User:Metacomet/Fourier

Proof of equivalence
$$ X(f) = {1 \over 2 \pi } X(\omega) |_{\omega = 2 \pi f}  = {1 \over 2 \pi } \left[ \int_{-\infty}^{\infty} x(t) \ e^{-i \omega t} \ dt \ \right]_{\omega = 2 \pi f }  $$


 * $$ = \int_{-\infty}^{\infty} x(t) \ e^{-j 2 \pi f t} \ dt \   $$

$$ \omega = 2 \pi f \, $$

$$ d \omega = 2 \pi \cdot df \, $$

$$ x(t) = \left[ \frac{1}{2 \pi} \int_{-\infty}^{\infty} X(\omega) \ e^{i \omega t} \ d \omega \ \right]_{\omega = 2 \pi f } $$


 * $$ =  \int_{-\infty}^{\infty}  \frac{1}{2 \pi}  X(\omega)|_{ \omega = 2 \pi f } \ e^{i 2 \pi f t} \cdot 2 \pi \cdot df \ $$


 * $$ = \int_{-\infty}^{\infty} X(f) \ e^{j 2 \pi f t}\, df \ $$


 * Q.E.D

New Proof of Equivalence

 * Start by defining:


 * $$ X_1(\omega) \equiv \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x(t) \ e^{-i \omega t}\, dt \ $$


 * Then:


 * $$ x(t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} X_1(\omega) \ e^{i \omega t}\, d \omega \ $$


 * Next, define:


 * $$ X_2(\omega) \equiv \sqrt{2 \pi} X_1(\omega)$$


 * Therefore:


 * $$ X_2(\omega) = \int_{-\infty}^{\infty} x(t) \ e^{-i \omega t}\, dt \ $$


 * and


 * $$ X_1(\omega) =  {  1 \over \sqrt{2 \pi}  }  X_2(\omega)$$


 * As a result:


 * $$ x(t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}  {  1 \over \sqrt{2 \pi}  }    X_2(\omega) \ e^{i \omega t}\, d \omega   =  \frac{1}{2 \pi} \int_{-\infty}^{\infty}   X_2(\omega) \ e^{i \omega t}\, d \omega \ $$


 * Next, define:


 * $$ X_4(\omega) \equiv   {  1 \over 2 \pi  }  X_2(\omega)$$


 * Therefore:


 * $$ X_4(\omega) = {  1 \over 2 \pi  }   \int_{-\infty}^{\infty} x(t) \ e^{-i \omega t}\, dt \ $$


 * and


 * $$ X_2(\omega) =   2 \pi    X_4(\omega)  \, $$


 * As a result:


 * $$ x(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}   2 \pi    X_4(\omega)    \ e^{i \omega t}\, d \omega   =  \int_{-\infty}^{\infty}   X_4(\omega) \ e^{i \omega t}\, d \omega \ $$


 * Next:


 * let $$ \omega = 2 \pi f \, $$


 * Therefore:


 * $$ x(t) = \lim_{ \gamma \to \infty } \int_{ \omega = -\gamma  }^{ \omega = \gamma }   X_4(\omega) \ e^{i \omega t}\, d \omega \ $$


 * $$  = \lim_{ \gamma \to \infty }  \int_{2 \pi f = -\gamma }^{ 2 \pi f  =  \gamma }   X_4(  2 \pi f  ) \ e^{i  2 \pi f  t}\, d ( 2 \pi f )   \ $$


 * $$ =  \lim_{ \gamma \to \infty }   \int_{  f = -\gamma / 2 \pi  }^{ f  =  \gamma / 2 \pi  }   X_4(  2 \pi f  ) \ e^{i  2 \pi f  t}\,  2 \pi df    \ $$


 * $$ =  \lim_{ \gamma \to \infty }   \left[ 2 \pi  \int_{  f = -\gamma / 2 \pi  }^{ f  =  \gamma / 2 \pi  }   X_4(  2 \pi f  ) \ e^{i  2 \pi f  t}\,  df  \right]  \ $$


 * $$ =  \lim_{ \gamma \to \infty }    \int_{  f = -\gamma  }^{ f  =  \gamma  }   X_4(  2 \pi f  ) \ e^{i  2 \pi f  t}\,  df    \ $$


 * $$ =  \int_{  -\infty  }^{   \infty  }   X_4(  2 \pi f  ) \ e^{i  2 \pi f  t}\,  df    \ $$


 * Finally, define:


 * $$ X_3(f) \equiv X_4(\omega) \bigg|_{\omega = 2 \pi f} $$


 * Then:


 * $$ X_3(f) = {1 \over 2 \pi }  X_2(\omega) \bigg|_{\omega = 2 \pi f}  $$


 * and


 * $$ x(t) =  \int_{  -\infty  }^{   \infty  }   X_3(  f  ) \ e^{i  2 \pi f  t}\,  df    \ $$