Fourier Series Examples

This document derives the Fourier Series coefficients for several functions. The functions shown here are fairly simple, but the concepts extend to more complex functions.

Even Pulse Function (Cosine Series)

Consider the periodic pulse function shown below. It is an even function with period T. The function is a pulse function with amplitude A, and pulse width T_p. The function can be defined over one period (centered around the origin) as:

$$x_T(t) = \left\{ {\matrix{
{A,\quad |t| \le {{{T_p}} \over 2}} \cr
{0,\quad |t| \gt {{{T_p}} \over 2}} \cr
} } \right.,\quad \quad - {T \over 2} < t \le {T \over 2}$$

Aside: the periodic pulse function

The periodic pulse function can be represented in functional form as Π_T(t/T_p). During one period (centered around the origin)

$$\Pi_T(t) = \left\{ {\matrix{
{1,\quad |t| \le {1 \over 2}} \cr
{0,\quad |t| \gt {1 \over 2}} \cr
} } \right.,\quad \quad - {T \over 2} < t \le {T \over 2}$$

Π_T(t) represents a periodic function with period T and pulse width ½. The pulse is scaled in time by T_p in the function Π_T(t/T_p) so:

$${\Pi _T}\left( {{t \over {{T_p}}}} \right) = \left\{ {\matrix{
{1,\quad |t| \le {{{T_p}} \over 2}} \cr
{0,\quad |t| > {{{T_p}} \over 2}} \cr } } \right.,\quad \quad - {T \over 2} < t \le {T \over 2}$$

This can be a bit hard to understand at first, but consider the sine function. The function sin(x/2) twice as slow as sin(x) (i.e., each oscillation is twice as wide). In the same way Π_T(t/2) is twice as wide (i.e., slow) as Π_T(t).

The Fourier Series representation is

$${x_T}(t) = {a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \left( {n{\omega _0}t} \right) + {b_n}\sin \left( {n{\omega _0}t} \right)} \right)} $$

Since the function is even there are only a_n terms.

$${x_T}(t) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {n{\omega _0}t} \right)} = \sum\limits_{n = 0}^\infty {{a_n}\cos \left( {n{\omega _0}t} \right)} $$

The average is easily found,

$$a_0=A\frac{T_p}T$$

The other terms follow from

$${a_n} = {2 \over T}\int\limits_T {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} ,\quad n \ne 0$$

Any interval of one period is allowed but the interval from -T/2 to T/2 is straightforward in this case.

$${a_n} = {2 \over T}\int\limits_{{-{T} \over 2}}^{{T \over 2}} {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} $$

Since x_T(t)=A between -T_p/2 to +T_p/2 and zero elsewhere the integral simplifies and can be solved

$$\begin{align}
{a_n} &= {2 \over T}\int\limits_{ - {{{T_p}} \over 2}}^{ + {{{T_p}} \over 2}} {A\cos \left( {n{\omega _0}t} \right)dt} \cr
&= \left. {{2 \over T}{A \over {n{\omega _0}}}\sin \left( {n{\omega _0}t} \right)} \right|_{ - {{{T_p}} \over 2}}^{ + {{{T_p}} \over 2}} \cr
&= {2 \over T}{A \over {n{\omega _0}}}\left( {\sin \left( { + n{\omega _0}{{{T_p}} \over 2}} \right) - \sin \left( { - n{\omega _0}{{{T_p}} \over 2}} \right)} \right) \cr\end{align} $$

Since sine is an odd function, sin(a)-sin(-a)=2sin(a), and using the fact that ω₀=2π/T and

$${a_n} = {4 \over T}{A \over {n{\omega _0}}}\sin \left( {n{\omega _0}{{{t_p}} \over 2}} \right) = 4{A \over {n2\pi }}\sin \left( {n\pi {{{t_p}} \over T}} \right) = 2{A \over {n\pi }}\sin \left( {n\pi {{{t_p}} \over T}} \right)$$

This result is further explored in two examples.

