# Real and imaginary of F(ω)

1. May 17, 2014

### 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?

Last edited: May 17, 2014
2. May 17, 2014

### Simon Bridge

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

3. May 17, 2014

### Jhenrique

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

4. May 17, 2014

### HallsofIvy

Use the fact that $e^{-i\omega t}= cos(\omega t)- i sin(\omega t)$.

5. May 17, 2014

### 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).

6. May 17, 2014

### Simon Bridge

... 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.

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?

7. May 17, 2014

### 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)).

8. May 17, 2014

### Simon Bridge

9. May 18, 2014

### Jhenrique

$$\\f(x) = \int_{0}^{\infty} A(\omega) \cos(x \omega) + B(\omega) \sin(x \omega) d\omega \\ = \frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)(\cos(\omega t) \cos(\omega x) + \sin(\omega t) \sin(\omega x))dt d\omega \\ = \frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)\cos(\omega(x-t))dt d\omega \\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)(\exp(i \omega (x-t)) + \exp(-i \omega(x-t)))dt d\omega \\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega (x-t))dt d\omega + \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(-i \omega(x-t)) dt d\omega \\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega (x-t))dt d\omega + \frac{1}{2 \pi} \int_{-infty}^{0} \int_{-\infty}^{\infty}f(t) \exp(i \omega(x-t)) dt d\omega \\ = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega(x-t))dt d\omega \\ = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(t) \exp(-i \omega t) dt \right) \exp(i \omega x) d \omega \\ = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \left( \hat f(\omega) \right) \exp(i \omega x) d \omega$$

10. May 18, 2014

### Jhenrique

So...?

11. May 18, 2014

### 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.