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In "High-speed Digital Design - A handbook of black magic" p. 2,

The knee frequency for any digital signal is related to the rise (and fall) time of its digital edges, but not to its clock rate:

$$ F_{knee}=\frac{0.5}{T_{r}} $$

where

\$F_{knee}\$ = frequency below which most energy in digital pulses concentrates

\$T_{r}\$ = pulse rise time

What's the math and theory behind this equation? I know the max. frequency in a clocked circuit must greater than the clock rate, but why 0.5?

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  • \$\begingroup\$ Here's some on the math behind it on SI-LIST, freelists.org/post/si-list/3dB-or-Knee-Frequency,3 It's basically a conservative choice you use to consider what frequency band you will find "most" of your energy in for your signal. \$\endgroup\$ – Some Hardware Guy Nov 24 '14 at 16:21
  • \$\begingroup\$ I know the 3dB frequency, it's actually RC like filter response's half-power frequency. But the link doesn't talk 0.5 one in detail. \$\endgroup\$ – diverger Nov 24 '14 at 16:29
  • \$\begingroup\$ I think the 0.5 is just a way to be more conservative with your numbers. Here's someone who did a power point on it for Kansas state google.com/… \$\endgroup\$ – Some Hardware Guy Nov 24 '14 at 16:33
  • \$\begingroup\$ I found a useful link: referencedesigner.com/books/si/Knee-frequency.php Hope it can help you. \$\endgroup\$ – Hosea Oct 18 '17 at 10:28
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If you take a random digital signal with Tr as rise/fall time and plot the power versus frequency, you will see that the power falls off by 20dB/decade until about Fknee, where the power drops off by 40 dB/decade.

If you use 10%-90% rise/fall time definition, at Fknee you will have 6.8dB additional loss compared to the 20db/decade.

If you go for 3dB additional loss, the magic constant "0.5" becomes the more often used "0.35", so Fknee = 0.35/Tr.

If you go for 3dB additional loss and use the 20%-80% risetime definition, the constant becomes "0.22" instead.

Great if someone can find a Matlab/Scilab script to plot and show this :-)

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  • \$\begingroup\$ All your cases, are to change \$T_{r}\$, so why not choose "0.35", but "0.5", is it special than 0.35:)? \$\endgroup\$ – diverger Nov 24 '14 at 16:45
  • \$\begingroup\$ Very good question. The Howard Johnson book uses "0.5", most other places you see "0.35". In reality it does not matter much. It's mostly used to get a rough idea of the bandwidth needed to measure a signal. So if you have modern digital stuff with 200ps risetime you know you need about 0.35/.2 to 0.5/.2 or roughly 2 GHz BW for your scope - more is better, but that should be just about enough :-) \$\endgroup\$ – Rolf Ostergaard Nov 24 '14 at 16:52
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If you had a signal consisting of only a 0.5ns rise and 0.5ns fall then quite rightly you'd say "this signal has a period of 1ns". You'd probably then conclude that your signal (with a period of 1ns) has a frequency of 1GHz.

You could then assume that if rise times and fall times are equal F = 0.5/ rise (or fall) time i.e. F = 0.5/0.5ns = 1GHz.

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  • \$\begingroup\$ Just a triangle wave with equal rising/falling edge, right? So, it's a estimated value? \$\endgroup\$ – diverger Nov 24 '14 at 16:24
  • \$\begingroup\$ @diverger no, not an estimated value but a precise value and yes a triangle wave. \$\endgroup\$ – Andy aka Nov 24 '14 at 18:12
  • \$\begingroup\$ @Andy aka, sorry to downvote. But Fknee (as per the definition in the Howard Johnson book) has nothing to do with the fundamental frequency of the signal. \$\endgroup\$ – Rolf Ostergaard Nov 29 '14 at 6:39
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Lets look on perfect sinusoidal shape and on rise time definition. Rise time is between 10% and 90% of slope. In high speed signals rise time includes highest frequency of signal.

In typical sinusoidal shape rise time is about 33% to of period. If we know rise time of signal we can calculate its period (period of highest frequency in this signal): period= 3*rise_time.

F_knee = 1/(3*rise_time) = 0.33/rise_time

So F_knee is just a highest frequency in digital signal calculated from rise time or from fall time (use shorter one).

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