Aperiodic Functions: From Fourier Series to Fourier Transform

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Thus far, only periodic functions have been considered though most functions are not periodic. To move from periodic functions (with period T) to aperiodic functions we simply let the period get very large, i.e., T→∞. Though this seems straightforward in concept, it fundamentally changes the nature of the transformation from time domain to frequency domain. Throughout this section we will work exclusively with the Exponential Fourier Series (which will lead to the Fourier Transform).

Behavior as T Increases

In the previous document, the Fourier Series of the pulse function was derived and discussed. To recap, the periodic pulse function ΠT(t/Tp) has Fourier Series coefficients

$${c_n} = {1 \over T}\int_T {{\Pi _t}\left( {{t \over {{T_p}}}} \right){e^{ - jn{\omega _0}t}}dt}$$ $$ {c_n} = {{{T_p}} \over T}{\mathop{\rm sinc}\nolimits} \left( {{{n{T_p}} \over T}} \right)$$

Recall that this expression uses the engineering form of the "sinc()" function (i.e., ${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( {\pi x} \right)} \over {\pi x}}$), and not the definition favored by mathematicians (${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( x \right)} \over x}$).


To understand what happens as the period T increases (with Tp=1.5), consider the images below.


T = 2    ...4   ...8   ...16  

The graph on the left side of the image shows the periodic pulse function (a function of time). The width of the pulse is 1.5 seconds and, the period of the function is chosen with the buttons at the bottom of the image. There are two horizontal scales; the top one shows time in seconds, and the bottom one shows the location of the times -2T, -T, -T/2, 0, T/2, T and 2T. The right graph shows the Fourier Series coefficients cn. Note that there are also two scales on the horizontal axis. The bottom scale is the integer index, n. The top scale is a frequency given by ω=n·ω0 where ω0=2·π/T is the frequency of the original periodic function. The term n·ω0 is, therefore, the frequency of the nth harmonic (i.e., it goes through n oscillations in one period, T). On this top scale, the spacing between the cn coefficients is ω0.

As T increases, there are two important features to note:

If T were to increase to become arbitrarily large, the magnitude of the Fourier Series coefficients goes to zero. We can keep this from happening by multiplying cn by T.

$$ {c_n} = {{{T_p}} \over T}{\mathop{\rm sinc}\nolimits} \left( {{{n{T_p}} \over T}} \right)$$

The result is shown below (where the right hand graph is now T·cn. Now as T increases (Tp=1.5), the height of c0 remains constant. However the coefficients, cn, continue to get closer together according to the ω scale (since the spacing is given by ω0=2·π/T).


T = 2    ...4   ...8   ...16  

Behavior as T→∞

Now imagine what would happen as T increases further. The overall magnitude of the coefficients is determined by the sinc() function, but the coefficients get closer and closer together (the spacing is ω0) until, as T→∞, the coefficients describe a continuous function. The value of n·ω0 can now take on any value (because ω0 is very small, and n can be any integer value). If we now call this quantity by the name ω = n·ω0 and call the value Δω = ω0. As T→∞, we replace the discrete quantity T·cn by a continuous quantity X(ω). We call this transformation from a continuous function of time, x(t), to a continuous function of frequency, X(ω), the Fourier Transform. The analysis equation for the Fourier Transform follows directly from that of the Fourier Series as T→∞.

$$\eqalign{
& T{c_n} = \int_T {x(t){e^{ - jn \cdot {\omega _0}t}}dt} \quad \quad T \to \infty \cr
& X(\omega ) = \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt} \cr} $$

Likewise, if we start from the Synthesis equation of the Fourier Series,

$$x(t) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn \cdot {\omega _0}t}}} $$

Multiply and divide the right side by T, then replace 1/T by ω0/(2*π)

$$x(t) = \sum\limits_{n = - \infty }^{ + \infty } {T{c_n}{e^{jn \cdot {\omega _0}t}}{1 \over T}} = \sum\limits_{n = - \infty }^{ + \infty } {T{c_n}{e^{jn \cdot {\omega _0}t}}{{{\omega _0}} \over {2\pi }}} $$

Make the substitutions as described in the previous paragraph (T·cn=X(ω), ω0=Δω)

$$x(t) = {1 \over {2\pi }}\sum\limits_{n = - \infty }^{ + \infty } {X(\omega ){e^{j\omega t}}\Delta \omega } $$

In the limit as T→∞ and Δω→0 this becomes an integral

$$x(t) = {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {X(\omega ){e^{j\omega t}}d\omega } $$

Example: Fourier Transform of pulse function, Π(t/Tp)

Using the definition given above, the Fourier Transform of the pulse function Π(t/Tp) is given by

$$\eqalign{
& x(t) = \Pi \left( {{t \over {{T_p}}}} \right) \cr
& X(\omega ) = \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt} = \int\limits_{ - {{{T_p}} \over 2}}^{ + {{{T_p}} \over 2}} {{e^{ - j\omega t}}dt} = - {1 \over {j\omega }}\left( {{e^{ - \omega {{{T_p}} \over 2}}} - {e^{ + \omega {{{T_p}} \over 2}}}} \right) \cr & \ \ \ \ \ \ \ \ \ = - {1 \over {j\omega }}\left( { - 2j\sin \left( {\omega {{{T_p}} \over 2}} \right)} \right) = {2 \over \omega }\sin \left( {\omega {{{T_p}} \over 2}} \right) \cr} $$

To get this into a form using the sinc() function multiply and divide by Tp and (and, later, by π), then rearrange

$$\eqalign{
& X(\omega ) = {2 \over \omega }{{{T_p}} \over {{T_p}}}\sin \left( {\omega {{{T_p}} \over 2}} \right) = {{{T_p}} \over {\omega {{{T_p}} \over 2}}}\sin \left( {\omega {{{T_p}} \over 2}} \right) = {{{T_p}} \over {\pi \omega {{{T_p}} \over {2\pi }}}}\sin \left( {\pi \omega {{{T_p}} \over {2\pi }}} \right) \cr
& X(\omega ) = {T_p}{\mathop{\rm sinc}\nolimits} \left( {\omega {{{T_p}} \over {2\pi }}} \right) \cr} $$

Thus, the pulse function, x(t)=Π(t/Tp) and the sinc() function, X(ω)=Tpsinc(ωTp/(2π)) form a Fourier Transform pair.

Onward to the Fourier Transform

If you are interested in the Fourier Transform, go here.

References

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