Applying a Periodic Input Given a Transfer Function

Discussion in 'Homework Help' started by chet47, Mar 9, 2012.

  1. chet47

    Thread Starter New Member

    Mar 4, 2012
    For my homework, I need to find the transfer function of a circuit and find the response of a periodic input. I found the transfer function to be:
    H(s) = Vo(s)/Vi(s) = 1000/(s+10)(s+100) = 1000/(s^2+110s+1000)

    I found the periodic input to be: u(t) - u(t - pi/2)
    because the plot shows Vi(t) as 1 from t = 0 to pi/2, and Vi(t) as 0 from pi/2 to pi, and so on

    I then rearranged H(s), so that
    Vo1(s) = (1000/s^2+110s+1000) * Vi1(s) (1st response from periodic input)
    Vo2(s) = (1000/s^2+110s+1000) * Vi2(s) (2nd response from periodic input)

    Now, the responses should look something like:
    Vo1 = (A + Be^-10t + Ce^-100t)u(t)
    Vo2 = -(A + Be^-10(t-pi/2) + Ce^-100(t-pi/2))u(t-pi/2)

    Now, is this the equation to find A, B, and C?
    A/s + B/(s+10) + C/(s+100) (for 1st response)

    I'm also confused about how to find A, B, and C for the second response, because I have:

  2. mwalden824


    Mar 6, 2011
    Your first question is yes, that is the equation to find A, B and C for the 1st response.

    The second question, you do it the same way using partial fraction expansion. When you set it equal to the expanded terms with the A, B, and C coefficients, you will factor out the exponential and there will be an exponential term on each side that will effectively cancel so you can solve for the coefficients just as you did in the first response.

    Here is a website that describes it better than I can...

    Scroll down to the very bottom of that website, and it has a very nice symbolic description of what to do when there is an exponential in the numerator.


    P.S. You also need to multiply [1000e^-s(pi/2)]/(s+10)(s+100) * (1/s) because the input for the second response is still a step function, just time-shifted.
    Last edited: Mar 10, 2012
  3. chet47

    Thread Starter New Member

    Mar 4, 2012
    Thanks for the help!
    This is what I found for the first response: A=1;B=-10/9;C=1/9
    and the second response: A=e^-(pi/2)s B=(-10/9)e^-(pi/2)s C=(1/9)e^-(pi/2)s

    So, now the responses look like:
    Vo1 = (1 + (-10/9)e^-10t + (1/9)e^-100t)u(t)
    Vo2 = [-1+(10/9)e^-110(t-pi/2)-(1/9)e^-1000(t-pi/2))e^-(pi/2)s] u(t-pi/2)

    Do you know how I could plot these responses of the system in matlab?