# Find y(t), having u(t) and transfer function

Discussion in 'Homework Help' started by ellosma, Jan 9, 2014.

1. ### ellosma Thread Starter New Member

Jan 9, 2014
8
0
I have a exercise and i can't do a part of it. I've a transfert function G(s) = 1/(s+1) and u(t), the entering signal:

u(t) is
1+2t for 0<=t<1
t-1 for t>=1

I have to find y(t), the outcoming signal of the system.

I find, correctly, the signal y(t) about the first part of u(t), 1+2t:
y1(t)=2t + e^{-t} -1

First i use laplace transform on u(t), for have U(s) , only the first part
Second i know that Y(s)= G(s)U(s) so, i calculate Y(s) because i've G(s) and U(s)
Antitransform Y(s) for have y(t)

But now i don't know how to find y(t) for the second part, t-1. Could someone tell me how to find it, i'd like to comprend because it's not an homework but an exercise that i do for preparing me to an exam. Thanks!

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,522
717
Do the Laplace transform of t-1, the interval of integration is from 1 to infinity.

3. ### WBahn Moderator

Mar 31, 2012
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The "best" way to do it, unless it introduces too much complexity which it may or may not, is to take the Laplace transform of your u(t). Not the first part of it, but the whole thing.

So use unit step functions to write a single equation for u(t) and then take the Laplace transform of that one equation and proceed from there.

The alternative is to treat your signal as two different signals, one from 0 < t < 1 and a second from 1 < t <∞. Keep in mind that to do this, you will need to determine two sets of initial conditions -- one at t=0 and one at t=1.

A third way it to take the inverse Laplace transform of H(s) to get h(t) and then use convolution.

4. ### ellosma Thread Starter New Member

Jan 9, 2014
8
0
For solve the problem in the first way, that you say it's the best, how i could write the intere laplace transformation of the first and second part ?

I write u(t) as : 1+2t - (1+2t)*1(t-1) + (t-1)*1(t-1) and now i should apply the laplace transform for have U(s). But i don't know if it's correct :/ really Thanks for the help, i'm going crazy with this ex!

5. ### rsashwinkumar New Member

Feb 15, 2011
23
2
First i suggest you to plot your u(t) in time domain; Differentiate it w.r.t. time, twice and you will finally end up with some impulses. Then apply differentiation property and find Laplace transform of u(t);

Now you can easily determine the inverse laplace of U(s)*1/(s+1);

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Last edited: Jan 10, 2014
6. ### WBahn Moderator

Mar 31, 2012
23,858
7,382
Please. The first part of your post is just fine -- great, in fact. But then, after giving a good suggestion on a possible path to take, you then proceed to just give out the answer to that part instead of letting the OP struggle through it on their own for awhile first. That is frowned upon.

Last edited: Jan 10, 2014
7. ### rsashwinkumar New Member

Feb 15, 2011
23
2
Point noted.! Edited the post..

8. ### WBahn Moderator

Mar 31, 2012
23,858
7,382
Thanks! I've edited my post to also remove the answer from it.

9. ### ellosma Thread Starter New Member

Jan 9, 2014
8
0
Really Thanks, now i'm oggi to translate some of your advice because i'm italian and not so good in english. I've the complet solution of this exercise, but the first part i solved ti so i don't look the solution. But for the second part, in that i've problems , my book find the second part of y(t) writing
y(t) = 2t - 1 + e^{-t} - [ t - 1 + 2 -2e^{-(t-1)}]

I don't know how is the espression between [....] ?! Sorry for the question and really Thanks for the help , i'd like to comprend and after i'll do some ochre ex like this !!

10. ### WBahn Moderator

Mar 31, 2012
23,858
7,382
Please show your work as far as you were able to get. Then we can see exactly where and what concept is giving you grief and can help you understand that concept better. That will also help with the language issue because it will better focus on onto the same point of the process.

11. ### ellosma Thread Starter New Member

Jan 9, 2014
8
0

In the post i put the photo of that i wrote. I don't put the photo of the Two inverse laplace transform that i do but if you want i can send also them. The result of the first part is correct but not the second.. Thanks and sorry for my bad english!

12. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,522
717
Write out how you did second part on paper and post the picture of the paper.

13. ### ellosma Thread Starter New Member

Jan 9, 2014
8
0

I know that the error is that i study the first part, and the second as Two indipendent part but i should consider the second as the joined with the y(t) of the first,but i don't know how

Jan 9, 2014
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0

Apr 5, 2008
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16. ### ellosma Thread Starter New Member

Jan 9, 2014
8
0
Ok ,Sorry i posted the image on dropbox but now it should be ok

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17. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,522
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t->1/s^2 when the integral is from 0 to infinity. You don't have integral from 0 to infinity. You have integral from 1 to infinity. That is why this is wrong.

1->1/s when the integral is from 0 to infinity. You don't have integal from 0 to infinity. You have integral from 1 to infinity. That is why this is also wrong.

18. ### ellosma Thread Starter New Member

Jan 9, 2014
8
0
I know i'm a stalker, i'm really Sorry, but i don't know how to solve the problem. May e i had to use the convolution? how to join the y(t) of the first part with the u(t) of the second?

19. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,522
717
Do the integral.
u(t)=t-1, for $t\geq1$
$U(s)=\int ^\infty_1 (t-1)e^{-st}dt$

$U(s)=\int ^\infty_1 te^{-st}dt-\int ^\infty_1 1e^{-st}dt$

$U1(s)=\int ^\infty_1 te^{-st}dt$

$U2(s)=\int ^\infty_1 1e^{-st}dt$

Consult Table of Integrals.
Plug in infinity and 1 to get an actual answer for U1(s) and U2(s).
U(s)=U1(s)-U2(s)