Transfer functions at 0 frequency

Discussion in 'General Electronics Chat' started by midnightblack, May 31, 2012.

  1. midnightblack

    Thread Starter Member

    Feb 29, 2012
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    0
    Hi.

    I was wondering how do you work out the gain at 0 frequencies for transfer functions.

    Say I have a transfer function

    Y/X= (jωRC)/(1+jωRC)

    The high frequency asymptote is at 1/1=1
    The low frequency asymptote is at jωRC/1=jωRC

    When ω->0, jωRC->zero. This I understand.

    But when putting on a bode plot, we can't take the logarithm of zero?

    So how do you find out where it touches the y axis where ω=0?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You can't represent that particular condition on a Bode plot.

    Why do you want to know this?
     
  3. Papabravo

    Expert

    Feb 24, 2006
    10,154
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    Zero frequency does not exist on a Bode plot because the horizontal axis is logarithmic as you have already noted. To get and answer that makes more sense you should multiply the numerator and denominator of the transfer function by the complex conjugate of the denominator. With a real number in the denominator the limit as ω approaches 0 should be more obvious.

    I'll try it and get back to you.

    I was thinking of using L'hopital's rule but it does not apply here.
     
    Last edited: May 31, 2012
  4. steveb

    Senior Member

    Jul 3, 2008
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    While the zero frequency is not on the Bode plot, one can easily identify the response at zero if one knows how to read a Bode plot. The only caveat is that you must plot down below the lowest poles and zeros. If there are poles and/or zeros at a frequency of zero, you can still figure out the response, but obviously you can't plot down that far.

    Keep in mind that pole/zero cancellations, or even nearly canceled pole/zero occurrences are not obvious on a Bode plot.

    The whole idea of the Bode plot is to be able to figure out the general response information from the plot itself. This means that all relevant information is there. Note that pole/zero cancellations are not discernible, but that is extraneous information when it comes to the actual magnitude and phase responses.

    For example, lets say the system has an integrator that places a pole at zero. It will be clear that the magnitude response is rising at 20 dB/decade as you move towards zero. The phase response will be approaching -90 degrees.

    If there is a zero at zero, then you will see decreasing magnitude at you approach zero and the phase will approach 90 degrees.

    There can be multiple poles and or zeros at zero frequency, and then it is the net number of poles/zeros (after all cancellations) that determine the response. You might see plus and minus 20, 40 60 db/decade slopes on the plots, and integer multiples of 90 degrees.

    If there are no poles and no zeros at zero frequency, then the phase will go to zero and the magnitude will stay flat, at whatever value shown at the low frequency end of the plot.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The function mentioned in post #1 doesn't present any unusual challenges.

    The OP points out quite rightly that the high frequency assymptote is to unity - or 0 dB. Moving in frequency below the pole breakpoint [-3dB @ ω=1/(RC)] the gain must decrease along an assymptote of -20dB per decade.

    In any event, if in doubt, one can cross-check the gain and phase at any frequency ω on the constructed Bode plot by doing a direct calculation of magnitude & phase from the known transfer function. This was implicit in the OP's comment about the high frequency gain value well above the pole frequency approaching unity value or 0dB.
     
  6. midnightblack

    Thread Starter Member

    Feb 29, 2012
    31
    0
    The mistake I was making was that I incorrectly read axes on a question. ω was actually going down as 0.1,0.001,0.0001. It seemed as though the bode plot touched the y axis on this particular diagram.

    But since I made that mistake, I understand it better anyways.

    Anyway, thank you for your replies. It helped me understand this more.
     
    Last edited: Jun 2, 2012
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