Functions of period 2 2 2 π π π

2 Functions of period $2 π$

We now discuss how to represent periodic non-sinusoidal functions $f (t)$ of period $2 π$ in terms of sinusoids, i.e. how to obtain Fourier series representations. As already discussed we expect such Fourier series to contain harmonics of frequency $\frac{n}{2 π}$ $(n = 1, 2, 3, \dots)$ and, if the periodic function has a non-zero average value, a constant term.

Thus we seek a Fourier series representation of the general form

$f (t) = \frac{a_{0}}{2} + a_{1} cos t + a_{2} cos 2 t + \dots + b_{1} sin t + b_{2} sin 2 t + \dots$

The reason for labelling the constant term as $\frac{a_{0}}{2}$ will be discussed later. The amplitudes $a_{1}, a_{2}, \dots$ $b_{1}, b_{2}, \dots$ of the sinusoids are called Fourier coefficients .

Obtaining the Fourier coefficients for a given periodic function $f (t)$ is our main task and is referred to as Fourier Analysis. Before embarking on such an analysis it is instructive to establish, at least qualitatively, the plausibility of approximating a function by a few terms of its Fourier series.

Task!

Consider the square wave of period $2 π$ one period of which

is shown in Figure 10.

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Write down the analytic description of this function,
State whether you expect the Fourier series of this function to contain a constant term,
List any other possible features of the Fourier series that you might expect from the graph of the square-wave function.

We have $\begin{array}{rcl} f (t) & = & \{\begin{matrix} 4 & - \frac{π}{2} < t < \frac{π}{2} \\ 0 & - π < t < - \frac{π}{2}, & \frac{π}{2} < t < π \end{matrix} \\ f (t + 2 π) & = & f (t) \end{array}$
The Fourier series will contain a constant term since the square wave here is non-negative and cannot therefore have a zero average value. This constant term is often referred to as the d.
(direct current) term by engineers.
(c) Since the square wave is an even function (i.e. the graph has symmetry about the $y$ axis) then its Fourier series will contain cosine terms but not sine terms because only the cosines are even functions. (Well done if you spotted this at this early stage!)

It is possible to show, and we will do so later, that the Fourier series representation of this square wave is

$2 + \frac{8}{π} \{cos t - \frac{1}{3} cos 3 t + \frac{1}{5} cos 5 t - \frac{1}{7} cos 7 t + \dots\}$

i.e. the Fourier coefficients are

\begin{array}{rcl} \frac{a_{0}}{2} = 2, a_{1} = \frac{8}{π}, a_{2} = 0, a_{3} = - \frac{8}{3 π}, a_{4} = 0, a_{5} = \frac{8}{5 π}, \dots \end{array}

Note, as well as the presence of the constant term and of the cosine (but not sine) terms, that only odd harmonics are present i.e. sinusoids of period $2 π, \frac{2 π}{3}, \frac{2 π}{5}, \frac{2 π}{7}, \dots$ or of frequency $1, 3, 5, 7, \dots$ times the fundamental frequency $\frac{1}{2 π} .$

We now show in Figure 8 graphs of

the square wave
the first two terms of the Fourier series representing the square wave
the first three terms of the Fourier series representing the square wave
the first four terms of the Fourier series representing the square wave
the first five terms of the Fourier series representing the square wave

Note: We show the graphs for $0 < t < π$ only since the square wave and its Fourier series are even.

Figure 8

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We can clearly see from Figure 8 that as the number of terms is increased the graph of the Fourier series gradually approaches that of the original square wave - the ripples increase in number but decrease in amplitude. (The behaviour near the discontinuity , at $t = \frac{π}{2}$ , is slightly more complicated and it is possible to show that however many terms are taken in the Fourier series, some “overshoot” will always occur. This effect, which we do not discuss further, is known as the Gibbs Phenomenon.)

2.1 Orthogonality properties of sinusoids

As stated earlier, a periodic function $f (t)$ with period $2 π$ has a Fourier series representation

\begin{array}{rcl} f (t) & = & \frac{a_{0}}{2} + a_{1} cos t + a_{2} cos 2 t + \dots + b_{1} sin t + b_{2} sin 2 t + \dots, \\ = & \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} cos n t + b_{n} sin n t) (3) \end{array}

To determine the Fourier coefficients $a_{n}, b_{n}$ and the constant term $\frac{a_{0}}{2}$ use has to be made of certain integrals involving sinusoids, the integrals being over a range $α, α + 2 π$ , where $α$ is any number. (We will normally choose $α = - π$ .)

Task!

Find $\int_{- π}^{π} sin n t d t$ and $\int_{- π}^{π} cos n t d t$ where $n$ is an integer.

