Problem with Laplace transform unit step function questions

[enger]

Homework Statement

L{2t u(t-1)}

Homework Equations

L{g(t) u(t-c)} = e^-cs L{g(t+c)}

The Attempt at a Solution

L{2t u(t-1)}=e^-s L{2(t+1)}
L{2(t+1)}=2/s^2+2/s
L{2t u(t-1)} = e^-s {2/s^2 + 2/s}


i think the whole attempt is wrong , I'm getting confused in this type of questions , i would appreciate if someone could explain the u(t-a) function & solve this problem...

 

[Char. Limit]

I dunno, looks like you have it right to me.

 

[hunt_mat]

By step unit function, I assume you mean that the function that is 1 on the interval [a,b] and 0 elsewhere? Just write down the definition:

[tex]
\int_{0}^{\infty}\chi_{[a,b]}(x)e^{-sx}dx=\int_{a}^{b}e^{-sx}dx
[/tex]

I presume you can do the rest from here.

 

[LCKurtz]

I dunno, looks like you have it right to me.

Looks right to me too. What is PF SAS?

 

[Char. Limit]

Looks right to me too. What is PF SAS?

Oh, not much. Just a small group of people, unrelated to PF, dedicated to fighting crackpots who appear on this site. Like the .9999... does not equal 1 people.

 

[LCKurtz]

What does the acronym SAS stand for? And if you really want to engage the crackpots, visit usenet's sci.math. But, of course, a real crank is immune to correction.

 

[Char. Limit]

Special Anti-crackpot Service. The name is modeled after the British SAS.

 

[vela]

i think the whole attempt is wrong , I'm getting confused in this type of questions , i would appreciate if someone could explain the u(t-a) function & solve this problem...

The unit step function u(t) is 0 if t<0 and 1 if t≥0. That's it.

If you have u(t-a), it's 0 if t<a and 1 if t≥a.

Now consider g(t)u(t-a). When t<a, g(t)u(t-a)=g(t)×0=0. When t≥a, g(t)u(t-a)=g(t)×1=g(t). So g(t)u(t-a) is just a convenient way of writing

[tex]g(t)u(t-a) = \left\{ \begin{array}{cr}0 & t<a \\ g(t) & t\ge a\end{array}\right.[/tex]

 

[enger]

is there a good reference with examples , i need to understand this part a little more.

the (t-a) is the shift , so i shouldn't be expanding it in the problem i solved above !

 

[vela]

I have no idea what you're asking. Could you elaborate?

 

[enger]

I have no idea what you're asking. Could you elaborate?

in this step

L{2(t+1)}=2/s^2+2/s

i expanded the equation to (2t+2) then used Laplace transform for t^n & constant , is this right or wrong , since i understand that U(t-1) is the shift of the equation

 

[vela]

Yes, it is. The shift isn't involved in that step. You're finding the Laplace transform of an unshifted 2(t+1). When you combine that with the exponential factor, you get the transform of the truncated 2t.

Another way you can do the problem is to use t=t-1+1 to write

2t u(t-1) = 2 [(t-1)+1] u(t-1)

Now everything is in terms of t-1, so you should be able to see it's just the function 2(t+1)u(t) shifted to the right by 1. That's why you calculate the Laplace transform of 2(t+1), unshifted, and then slap the exponential factor on, which indicates the shift.

 

[enger]

Yes, it is. The shift isn't involved in that step. You're finding the Laplace transform of an unshifted 2(t+1). When you combine that with the exponential factor, you get the transform of the truncated 2t.

Another way you can do the problem is to use t=t-1+1 to write

2t u(t-1) = 2 [(t-1)+1] u(t-1)

Now everything is in terms of t-1, so you should be able to see it's just the function 2(t+1)u(t) shifted to the right by 1. That's why you calculate the Laplace transform of 2(t+1), unshifted, and then slap the exponential factor on, which indicates the shift.

so the answer i added for the problem is right?

also is there a reference with examples?

i get confused when solving problems like l{u(t-pi)sin2t)

 

[vela]

Yes, as the others noted, your answer is correct.

I don't know of any references offhand. Electrical engineering textbooks on linear system analysis probably cover this material.

If it helps, just write everything out step by step. For u(t-pi) sin 2t, you have g(t)=sin 2t and c=pi. So g(t+c) = g(t+pi) = sin 2(t+pi) = ... and so on.

Or use the trick where you say t=(t-pi)+pi. So sin 2t = sin [2(t-pi)+2pi] = ...

 

[LCKurtz]

so the answer i added for the problem is right?

also is there a reference with examples?

i get confused when solving problems like l{u(t-pi)sin2t)

Maybe you just need to think about the proof for arbitrary f(t):

[tex]L(f(t)u(t-a)) = \int_0^\infty e^{-st}f(t)u(t-a)\, dt = \int_a^\infty e^{-st}f(t)\cdot 1\, dt[/tex]

Now let w = t-a, dw = dt

[tex] = \int_0^\infty e^{-s(w+a)}f(w+a)\, dw = e^{-as}\int_0^\infty e^{-sw}f(w+a)\, dw= e^{-as}\int_0^\infty e^{-st}f(t+a)\, dt=e^{-as}L(f(t+a))[/tex]