anonymous
  • anonymous
Show the procedure for finding the transform of f(t)=cos5t-sin5t
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Fourier?
anonymous
  • anonymous
It doesn't say but thats what I'm guessing since thats what we've covered.
anonymous
  • anonymous
You haven't covered Laplace?

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anonymous
  • anonymous
Series and Sequences are my absolute weakness.
anonymous
  • anonymous
I have covered LaPlace as well.
anonymous
  • anonymous
hmm
anonymous
  • anonymous
That was our last week of my class
anonymous
  • anonymous
Its a short answer type question though so I'm not sure what it means by transform.. the only 2 I can recall are fourier and laplace
anonymous
  • anonymous
When's it due?
anonymous
  • anonymous
3 hours
anonymous
  • anonymous
ouch
anonymous
  • anonymous
Yeah I have like 20 more questions, but they are much more simple.
anonymous
  • anonymous
The answer's going to depend on whether it's Fourier or Laplace, that's the problem.
anonymous
  • anonymous
Well wouldn't fourier be a series? Which would give me an interval?
anonymous
  • anonymous
I have to leave for a bit - bad timing. I'll try and be back before your three hours is up.
anonymous
  • anonymous
Cool thanks dude. No problem
anonymous
  • anonymous
kk
anonymous
  • anonymous
I can do it quickly if it's Laplace transform.
anonymous
  • anonymous
I haven't done any Fourier transform in a while.
anonymous
  • anonymous
I think its LaPlace
anonymous
  • anonymous
I don't think we did Fourier transforms, we did Fourier Series.
anonymous
  • anonymous
They are different right?
anonymous
  • anonymous
yes, they are.
anonymous
  • anonymous
Ok well I guess its a laplace because I haven't done any transform other than LaPlace
anonymous
  • anonymous
It would be easy then. do you use a Laplace transform table? or should I solve using the definition of Laplace transform?
anonymous
  • anonymous
We use transform tables usually....
anonymous
  • anonymous
that's good.. we will just use two formulas, which are: \[1) L[\sin kt]={k \over s^2+k^2}\] \[2) L[\cos kt]= {s \over s^2+k^2}\] where L[f(t)] denotes the Laplace transform of f(t).
anonymous
  • anonymous
do you think you can apply these two formulas?
anonymous
  • anonymous
Hmm Let me see... I'm really not fluent in using the table as it was our last topic discussed in the class.
anonymous
  • anonymous
Wouldn't I just substitute K for the cosine and sine values?
anonymous
  • anonymous
yeah k in this case is what?
anonymous
  • anonymous
5
anonymous
  • anonymous
OK.. show me what you get.
anonymous
  • anonymous
\[5-s/s^2+25\]
anonymous
  • anonymous
but I can factor the bottom if I wanted
anonymous
  • anonymous
are you sure about factorizing the bottom?
anonymous
  • anonymous
Nevermind I can;t
anonymous
  • anonymous
f(t)=5-s/s^2+25 final solution.
anonymous
  • anonymous
1) well what is the Laplace transform of f(t)? 2) is it 5-s or s-5?
anonymous
  • anonymous
s-5 I read it inccorectly
anonymous
  • anonymous
ok.. the notation that's usually used to denote a function after being transformed is uppercase letter, in our case it will be: \[F(s)= {s-5 \over s^2+25}\]
anonymous
  • anonymous
Oh right... because you're transforming it in terms of f(t) and s... I recall that now.
anonymous
  • anonymous
if you're asking about how you can get the formulas (which is the most important part of the answer), you can just use the definition and integrate for the first one: \[\int\limits_{0}^{\infty}{\cos (kt)} e ^{-st} dt\] the result of this integral (can be integrated using integration by parts) is, by definition, the Laplace transform of cos (kt).
anonymous
  • anonymous
Right which will help me build a transform table if I didn't have one available?
anonymous
  • anonymous
Exactly.
anonymous
  • anonymous
Sweet tthanks a lot :)
anonymous
  • anonymous
You're welcome!

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