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The Inverse Fourier Transform

Module by: Don Johnson. E-mail the author

Summary: This module explains the inverse Fourier transform.

The Fourier transform relates a signal's time and frequency domain representations to each other. The direct Fourier transform (or simply the Fourier transform) calculates a signal's frequency domain representation from its time-domain variant (Equation 1). The inverse Fourier transform (Equation 2) finds the time-domain representation from the frequency domain. Rather than explicitly writing the required integral, we often symbolically express these transform calculations as ℱ⁢s ℱ s and ℱ-1⁢S ℱ S , respectively.

ℱ⁢s=S⁢f=∫−∞∞s⁢t⁢e(−i)⁢2⁢π⁢f⁢tdt ℱ s S f t s t 2 f t (1)
ℱ-1⁢S=s⁢t=∫−∞∞S⁢f⁢e(i)⁢2⁢π⁢f⁢tdf ℱ S s t f S f 2 f t (2)
We must have s⁢t=ℱ-1⁢ℱ⁢s⁢t s t ℱ ℱ s t and S⁢f=ℱ⁢ℱ-1⁢S⁢f S f ℱ ℱ S f , and these results are indeed valid with minor exceptions.

Note:

Recall that the Fourier series for a square wave gives a value for the signal at the discontinuities equal to the average value of the jump. This value may differ from how the signal is defined in the time domain, but being unequal at a point is indeed minor.
Showing that you "get back to where you started" is difficult from an analytic viewpoint, and we won't try here. Note that the direct and inverse transforms differ only in the sign of the exponent.

Exercise 1

The differing exponent signs means that some curious results occur when we use the wrong sign. What is ℱ⁢S⁢f ℱ S f ? In other words, use the wrong exponent sign in evaluating the inverse Fourier transform.

Solution

ℱ⁢S⁢f=∫−∞∞S⁢f⁢e(−i)⁢2⁢π⁢f⁢tdf=∫−∞∞S⁢f⁢e(i)⁢2⁢π⁢f⁢(−t)df=s⁢−t ℱ S f f S f 2 f t f S f 2 f t s t

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