laplace transforms help

Discussion in 'Homework Help' started by t014y, Jun 8, 2010.

  1. t014y

    Thread Starter New Member

    Jun 8, 2010
    1
    0
    I'm studying for test tomorrow and i feel that i know how to do this but the numbers are not "nice" number so i wanna check it see if they are right. it 3 problems and I'm very sure the first one is right. from these i can check to see if the others are right. I'm going to put the question, circuit and my work for the answer (the circuits and graphs are attached.) thank you for you help.
    (i'm sorry if i uploaded a bunch of stuff)

    ---------------------------------------------------------------------------

    1) Sketch the given function and find it's Laplace transform.
    Code ( (Unknown Language)):
    1. f(t) = (8t-8)[u(t-1)-u(t-2)] + (24-8t)[u(t-2)-u(t-4)] + (8t-40)[u(t-4)-u(t-5)]
    this is what i got:
    (graph is in attachment not sure how it shows up)
    Code ( (Unknown Language)):
    1. L{(6t-8)u(t-1) - (8t-8)u(t-2) + (24-8)u(t-2) - (24-8t)u(t-4) + (8t-40)u(t-4) - (8t-40)u(t-5)}
    2.  
    3. L{8(t-1)u(t-1) + (24-8t-8t+8)u(t-2) + (8t-40-24+8t)u(t-4) - 8(t-5)u(t-5)}
    4.  
    5. (8e^-s)/s^2 + L{(-16t + 32)u(t-2) + (16t - 64)u(t-4)} - (8e^-5s)/s^2
    6.  
    7. (8e^-s - 16e^-2s + 16e^-4s - 8e^-5s)/s^2
    -------------------------------------------------------------------------

    2)the switch in the circuit has been closed for a long time. at t=0 the switch is opened. find v0(t) for t >= 0 using the Laplace transform.
    (pic is 2nd in attachment )

    first i found initial conditions across the capacitor and inductor.
    v = 0
    i = .06A

    then did the transformation to s domain using a voltage source from the inductor. (3rd in attachment)

    used voltage divider to find
    Code ( (Unknown Language)):
    1. (12s + 48.72)/(s^2+4.06s +8.33333)
    then partial fraction decomposition (PFD) yields the numerator to be
    Code ( (Unknown Language)):
    1. 6.23 <-94.024°
    in polar

    making the answer:
    Code ( (Unknown Language)):
    1. 12.46*e^(-2.03t)*cos(28.796t-94.024)u(t)
    -------------------------------------------------------------------------
    last one
    -------------------------------------------------------------------------
    3)there is no energy stored in the capacitors at the time the switch is closed. find I1(s), V1(s),V2(s), i1(t), v1(t), and v2(t)
    (circuit is 4th in attachment)

    first i added the parallel capacitors giving one of 40uF

    then i converted to s domain (5th in attachment)

    then i found I1(s) to be
    Code ( (Unknown Language)):
    1. .02/(s+20000)
    and i1(t)
    Code ( (Unknown Language)):
    1. .02e^-20000t
    then used I1(s) to find V1 and V2
    Code ( (Unknown Language)):
    1. V1(s) = .02/(s^2+s/2000)
    2. V2(s) = .02/(s^2 + s/500)
    then PFD on V1 and V2
    Code ( (Unknown Language)):
    1. V1(s) = 40/s + 80000/s+.0005
    2. V2(s) = 10/s + 5000/x+.002
    giving v1(t) and v2(t)
    Code ( (Unknown Language)):
    1. v1(t) = 40t + 80000e^-.0050t
    2. v2(t) = 10t + 5000e^-.002t
    -------------------------------------------------------------------------
    thank you :)
     
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