finding impulse response of the following holding circuit

Discussion in 'Homework Help' started by Amar_903, Sep 30, 2010.

  1. Amar_903

    Thread Starter New Member

    Sep 30, 2010
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    there is the problem
    find the impulse response of the following holding circuit:

    image is in the attached file
     
  2. Georacer

    Moderator

    Nov 25, 2009
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    The policy of this forum is that you, the Original Poster, must show his efforts (even if they failed) on solving the exercise first, in order to get help from the rest of the community.
    So please post your work and/or thoughts so far and we will be glad to help you!
     
  3. Amar_903

    Thread Starter New Member

    Sep 30, 2010
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    sorry i was not aware of this fact,

    i have solved some problems of impulse response in which we have provided with the input function, we change it in delta input and take laplace transform then multiply with the transfer function and take inverse laplace, but in this no input function is given, so is it any other method to solve this,

    i studied on the internet tht we can find impulse response by finding the step response of heavyside function, but m unable to understand this completely, plz do explain now
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    My recollection in this matter is that the Laplace Transform of the response of a system excited by an impulse is identical to the [Laplace] transfer function.

    If you are able to write the Laplace form of the transfer function [H(s)] then the impulse response f(t) will simply be

    f(t) = Inverse Laplace transform of H(s)
     
  5. Georacer

    Moderator

    Nov 25, 2009
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    Do you need to draw the response or get a formula for it? If a drawing is what you need, the solution is quite obvious:

    Suppose you give a dirac spike δ at the input at time 0. A the point g, at time 0 that dirac δ will arrive and at time t0 the negative dirac -δ will arrive. Finally, those two are integrated. Remember that the integral of a dirac spike is 1? Consequently the final graph will consist of a step from 0 to 1 at time 0, and a step down from 1 to 0 at time t0.
    Simple as that.
    P.S. I will look at the Trasfer Function too.
     
  6. Amar_903

    Thread Starter New Member

    Sep 30, 2010
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    i need a formula for it, t_n_k thanks for ur concern, but plz expalin more,
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Georacer already gave you some good clues .....

    So more info ....

    y(t)=\int{g(t) dt}

    so - [ignoring the interval from t=-∞ to 0]

    Y(s)=\frac{G(s)}{s}

    and

    g(t)=x(t)-x(t-t_0)u(t-t_0)

    where u(t-t0) is the delayed (Heavyside) step function.

    so

    G(s)=X(s)-e^{-t_{0} s}X(s)

    You need to find the overall transfer function H(s) where

    H(s)=\frac{Y(s)}{X(s)}
     
  8. Amar_903

    Thread Starter New Member

    Sep 30, 2010
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    thanks i understand that quite good, but our prof given us the solution which is attached i think he cut down many steps.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If you think about the solution your prof gave it is precisely what Georacer proposed to you in his last post.

    Clearly the prof didn't require you to set up the Laplace form of the transfer function.
     
  10. Georacer

    Moderator

    Nov 25, 2009
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    Here is a graph of the exercise. In the upper part there is the input of the integrator, in the lower is the integral, in case it's not obvious yet.
    impulses.png
     
  11. Amar_903

    Thread Starter New Member

    Sep 30, 2010
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    thanks sir, i got it
     
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