> inductor network you propose, sensed by a "perfect" high impedance probe at the top of the inductor, would see the following:
1. At very low frequencies, the capacitor is essentially an open circuit, the inductor a short circuit, so the output is nearly zero
2. At very high frequencies, the capacitor is essentially a short circuit, the inductor an open circuit, so the output is nearly identical to the input
3. At the frequency where the capacitor and inductor have nearly equal impedances, which is what gives you that LC equation, you would expect the signal at the junction of the two to be about half the input. And so it would be, excepting that the cap and inductor interact by resonance.
4. The resonance is caused by the L and C exchanging energy from their energy storage mechanisms. It causes a signal peak right at the resonant frequency.
5. How high the voltage at resonance gets is determined by the loading on the L and C. If L and C are perfect (i.e. no resistance to eat up some of the energy) the voltage at resonance goes to infinity. There is always some resistance in the real world, so the voltage is not infinite, but it can be many times the incoming source voltage. Or hardly higher at all, if there is a lot of loading in the inductor's inevitable internal resistance or the loading of whatever is attached across the inductor to look at the signal.

[fuzzy645]

1. Go read about impedances, on the web, and in the various discussions here, using the search function.
2. Capacitive impedance decreases with increasing frequency.
3. Inductive impedance increases with increasing frequency.

With that as background, a "perfect" low impedance source driving the capacitor -> inductor network you propose, sensed by a "perfect" high impedance probe at the top of the inductor, would see the following:
1. At very low frequencies, the capacitor is essentially an open circuit, the inductor a short circuit, so the output is nearly zero
2. At very high frequencies, the capacitor is essentially a short circuit, the inductor an open circuit, so the output is nearly identical to the input
3. At the frequency where the capacitor and inductor have nearly equal impedances, which is what gives you that LC equation, you would expect the signal at the junction of the two to be about half the input. And so it would be, excepting that the cap and inductor interact by resonance.
4. The resonance is caused by the L and C exchanging energy from their energy storage mechanisms. It causes a signal peak right at the resonant frequency.
5. How high the voltage at resonance gets is determined by the loading on the L and C. If L and C are perfect (i.e. no resistance to eat up some of the energy) the voltage at resonance goes to infinity. There is always some resistance in the real world, so the voltage is not infinite, but it can be many times the incoming source voltage. Or hardly higher at all, if there is a lot of loading in the inductor's inevitable internal resistance or the loading of whatever is attached across the inductor to look at the signal.

RG -  I appreciate the detailed reply you provided. I must identify myself as a major NOOB  and therefore understood very little of what you said....straight over my head.  Again, I appreciate the time you have taken.  I will certainly take you up on your advice read up on the concept of impedance.

Maybe I will rephrase my question in simple terms (which is all I know).  Hypothetically, lets say you fed a basic Tele with the bridge pickup selected measuring around 6K as input into this circuit.   Imagine the guitar has no volume pot, no tone pot, just the guitar pickup directly into this LC network and then straight out to the jack.  It is going to either cut (or perhaps resonate) some frequency for sure.  If we had a switch to listen on vs. off, the human ear would clearly hear some tonal difference (good or bad).     I completely understand that the idea of calculating such frequencies via a formula is likely flawed as their are many other variables if we wanted to be super accurate (such as input impedance etc...).  That being said, shouldn't this formula provided at least give us a "ballpark" figure?  It seems to me (and my NOOB ignorance) that "ballpark" figures are in fact useful in this engineering business.  For example, the terms "low impedance" and "high impedance" are used to make decisions - and the terms "low" and "high" are by definition approximations (bigger than a bread box type of thing)

So I guess my questions are:

1. does the aforementiononed formula at least give us a reasonable approximation of the frequency (bigger than a breadbox kind of thing)?
2.  if we then allow this # to be our "resonant" frequency (however flawed), what does this  mean in musical terms?  Would that frequency be more "pronounced" or rather would it dip? I would assume we can answer this question even if  turns out the formula is totally flawed.
3.  am I completely barking up the wrong tree (as a NOOB would do  )

Thanks again for your time.

[PRR]

Worth repeating:

1. Go read about impedances, on the web, and in the various discussions here, using the search function. Or good old books!
2. Capacitive impedance decreases with increasing frequency.
3. Inductive impedance increases with increasing frequency.


> It "cuts" everything else above/below the resonant frequency (bleeds to ground) and leaves the resonant frequency intact.  Sound right?

A series-resonant tank has "zero" impedance at resonance and higher impedance elsewhere; ultimately infinite impedance at zero and infinite frequency.

But draw the WHOLE circuit. Including source and load.

In this case the load is a ~~1Meg guitar amp input, and after checking I found it has little effect.

The source: if you "load" a _zero_ impedance ideal source with ANYthing, it is unaffected. But there are no zero-Z sources. Guitar has significant source impedance. The naked pickup is 5K+5H (and that 5H would interact with your 1.5H inductor). Nearly all axes have volume controls, and when not full-up or full-down the output is dominated by pot resistance. So there is a chance a "zero" impedance tank will be able to do something.

In fact the real-world series tank never goes to zero. All parts have parasitic resistance, and in audio coils it is often very signifcant. A small 1.5H coil may have 1K of pure series resistance. That's the lowest the whole tank will go.

Take the guitar as 50K. Start from DC toward bass. Assume coil Z is small at first. 0.047 is infinite at DC then drops. When does 0.047uFd become similar to 50K, and start to load-down the guitar? I like the Reactance Chart for this. I squint 65Hz. Response flat from DC to 65Hz then falling. At some high frequency the cap is near-zero but the coil impedance rises above 50K, and load-down stops. I squint 50K+1.5H as 7KHz.

So response is flat DC to 65Hz then drops, then rises and is flat again above 7KHz.

How does it drop? Well, when not near L-C resonance, it drops like a simple R-C and rises like a simple L-R. 6db/oct.

What happens in between? The L and C cross near 600Hz (yes, geo-mean of 65Hz and 7KHz).  For a perfect tank the drop is an infinite null. In many practical systems, the coil resistance dominates and limits the dip. Ass-uming 50K source and 1K in the coil, about a 50:1 dip.