BW = .35/tr
Why doesn't the voltage step size of the signal influence the BW calculation?
A 0-10V signal with a rise time of 10nS has the same BW of a 0-1V with a 10nS rise time. This seem counter intuitive since the 10V signal will rise to 1V in 1nS. Which is much faster?
Thanks
the bandwidth calculation is done by comparing the maximum of the magnitude of the fourier transform, to the place on the spectrum in which its max/sqrt(2)...
its by virtue of the definition. the 10V step you describe will have more energy concentrated at w=0, but it falls off as you go higher in frequency at the same rate as the 1V signal.
john
snipped-for-privacy@snet.net wrote on 04-06-07 19:08 :
Have you tested such a circuit? The relationship is based on similarity to a single-pole response. If the equivalence is accurate, then the circuit will always consume a finite number of time constants ("R-C") to change the output from 10% to 90% of a final value, as long as equivalent R and C do not change.
For your 1v/1nS calculation with the 10V signal, what is the final voltage value? Is that really the same as the 1V/10nS example?
Google:
Results 1 - 10 of about 116,000,000 for step response. (0.05 seconds)
Realize that most simple (no nonlinear elements) circuitry follows superposition. The bandwidth of each of the 10 one volt steps is the same as any other. A ten volt step is just 10 one volt added together. You do, however need 100 times the power to drive a 10 volt step compared to a one volt step.
Bill
Salmon Egg wrote on 10-06-07 13:11 :
Uh-huh... and that means what, exactly, to me?
Beats me.
Bioll
-- Support the troops. Impeach Bush. Oh, I forgot about Cheney.
Salmon Egg wrote on 10-06-07 18:08 :
Not a surprise, dimbulb.
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