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How do I find Bandwidth from looking at Graph?

  1. Dec 18, 2012 #1
  2. jcsd
  3. Dec 18, 2012 #2
    You have to convert the magnitude scale into dB

    dB = 20*log10(magnitude)

    Look at where it is -3 dB. That is your cutoff frequency (fc).

    Your bandwidth in this case will be 2*fc. Read more on http://en.wikipedia.org/wiki/Bandwidth_(signal_processing [Broken])

    EDIT: Note however, that it is only a convention, that the cutoff frequency is -3 dB. It is something we have defined.
    Last edited by a moderator: May 6, 2017
  4. Dec 18, 2012 #3
  5. Dec 18, 2012 #4
    I thought BW is fc in this case. In this case, it looks like the -3dB is about 100KHz. The bandwidth is 100KHz as it is from 0Hz to 100KHz. I don't think it is 2fc = 200KHz.
  6. Dec 18, 2012 #5
    Yes, I think you are right. The circuit for this graph is a non-inverting amplifier made from a 741 op-amp. Op-amp performance is limited at high frequency; it behaves as a low-pass filter.

    "In case of a low-pass filter or baseband signal, the bandwidth is equal to its upper cutoff frequency."
  7. Dec 18, 2012 #6
    EDIT: I see, the article says it all :) My fault.

    You might be right, but I seem to remember that we have to include the imaginary components also.

    The frequency response is symmetrical around f = 0, for this case. And using modulation, the signal would be positioned at a frequency f = f0 (where f0 is the carrier wave), with a symmetrical shape like the picture around this frequency. The bandwidth of the signal would then be 2*fc.
  8. Dec 18, 2012 #7


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    Staff: Mentor

    In my experience, for a LPF like that, the bandwidth is just fc. If it were a BP filter, then the bandwidth extends to both -3dB frequencies.
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