Unit Step Function and Laplace Transforms

[Spectre5]

Hey

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

Note:
{\cal L} = Laplace
{\cal U} = Unit Step Function


{\cal L} \{ \cos(2t) \,\,\, {\cal U} (t - \pi)\}

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

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

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

=\frac{se^{-\pi s}}{s^2+4}


Can anyone help me with this problem:

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

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

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

=\frac{s \, e^{-\frac{\pi s}{2}}}{s^2 + 1}

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

 

[Data]

First one is correct.

{\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}

 

[Spectre5]

oh, I edited my post as you posted that...so is the books answer wrong then?

 

[Data]

your answer is correct...

Sounds like the author of the solutions manual was having a bad day.

 

[Spectre5]

ok, thanks

 

[Spectre5]

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):

http://umath.nuk.edu.tw/~scnet/CourseModule/Class/DE/download/04_DE_Huang.pdf

Then go to PAGE 20

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

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

=e^{-as}{\cal L}\{ f(t) \}

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!

 

[StatusX]

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.

 

[Spectre5]

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

 

[Data]

It's not the Laplace transform of anything, the way it's written (I assume you missed a minus sign! ).

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

\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.

The t, v, and \asymp are just "dummy variables."

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

{\cal L}\{f(t - a)\} = {\cal L}\{f(t)\}

for every real a. What they really mean when they look at {\cal L}\{f(t-a)\}, of course, is {\cal L}\{g(t)\} where g(t) = f(t-a).

 

[Spectre5]

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

 

[Spectre5]

So is this true:

{\cal L}\{ f(t + a) \} = {\cal L}\{ f(t) \}

For any real number a?

 

[Data]

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?

It's right, because \cos is 2\pi-periodic. Otherwise you would be right.

 

[StatusX]

I'm sorry, I'm probably too tired to be trying to help.

 

[Data]

So is this true:

{\cal L}\{f(t - a)\} = {\cal L}\{f(t)\}

For any real number a?

Yes. But that isn't what your textbook means when it says {\cal L}\{f(t - a)\}. As I said, it has many "errors" in it as a result.

Recall the definition of the Laplace transform,

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

and thus

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

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

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

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

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

which is not the same thing (note we now have just dt, and the exponent on the e has changed!).

 

[Data]

Good. I should have pointed out the mathematical resolution to this apparent conundrum (no, current math isn't ultimately totally inconsistent!):

instead define

{\cal L}\{u\} = \int_0^\infty u e^{-st} \ dt

while the difference doesn't look too big, it's pretty important... when you want {\cal L}\{u\} for any u, just throw it into that position in the integral.

 

[Spectre5]

ok, thanks