Laplace Transform Question

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• Aug 25th 2010, 05:33 PM
Khonics89
Laplace Transform Question
Response of an dampened system to a single square wave.
y'' + 4y = r(t); y(0) = 0; y0(0) = 0 :
where r(t) = 1 if 0 < a < t < b and 0 otherwise.

Find the condition on a and b such that the system is at rest when
t > b

I found the equation for this system, and it goes something like this

I change r(t)=H(t)=1/s

Then I let L(y(t))=F(s) Laplace Transformation

Then F(s)=1/s*(s^2+4) I manipulated this to get

1/4s -1/4 (s/(s^2+4)) then changing it back to in terms of t yields

1/4 -1/4 cos(2t)

So I don't know how to go about in finding out the conditions for a,b !!

• Aug 25th 2010, 09:52 PM
CaptainBlack
Quote:

Originally Posted by Khonics89
Response of an dampened system to a single square wave.
y'' + 4y = r(t); y(0) = 0; y0(0) = 0 :
where r(t) = 1 if 0 < a < t < b and 0 otherwise.

Find the condition on a and b such that the system is at rest when
t > b

I found the equation for this system, and it goes something like this

I change r(t)=H(t)=1/s

Then I let L(y(t))=F(s) Laplace Transformation

Then F(s)=1/s*(s^2+4) I manipulated this to get

1/4s -1/4 (s/(s^2+4)) then changing it back to in terms of t yields

1/4 -1/4 cos(2t)

So I don't know how to go about in finding out the conditions for a,b !!

Why do we not see the LT of a square pulse from a to b any where in this?

CB
• Aug 25th 2010, 10:44 PM
Khonics89
Huh??

I'm confused!

I did this if 0<a<t<b then 0<a<t

and b<t ??

Poster above, can you please be more clear.

Thanks
• Aug 26th 2010, 12:15 AM
CaptainBlack
Quote:

Originally Posted by Khonics89
Huh??

I'm confused!

I did this if 0<a<t<b then 0<a<t

and b<t ??

Poster above, can you please be more clear.

Thanks

What do you mean by that, you have:

$r(x)=\begin{cases}
1&, 0 0&, \text{otherwise}
\end{cases}$

and:

$y''+4y=r(t)$,

with initial conditions $y(0)=0$, $y'(0)=0$

Take Laplace transforms:

$s^2Y(s)+4Y(s)=R(s)$

where $R(s)$ is the LT of $r(t)$ and is in every table of LTs. Then:

$Y(s)=\dfrac{R(s)}{s^2+4}$

which you will then have to find the ILT of (hopefully by table look-up)

CB
• Aug 26th 2010, 12:22 AM
CaptainBlack
Quote:

Originally Posted by Khonics89
Response of an dampened system to a single square wave.
y'' + 4y = r(t); y(0) = 0; y0(0) = 0 :
where r(t) = 1 if 0 < a < t < b and 0 otherwise.

Find the condition on a and b such that the system is at rest when
t > b

I found the equation for this system, and it goes something like this

I change r(t)=H(t)=1/s

Then I let L(y(t))=F(s) Laplace Transformation

Then F(s)=1/s*(s^2+4) I manipulated this to get

1/4s -1/4 (s/(s^2+4)) then changing it back to in terms of t yields

1/4 -1/4 cos(2t)

So I don't know how to go about in finding out the conditions for a,b !!

What do you think the material above highlighted in red means?

1. That is not how you write out that 1/s is the LT of r(t)

2. 1/s is not the LT of r(t), you need to look up the required LT in a LT table.

CB
• Aug 26th 2010, 07:51 AM
Khonics89
Thanks for the clarification.

I arrived at something like this...

r(t) in terms of heaviside funtion is H(t-a)-H(t-b)

If that's the case then F(s)=1/s(s^2+4)(e^-as-e^-bs)

I'll go find the ILT.. then hopefully get the conditions for a,b ?

What do you think?