Example 1: Special case, Duty Cycle = 50%

Consider the case when the duty cycle is 50% (this means that the function is high 50% of the time, or T_p=T/2), A=1, and T=2. In this case a₀=average=0.5 and for n≠0:

$$\displaylines{
{a_n} = 2{A \over {n\pi }}\sin \left( {n\pi {{{t_p}} \over T}} \right) = 2{A \over {n\pi }}\sin \left( {{{n\pi } \over 2}} \right) \cr
n = 0,\,1,\,2,\,3,\,4,\,5,...\quad \sin \left( {{{n\pi } \over 2}} \right) = 0,1,0, - 1,0,1,... = \left\{ {\matrix{
{ - {1^{{{n - 1} \over 2}}},\quad n\;odd} \cr
{0,\quad n\;even} \cr } } \right. \cr
{a_n} = \left\{ {\matrix{
{2{A \over {n\pi }}\left( { - {1^{{{n - 1} \over 2}}}} \right),\quad n\;odd} \cr
{\quad 0,\quad n\;even,\;n \ne 0\;} \cr } } \right. \cr} $$

The values for a_n are given in the table below. Note: this example was used on the page introducing the Fourier Series. Note also, that in this case a_n (except for n=0) is zero for even n, and decreases as 1/n as n increases.

n	a_n
0	0.5
1	0.6366
2	0
3	-0.2122
4	0
5	0.1273
6	0
7	-0.0909

Average + 1^st harmonic up to 3^rd harmonic ...5th harmonic ...7^th ...21^st

The graph shows the function x_T(t) (blue) and the partial Fourier Sum (from n=0 to n=N) (red)

$$\sum\limits_{n = 0}^N {{a_n}\cos ({\omega _0}t)} $$

as well as the highest frequency harmonic, $a_Ncos(N\omega_0t)$ (dotted magenta). Lower frequency harmonics in the summation are thin dotted blue lines (but harmonics with $a_n0$ are not shown. You can change n by clicking the buttons. As before, note:

As you add sine waves of increasingly higher frequency, the approximation improves.
The addition of higher frequencies better approximates the rapid changes, or details, (i.e., the discontinuity) of the original function (in this case, the square wave).
Gibb's overshoot exists on either side of the discontinuity.
Because of the symmetry of the waveform, only odd harmonics (1, 3, 5, ...) are needed to approximate the function. The reasons for this are discussed below
The rightmost button shows the sum of all harmonics up to the 21st harmonic, but not all of the individual sinusoids are explicitly shown on the plot. In particular harmonics between 7 and 21 are not shown.

Example 2: Special case, Duty Cycle = 40%

Now consider the case when the duty cycle is 40%, A=1, and T=2. In this case a₀=average=0.4 and for n≠0:

$${a_n} = {4 \over T}{A \over {n{\omega _0}}}\sin \left( {n{\omega _0}{{{t_p}} \over 2}} \right) = 2{A \over {n\pi }}\sin \left( {0.4n\pi } \right)$$

The values for a_n are given in the table below (note: this example was used on the previous page).

n	a_n
0	0.4
1	0.6055
2	0.1871
3	-0.1247
4	-0.1514
5	-0.0000
6	0.1009
7	0.0535

Average + 1^st harmonic up to 2^nd harmonic ...3^rd harmonic ...4^th ...21^st

The graph shows the function x_T(t) (blue) and the partial Fourier Sum (from n=0 to n=N) (red)

$$\sum\limits_{n = 0}^N {{a_n}\cos ({\omega _0}t)} $$

Note that because this example is similar to the previous one, the coefficients are similar, but they are no longer equal to zero for n even.

Even Square Wave (Exploiting Symmetry)

In problems with even and odd functions, we can exploit the inherent symmetry to simplify the integral. We will exploit other symmetries later. Consider the problem above. We have an expression for a_n, n≠0

$${a_n} = {2 \over T}\int\limits_{ - {T \over 2}}^{{T \over 2}} {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} $$

If x_T(t) is even, then the product x_T(t)·cos(n·ω₀t) is even (the product of two even functions is even). We can then use the fact that for an even function, e(t),

$$\int\limits_{ - a}^{ + a} {e(t)dt} = 2\int\limits_0^a {e(t)dt} $$

$${a_n} = {4 \over T}\int\limits_0^{{T \over 2}} {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} $$

which generates the same answer as before. This will often be simpler to evaluate than the original integral because one of the limits of integration is zero.

