Laplace transform with heaviside step function

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Oct 6th 2010, 08:17 AM
BobP

Reply to Ackbeet @ 13
Sorry, I missed a 2 out of my expression.
I should have put 1 - H(t-1) + 2H(t-2).
The 1 I admit is lazy, H(t) would be better, but since the Laplace transform is defined for zero to infinity they are in effect the same aren't they ? (They have the same LT).
Can you confirm that x(t) is defined as being equal to 1 for 0<=t<=1, zero for 1<t<2 and 2 for t>=2 ?
Oct 6th 2010, 08:25 AM
Ackbeet

The OP has the definition of x(t).

Quote:

Can you confirm that x(t) is defined as being equal to 1 for 0<=t<=1, zero for 1<t<2 and 2 for t>=2 ?

Incorrect for this problem. It is defined by

$\displaystyle x(t)=\begin{cases} 0,\quad t<0\\ 1,\quad 0\le t<1\\ 2,\quad 2\le t\end{cases}.$ At least, that's my interpretation of this line of the OP:

Quote:

where x(t) = 1, 0≤t≤1;
2, t≥2;
0 elsewhere
Oct 6th 2010, 08:40 AM
coley0412

well...as mentioned previously - this is the part where i get confused.

s+e^-s/s(s^2+2s+2)

=L^-1(1/s)+(L^-1(1/(s+1) + L^-1(1/(s+1)))e^-s

= 1 + 2e^-t(t-1)H(t-1)
Oct 6th 2010, 08:50 AM
BobP

Reply to Ackbeet @ 17

So x(t) is not defined for values of t between 1 and 2 ?
Oct 6th 2010, 09:05 AM
Ackbeet

Reply to BobP:

Oh, I see what you're getting at. Silly me. Brain not working correctly! The function is 0 in that range. So we have

$\displaystyle x(t)=\begin{cases} 0,\quad t<0\\ 1,\quad 0\le t<1\\ 0,\quad 1\le t<2\\ 2,\quad 2\le t\end{cases}.$

To get this function, we could certainly use your expression, but I do think the $H(t)$ is better. It's more explicitly correct, anyway. So we get

$x(t)=H(t)-H(t-1)+2H(t-2).$

Are we on the same page now?
Oct 6th 2010, 09:16 AM
BobP

Yes.
Oct 6th 2010, 09:26 AM
Ackbeet

So the implications of my last post are that the Laplace Transform of the ODE is

$\displaystyle s^{2}Y(s)-1+2sY(s)+2Y(s)=\frac{1}{s}- \frac{e^{-s}}{s}+\frac{2e^{-2s}}{s}.$ From this we get

$\displaystyle Y(s)(s^{2}+2s+2)=\frac{1+s-e^{-s}+2e^{-2s}}{s}.$ Therefore,

$\displaystyle Y(s)=\frac{1+s-e^{-s}+2e^{-2s}}{s(s^{2}+2s+2)}.$

As you say, you're a little confused in computing inverse LT's. Probably the first thing I would do is complete the square in the denominator (the quadratic only factors over the complex numbers):

$\displaystyle Y(s)=\frac{1+s-e^{-s}+2e^{-2s}}{s(s^{2}+2s+1+1)}=\frac{1+s-e^{-s}+2e^{-2s}}{s((s+1)^{2}+1)}.$

Now, at this point, I would probably break up the fraction into four fractions like so:

$\displaystyle Y(s)=\frac{1}{s((s+1)^{2}+1)}+\frac{1}{(s+1)^{2}+1 }-\frac{e^{-s}}{s((s+1)^{2}+1)}+\frac{2e^{-2s}}{s((s+1)^{2}+1)}.$

The second fraction there you can invert directly using the table here. That same link will help you a great deal in computing the rest of them. The first term I would use the technique of partial fractions on. That is, you can write

$\displaystyle \frac{1}{s((s+1)^{2}+1)}=\frac{A}{s}+\frac{Bs+C}{( s+1)^{2}+1}.$ Get a common denominator on the RHS and set the numerators equal.

For the last two terms, the exponentials correspond to time shifting, along with the Heaviside function again. Again, consult the table there, and see if you can see the pattern.
Oct 6th 2010, 11:51 AM
coley0412

ok, for the first and second terms I get:
A=1/2; B=-1/2; C=-1

e^-tsin(t)+1/2-1/2*e^tcos(t) (I THINK)

still working on the second two terms...
Oct 7th 2010, 06:39 PM
Ackbeet

I'd agree with your constants A, B, C. That gives

$\displaystyle \frac{1}{s((s+1)^{2}+1)}=\frac{1}{2s}+\frac{-\frac{1}{2}s-1}{(s+1)^{2}+1}=1)}=\frac{1}{2s}+\frac{-\frac{1}{2}(s+1)+\frac{1}{2}-1}{(s+1)^{2}+1}=\frac{1}{2s}+\frac{-\frac{1}{2}(s+1)-\frac{1}{2}}{(s+1)^{2}+1},$

with inverse LT of

$\displaystyle\frac{1}{2}\,H(t)+e^{-t}\mathcal{L}^{-1}\left[\frac{-\frac{1}{2}\,s-\frac{1}{2}}{s^{2}+1}\right] =\frac{1}{2}\,H(t)-\frac{1}{2}\,e^{-t}\mathcal{L}^{-1}\left[\frac{s+1}{s^{2}+1}\right] =\frac{1}{2}\,H(t)-\frac{1}{2}\,e^{-t}\left(\mathcal{L}^{-1}\left[\frac{s}{s^{2}+1}\right]+\mathcal{L}^{-1}\left[\frac{1}{s^{2}+1}\right]\right)$

$\displaystyle =\frac{1}{2}\,H(t)-\frac{1}{2}\,e^{-t}H(t)\left(\cos(t)+\sin(t)\right).$

$\displaystyle =\frac{1}{2}\,H(t)\left(1-e^{-t}\left(\cos(t)+\sin(t)\right)\right).$

This isn't quite what you got, but it's very close.
Oct 8th 2010, 04:55 AM
coley0412

ok and now i add the inverse transform of the last two fractions to this?
Oct 8th 2010, 07:03 AM
Ackbeet

Right. The inverse LT is a linear operation just like the LT is.
Jul 13th 2012, 04:52 PM
misspositivity10

Re: Laplace transform with heaviside step function

excuse me, but can someone PLEASE tell me what's mean?
Jul 14th 2012, 12:46 AM
tom@ballooncalculus

Re: Laplace transform with heaviside step function

I believe this is a bug due to some previous latex rules allowing for line breaks. (Which is all you want to see at those points.)

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