Question

Show the procedure for finding the transform of f(t)=cos5t-sin5t

Answer 1

Fourier?

Answer 2

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

Answer 3

You haven't covered Laplace?

Answer 4

Series and Sequences are my absolute weakness.

Answer 5

I have covered LaPlace as well.

Answer 6

hmm

Answer 7

That was our last week of my class

Answer 8

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

Answer 9

When's it due?

Answer 10

3 hours

Answer 11

ouch

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Cool thanks dude. No problem

Answer 17

kk

Answer 18

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

Answer 19

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

Answer 20

I think its LaPlace

Answer 21

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

Answer 22

They are different right?

Answer 23

yes, they are.

Answer 24

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

Answer 25

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

Answer 26

We use transform tables usually....

Answer 27

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).

Answer 28

do you think you can apply these two formulas?

Answer 29

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

Answer 30

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

Answer 31

yeah k in this case is what?

Answer 32

5

Answer 33

OK.. show me what you get.

Answer 34

\[5-s/s^2+25\]

Answer 35

but I can factor the bottom if I wanted

Answer 36

are you sure about factorizing the bottom?

Answer 37

Nevermind I can;t

Answer 38

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

Answer 39

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

Answer 40

s-5 I read it inccorectly

Answer 41

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

Answer 42

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

Answer 43

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).

Answer 44

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

Answer 45

Exactly.

Answer 46

Sweet tthanks a lot :)

Answer 47

You're welcome!