Even Square Wave (Exponential Series)

Consider, again, the pulse function. We can also represent x_T(t) by the Exponential Fourier Series

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

We find the c_n

$${c_n} = \int\limits_T {{x_T}(t){e^{ - jn{\omega _0}t}}dt} $$

As before the integral is from -T/2 to +T/2 and make use of the facts that the function is constant for |t|<T_p/2 and zero elsewhere, and the T·ω₀=2*·π.

$$\begin{align}
{c_n} &= {1 \over T}\int\limits_{ - {T \over 2}}^{ + {T \over 2}} {{x_T}(t){e^{ - jn{\omega _0}t}}dt} = {1 \over T}\int\limits_{ - {{{T_p}} \over 2}}^{ + {{{T_p}} \over 2}} {A{e^{ - jn{\omega _0}t}}dt} \cr
&= \left. {{A \over { - Tjn{\omega _0}}}{e^{ - jn{\omega _0}t}}} \right|_{ - {{{T_p}} \over 2}}^{ + {{{T_p}} \over 2}} = {A \over { - j2\pi n}}\left( {{e^{ - jn{\omega _0}{{{T_p}} \over 2}}} - {e^{ + jn{\omega _0}{{{T_p}} \over 2}}}} \right) \cr\end{align} $$

Euler's identities dictate that e^+jθ-e^-jθ=2jsin(θ) so e^-jθ-e^+jθ=-2jsin(θ).

and $${c_n} = {{ - 2jA} \over { - j2\pi n}}\sin \left( {n{\omega _0}{{{T_p}} \over 2}} \right) = {A \over {\pi n}}\sin \left( {n{\omega _0}{{{T_p}} \over 2}} \right)$$

Note that, as expected, c₀=a₀ and c_n=a_n/2, (n≠0) (since this is an even function b_n=0).

Even Triangle Wave (Cosine Series)

Consider the triangle wave

The average value (i.e., the 0^th Fourier Series Coefficients) is a₀=0. For n>0 other coefficients the even symmetry of the function is exploited to give

$${a_n} = {2 \over T}\int\limits_T {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} = {2 \over T}\int\limits_{ - {T \over 2}}^{ + {T \over 2}} {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} = {4 \over T}\int\limits_0^{ + {T \over 2}} {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} $$

Between t=0 and t=T/2 the function is defined by x_T(t)=A-4At/T so

$${a_n} = {4 \over T}\int\limits_0^{ + {T \over 2}} {\left( {A - {{4A} \over T}t} \right)\cos \left( {n{\omega _0}t} \right)dt} = {{4A} \over T}\left( {\int\limits_0^{ + {T \over 2}} {\cos \left( {n{\omega _0}t} \right)dt} - {4 \over T}\int\limits_0^{ + {T \over 2}} {t\cos \left( {n{\omega _0}t} \right)dt} } \right)$$

Perform the integrations (either by hand using integration by parts, or with a table of integrals, or by computer) and use the fact that ω₀·T=2·π

$${a_n} = {{4A} \over T}\left( {{{T\sin (\pi n)} \over {2\pi n}} + {4 \over T}{{{T^2}{\mkern 1mu} \left( {2{\mkern 1mu} \sin {{\left( {{{\pi {\mkern 1mu} n} \over 2}} \right)}^2} - \pi {\mkern 1mu} n{\mkern 1mu} \sin \left( {\pi {\mkern 1mu} n} \right)} \right)} \over {4{\mkern 1mu} {\pi ^2}{\mkern 1mu} {n^2}}}} \right)$$

Since sin(π·n)=0 this simplifies to

$${a_n} = {{4A} \over T}{4 \over T}{{{T^2}{\mkern 1mu} 2{\mkern 1mu} \sin {{\left( {{{\pi {\mkern 1mu} n} \over 2}} \right)}^2}} \over {4{\mkern 1mu} {\pi ^2}{\mkern 1mu} {n^2}}} = {{8A\sin {{\left( {{{\pi {\mkern 1mu} n} \over 2}} \right)}^2}} \over {{\pi ^2}{\mkern 1mu} {n^2}}}$$

This answer is correct, but noting that

$$n = 0,1,2,3,4,5,6,7,...\quad \quad \sin {\left( {{{\pi {\mkern 1mu} n} \over 2}} \right)^2} = 0,1,0,1,0,1,0,... = {{{ 1 -{\left( { - 1} \right)}^n}} \over 2}$$

yields an even simpler result

$${a_n} = \left\{ {\matrix{
{4A{{1 - {{\left( { - 1} \right)}^n}} \over {{\pi ^2}{\mkern 1mu} {n^2}}},\quad {\rm{n odd}}} \cr
{0,\quad {\rm{n even}}} \cr } } \right.$$

Example 3: Triangle wave

If x_T(t) is a triangle wave with A=1, the values for a_n are given in the table below (note: this example was used on the previous page).

n	a_n
0	0
1	0.8106
2	0
3	0.0901
4	0
5	0.0324
6	0
7	0.0165

Average + 1^st harmonic up to 3^rd harmonic ...5^th harmonic ...7^th ...9^th

Note: this is similar, but not identical, to the triangle wave seen earlier.

