Is the Hypothesis of Real and Imaginary Components for F(ω) True?

[Jhenrique]

I never see the following hypothesis but I believe that they are true...

##\text{Re}(\hat f (\omega)) = a(\omega)##

##\text{Im}(\hat f (\omega)) = b(\omega)##


where:

##f(t) = \int_{-\infty}^{+\infty}\hat f(\omega) \exp(i \omega t) d\omega = \int_{0}^{\infty} a(\omega) \cos(\omega t) + b(\omega) \sin(\omega t) d\omega##


##\hat f (\omega) = \int_{-\infty}^{+\infty}f(t) \exp(-i \omega t) dt##


##a(\omega) = \frac{1}{\pi}\int_{-\infty}^{+\infty} f(t) \cos(\omega t) dt##

##b(\omega) = \frac{1}{\pi}\int_{-\infty}^{+\infty} f(t) \sin(\omega t) dt##


So, the two first equations are true?

 

[Simon Bridge]

You are missing the definition of ##\hat f(\omega)##.
From there you should be able to prove (or disprove) the relationships yourself.

 

[Jhenrique]

You are missing the definition of ##\hat f(\omega)##.
From there you should be able to prove (or disprove) the relationships yourself.

Is unmistakable that ##\hat f## represents the Fourier transform of ##f## !

 

[HallsofIvy]

I never see the following hypothesis but I believe that they are true...

##\text{Re}(\hat f (\omega)) = a(\omega)##

##\text{Im}(\hat f (\omega)) = b(\omega)##

where:

##\hat f (\omega) = \int_{-\infty}^{+\infty}f(t) \exp(-i \omega t) dt##

Use the fact that [itex]e^{-i\omega t}= cos(\omega t)- i sin(\omega t)[/itex].

##a(\omega) = \frac{1}{\pi}\int_{-\infty}^{+\infty} f(t) \cos(\omega t) dt##

##b(\omega) = \frac{1}{\pi}\int_{-\infty}^{+\infty} f(t) \sin(\omega t) dt##

So, the two first equations are true?

 

[Jhenrique]

An phasor in the complex form is ##A \exp(i(\omega t + \varphi)) = A \exp(i \varphi) \exp(i \omega t)##, the summation of phasors wrt angular frequency is ##\sum A(\omega) \exp(i \varphi(\omega)) \exp(i \omega t) \Delta \omega = \sum \hat f(\omega) \exp(i \omega t) \Delta \omega##. So, becomes clear that the ##\text{Abs}(\hat f(\omega)) = A(\omega)## and ##\text{Arg}(\hat f(\omega)) = \varphi(\omega)##.

What I want mean is that I don't understand the relation that a(ω) and b(ω) has with f(ω).

 

[Simon Bridge]

Is unmistakable that ##\hat f## represents the Fourier transform of ##f## !

... but you edited post #1 to include that anyway - thank you ;)
So your next step was to relate the sine and cosine form to the exponential in the Fourier transform re post #4.

What I want mean is that I don't understand the relation that a(ω) and b(ω) has with f(ω).

As in post #4. ##e^{-i\omega t}=\cos\omega t - i\sin\omega t##
Make the substitution in the Fourier transformation definition... which you gave as:
$$\hat f(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\;dt$$ ... and follow your nose.

Did you try that?

 

[Jhenrique]

Yeah, but I don't see in none place a direct connection between a(ω) and b(ω) with Re(f(ω)) and Im(f(ω)).

 

[Simon Bridge]

Please show your working.

 

[Jhenrique]

$$

\\f(x) = \int_{0}^{\infty} A(\omega) \cos(x \omega) + B(\omega) \sin(x \omega) d\omega

\\ = \frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)(\cos(\omega t) \cos(\omega x) + \sin(\omega t) \sin(\omega x))dt d\omega

\\ = \frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)\cos(\omega(x-t))dt d\omega

\\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)(\exp(i \omega (x-t)) + \exp(-i \omega(x-t)))dt d\omega

\\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega (x-t))dt d\omega + \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(-i \omega(x-t)) dt d\omega

\\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega (x-t))dt d\omega + \frac{1}{2 \pi} \int_{-infty}^{0} \int_{-\infty}^{\infty}f(t) \exp(i \omega(x-t)) dt d\omega

\\ = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega(x-t))dt d\omega

\\ = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(t) \exp(-i \omega t) dt \right) \exp(i \omega x) d \omega

\\ = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \left( \hat f(\omega) \right) \exp(i \omega x) d \omega

$$

 

[Jhenrique]

So...?

 

[Simon Bridge]

You started at the wrong place.

Start from your stated definition for ##\hat f## ... the one with the exponential in it. That is your first line.

Your second line should use the substitution for the exponential in terms of sine and cosine.