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Don Pearce wrote: The frequency of the ring is determined by the LC circuit that is being struck by that edge. ** So a linear circuit with a resonance creates a new frequency ? Does not do so with a sine sweep. You bet it does! You get close to the resonance and it kicks off... you don't need to be right on the resonance, you only need to be in the ballpark. This is the basic principle that makes bandpass speaker enclosures do what they do. You don't need to put that one bass note into it... any bass going into it will come out as that one note. --scott -- "C'est un Nagra. C'est suisse, et tres, tres precis." |
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