Fourier Series from Fourier Transform

Contents

Intro - Calculating Fourier Series Coefficients without Integration

We derived the Fourier Transform as an extension of the Fourier Series to non-periodic function. Then we developed methods to find the Fourier Transform using tables of functions and properties, so as to avoid integration. Now we can come full circle and use these methods to calculate the Fourier Series of a aperiodic function from a Fourier Transform of one period of the function. In other words, we will calculate the Fourier Series coefficients without integration! We start with a description of the relationship between the Fourier Transform of a function and the Fourier Series of its periodic extension.

Relationship between Fourier Transform of x(t) and Fourier Series of xT(t)

Consider an aperiodic function, x(t), of finite extent (i.e., it is only non-zero for a finite interval of time). In the diagram below this function is a rectangular pulse.

The periodic extension of x(t) is called xT(t), and is just x(t) replicated every T seconds, such that it is periodic with period T (i.e., xT(t+nT)=x(t), with n an integer).

The question to be answered is: how are the Fourier Transform of x(t) and the Fourier Series for xT(t) related?

The Fourier Transform of x(t) is X(ω) and is given by:

$$X(\omega ) = \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt} $$

The Fourier Series coeffients, cn, of xT(T) are given by:

$${c_n} = {1 \over T}\int_T {{x_T}(t){e^{ - jn{\omega _0}t}}dt} = {1 \over T}\int\limits_{ - {T \over 2}}^{ + {T \over 2}} {{x_T}(t){e^{ - jn{\omega _0}t}}dt} $$

If the function x(t) is only non-zero in the inerval (-T/2, +T/2) we can rewrite the Fourier Transform as

$$X(\omega ) = \int\limits_{ - {T \over 2}}^{ + {T \over 2}} {x(t){e^{ - j\omega t}}dt} $$

Comparing the equations and noting the xT(t)=x(t) over the inverval of integration, we can see that the relationship between the Fourier series coefficients, cn, and the Fourier Transform, X(ω), is given by

$${c_n} = {1 \over T}X(n{\omega _0})$$

In the particular example of the rectangular pulse we know

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

For the periodic pulse with height =1, we know

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

We could also get cn directly from X(ω):

$${1 \over T}X(n{\omega _0}) = {1 \over T}\left( {{T_p}{\mathop{\rm sinc}\nolimits} \left( {{{n{\omega _0}{T_p}} \over {2\pi }}} \right)} \right) = {{{T_p}} \over T}{\mathop{\rm sinc}\nolimits} \left( {{{n{{2\pi } \over T}{T_p}} \over {2\pi }}} \right) = {{{T_p}} \over T}{\mathop{\rm sinc}\nolimits} \left( {{{n{T_p}} \over T}} \right) = {c_n}$$
Aside: Relationship between Fourier Series and Fourier Transform for Periodic Function

The relationship between the Fourier Transform and Fourier Series representation of a periodic function was derived earlier and is repeated here. If xT(t) has a Fourier Series representation

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

then its Fourier Transform is

$${X_T}\left( \omega \right) = 2\pi \sum\limits_{n = - \infty }^{ + \infty } {{c_n}\delta (\omega - n{\omega _0})} $$
Key Concept: Relationship between Fourier Series and Fourier Transform

If xT(t) is the periodic extension of x(t) (with period T) then the Fourier Series coefficients, cn, of xT(t) and Fourier Transform, X(ω), of x(t) are related by:

$${c_n} = {1 \over T}X(n{\omega _0})$$

Note: The Fourier Transform of xT(t) is given by: ${X_T}\left( \omega \right) = 2\pi \sum\limits_{n = - \infty }^{ + \infty} {{c_n}\delta (\omega - n{\omega _0})} $.


These relationships are spelled out on a one-page pdf.

The Periodic Rectangular Pulse

This is the example given above.

Find the Fourier Series representation of the periodic pulse train xT(t)=ΠT(t/Tp). Since xT(t) is the periodic extension of x(t)=Π(t/Tp), and we know from a Fourier Transform table (or from previous work)

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

then

$${c_n} = {1 \over T}X(n{\omega _0}) = {1 \over T}\left( {{T_p}{\mathop{\rm sinc}\nolimits} \left( {{{n{\omega _0}{T_p}} \over {2\pi }}} \right)} \right) = {{{T_p}} \over T}{\mathop{\rm sinc}\nolimits} \left( {{{n{{2\pi } \over T}{T_p}} \over {2\pi }}} \right) = {{{T_p}} \over T}{\mathop{\rm sinc}\nolimits} \left( {{{n{T_p}} \over T}} \right)$$

The Periodic Triangular Pulse

Find the Fourier Series representation of the periodic triangular pulse xT(t)=ΛT(t/Tp).

