Waveform in relation to Resonance

Thread Starter

Guinness

Joined Dec 31, 2009
81
Hi,

I was wondering, does the waveform have to be a pure sinewave for the following formula's to work?







Or could you use any repeating waveform, like a square, triangle, or a close but not really there sinewave?
 
Strictly speaking, yes, only for sine waves. But Fourier tells us that all periodic waveforms can be made from a harmonic series of sine waves*, so if it helps you could think of a square wave, say, as a dominant root sine wave with an ever-diminishing series of odd (sine wave) harmonics.

The best understanding comes from looking at these relationships in the time domain**, but that's not an easy path. Try a few LRC SPICE sims (don't forget the R or SPICE may fall over) with various waveform inputs and probe the input currents and output voltages, that should give a bit of a feel for it.

*Assume that Gibb was just being awkward at this point.
** Nature doesn't do frequencies, that's a convenient construct, it's all really down to rates of change.
 

studiot

Joined Nov 9, 2007
4,998
I don't know what your level of maths is, but the equations you refer to come from solution of the differential equation governing voltage in a cicuit, in relation to the inductance, capacitance and resistance.

We equate the forcing or driving voltage to these using Lenz law, Ohms law and Coulombs law and solve.

The forcing voltage can be a sine wave, square wave or whatever, each leading to a different solution.
 

Thread Starter

Guinness

Joined Dec 31, 2009
81
I don't know what your level of maths is, but the equations you refer to come from solution of the differential equation governing voltage in a cicuit, in relation to the inductance, capacitance and resistance.
Well lets just say, im not going to be a world leading maths teacher, but im ok to understand how to use the equations used in electronics, just about.

The forcing voltage can be a sine wave, square wave or whatever, each leading to a different solution
Thats the bit im trying to understand, im getting that there is a resonant frequency, even if I use a none perfect sinewave. But it will be a different frequency than what you would get if it was a sinewave?

If thats the case, then I guess the best way would be for me to make a simple circuit and scope it out.
Which I think is what Holdstock was saying, sorry I got a bit lost in your post.
 

studiot

Joined Nov 9, 2007
4,998
I've attached a worksheet showing how your formulae are derived from the differential equation for a sinusoidal voltage, as outlined in my first post.

If you replace the Vg = Vsin(2∏fT) with your chosen alternate function you can get an exact solution by solving the differential equation.

I warn you this may be difficult which is one reason we use sine functions. We generally use Laplace transforms for this purpose.
It can be easier if you can express your function as a sine series as this leads to a Fourier series solution or even a sine integral as you can use the Fast Fourier Transform.

Hope this helps.
 

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Thread Starter

Guinness

Joined Dec 31, 2009
81
studiot - thank you very much for the worksheet you attached. At first glance it doesn't seem to bad to understand.

Cheers for the info.
 

studiot

Joined Nov 9, 2007
4,998
Hopefully you can see how you assemble Ohms law, Lenz law and Coulombs law to form a (differential) equation and solve it.
 

KL7AJ

Joined Nov 4, 2008
2,229
Hi,

I was wondering, does the waveform have to be a pure sinewave for the following formula's to work?







Or could you use any repeating waveform, like a square, triangle, or a close but not really there sinewave?

The formula only works for a sine wave.

However, ANY repetitive waveform can be decomposed into individual sine waves and harmonics, each of which can be described by the resonance formula.

Hope this helps.

eric
 

studiot

Joined Nov 9, 2007
4,998
The formula only works for a sine wave.
Actually this is only partly true.

If you pass into (excite) a resonant circuit another waveshape at (near) the resonant frequency as calculated above you will see sine waves out.

This is inherent in the solution of the second order differential equation I outlined above.
 
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