Note:

As you add sine waves of increasingly higher frequency, the approximation gets better and better, and these higher frequencies better approximate the details, (i.e., the change in slope) in the original function.
The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n² (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). Conceptually, this occurs because the triangle wave looks much more like the 1st harmonic, so the contributions of the higher harmonics are less. Even with only the 1st few harmonics we have a very good approximation to the original function.
There is no discontinuity, so no Gibb's overshoot.
As before, only odd harmonics (1, 3, 5, ...) are needed to approximate the function; this is because of the symmetry of the function.

Odd Function (Sawtooth Wave)

Thus far, the functions considered have all been even. The diagram below shows an odd function.

In this case, a Fourier Sine Series is appropriate

$${x_T}(t) = \sum\limits_{n = 1}^\infty {{b_n}\sin \left( {n{\omega _0}t} \right)} \quad \quad {b_n} = {2 \over T}\int\limits_T {{x_T}\left( t \right)\sin \left( {n{\omega _0}t} \right)dt} $$

It is easiest to integrate from -T/2 to +T/2. Over this interval $x_T(t)=2At/T$.

$${b_n} = {2 \over T}\int\limits_{ - {T \over 2}}^{ + {T \over 2}} {{x_T}\left( t \right)\sin \left( {n{\omega _0}t} \right)dt} = {2 \over T}\int\limits_{ - {T \over 2}}^{ + {T \over 2}} {{{2At} \over T}\sin \left( {n{\omega _0}t} \right)dt} $$

Performing the integration (and using the fact that ω₀·T=2·π) the integral yields

$${b_n} = {2 \over T}{{A{\mkern 1mu} T{\mkern 1mu} \left( {\sin \left( {\pi {\mkern 1mu} n} \right) - \pi {\mkern 1mu} n{\mkern 1mu} \cos \left( {\pi {\mkern 1mu} n} \right)} \right)} \over {{\pi ^2}{\mkern 1mu} {n^2}}}$$

Using two simplification, sin(π·n)=0 and cos(π·n)=(-1)ⁿ gives

$${b_n} = - {{2A{\mkern 1mu} } \over {\pi n}}{\left( { - 1} \right)^n}$$

Aside: using symmetry

In this case since x_T(t) is odd and is multiplied by another odd function (sin(n·ω₀t)), their product is even and the integral can be rewritten as:

$${b_n} = {2 \over T}\int\limits_{ - {T \over 2}}^{ + {T \over 2}} {{x_T}\left( t \right)\sin \left( {n{\omega _0}t} \right)dt} = {b_n} = {4 \over T}\int\limits_0^{ + {T \over 2}} {{x_T}\left( t \right)\sin \left( {n{\omega _0}t} \right)dt} $$

Example 4: Odd Sawtooth Wave

If x_T(t) is a sawtooth wave with A=1, the values for b_n are given in the table below

n	b_n
1	0.6366
2	-0.3183
3	0.2122
4	-0.1592
5	0.1273
6	-0.1061
7	0.0909

Average + 1^st harmonic up to 2^nd harmonic ...3^rd ...4^th ...5^th ...20^th

Note: this is similar, but not identical, to the sawtooth wave seen earlier.

Note:

As you add sine waves of increasingly higher frequency, the approximation gets better and better, and these higher frequencies better approximate the details, (i.e., the change in slope) in the original function.
Since this function doesn't look as much like a sinusoid as the triangle wave, the coefficients decrease less rapidly (as 1/n instead of 1/n²
There is Gibb's overshoot caused by the discontinuity.

