online course - (Laplace Transform problem)

Discussion in 'Homework Help' started by notoriusjt2, Mar 2, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    I am in an online course and the questions/examples from the book are quite terrible. I have read the entire chapter and this is the first question from my chapter quiz. Could someone please post the basic steps(not the answer) needed to solve this problem as I am honstley clueless. I will then use those steps to try to solve the problem. Thank You
    [​IMG]
     
  2. loosewire

    AAC Fanatic!

    Apr 25, 2008
    1,584
    435
    Look up and study in wikipedia,a lot of information on your
    study.
     
  3. dachikid

    Member

    Oct 19, 2007
    16
    0
    You might want to start by determining the impedance of the circuit withn the time domain. From there it's just a hop-skip and a jump away to the Laplace transform.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    While it might be informative there's no need to do that.

    This is the process I would adopt.

    1. Convert all components to the 's' domain.
    An Inductor L becomes Ls
    A Capacitor C becomes 1/(Cs)
    A Resistor R stays as R
    2. Treat the circuit as if it is a series-parallel network and solve it using techniques you have applied to series-parallel resistive networks in the past.

    So then, that part of the circuit comprising the (1/2)F (or 2/s) capacitor in parallel with the 1Ω resistor would become

    (2/s)/(1+2/s) [Remember two resistors in parallel - R1*R2/(R1+R2)]

    To this you would add the (4/3)H or (4s/3) inductor term.

    So that entire branch is equal to [4s/3 + (2/s)/(1+2/s)]. That branch is in parallel with the lone (3/2)F or (2/3s) capacitor.

    It's just more fiddly than manipulating purely resistive terms because the algebra is complicated by the terms in 's' - but it's still just a process of careful algebraic manipulation.

    Have a go at doing the problem - even if you go wrong, someone can then help you.
     
    Last edited: Mar 2, 2010
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