resonant frequency

Because an arbitrary wave can be broken into a sum of sine waves at different amplitudes and relative phases, it's easier to imagine the answer that way. A sine wave at one frequency is better approximated by a ... sine wave at that frequency ... as no surprise to anyone. So you can analyze it entirely at that single frequency and be done with it.

In the case of a square wave, what you need to know is what collection of sine waves can be conjured up to form it. Mathematically, this requires an infinite number of sine waves in various amplitude combinations at odd multiples of the fundamental frequency, if memory serves. Reality is different from this, of course, and a practical square wave is never truly actually equal to the mathematical picture of one. But you can analyze the circuit in terms of each continuent frequency that makes up the input square wave and then add up the results at the output to see the resulting wave and that works well.

Or, you can analyze the system by applying the filter description in the Fourier equation and working the integral to see what comes out that way.

Just quickly from scratch paper, I think a square wave going between

-1 and 1 works out to something like the following:

SUM [ (5/4)*(1/n)*SIN(n*w)/SQRT(1+(n/5)^2) ], n=1,3,5,7, ...

with w=2*PI*f

'n' is the odd multiple of the fundamental frequency.

Try it out and see.

Anyway, you are allowed to treat each frequency independently, analyze them individually, and then recombine them through superposition (simple addition back to a composite wave.)

Jon

To a degree, yes, because a square wave at the resonant frequency includes among its components a sine wave at the resonant frequency. It's not quite the same as a sinusoid of the same amplitude, because in the case of the square wave there's energy at the frequencies of the other components (which are multiples - in this case, odd multiples

- of the base frequency).

Any regular, non-sinusoidal wave is the equivalent of a series of pure sine waves at frequencies which are multiples of the base frequency, and so there's going to be SOMETHING going on at that frequency. It's just that the pure sine wave at a given rate is the only waveform where ALL of the energy is at that frequency.

Bob M.

What happens to a signal of which a certain frequency is amplified(say an ideal resonant "curve")?

If its just a pure sine way then that sine way gets amplified but if its a square wave, which contains all frequencies, then only some of the frequencies get amplified.

What happen is that the resonance sorta picks out the frequency of the signal. This is a sine wave that we get. Its not perfect but as the resonance becomes sharper we get more of a pure sine wave and less of the original signal. So think of the signal as "morphing" to a sine wave at the resonant frequency as the resonance gets stronger.

Mathematically we can think of S(w) as the signal in the frequency domain and then we are multiplying by a function like exp(-(w-w0)^2/q).

so we get

S(w)*exp(-(w-w0)^2/q)

No matter what S(w) is, as q->0 we get get a dirac(w-w0). When converted back into the time domain this is just A*sin(w0*t).

What you can think of is that the resonance "shrinks" as a band pass filter. What happens to a signal when you do this? You remove the lower frequencies and the higher frequencies. What happens when you do this to a square wave? Removing the higher frequencies makes it more into a sine like wave...

Better to look at what actually happens:

The fourier series of a square wave is

sum(sin(k*w*t)/k)

(doesn't matter about constants or harmonicity because we just want the general behavor. i.e., it doesn't matter if k is even or not in the sum above(using a sawtooth results in the same logic_).

Now what is this in the frequency domain? Its just a sum of impulses with frequencies kw.

If we bandpass that it means we are removing the lower and higher frequencies from the sum.

Essentially resulting in something like

sum(sin(kwt)/k,k=K1..K2)

(actually its a convolution with our filter but in the limit it works out to be something like this) where K1 and K2 get closer together and center around our resonant frequency.

Since the signal here contains all frequencies it will pick out one and be ok(although the signal could be quite complex).

But if you just have a pure sine wave you then might "miss" the sine wave and not amplify it at.

Basically the point is that all you have to do is think of a band pass for a resonance curve in this case.

If your interested in seeing how it works then just take a function, transform it in the frequency domain, filter it and then take it back into the time domain.

You can see that this type of analysis applies to all functions with continuous or discrete transforms. Just take your signal's fourier representation and convolve it with your filter. If you are looking at an "idea" resonance curve then it simplifies a great deal and essentially picks out the frequency at resonance. If its not a perfect filter then it picks out more frequencies around the resonance and so it can get quite complicated(looking nothing like the original single).

Theoretically if you have a pure sine wave and an ideal resonance curve you will not get any signal unless the resonance frequency is the frequency of the sine wave. This also happens in the square save if your resonance frequency is not an odd multiple the fundamental frequency. In real life things are rarely ideal and you cannot have an ideal resonance curve.

Anyways...

Jon

No. The square wave has harmonics, and they can affect the output. However. the effects of the odd harmonics might not be a problem in the case of a high-Q circuit. A high-Q circuit rejects/attenuates frequencies other than the resonant frequency (more so than a low-Q circuit).

yes, to a lesser extent so will a square wave at 1/3 the resoonant frequency or 1/5, 1/7 .....

Bye. Jasen