# 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.

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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:

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...

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.

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