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Derivation of the Fourier Transform

Module by: Carlos E. Davila. E-mail the author

Summary: The Fourier transform is derived from the Fourier series.

Let's begin by writing down the formula for the complex form of the Fourier Series:

x ( t ) = ∑ n = - ∞ ∞ c n e j n Ω 0 t x ( t ) = ∑ n = - ∞ ∞ c n e j n Ω 0 t
(1)

as well as the corresponding Fourier Series coefficients:

c n = 1 T ∫ t 0 t 0 + T x ( t ) e - j n Ω 0 t d t c n = 1 T ∫ t 0 t 0 + T x ( t ) e - j n Ω 0 t d t
(2)

As was mentioned in Chapter 2, as the period TT gets large, the Fourier Series coefficients represent more closely spaced frequencies. Lets take the limit as the period TT goes to infinity. We first note that the fundamental frequency approaches a differential

Ω 0 = 2 π T → d Ω Ω 0 = 2 π T → d Ω
(3)

consequently

1 T = Ω 0 2 π → d Ω 2 π 1 T = Ω 0 2 π → d Ω 2 π
(4)

The nnth harmonic, nΩ0nΩ0, in the limit approaches the frequency variable ΩΩ

n Ω 0 → Ω n Ω 0 → Ω
(5)

From equation Equation 2, we have

c n T → ∫ - ∞ ∞ x ( t ) e - j Ω t d t c n T → ∫ - ∞ ∞ x ( t ) e - j Ω t d t
(6)

The right hand side of Equation 6 is called the Fourier Transform of x(t)x(t):

X ( j Ω ) ≡ ∫ - ∞ ∞ x ( t ) e - j Ω t d t X ( j Ω ) ≡ ∫ - ∞ ∞ x ( t ) e - j Ω t d t
(7)

Now, using Equation 6, Equation 4, and Equation 5 in equation Equation 1 gives

x ( t ) = 1 2 π ∫ - ∞ ∞ X ( j Ω ) e j Ω t d Ω x ( t ) = 1 2 π ∫ - ∞ ∞ X ( j Ω ) e j Ω t d Ω
(8)

which corresponds to the inverse Fourier Transform. Equations Equation 7 and Equation 8 represent what is known as a transform pair. The following notation is used to denote a Fourier Transform pair

x ( t ) ↔ X ( j Ω ) x ( t ) ↔ X ( j Ω )
(9)

We say that x(t)x(t) is a time domain signal while X(jΩ)X(jΩ) is a frequency domain signal. Some additional notation which is sometimes used is

X ( j Ω ) = F x ( t ) X ( j Ω ) = F x ( t )
(10)

and

x ( t ) = F - 1 X ( j Ω ) x ( t ) = F - 1 X ( j Ω )
(11)

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