## anonymous 5 years ago Show the procedure for finding the transform of f(t)=cos5t-sin5t

1. anonymous

Fourier?

2. anonymous

It doesn't say but thats what I'm guessing since thats what we've covered.

3. anonymous

You haven't covered Laplace?

4. anonymous

Series and Sequences are my absolute weakness.

5. anonymous

I have covered LaPlace as well.

6. anonymous

hmm

7. anonymous

That was our last week of my class

8. 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

9. anonymous

When's it due?

10. anonymous

3 hours

11. anonymous

ouch

12. anonymous

Yeah I have like 20 more questions, but they are much more simple.

13. anonymous

The answer's going to depend on whether it's Fourier or Laplace, that's the problem.

14. anonymous

Well wouldn't fourier be a series? Which would give me an interval?

15. anonymous

I have to leave for a bit - bad timing. I'll try and be back before your three hours is up.

16. anonymous

Cool thanks dude. No problem

17. anonymous

kk

18. anonymous

I can do it quickly if it's Laplace transform.

19. anonymous

I haven't done any Fourier transform in a while.

20. anonymous

I think its LaPlace

21. anonymous

I don't think we did Fourier transforms, we did Fourier Series.

22. anonymous

They are different right?

23. anonymous

yes, they are.

24. anonymous

Ok well I guess its a laplace because I haven't done any transform other than LaPlace

25. anonymous

It would be easy then. do you use a Laplace transform table? or should I solve using the definition of Laplace transform?

26. anonymous

We use transform tables usually....

27. 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).

28. anonymous

do you think you can apply these two formulas?

29. anonymous

Hmm Let me see... I'm really not fluent in using the table as it was our last topic discussed in the class.

30. anonymous

Wouldn't I just substitute K for the cosine and sine values?

31. anonymous

yeah k in this case is what?

32. anonymous

5

33. anonymous

OK.. show me what you get.

34. anonymous

$5-s/s^2+25$

35. anonymous

but I can factor the bottom if I wanted

36. anonymous

are you sure about factorizing the bottom?

37. anonymous

Nevermind I can;t

38. anonymous

f(t)=5-s/s^2+25 final solution.

39. anonymous

1) well what is the Laplace transform of f(t)? 2) is it 5-s or s-5?

40. anonymous

41. anonymous

ok.. the notation that's usually used to denote a function after being transformed is uppercase letter, in our case it will be: $F(s)= {s-5 \over s^2+25}$

42. anonymous

Oh right... because you're transforming it in terms of f(t) and s... I recall that now.

43. 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).

44. anonymous

Right which will help me build a transform table if I didn't have one available?

45. anonymous

Exactly.

46. anonymous

Sweet tthanks a lot :)

47. anonymous

You're welcome!