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 !!
Thanks in advance!
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
A couple of comments:
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?
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.
@ 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
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?
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!