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I had a question with regards to the decay rate of filters and how it responds to transmission speed.

Suppose we have implemented a simple BFSK system with two asynchronous receivers and we have a signal applied to one of the filters(doesn't matter which). Before another signal can be applied it seems the initial signals filter response needs to be sufficiently decayed before another signal can be transmitted.

How is this required time calculated? Are the filters drained of energy after each received transmission?

Thank you for the information.

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Decay time depends on the dynamic range of decay to be measured. Normally 90 to 10% rise or fall time, t is related to f -3dB bandwidth by t=0.35 * f-3dB.

If one considers a lower threshold for decay, then the decay time increases significantly.

When we have a BPF Q=fc/Δf(-3dB). Thus the envelope decay time is equivalent to behaves like the LPF response with a BW = 1/2Δf(-3dB)

Consider applying a DC step voltage to a series R+ L//C parallel resonant circuit. Let L= 1H, C= 250uF fo~10 Hz Z(10Hz)= 62.8 Ohms thus for a Q of 10 one chooses Rs=628 Ohms and the envelope decay bandwidth is the single side bandwidth of 5Hz . enter image description here

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A q=5 circuit has almost exactly 6dB/cycle (6.28 radians) of ringing.

For 1% inter-symbol-interference, you need 40dB/6dB = 7cycles of delay so the stored energy supports a voltage that is 99% attenuated.

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  • \$\begingroup\$ Q factor? Does this apply to all filter types? \$\endgroup\$ – FourierFlux Jan 1 '18 at 5:39

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