Thanks
• Aug 26th 2010, 07:54 AM
Ackbeet
Careful with your parentheses. That looks right. That is,

$\displaystyle{F(s)=\frac{e^{-as}-e^{-bs}}{s(s^{2}+4)}.}$
• Aug 26th 2010, 08:07 AM
Khonics89
Yeah, sorry I don't know how to use Latex on this website.

Just a quick note, after I find the inverse laplace using t shifting etc.. I should be able to say what my a and b's should be right?
• Aug 26th 2010, 08:10 AM
Ackbeet

1. Your system isn't actually damped, because there's no first derivative in the original DE. That's just a side observation.
2. In order to find a condition on a and b, you should try to form a mathematical equation from the statement "...system is at rest when
t > b." It strikes me that "being at rest" says nothing about where the system is (i.e., y, assuming y is a coordinate), but, instead, about what?
• Aug 26th 2010, 06:39 PM
Khonics89
I found my y(t) function which is

{H(t-a)-H(t-b)+cos(2(t-b))H(t-b)-cos(2(t-a))H(t-a)}/4

So now to find the conditions, would mean y'=0 ? but wouldn't for t>b my heavi side function would equate to 0 ? hence my y(t)=0 ???
• Aug 26th 2010, 06:54 PM
Ackbeet
Take the derivative and see what you get. The derivative of the Heaviside step function is the Dirac Delta function. All you really need to know about it for this application is that it's zero pretty much everywhere except where its argument is zero. So you can basically cancel any terms that are multiplied by this function.
• Aug 31st 2010, 07:07 PM
Khonics89
@ Ackbeet : Then I get a situation where for all t>b I get my delta(t-a) or Delta(t-b) = constant...

My equation for y'(t)=1/4{d(t-a)-d(t-b)-2H(t-b)sin(2t-s2b)+d(t-b)cos(2t-2b)+2H(t-a)sin(2t-2a)-d(t-a)cos(2t-2a)

now I know that y d(t-a)=d(t-b)=K K is any integer, also that H(t-b)=0 if t>b refer above my limts..

hence I get

d(t-b)cos(2t-2b)-d(t-a)cos(2t-2a)=0

I know my cosine function will only give me t=pi/4+a or b

but then again my delta function will give me d(t-b)=0 hence t-b=0 only if t=/=b, same applies with a.

So my question, what are the conditions for a,b such that the system is at rest when t>b..

is it when a=t-pi/4 or when b=t-pi/4 ??

or simply every where except t=/= a,b ?

Thanks
• Sep 1st 2010, 03:07 AM
Ackbeet
Hmm. I'm not sure I follow all that reasoning there. I agree with your derivative. So, we're looking at t > b. That means d(t-b) = d(t-a) = 0. Hence, in this region, the derivative simplifies down to

y'(t)=1/4{2H(t-a)sin(2t-2a)-2H(t-b)sin(2t-2b)}, t>b. (Incidentally, you forgot the curly brace on the end. *ahem*). One slight simplification more:

y'(t)=1/2{H(t-a)sin(2t-2a)-H(t-b)sin(2t-2b)}, t>b. Now, if t>b, then t>a. Hence, both Heaviside step functions are unity. Thus, we write

y'(t)=1/2{sin(2t-2a)-sin(2t-2b)}, t>b.

We need this to be zero for all t>b. How do you suppose we can do that?
• Sep 1st 2010, 04:50 AM
Khonics89
Well I know y'(t)=0 because we want the system to be at rest.. which leaves us with

sin(2(t-a))-sin(2(t-b))=0

I can use this identity sina-sinb=2sin(a+b/2)cos(a-b/2)
List of trigonometric identities - Wikipedia, the free encyclopedia Sum to product rule

that gives me

sin(2t-a-b)cos(b-a)=0

I can say b=a or 2t-a-b=0 which gives me t=a+b/2 ?

whch is greater than b!
• Sep 1st 2010, 05:25 AM
Ackbeet
Your result should be independent of t. Why not approach it this way:

sin(2(t-a)) = sin(2(t-b)).

One way to guarantee that this relationship holds is if the arguments of the two sine functions differ by what? Where does that take you?
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