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This module shows how to derive the scintillating and useful Fourier transform.

Fourier series clearly open the frequency domain as an interesting and useful way of determining how circuits andsystems respond to periodic input signals. Can we use similar techniques for nonperiodic signals? What isthe response of the filter to a single pulse? Addressing these issues requires us to find the Fourier spectrum of all signals,both periodic and nonperiodic ones. We need a definition for the Fourier spectrum of a signal, periodic or not. This spectrum is calculated by what is known as the Fourier transform .

Let s T t be a periodic signal having period T . We want to consider what happens to this signal's spectrum as welet the period become longer and longer. We denote the spectrum for any assumed value of the period by c k T . We calculate the spectrum according to the familiar formula

c k T 1 T t T 2 T 2 s T t 2 k t T
where we have used a symmetric placement of the integration interval about the origin for subsequent derivationalconvenience. Let f be a fixed frequency equaling k T ; we vary the frequency index k proportionally as we increase the period. Define
S T f T c k T t T 2 T 2 s T t 2 f t
making the corresponding Fourier series
s T t k S T f 2 f t 1 T
As the period increases, the spectral lines become closer together, becoming a continuum. Therefore,
T s T t s t f S f 2 f t
with
S f t s t 2 f t
S f is the Fourier transform of s t (the Fourier transform is symbolically denoted by the uppercase version of the signal's symbol) and is definedfor any signal for which the integral ( [link] ) converges.

Let's calculate the Fourier transform of the pulse signal , p t . P f t p t 2 f t t 0 Δ 2 f t 1 2 f 2 f Δ 1 P f f Δ f Δ f Note how closely this result resembles the expression for Fourier series coefficients of the periodic pulse signal .

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Spectrum

The upper plot shows the magnitude of the Fourier seriesspectrum for the case of T 1 with the Fourier transform of p t shown as a dashed line. For the bottom panel, we expanded the period to T 5 , keeping the pulse's duration fixed at 0.2, and computed itsFourier series coefficients.

[link] shows how increasing the period does indeed lead to a continuum of coefficients, and thatthe Fourier transform does correspond to what the continuum becomes. The quantity t t has a special name, the sinc (pronounced "sink") function, and is denoted by sinc t . Thus, the magnitude of the pulse's Fourier transform equals Δ sinc f Δ .

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

s S f t s t 2 f t
S s t f S f 2 f t
We must have s t s t and S f S f , and these results are indeed valid with minor exceptions.
Recall that the Fourier series for a square wave gives a valuefor the signal at the discontinuities equal to the average value of the jump. This value may differ from how the signalis 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 thedirect and inverse transforms differ only in the sign of the exponent.

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Source:  OpenStax, Fundamentals of electrical engineering i. OpenStax CNX. Aug 06, 2008 Download for free at http://legacy.cnx.org/content/col10040/1.9
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