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
    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
    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
    There is a rule of thumb that suggests


    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
    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.