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Homework Help: Unit Step Function and Laplace Transforms

  1. Apr 2, 2005 #1

    I was wondering if someone would check my work on this problem:

    [tex]{\cal L}[/tex] = Laplace
    [tex]{\cal U}[/tex] = Unit Step Function

    [tex]{\cal L} \{ \cos(2t) \,\,\, {\cal U} (t - \pi)\}[/tex]

    [tex]=e^{-\pi s} {\cal L} \{ \cos(2(t + \pi)) \}[/tex]

    [tex]=e^{-\pi s} {\cal L} \{ \cos(2t + 2\pi) \}[/tex]

    [tex]=e^{-\pi s} {\cal L} \{ \cos(2t) \}[/tex]

    [tex]=\frac{se^{-\pi s}}{s^2+4}[/tex]

    Can anyone help me with this problem:

    [tex]{\cal L} \{ \sin(t) \,\,\, {\cal U} (t - \pi/2)\}[/tex]

    [tex]=e^{-\frac{\pi s}{2}} {\cal L} \{ \sin(t + \pi / 2) \}[/tex]

    [tex]=e^{-\frac{\pi s}{2}} {\cal L} \{ \cos(t) \}[/tex]

    [tex]=\frac{s \, e^{-\frac{\pi s}{2}}}{s^2 + 1}[/tex]

    But this is different from the books published answer (in the exponential, they do not have it divived by two, they just have e^(-pi*s)...but the rest of it the same....also the solution manual did the problem differently, it changed the sin(t) to cos(t - pi/2)...but then wouldn't that just convert back to sin(t)? Anyways...it skips a lot of work and just ends up with the answer after that step
    Last edited: Apr 2, 2005
  2. jcsd
  3. Apr 2, 2005 #2
    First one is correct.

    [tex] {\cal L}\{ \cos(t - \pi / 2) u(t - \pi / 2) \} = e^{-\frac{\pi}{2}s}{\cal L}\{\cos \left( (t+\pi /2) - \pi /2\right)\} = e^{-\frac{\pi}{2}s}{\cal L}\{\cos{t}\} = \frac{e^{-\frac{\pi}{2}s}s}{s^2+1}[/tex]
  4. Apr 3, 2005 #3
    oh, I edited my post as you posted that...so is the books answer wrong then?
  5. Apr 3, 2005 #4
    your answer is correct...

    Sounds like the author of the solutions manual was having a bad day.
    Last edited: Apr 3, 2005
  6. Apr 3, 2005 #5
    ok, thanks
  7. Apr 3, 2005 #6
    ok, I am confused on the end of the proof of the Second Translation Theorem

    If you go to this link (just a link I found that has the proof so I don't have to type it all in LaTeX):


    Then go to PAGE 20

    That is the page with the end of the proof.,..what I don't understand is this part:

    [tex]=e^{-as}\int_0^\infty{e^{sv} \, f(v) \, dv}[/tex]

    [tex]=e^{-as}{\cal L}\{ f(t) \}[/tex]

    Isn't that integral the definition of the Laplace transform of f(v), not f(t)?? I don't understand why it is the Laplace of t instead of the transformed variable v.

    Thanks for any help!
    Last edited by a moderator: Apr 21, 2017
  8. Apr 3, 2005 #7


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    Actually I think the first one is wrong. When you say L{cos(2t+2pi)} you implicitly mean L{f(t)} where f(t) = cos(2t+2pi) for t>0 and 0 for t<0. This is not the same as g(t) = cos(2t) for t>0 and 0 for t<0. This could be obtained by doing another translation. Do you have an answer for this problem?

    Regarding your last post, v and t are just dummy variables and can be interchanged. I'm not exactly sure why they used v in the first place, but maybe they explain it somewhere else.
  9. Apr 3, 2005 #8
    it is explained as to why v is there, it was used in a transformation:

    v = t - a

    This was done to make it to correct for of a Laplace transformation
  10. Apr 3, 2005 #9
    It's not the Laplace transform of anything, the way it's written (I assume you missed a minus sign! :smile:).

    Once corrected, it's really the Laplace transform of the function [itex]f[/itex]. The argument is irrelevant. Note that

    [tex]\int_0^\infty f(t)e^{-st} \ dt = \int_0^\infty e^{-sv}f(v) \ dv = \int_0^\infty e^{-s\asymp}f(\asymp}) \ d\asymp.[/tex]

    The [itex]t[/itex], [itex]v[/itex], and [itex]\asymp[/itex] are just "dummy variables."

    Your textbook undoubtedly actually has many mistakes in it, because of this. By their own definition,

    [tex]{\cal L}\{f(t - a)\} = {\cal L}\{f(t)\}[/tex]

    for every real [itex]a[/itex]. What they really mean when they look at [itex]{\cal L}\{f(t-a)\}[/itex], of course, is [itex]{\cal L}\{g(t)\}[/itex] where [itex]g(t) = f(t-a)[/itex].
  11. Apr 3, 2005 #10
    As for your comment about my first problem posted, I had already accounted for the Unit Step function when I pulled out the exponential and plugged in t - pi into the cos function
  12. Apr 3, 2005 #11
    So is this true:

    [tex]{\cal L}\{ f(t + a) \} = {\cal L}\{ f(t) \}[/tex]

    For any real number a?
  13. Apr 3, 2005 #12
    It's right, because [itex]\cos[/itex] is [itex]2\pi[/itex]-periodic. Otherwise you would be right.
  14. Apr 3, 2005 #13


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    I'm sorry, I'm probably too tired to be trying to help.
  15. Apr 3, 2005 #14
    Yes. But that isn't what your textbook means when it says [itex]{\cal L}\{f(t - a)\}[/itex]. As I said, it has many "errors" in it as a result.

    Recall the definition of the Laplace transform,

    [tex]{\cal L}\{f(t)\} = \int_0^\infty f(t)e^{-st} \ dt.[/tex]

    and thus

    [tex]{\cal L}\{f(t - a)\} = \int_0^\infty f(t-a)e^{-s(t-a)} \ d(t-a).[/tex]

    This integral is the same one as in the definition (note that it is now [itex]d(t-a)[/itex]!!!), with the dummy variable changed from [itex]t[/itex] to [itex]t-a[/itex].

    This is NOT what your textbook means when it says [itex]{\cal L}\{f(t - a)\}[/itex], though, and this is why they have made errors (unless they have some cleverly hidden fine print somewhere :wink:).

    What your textbook actually means by [itex]{\cal L}\{f(t - a)\}[/itex] is

    [tex]\int_0^\infty f(t-a)e^{-st} \ dt,[/tex]

    which is not the same thing (note we now have just [itex]dt[/itex], and the exponent on the [itex]e[/itex] has changed!).
    Last edited: Apr 3, 2005
  16. Apr 3, 2005 #15
    Good. I should have pointed out the mathematical resolution to this apparent conundrum (no, current math isn't ultimately totally inconsistent!):

    instead define

    [tex]{\cal L}\{u\} = \int_0^\infty u e^{-st} \ dt[/tex]

    while the difference doesn't look too big, it's pretty important... when you want [itex]{\cal L}\{u\}[/itex] for any [itex]u[/itex], just throw it into that position in the integral.
  17. Apr 3, 2005 #16
    ok, thanks
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