# Rise Time vs Harmonics Reduction

Discussion in 'Homework Help' started by Bill B, Sep 28, 2014.

1. ### Bill B Thread Starter Active Member

Nov 29, 2009
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0
I have a problem I have been trying to figure out for several days now.

If the peak signal energy for a 10 MHz signal with a rise/fall time of 1 nS begins to drop at
40 dB/decade above 320 MHz, how much will the peak energy around 800‐900 MHz be reduced
when the rise/fall time is increased to 10 nS?

This is one of those questions that really wasn't covered in the class, but is related to the lesson. I just can't see the connection. I'm not looking for the answer, I'm looking some direction on how to work the problem. Any help would be greatly appreciated.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,071
Are you taking a DSP class? You need to be able to do FFTs of waveforms to answer this question (or have access to a Spectrum Analyzer).

Or you can use this simplification:

Last edited: Sep 29, 2014
3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
There is a rule of thumb that suggests

Bandwidth=K/tr

Where K is an heuristic value and tr is the rise time.

If a rise time of 1 nsec equates to a 320MHz bandwidth then a rise time of 10 nsec presumably gives a ten-fold bandwidth reduction to 32MHz.

From 32MHz to 900MHz at a 40dB/decade signal attenuation one would guess the attenuation would be about 1.45[decades]*40 dB or 58 dB.

For a first order system [the usual assumption for the rule of thumb] the drop-off per decade would presumably be -20dB which is somewhat inconsistent with the claimed -40dB / decade drop off.

Check this out ....

http://www.k-state.edu/ksuedl/publications/Technote 2 - Bandwidth and Risetime.pdf

Playing around with a pure double [and equal] pole 2nd order system the bandwidth looks to be roughly 530MHz per nanosecond of step input rise-time. The roll off would be at -40dB / decade with a double pole system.

Last edited: Sep 29, 2014
4. ### Bill B Thread Starter Active Member

Nov 29, 2009
61
0
Thanks so much for the help. I was mistaken in thinking this was a harmonics related question. It is a filtering question. The link you both provided (it was the same link) was very helpful. t n k actually provided the solution (58 dB). Working back from that solution I was able to figure out how it was derived and I now have a clear understanding of the problem. Thanks again. You were both a huge help.

Bill B.