Finding Laplace transforms

Thread Starter

Kayne

Joined Mar 19, 2009
105
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
 

Attachments

Ghar

Joined Mar 8, 2010
655
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:

Thread Starter

Kayne

Joined Mar 19, 2009
105
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.

Thanks for your help.
 

Ghar

Joined Mar 8, 2010
655
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.
 
Top