Functions that are neither even nor odd

So far, all of the functions considered have been either even or odd, but most functions are neither. This presents no conceptual difficult, but may require more integrations. For example if the function x_T(t) looks like the one below

Since this has no obvious symmetries, a simple Sine or Cosine Series does not suffice. For the Trigonometric Fourier Series, this requires three integrals

$$\begin{align}
{x_T}(t) &= {a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \left( {n{\omega _0}t} \right) + {b_n}\sin \left( {n{\omega _0}t} \right)} \right)} \cr
{a_0} &= {2 \over T}\int\limits_T {{x_T}\left( t \right)dt} \cr
{a_n} &= {2 \over T}\int\limits_T {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt} ,\quad n \ne 0 \cr
{b_n} &= {2 \over T}\int\limits_T {{x_T}\left( t \right)\sin \left( {n{\omega _0}t} \right)dt} \cr\end{align} $$

However, an exponential series requires only a single integral

$$\displaylines{
{x_T}(t) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn{\omega _0}t}}} \cr
{c_n} = {1 \over T}\int\limits_T {{x_T}\left( t \right){e^{ - jn{\omega _0}t}}dt} \cr} $$

For this reason, among others, the Exponential Fourier Series is often easier to work with, though it lacks the straightforward visualization afforded by the Trigonometric Fourier Series.

Example 5: Neither Even nor Odd

In this case, but not in general, we can easily find the Fourier Series coefficients by realizing that this function is just the sum of the square wave (with 50% duty cycle) and the sawtooth so

n	a_n	b_n
0	0.5	----
1	0.6366	0.6366
2	0	-0.3183
3	-0.2122	0.2122
4	0	-0.1592
5	0.1273	0.1273
6	0	-0.1061
7	-0.0909	0.0909

From the relationship between the Trigonometric and Exponential Fourier Series

$$c_0=a_0 \ and \ c_n = \frac{a_n}2 - j\frac{b_n}2 for n \neq 0, \quad with \ c_{-n}=c_n^*$$

n	c_n
-7	-0.0455 + 0.0455j
-6	-0.0531j
-5	0.0637 + 0.0637j
-4	-0.0796j
-3	-0.1061 + 0.1061j
-2	-0.1592j
-1	0.3183 + 0.3183j
0	0.5
1	0.3183 - 0.3183j
2	0.1592j
3	-0.1061 - 0.1061j
4	0.0796j
5	0.0637 - 0.0637j
6	0.0531j
7	-0.0455 - 0.0455j

Average + 1^st harmonic up to 2^nd harmonic ...3^rd ...4^th ...5^th ...20^th

Note: this is similar, but not identical, to the sawtooth wave seen earlier.

Note:

As you add sine waves of increasingly higher frequency, the approximation gets better and better, and these higher frequencies better approximate the details, (i.e., the change in slope) in the original function.
There is Gibb's overshoot caused by the discontinuities.

Effect of Function Symmetry on Coefficients

If the function x_T(t) has certain symmetries, we can simplify the calculation of the coefficients.

Symmetry Trigonometric Series and Symmetry

Symmetry	Simplification
x_T(t) is even	$$\displaylines{ {a_0} = average \cr {a_n} = {4 \over T}\int\limits_0^{ + {T \over 2}} {{x_T}(t)\cos \left( {n{\omega _0}t} \right)dt} , \quad n \neq 0 \cr {b_n} = 0 \cr} $$
x_T(t) is odd	$$\displaylines{ {a_n} = 0 \cr {b_n} = {4 \over T}\int\limits_0^{ + {T \over 2}} {{x_T}(t)\sin \left( {n{\omega _0}t} \right)dt} \cr} $$
x_T(t) has half-wave symmetry A periodic function has half wave symmetry if f(t-T/2)=-f(t). In other words, if you shift the function by half of a period, then the resulting function is the opposite the original function. The triangle wave has half-wave symmetry. See below for clarification. A function can have half-wave symmetry without being either even or odd.	$$\displaylines{ {a_n} = {b_n} = 0,\quad n\;even \cr {a_n} = {4 \over T}\int\limits_0^{ + {T \over 2}} {{x_T}(t)\cos \left( {n{\omega _0}t} \right)dt} ,\quad n\;odd \cr {b_n} = {4 \over T}\int\limits_0^{ + {T \over 2}} {{x_T}(t)\sin \left( {n{\omega _0}t} \right)dt} ,\quad n\;odd \cr} $$

The first two symmetries are were discussed previously in the discussions of the pulse function (x_T(t) is even) and the sawtooth wave (x_T(t) is odd).

