Difference between Forward and Backward Fourier Transforms?

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The forward and backward Fourier transforms differ primarily in the sign of the exponent in their equations. The forward transform is defined as F(k) = ∫ f(x)e^{-i 2π k x} dx, while the inverse transform is f(x) = ∫ F(k)e^{i 2π k x} dk. The introduction of the minus sign in the forward transform allows for the generalization of Fourier transforms across all types of functions, ensuring the reciprocal property holds. If both transforms used the same sign, the reciprocal property would fail for odd functions or those that are neither even nor odd. Thus, the minus sign is essential for maintaining the integrity of the Fourier transform across various signal types.
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What is the difference between forward and backward Fourier transforms? I'm look:

<br /> F(k) = \int_{-\infty}^{\infty} f(x)\ e^{- i 2\pi k x }\,dx<br />

<br /> f(x) = \int_{-\infty}^{\infty} F(k)\ e^{ i 2\pi k x }\,dk<br />

If I swap the x and the k in the second equation, the transforms are then:

<br /> F(k) = \int_{-\infty}^{\infty} f(x)\ e^{- i 2\pi k x }\,dx<br />

<br /> F(k) = \int_{-\infty}^{\infty} f(x)\ e^{ i 2\pi x k }\,dx<br />

and the only difference is the minus sign in the exponent. What gives? Why aren't the forward and backwards transforms identical?
 
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Actually minus sign is introduced for generality.

Fourier transforms have reciprocal (reversible) property i.e., if you take Fourier transform on a signal two times you will end up with original signal. So if f(x) [/itex] is even function then both forward and inverse transforms can have the same sign and reciprocal property is valid. But if f(x) [/itex] is odd or neither even nor odd, then reciprocal property is lost (i.e, if you take FT two times you will end up with minus of original function) if you use same sign in forward and inverse transforms. &lt;br /&gt; &lt;br /&gt; In order to generalize Fourier transform for any signal (even, odd, neither of two) and to preserve reciprocal property minus sign is introduced.
 

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