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Observe a filter response to a unit step?

  1. Aug 14, 2012 #1
    Hi guys, what does it mean to observe a filter response to a unit step?

    The book gives a transfer function and I've recreated it in Proteus ISIS. Now it asks me to simulate a square wave as an input (i used a pulse generator), and then observe a filter response to a unit step, describing the results.

    I'm at a loss here - what exactly am I supposed to be describing.

    The output from my ISIS design is a square wave, but that's about it - what am I supposed to describe about that?

    Am I on the right track here or am I supposed to be doing something else?
     
  2. jcsd
  3. Aug 14, 2012 #2

    NascentOxygen

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    Hi Jaded1. A step response is the response to a transition of the input, a step up or a step down. But yes, you can use the edge of a pulse, but just make sure the pulse is wide enough that the filter's response to the rise has died out before the filter is subjected to the fall, or the responses will overlap and give a misleading appearance.

    Normally, when you subject a system's input to a sudden rise, the output tries its best to follow suit, but doesn't rise as steeply and sometimes shows an overshoot or ringing, yet neither of these imperfections were to be found on the input. So if this were a real circuit you would expand the settings on the oscilloscope — as though you are using a magnifying glass — to get a better look at the rise and corners on that output waveform. You'll find it's nowhere near as square as you think. :smile:
     
  4. Aug 14, 2012 #3
    Oh I see, so basically what you explained about the rising/falling slopes is what the book was implying when it says 'describe the filter response to a unit step'.

    I honestly wouldn't have know because this is what I got in ISIS:

    http://i.imgur.com/nRCKQ.jpg

    As you can see, it's a proper sqaure wave output.
     
  5. Aug 14, 2012 #4

    NascentOxygen

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    If you can't get an output different from the input, then there is something wrong with the calculations for your filter. It's not filtering!
     
  6. Aug 14, 2012 #5
    I created the following:

    OPsLow3order.png

    in ISIS. But you're right, my input/output are the same :frown: If I understand correctly what you explained above, I should be a getting a step response with a sloping rise and then a straight line ahead once it reaches the peak.
     
    Last edited: Aug 14, 2012
  7. Aug 14, 2012 #6

    NascentOxygen

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    If o/p equals i/p, then you've accidentally set the R's to 0 or very low, or C's to 0 or way too small. Though I can't be sure that you are using the oscilloscope function to best effect; rounded corners will always look square if you don't stretch out the time axis to display the rounding.

    Might be more straightforward to obtain its frequency response, and note where the gain starts to fall away. Does it agree with the designed response?
     
  8. Aug 15, 2012 #7

    NascentOxygen

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    Your input square wave is probably way too low in frequency. Keep increasing by factors of x10 the frequency of that input square wave until the output amplitude starts to fall and see how the corners look. If you designed the filter to have a bandwidth of 2kHz, give it a square wave of 2kHz and see what the output looks like.
     
  9. Aug 15, 2012 #8
    Yep thanks :smile:

    Sloping rise, overshoot, comes down to 1V and then remains constant at 1V until the fall, which is a sloping fall (not steep) and there is a dip before it comes back upto the 0V.
     
  10. Aug 15, 2012 #9

    uart

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    Hi Jaded. Your square wave is at a very low frequency of 1Hz, but many circuit could have transient responses (to the step) that are sub millisecond or even sub micro-second. You are probably looking on the wrong time scale for your circuit.

    Take a close look at your output trace at say 500ms. See how it undershoots by about 25 to 30 mV. This looks like a vertical line on your trace, but if you used say a 1kHz square wave so that the time scale was expanded, then you'd see some "ringing" (damped oscillations). This is the type of thing the book is asking you to describe.
     
  11. Aug 15, 2012 #10
    Thanks uart. Tried that and it does show on the analogue response differently to what I was doing before.
     
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