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Effect of zeros on impulse response

  1. Oct 5, 2013 #1
    The transfer function of a passive bandpass filter has one zero and two poles.

    The filter is:
    Signal -> L -> C -> R -> Gnd,
    where the signal is the input and the voltage across R is the output.

    [itex]H(s) = \frac{sRC}{s^2LC+sRC+1}[/itex]

    Initial value theorem states that it's impulse response has an initial value of R/L. How can this be? The system filters out high frequencies, so it should not be able to change its voltage instantaneously in response to an impulse.
     
  2. jcsd
  3. Oct 6, 2013 #2

    meBigGuy

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    Divide numerator and denominator by s and re-compute the initial value.
     
  4. Oct 6, 2013 #3
  5. Oct 6, 2013 #4

    meBigGuy

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    You are right. Has me stumped at the moment.

    The step response (Vin = 1/s) would give zero. Strange.
     
  6. Oct 6, 2013 #5

    meBigGuy

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    OK --- Since the impulse is infinite, there is finite current in the inductor at time 0.

    It has to do with the integral of the dirac delta function, which is the heaviside step function.
    If you go to the wikipedia page for rl circuit http://en.wikipedia.org/wiki/RL_circuit and look at the equation for the resistor current in the impulse response section, you can figure it out. The circuit topology is different, but by the logic you (and I) were applying you would expect the resistor current to be 0 at time 0, which it isn't.
     
  7. Oct 6, 2013 #6
    I don't see what was wrong with our logic, though. The current can't change instantaneously through the inductor. Prior to the impulse, the current is zero. Once the impulse occurs, it shouldn't be able to jump. It follows that the current in the inductor is equal to the current in the resistor, and thus the voltage should be zero.

    I see the math, and I've worked it myself both in this example and in the original and sure enough, there is a step function in the response. Does it make sense to you?
     
  8. Oct 7, 2013 #7

    meBigGuy

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    Yes. It makes sense that the sort of impulse that would have a laplace transform of 1 would be able to cause a current step function in an inductor. It doesn't exist in nature.

    An impulse contains all frequencies = 1, and contains infinite energy. It the limit you get a step function across an integrator.
     
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