Laplace and Z Transformations Help

Discussion in 'Math' started by crazyengineer, Jun 7, 2011.

  1. crazyengineer

    Thread Starter Member

    Dec 29, 2010
    156
    2
    Hello everyone

    I need some help understanding the mistakes I made in the following problems

    1) In the following Laplace Transformation problem
    e^(-2t) [u(t)-u(t-5)]
    originally I thought the answer was
    [1/(s+2)]-(e^(-5s))/(s+2)
    but the actual answer is
    [1/(s+2)](1-e^(-5(s+2)))

    2) Can anyone explain why the zero transformation of z/(z-2)^2 is n2^(n-1)u[-n-1] when |z|<1?
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    For this first one, try the following ...

    e^(-2t) u(t) - e^(-2t) u(t-5)

    e^(-2t) u(t) - e^(-10) e^(-2t+10) u(t-5)

    e^(-2t) u(t) - e^(-10) e^(-2(t-5)) u(t-5)

    Now take the inverse transform using the time shifting property on the second term.

    1/(s+2) - e^(-10) e^(-5s) /(s+2)

    Simplify

    (1/(s+2))*(1 - e^(-5s-10) )

    (1/(s+2))*(1 - e^(-5(s+2)) )
     
    Last edited: Jun 7, 2011
  3. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I'm not sure what a zero transformation is. However, if I take the z-transform of n2^(n-1)u[-n-1] I get -z/(z-2)^2 with region of convergence |z|<2, which differs from your answer by a negative sign. So, I'm not sure if I'm understanding the question, or if I am, I may be making a mistake in the calculation.
     
    Last edited: Jun 7, 2011
  4. crazyengineer

    Thread Starter Member

    Dec 29, 2010
    156
    2
    for 1) how did you get e^(-10) e^(-2t+10) in e^(-2t) u(t) - e^(-10) e^(-2t+10) u(t-5)?

    for 2) I meant z transformation. The problems came from my schuam outline book on signals and systems
     
  5. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    So, for the first one, e^(-2t)=e^(0) e^(-2t)=e^(-10+10) e(-2t)=e^(-10) e^(+10) e^(-2t)=e^(-10) e^(-2t+10)

    So just play around with exponents via allowed rules.

    For the second one, I took the z-transform and got the same answer you posted, but with a minus sign. I can post this if you want, and you can try to figure out if I made a mistake or not.
     
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