# Whether Fourier transform formula be considered as Convolution or Correlation?

The expression for Fourier transform is given as $$F(\omega) = \int\limits_{-\infty}^{+\infty} f(t) \ e^{-j \omega t} \ dt \tag{1}$$

Now, let one function be $f(x)$ and other be $e^{j\omega t}$ then their convolution at $p(0)$ is given as

$$p(0) = \int\limits_{-\infty}^{+\infty} f(t) \ e^{-j \omega t} \ dt \tag{2}$$ same as in $(1)$.

Also, if we split the exponential term into sinusoidal,the expression $(1)$ becomes

$$F(\omega) = \int\limits_{-\infty}^{+\infty} (f(t) \cos{\omega t} -j f(t)\sin{\omega t})\ dt \tag{3}$$

Let $f(t)$ and $\cos{\omega t}$ be 2 functions. So their correlation at value $q(0)$ is given as

$$q(0) = \int\limits_{-\infty}^{+\infty} (f(t) \cos{\omega t}) \ dt \tag{4}$$ same as in $(3)$. (we can do with f(t) and sine function also.)

So I want to know

1. Whether Fourier transform formula be considered as Convolution or Correlation?

2. Also, what is meaning of $p(0)$ and $q(0)$ terms ?

Correlation and convolution are basically the same operations. You can express the cross-correlation of two functions $f(t)$ and $g(t)$ by a convolution:

$$R_{fg}(\tau)=f(\tau)\star g^*(-\tau)$$

where $\star$ denotes convolution, and $*$ denotes complex conjugation.

If you evaluate the cross-correlation at $\tau=0$ you get the inner product of $f(t)$ and $g(t)$, and that's exactly what the Fourier transform is: it is the projection of $f(t)$ on the complex exponential $e^{j\omega t}$.

1. It is actually an inner product. Due to the equivalence of convolution and correlation, it can be seen as both, evaluated at $\tau=0$.