Half-wave symmetry is depicted the diagram below.

The top function, x_T1(t), is odd (x_T1(t)=-x_T1(-t)), but does not have half-wave symmetry. The bottom function, x_T2(t) is nether even nor odd, but since x_T2(t)=-x_T2(t-T/2), it has halfwave symmetry. To visualize this imagine shifting the function by half a period (T/2); for half-wave symmetry the shifted function should be the mirror image of the original function (about the horizontal axis) as shown below

The reason the coefficients of the even harmonics are zero can be understood in the context of the diagram below. The top graph shows a function, x_T(t) with half-wave symmetry along with the first four harmonics of the Fourier Series (only sines are needed because x_T(t) is odd). The bottom graph shows the harmonics multiplied by x_T(t).

Now imagine integrating the product terms from -T/2 to +T/2. The odd terms (from the 1st (red) and 3rd (magenta) harmonics) will have a positive result (because they are above zero more than they are below zero). The even terms (green and cyan) will integrate to zero (because they are equally above and below zero). Though this is a simple example, the concept applies for more complicated functions, and for higher harmonics.

The only function discussed with half-wave symmetry was the triangle wave and indeed the coefficients with even indices are equal to zero (as are all of the b_n terms because of the even symmetry). The square wave with 50% duty cycle would have half wave symmetry if it were centered around zero (i.e., centered on the horizontal axis). In that case the a₀ term would be zero and we have already shown that all the terms with even indices are zero, as expected.

Simplifications can also be made based on quarter-wave symmetry, but these are not discussed here.

A periodic function has quarter wave symmetry if it has half wave symmetry and it is either even or odd around its two half-cycles.

Exponential Series and Symmetry

Since the coefficients c_n of the Exponential Fourier Series are related to the Trigonometric Series by

$$\displaylines{
{c_0} = {a_0} \cr
{c_n} = {{{a_n}} \over 2} - j{{{b_n}} \over 2}for\;n \ne 0 \cr
{c_{ - n}} = c_n^* \cr} $$

(assuming x_T(t) is real) we can use the symmetry properties of the Trigonometric Series to find a_n and b_n and hence c_n.

However, in addition, the coefficients of c_n contain some symmetries of their own. In particular,

The magnitude of the c_n terms are even with respect to n: |c_-n|=|c_n|.
The angle of the c_n terms are odd with respect to n: ∠c_-n=-∠c_n.
The real part of c_n is even (Re(c_-n) = Re(c_n)) and the imaginary part is odd (Im(c_-n) =-Im(c_n))
If x_T(t) is even, then b_n=0 and c_n is even and real.
If x_T(t) is odd, then a_n=0 and c_n is odd and imaginary.

Some Comments about the Pulse Function

Let's examine the Fourier Series representation of the periodic rectangular pulse function, Π_T(t/T_p), more carefully.

Since the function is even, we expect the coefficients of the Exponential Fourier Series to be real and even(from symmetry properties). Furthermore, we have already calculated the coefficients of the Trigonometric Series, and could easily calculate those of the Exponential Series. However, let us do it from first principles. The Exponential Fourier Series coefficients are given by

$$\displaylines{
{\Pi _t}\left( {{t \over {{T_p}}}} \right) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn{\omega _0}t}}} \cr
with \cr {c_n} = {1 \over T}\int_T {{\Pi _t}\left( {{t \over {{T_p}}}} \right){e^{ - jn{\omega _0}t}}dt} \cr } $$

We can change the limits of integration to -T_p/2 and +T_p/2 (since the function is zero elsewhere) and proceed (the function is one in that interval, so we can drop it). We also make use of the fact the ω₀=2π/T and Euler's identity for sine.

$$\begin{align} {c_n} &= {1 \over T}\int\limits_{ - {T \over 2}}^{ + {T \over 2}} {{\Pi _t}\left( {{t \over {{T_p}}}} \right){e^{ - jn{\omega _0}t}}dt} = {1 \over T}\int\limits_{ - {{{T_p}} \over 2}}^{ + {{{T_p}} \over 2}} {{e^{ - jn{\omega _0}t}}dt} \cr
&= {1 \over { - jn{\omega _0}T}}\left( {{e^{ - jn{\omega _0}{{{T_p}} \over 2}}} - {e^{ + jn{\omega _0}{{{T_p}} \over 2}}}} \right) = {1 \over { - jn2\pi }}\left( {{e^{ - j{{n\pi {T_p}} \over T}}} - {e^{ + j{{n\pi {T_p}} \over T}}}} \right)\quad \quad \quad {\omega _0} = {{2\pi } \over T} \cr
&= {1 \over { - jn2\pi }}2j\sin \left( { - {{n\pi {T_p}} \over T}} \right) = {1 \over {n\pi }}\sin \left( {{{n\pi {T_p}} \over T}} \right) \cr
&= {{{T_p}} \over T}{{\sin \left( {{{n\pi {T_p}} \over T}} \right)} \over {{{n\pi {T_p}} \over T}}} \end{align} $$