From the Fourier Transform table we know the transform, X(ω) of a single triangular pulse (x(t)=Λ(t/Tp)) is given by:

$$X(\omega ) = {T_p}{{\mathop{\rm sinc}\nolimits} ^2}\left( {{{\omega {T_p}} \over {2\pi }}} \right)$$

so

$${c_n} = {1 \over T}X(n{\omega _0}) = {1 \over T}\left( {{T_p}{{{\mathop{\rm sinc}\nolimits} }^2}\left( {{{n{\omega _0}{T_p}} \over {2\pi }}} \right)} \right) = {{{T_p}} \over T}{{\mathop{\rm sinc}\nolimits} ^2}\left( {{{n{{2\pi } \over T}{T_p}} \over {2\pi }}} \right) = {{{T_p}} \over T}{{\mathop{\rm sinc}\nolimits} ^2}\left( {{{n{T_p}} \over T}} \right)$$

The Triangle Function

Find the Fourier Series representation of the triangle wave, xT(t), shown.

While perhaps not obvious at first, the corresponding aperiodic triangle wave, x(t) is simply the sum of a triangle pulse (2·Λ(t/(T/2))) plus a negative pulse (-Π(t/T)), as shown:

$$x(t) = 2\Lambda \left( {{t \over {\left( {{T \over 2}} \right)}}} \right) - \Pi \left( {{t \over T}} \right)$$

We can find X(ω) from the Fourier Transform table, and then find cn.

$$\eqalign{
X\left( \omega \right) &= 2 \cdot {T \over 2}{{\mathop{\rm sinc}\nolimits} ^2}\left( {{{T\omega } \over {4\pi }}} \right) - T{\mathop{\rm sinc}\nolimits} \left( {{{T\omega } \over {2\pi }}} \right) \cr
{c_n} &= {1 \over T}X(n{\omega _0}) = {1 \over T}\left( {2 \cdot {T \over 2}{{{\mathop{\rm sinc}\nolimits} }^2}\left( {{{Tn{\omega _0}} \over {4\pi }}} \right) - T{\mathop{\rm sinc}\nolimits} \left( {{{Tn{\omega _0}} \over {2\pi }}} \right)} \right)\quad \quad \quad \quad T{\omega _0} = 2\pi \cr
&= {{\mathop{\rm sinc}\nolimits} ^2}\left( {{{2\pi n} \over {4\pi }}} \right) - {\mathop{\rm sinc}\nolimits} \left( {{{2\pi n} \over {2\pi }}} \right) = {{\mathop{\rm sinc}\nolimits} ^2}\left( {{n \over 2}} \right) - {\mathop{\rm sinc}\nolimits} \left( n \right) \cr} $$

We could stop here, but instead we will show that this is equivalent to a previous result obtained through integration. First we note that c0=0, and for all other values of n, sinc(n)=0, so we can ignore the second term.

$${c_n} = {{\mathop{\rm sinc}\nolimits} ^2}\left( {{n \over 2}} \right) = {{{{\sin }^2}\left( {\pi {n \over 2}} \right)} \over {{{\left( {\pi {n \over 2}} \right)}^2}}} = 4{{{{\sin }^2}\left( {\pi {n \over 2}} \right)} \over {{{\left( {\pi n} \right)}^2}}},\quad \quad \quad n \ne 0$$

By brute force integration and a lot of algebra, we get the result (in the example in the link A=1) in the form of a Fouries Cosine series

$${a_n} = 8{{{{\sin }^2}\left( {\pi {n \over 2}} \right)} \over {{{\left( {\pi n} \right)}^2}}}$$

Which agrees with our answer because we know 2Re(cn)=an

The Periodic Impulse Train

Find the Fourier Series representation of a periodic impulse train, ${x_T}\left( t \right) = \sum\limits_{n = - \infty }^{ + \infty } {\delta \left( {t - nT} \right)} $.

The aperiodic function is a single pulse at the origin, x(t)=δ(t); so X(ω)=1, and cn=X(nω0)/T=1/T.

A Half-Wave Rectified Sine Wave.

Find the Fourier Series representation of xT(t) if it is a half-wave rectified sine wave as shown.

We can get create the aperiodic version of the wave by multiplying a sine wave (with period T) byt a rectangular pulse of width T/2 delayed by a time T/4.

$$x\left( t \right) = \left( {\sin \left( {{{2\pi t} \over T}} \right)} \right) \cdot \left( {\Pi \left( {{{t - {T \over 4}} \over {{T \over 2}}}} \right)} \right)$$

We find the Fourier Transform of both functions from the Fourier Transform table (using the time shift property with the rectangular pulse), and convolve (recall that multiplication in time is convolution in frequency.

