# Signal integrity (rise time and BW)

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
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On Jun 4, 7:08 pm, snipped-for-privacy@snet.net wrote:

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
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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?
Results 1 - 10 of about 116,000,000 for step response. (0.05 seconds)
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On 6/9/07 8:49 PM, in article snipped-for-privacy@corp.supernews.com,

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
--
Iraq: About three Virginia Techs a month

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Salmon Egg wrote on 10-06-07 13:11 :

Uh-huh... and that means what, exactly, to me?
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On 6/10/07 1:59 PM, in article snipped-for-privacy@corp.supernews.com,

Beats me.
Bioll -- Support the troops. Impeach Bush. Oh, I forgot about Cheney.
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Salmon Egg wrote on 10-06-07 18:08 :

Not a surprise, dimbulb.
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