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resonant frequency

Discussion in 'Electronic Basics' started by [email protected], Feb 9, 2007.

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  1. Guest

    When a system is driven by a wave at its resonant frequency, how does
    the shape of the wave come into play? For instance, will a square
    wave at a resonant frequency have the same effect as a sine wave at
    that frequency?

    Thank you.
  2. 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

    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.)

  3. Bob Myers

    Bob Myers Guest

    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.
  4. 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


    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


    (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


    (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

    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.


  5. John Fields

    John Fields Guest

  6. 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).
  7. jasen

    jasen Guest

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

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