$$\eqalign{
x\left( t \right) &= \left( {\sin \left( {{{2\pi t} \over T}} \right)} \right) \cdot \left( {\Pi \left( {{{t - {T \over 4}} \over {{T \over 2}}}} \right)} \right) = {f_1}\left( t \right) \cdot {f_2}\left( t \right) \cr
{f_1}\left( t \right) &= \sin \left( {{{2\pi t} \over T}} \right)\quad \quad \quad {F_1}\left( \omega \right) = j\pi \left( {\delta \left( {\omega + {{2\pi } \over T}} \right) - \delta \left( {\omega - {{2\pi } \over T}} \right)} \right) \cr
{f_2}\left( t \right) &= \Pi \left( {{{t - {T \over 4}} \over {{T \over 2}}}} \right)\quad \quad \quad {F_2}\left( \omega \right) = {T \over 2}{\mathop{\rm sinc}\nolimits} \left( {{{T\omega } \over {4\pi }}} \right){e^{ - j\omega {T \over 4}}} \cr
X\left( \omega \right) &= {1 \over {2\pi }}{F_1}\left( \omega \right) * {F_2}\left( \omega \right) \cr
&= {1 \over {2\pi }}\left( {j\pi \left( {\delta \left( {\omega + {{2\pi } \over T}} \right) - \delta \left( {\omega - {{2\pi } \over T}} \right)} \right)} \right) * \left( {{T \over 2}{\mathop{\rm sinc}\nolimits} \left( {{{T\omega } \over {4\pi }}} \right){e^{ - j\omega {T \over 4}}}} \right) \cr
&= {{jT} \over 4}\left( {\delta \left( {\omega + {{2\pi } \over T}} \right) - \delta \left( {\omega - {{2\pi } \over T}} \right)} \right) * \left( {{\mathop{\rm sinc}\nolimits} \left( {{{T\omega } \over {4\pi }}} \right){e^{ - j\omega {T \over 4}}}} \right) \cr} $$

Recall that any function convolved with a shifted impulse is just a shifted version of that function: f(x)*δ(x-x0)=f(x-x0), so this simplifies to

$$X\left( \omega \right) = {{j T} \over 4}\left( {{\mathop{\rm sinc}\nolimits} \left( {{{T\left( {\omega + {{2\pi } \over T}} \right)} \over {4\pi }}} \right){e^{ - j\left( {\omega + {{2\pi } \over T}} \right){T \over 4}}} - {\mathop{\rm sinc}\nolimits} \left( {{{T\left( {\omega - {{2\pi } \over T}} \right)} \over {4\pi }}} \right){e^{ - j\left( {\omega - {{2\pi } \over T}} \right){T \over 4}}}} \right)$$

From this the Fourier Series coefficients are determined.

$${c_n} = {1 \over T}X(n{\omega _0}) = {j \over 4}\left( {{\mathop{\rm sinc}\nolimits} \left( {{{T\left( {n{\omega _0} + {{2\pi } \over T}} \right)} \over {4\pi }}} \right){e^{ - j\left( {n{\omega _0} + {{2\pi } \over T}} \right){T \over 4}}} - {\mathop{\rm sinc}\nolimits} \left( {{{T\left( {n{\omega _0} - {{2\pi } \over T}} \right)} \over {4\pi }}} \right){e^{ - j\left( {n{\omega _0} - {{2\pi } \over T}} \right){T \over 4}}}} \right)$$

Simplify by using the fact the ω0T=2π.

$$\eqalign{
{c_n} &= {j \over 4}\left( {{\mathop{\rm sinc}\nolimits} \left( {{{nT{\omega _0} + 2\pi } \over {4\pi }}} \right){e^{ - j\left( {{{n{\omega _0}T} \over 4} + {\pi \over 2}} \right)}} - {\mathop{\rm sinc}\nolimits} \left( {{{nT{\omega _0} - 2\pi } \over {4\pi }}} \right){e^{ - j\left( {{{n{\omega _0}T} \over 4} - {\pi \over 2}} \right)}}} \right) \cr
&= {j \over 4}\left( {{\mathop{\rm sinc}\nolimits} \left( {{{n2\pi + 2\pi } \over {4\pi }}} \right){e^{ - j\left( {{{n2\pi } \over 4} + {\pi \over 2}} \right)}} - {\mathop{\rm sinc}\nolimits} \left( {{{n2\pi - 2\pi } \over {4\pi }}} \right){e^{ - j\left( {{{n2\pi } \over 4} - {\pi \over 2}} \right)}}} \right) \cr
&= {j \over 4}\left( {{\mathop{\rm sinc}\nolimits} \left( {{{n + 1} \over 2}} \right){e^{ - j{{\left( {n + 1} \right)\pi } \over 2}}} - {\mathop{\rm sinc}\nolimits} \left( {{{n - 1} \over 2}} \right){e^{ - j{{\left( {n - 1} \right)\pi } \over 2}}}} \right) \cr} $$

Further simplification (not shown) yields:

$${c_n} = - {{{{\left( { - 1} \right)}^n} + 1} \over {2{\kern 1pt} \pi {\kern 1pt} \left( {{n^2} - 1} \right)}} = \left\{ {\matrix{
{0,\quad n\;odd} \cr
{{1 \over {\pi {\kern 1pt} \left( {1 - {n^2}} \right)}}} \cr } } \right.,\quad n\;even$$

with the values for c1 and c-1 determined by application of L'Hôpital's rule.


References

© Copyright 2005 to 2015 Erik Cheever    This page may be freely used for educational purposes.

Erik Cheever       Department of Engineering         Swarthmore College