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Lead-Lag Compensator: Band-Pass Filter

  1. Apr 12, 2012 #1
    My professor in Feedback Control Systems told me that a lead-lag compensator is basically a low pass filter and that you can't amplify high frequencies because it requires a lot of energy. Thus, you can't design a PID/lead-lag compensated transfer function which acts like a bandpass filter. Is this correct?

    Based on this assumption, I've come up with a theory: In my laboratory, I've built an infrared transmitter and receiver with an RLC as its bandpass filter. The receiver isn't working and no other engineers have been able to make it work. So according to what my professor told me, is this because the power supplied isn't sufficient? (If the voltage was any higher, the op-amps would burn up.)

    The circuit is shown in the website bellow;

    http://howcircuits.com/active-ir-motion-detector.html

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    I recently discovered something interesting that might support what my professor stated:

    "If you attempt to set ωCO1 to a higher frequency than ωCO2, the band-pass filter will block all frequencies, and no signal will get through."

    http://www.play-hookey.com/ac_theory/band_pass_filters.html [Broken]
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    Another proof that he is right:

    The lead compensator offers PD control. This causes it to speed up the response of a system.
    With a lead compensator high frequencies are amplified.

    The PI controller reduces high-frequency
    noise. As such, it can be used as a low-pass filter

    The lead-lag compensator combines the lead and the lag compensator.

    http://aerostudents.com/files/automaticFlightControl/adjustingSystemProperties.pdf

    So from this I assume it is just a compensated low-pass filter.
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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 12, 2012 #2

    jim hardy

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    that doesn't sound quite right.

    They have transfer function of form ## \frac {T1s +1 }{T2s+1} ##
    so can have ac gain depending on ratio of lead & lag times.

    The Bode plot shows how they'll respond to a frequency sweep.

    Here's an article on them:
    http://www.library.cmu.edu/ctms/ctms/extras/leadlag.htm [Broken]

    And here's a powerpoint show that has some nice graphics.
    It'll open with Microsoft's free Powerpoint viewer or OpenOffice.

    http://www.google.com/url?sa=t&rct=...w5WeCA&usg=AFQjCNGrLxwH3bEqljE0CW2RjzOBQwuHSg

    hope this helps. (I'm no expert)
     
    Last edited by a moderator: May 5, 2017
  4. Apr 12, 2012 #3
    Lead-leg compensators have different frequencies for the zero and the pole. Based on the difference of the two frequencies we get different static gains below the lower frequency and above the higher frequency.

    For example, a widely used PI compensator has a pole at the origin and a zero at a non-zero frequency. A lead-lag compensator cannot achieve zero steady-state error as it by definition has non zero pole location.

    Also, consider a lead-leg with zeros at 1 and 100000 radians and poles at 100 and 1000 radians. You will get a boost between the two zeros and attenuation elsewhere, which is pretty much a bandpass filter.

    The P, PI, PD, PID, lead and lag complement each other (some are special cases of other ones).
     
  5. Apr 12, 2012 #4
    So you are assuming that a high pass filter can be made with a lead-lag system?
    What about the energy requirements, is it very high?
     
  6. Apr 12, 2012 #5
    How about a lead compensator that has a zero at -1, a pole at -10 rad, and Kp = 0.1? (i like to use the Kp*(1+s/wz)/(1+s/wp) formula as it gives negative pole locations -> stable systems). Though any formula will work just fine.

    Now stack five of these systems together. The resulting system is a HPF with -100 dB attenuation at 1 rad and no attenuation at 10 rads (disregarding the curving effect). It's a high-pass filter.

    Regarding the energy inquiry, I do not really know the answer to that. Perhaps your teacher meant that to get high gains, one would need a physical system capable of producing high voltages/currents?
     
  7. Apr 12, 2012 #6

    jim hardy

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    connect a resistor and capacitor in parallel
    Place them as Zfeedback around an inverting op-amp

    Connect another resistor and capacitor in parallel
    place them as Zinput into same inverting opamp

    and you have built a lead-lag element.
     
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