Fourier sine and cosine tranformation, difficult problem, (for me)

[difficult]

Homework Statement

What are the Fourier sine and cosine transformations of exp(5t)?

Homework Equations

Fc (ω) = (√(2/∏))∫exp(5t)cos(ωt)dt , (between boundaries of infinity and zero)

The Attempt at a Solution

When I try to integrate by parts I just end up going round in circles.
Or, if I first convert cos(ωt) into 0.5(exp(iωt) + exp(-iωt)) and then multiply this with exp(5t),
calculating the boundary values then becomes difficult

 

[vela]

Were you really given e^+5t and not e^-5t?

 

[difficult]

Were you really given e^+5t and not e^-5t?

Oops sorry! I forgot to mention that negative, but I still can't do the problem.

 

[vela]

Either method you described will work. Can you show us what you've gotten so far?

 

[difficult]

Well for the Fourier cosine transformation I end up with
(1/√2∏) ( (exp((iω - 5)t)/(iω - 5)) + (exp((-iω - 5)t) / (-iω - 5)) ) evaluated over the range from 0 to ∞, but what do

exp((iω - 5)t) and exp((-iω - 5)t) equal at t = ∞? .

Similarly for the Fourier sine transformation.

If there are plenty worked solutions to Fourier sine and cosine transformation problems, where are they?
Please I am referring to worked solutions to problems finding the Fourier sine and cosine transformations of a difficult expression, not of some stupidly simple number like 1.

 

[difficult]

Apparently the ultimate solution is

Fc(ω) = (√(2/∏))(5/5squaredωsquared)

but how?

 

[difficult]

For the Fourier cosine transformation (and similarly for the sine) I got this far by firstly converting cos(ωt) into 0.5(exp(iωt) + exp(-iωt)) and then multiplying by exp(-5t),

 

[difficult]

Is there an altogether better approach to this problem?

 

[difficult]

I only know the simple answer to the problem I asked about, but not how to produce it, which is the real issue.
If you are not going to help me out with this particular problem, which I can assure you is becoming a serious distraction from the rest of the subject, could you please describe the best general approach to Fourier sine and cosine transformations? The only worked solutions I can find are pretty unhelpful, as they are just for the Fourier sine and cosine transformations of the number 1, which largely translates into finding the integral of sin(ωt) or cos(ωt).

 

[difficult]

Should I use frequency shifting?
The Fourier transform of f(t)cos(ωsub0t) being equal to 0.5(F(ω-ωsub0) + F(ω+ωsub0))
and the Fourier transform of exp(-5t) being 1 / (5 + iω)

is the Fourier Transformation of exp(-5t)cos(ωsub0t) equal to 1 / (5 + iωsub0)?
In that case the simple, final answer provided would be wrong.

 

[vela]

Well for the Fourier cosine transformation I end up with
(1/√2∏) ( (exp((iω - 5)t)/(iω - 5)) + (exp((-iω - 5)t) / (-iω - 5)) ) evaluated over the range from 0 to ∞, but what do

exp((iω - 5)t) and exp((-iω - 5)t) equal at t = ∞? .

Take, for example, your first term. The factor of e^iωt has a magnitude of 1, regardless of what t equals, so
$$\left|\frac{e^{(i\omega-5)t}}{i\omega-5}\right| = \left|\frac{e^{i\omega t}}{i\omega-5}\right| e^{-5t} = \frac{1}{\sqrt{\omega^2+5^2}}e^{-5t}.$$ When t goes to infinity, that quantity goes to 0, which means $\frac{e^{(i\omega-5)t}}{i\omega-5}$ goes to 0.

 

[difficult]

Thanks.
If |exp(iωt)| = 1 and |exp(-iωt)| = 1 for all t, how do sin(ωt) and cos(ωt) vary with t, when they are functions of exp(iωt) and exp(-iωt)?

 

[vela]

Because |exp(iωt)| and exp(iωt) aren't the same thing. Sine and cosine aren't functions of |exp(iωt)|.

 

[difficult]

Thanks again. Going off on a tangent for a moment, do exp(iωt) and exp(-iωt) each form a circle of unit radius on an argand diagram?

When I substituted the number 1 for |exp(iωt)| and again the number 1 for |exp(-iωt)|, I
ended up with Fc(ω) =(√(2/∏)) x (5 / (5squared PLUS ωsquared))

Apparently it should actually be Fc(ω) =(√(2/∏)) x (5 / (5squared TIMES ωsquared))

 

[vela]

Thanks again. Going off on a tangent for a moment, do exp(iωt) and exp(-iωt) each form a circle of unit radius on an argand diagram?

Yes.

When I substituted the number 1 for |exp(iωt)| and again the number 1 for |exp(-iωt)|, I ended up with Fc(ω) =(√(2/∏)) x (5 / (5squared PLUS ωsquared))

Apparently it should actually be Fc(ω) =(√(2/∏)) x (5 / (5squared TIMES ωsquared))

The plus sign is correct. Another way you can look at the integral
$$\int_0^\infty e^{-5t}\cos\omega t\,dt = \left.\int_0^\infty e^{-st}\cos\omega t\,dt\right|_{s=5}$$ is as the Laplace transform of cos ωt evaluated when s=5. If you look up the Laplace transform for cosine in a table, you'll see there's supposed to be a plus there.

 

[difficult]

Thanks.

 

[difficult]

Does |exp(-iωt)| = 1 aswell?

 

[vela]

What does Euler's formula tell you?

 

[difficult]

How should I manipulate Euler's formula ?

 

[difficult]

Incorporated this post in the next one...

 

[difficult]

More precisely, is it because

√ ((cosine squared x) + (sine squared x)) = 1?

 

[difficult]

Or is that not why |exp(-iωt)| = 1, if indeed it equals 1 at all?