Finding Laplace transforms

Discussion in 'Homework Help' started by Kayne, Mar 19, 2010.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Hi all,

    I have the equation G(s)=C(s)/M(s)=5e^-(ts)/s+5 and I have to find the open loop response c(t) with a unit step which is 1/s


    I have 2 attachments that are the answer to this question, if anyone can let me know if i am correct or incorrect that would be great.

    Thanks for your time
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    You made a typo in your post here: there shouldn't be a 't' in the transfer function.

    Edit:
    Mistakes of mine... correcting...

    Edit 2:

    Ok, sorry about that, maybe no one read it anyway.

    Your answer isn't right.
    First of all, this is the one sided Laplace meaning your time domain answer starts at t = 0, but in this case, it actually starts somewhere else. Exponentials in Laplace are time shifts in time domain.
    Look at "time shifting" in this table.
    http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems

    You can take out the e^-s term and do the problem ignoring it, then bring it back as a time shift as the last step.

    The partial fractions method you used doesn't work in this case. Try adding your A and B terms back together and you won't get your original function. It only works in certain situations and I don't remember which right now.
    You also made a mistake, you plugged in -5 but left e^-5, when it was e^-s originally. Then you still calculate e^5 and got 148...

    The answer I get is:

    c(t) = 1 - e^-5(t - 1), for t >= 1
     
    Last edited: Mar 19, 2010
  3. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Thanks for the reply, I should have read the question before submitting it so there were no problem. I am not sure if you can or not but are you able to show me the workings to you answer. I see how to use the time shift but the answer that i am getting is nothing like yours maybe i am doing something else incorrect. This is is what i can get out.

    [​IMG]

    where u(t) is the Heaviside step function, and the T is the constant 't' in my transfer function.

    Thanks for your help.
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    That's the same answer, I just assumed T = 1 and wrote it as t >= T (and omitting = 0 for t < T) rather than putting down the step function.

    I should learn to use Tex... not used to having it on forums.
     
    Kayne likes this.
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