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Design >> Analog Design >> connection between bandwidth and step response?
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Message started by Marc Murphy on Sep 13^th, 2005, 5:58am

Title: connection between bandwidth and step response?
Post by Marc Murphy on Sep 13^th, 2005, 5:58am

Hi all, I was wondering if there is a rigorous way of connecting the frequency response to the step response? I've seen it mentioned in The Art and Science of Analog Circuit Design where it links high frequeny parts of the freq. resp. to the early events and low to late.

They also say looking at the first 10 ns of the step response conveys information about from 1/10ns=100MHz and up. Can anyone elaborate on the links?

What is the risetime I can expect for a given 3dB bandwidth? Stuff like that...

Title: Re: connection between bandwidth and step response
Post by Jess Chen on Sep 13^th, 2005, 9:28am

To derive a rigorous relationship between rise time and bandwidth, I believe you must assume something about the system. If you assume the system is dominated by a single first order pole, the relationship is

bw = .35/tr

where rise time is the 10 to 90 % rise time and if memory serves, bw is in Hz.

There's also the uncertainty relationship from Fourier theory between bandwidth and the width of the impulse response. I suppose you could argue that if you integrate the impulse response you have the step response, such that the width of the impulse response equals the rise time.

-Jess

Title: Re: connection between bandwidth and step response
Post by Paul on Sep 14^th, 2005, 12:50am

Hi Marc,

from T.H.Lee's Design of CMOS RF Integrated Circuits:
"In general, the bandwidth-risetime product will range between 2 and 2.2 if the system is reasonably well-damped (or more precisely, if the impulse response is unipolar so that the step response is monotonic [...])." 2nd edition, chap 8.5.5.
Notice that this expression uses the bandwidth in rad/s, not in Hz. If you divide by 2Pi, you get the 0.35 value in Jess' post. Lee also has some math and intuitive explanations on this.

Paul

Title: Re: connection between bandwidth and step response
Post by Marc Murphy on Sep 14^th, 2005, 4:43am

Thanks Paul...I have yet to read that one...my textbook budget is tapped out right now, but I've got a ton of the analog circuit books.

Title: Re: connection between bandwidth and step response
Post by Jess Chen on Sep 14^th, 2005, 10:59am

The relationship I posted is actually fairly straightforward to derive. The normalized step response of a first order system is

f(t) = 1-exp(-t/tau)

Using the 10% and 90% points,

.1 = 1-exp(-t1/tau) and
.9 = 1-exp(-t2/tau).

Or

.9 = exp(-t1/tau) and
.1 = exp(-t2/tau).

Taking the natural log of both sides and solving for t2-t1 gives

trise = t2-t1 = tau*ln(9).

Since tau = 1/(2pi*bw), where bw is in Hz,

trise = ln(9)/(2pi*bw) = 0.35/(bw).

-Jess

Title: Re: connection between bandwidth and step response
Post by rfzingle on Sep 14^th, 2005, 4:04pm

It is important to understand that all these formulae are correct only for a first order system with single pole only. As soon as the system gets more complicated, all these equations fall apart. However it is failry simple to derive for each system. All what you need to do is find the laplace domain outptu response of your system for a step input. Then take the inverse laplace of that eqn to get the time domain. Once you have the time domain equn you can find the rise-fall time. Toms book does cover it for first two order system I beleive...

...

Title: Re: connection between bandwidth and step response
Post by wued on Sep 21^st, 2005, 5:32pm

rfzingle wrote on Sep 14^th, 2005, 4:04pm:

And all of these are just for small signal