The last step in the derivation is performed so we can use the sinc() function (pronounced like "sink"). This function comes up often in Fourier Analysis.

$${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( {\pi x} \right)} \over {\pi x}}$$

Note: unfortunately there are two common definitions of the "sinc()" function. Engineers tend to use ${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( {\pi x} \right)} \over {\pi x}}$; this definition is used in MATLAB. Mathematicians tend to use ${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( x \right)} \over x}$; this definition is used in Mathematica. We will use the former definition.

Using the "engineering" notation for "sinc()" the coefficients simplify to

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

Aside: the "sinc()" function

The sinc function has several important features:

sinc(x)=0 for all integer values of x except at x=0 where sinc(0)=1. This is because sin(π·n)=0 for all integer values of n. However at n=0 we have sin(π·n)/(π·n) which is zero divided by zero, but by L'Hôpital's rule get a value of 1.
The first zeros away from the origin occur when x=±1.
The function decays with an envelope of 1/(π·x) as x moves from the origin. This is because the sin() function has successive maxima with an amplitude of 1, and the sin function is divided by π·x.

The diagram below shows c_n vs n for several values of the duty cycle, T_p/T.

Duty cycle = 0.1 ...0.25 ...0.5 ...0.75 ...0.9

The graph on the left shows the time domain function. If you hit the middle button, you will see a square wave with a duty cycle of 0.5 (i.e., it is high 50% of the time). The period of the square wave is T=2·π;. The graph on the right shown the values of c_n vs n as red circles vs n (the lower of the two horizontal axes; ignore the top axis for now). The blue line goes through the horizontal axis whenever the argument of the sinc() function, n·T_p/T is an integer (except when n=0.). In particular the first crossing of the horizontal axis is given by n·T_p/T=1 or n=T/T_p (note this is not an integer values of T_p). There are several important features to note as T_p is varied.

As T_p decreases (along with the duty cycle, T_p/T), so does the value of c₀. This is to be expected because c₀ is just the average value of the function and this will decrease as the pulse width does.
As T_p decreases, the "width" of the sinc() function broadens. This tells us that as the function becomes more localized in time (i.e., narrower) it becomes less localized in frequency (broader). In other words, if a function happens very rapidly in time, the signal must contain high frequency coefficients to enable the rapid change.
Let us call the "width" of the sinc() function the width of the main lobe (i.e., between the first two zero crossings around ω=0), Δn=2·T/T_p. If we call the width of the pulse Δt=t_p then
$$\left( {\Delta n} \right) \cdot \left( {\Delta t} \right) = \left( {2{T \over {{T_p}}}} \right) \cdot \left( {{T_p}} \right) = 2T = {\rm{constant}}$$
This tells us explicitly that the product of width in frequency (i.e., Δn) multiplied by the width in time (Δt) is constant - if one is doubled, the other is halved. Or - as one gets more localized in time, it is less localized in frequency. We will discuss this more later.

References

Replace

Fourier Series Examples

Contents

Even Pulse Function (Cosine Series)

Aside: the periodic pulse function

Example 1: Special case, Duty Cycle = 50%

Example 2: Special case, Duty Cycle = 40%

Even Square Wave (Exploiting Symmetry)

Even Square Wave (Exponential Series)

Even Triangle Wave (Cosine Series)

Example 3: Triangle wave

Odd Function (Sawtooth Wave)

Aside: using symmetry

Example 4: Odd Sawtooth Wave

Functions that are neither even nor odd

Example 5: Neither Even nor Odd

Effect of Function Symmetry on Coefficients

Symmetry Trigonometric Series and Symmetry

Exponential Series and Symmetry

Some Comments about the Pulse Function

Aside: the "sinc()" function