In fact both integrals are zero for

\begin{array}{rcl} \int_{- π}^{π} sin n t d t & = & {[- \frac{1}{n} cos n t]}_{- π}^{π} = \frac{1}{n} (- cos n π + cos n π) = 0 n \neq 0 (4) \\ \int_{- π}^{π} cos n t d t & = & {[\frac{1}{n} sin n t]}_{- π}^{π} = 0 n \neq 0 (5) \end{array}

As special cases, if $n = 0$ the first integral is zero and the second integral has value $2 π$ .

N.B. Any integration range $α, α + 2 π$ , would give these same (zero) answers.

These integrals enable us to calculate the constant term in the Fourier series (3) as in the following task.

Task!

Integrate both sides of (3) from $- π$ to $π$ and use the results from the previous Task. Hence obtain an expression for $a_{0}$ .

We get for the left-hand side

$\int_{- π}^{π} f (t) d t$

(whose value clearly depends on the function $f (t)$ ).

Integrating the right-hand side term by term we get

$\frac{1}{2} \int_{- π}^{π} a_{0} d t + \sum_{n = 1}^{\infty} \{\int_{- π}^{π} a_{n} cos n t d t + \int_{- π}^{π} b_{n} sin n t d t\} = \frac{1}{2} {[a_{0} t]}_{- π}^{π} + \sum_{n = 1}^{\infty} {0 + 0}$

(using the integrals (4) and (5) shown above). Thus we get

$\int_{- π}^{π} f (t) d t = \frac{1}{2} (2 a_{0} π) or a_{0} = \frac{1}{π} \int_{- π}^{π} f (t) d t$ (6)

Key Point 1

The constant term in a trigonometric Fourier series for a function of period $2 π$ is

\frac{a_{0}}{2} = \frac{1}{2 π} \int_{- π}^{π} f (t) d t = average value of f (t) over 1 period.

This result ties in with our earlier discussion on the significance of the constant term. Clearly a signal whose average value is zero will have no constant term in its Fourier series. The following square wave (Figure 9) is an example.

Figure 9

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We now obtain further integrals, known as orthogonality properties, which enable us to find the remaining Fourier coefficients i.e. the amplitudes $a_{n}$ and $b_{n}$ ( $n = 1, 2, 3, \dots$ ) of the sinusoids.

Task!

Using the standard trigonometric identity that

$sin n t cos m t \equiv \frac{1}{2} \{sin (n + m) t + sin (n - m) t\}$

evaluate $\int_{- π}^{π} sin n t cos m t d t$ where $n$ and $m$ are any integers.

We get

\begin{array}{rcl} \int_{- π}^{π} sin n t cos m t d t = \frac{1}{2} \{\int_{- π}^{π} sin (n + m) t d t + \int_{- π}^{π} sin (n - m) t d t\} = \frac{1}{2} {0 + 0} = 0 \end{array}

using the results (4) and (5) since $n + m$ and $n - m$ are also integers.

This result holds for any interval of $2 π$ .

Key Point 2

Orthogonality Relation

For any integers $m, n$ , including the case $m = n$ ,

\int_{- π}^{π} sin n t cos m t d t = 0

We shall use this result shortly but need a few more integrals first.

Consider next

$\int_{- π}^{π} cos n t cos m t d t where m and n are integers.$

Using another trigonometric identity we have, for the case $n \neq m$ ,

\begin{array}{rcl} \int_{- π}^{π} cos n t cos m t d t & = & \frac{1}{2} \int_{- π}^{π} {cos (n + m) t + cos (n - m) t} d t \\ = & \frac{1}{2} {0 + 0} = 0 using the integrals (4) and (5). \end{array}

For the case $n = m$ we must get a non-zero answer since ${cos}^{2} n t$ is non-negative. In this case:

\begin{array}{rcl} \int_{- π}^{π} {cos}^{2} n t d t & = & \frac{1}{2} \int_{- π}^{π} (1 + cos 2 n t) d t \\ = & \frac{1}{2} {[t + \frac{1}{2 n} sin 2 n t]}_{- π}^{π} = π (provided n \neq 0) \end{array}

For the case $n = m = 0$ we have $\int_{- π}^{π} cos n t cos m t d t = 2 π$

Task!

Proceeding in a similar way to the above, evaluate

$\int_{- π}^{π} sin n t sin m t d t$

for integers $m$ and $n$ .

Again consider separately the three cases:

$n \neq m,$
$n = m \neq 0$ and
$n = m = 0.$

Using the identity $sin n t sin m t \equiv \frac{1}{2} \{cos (n - m) t - cos (n + m) t\}$ and integrating the right-hand side terms, we get, using (4) and (5)
$\int_{- π}^{π} sin n t sin m t d t = 0 n, m integers n \neq m$
Using the identity $cos 2 θ = 1 - 2 {sin}^{2} θ$ with $θ = n t$ gives for $n = m \neq 0$ $\begin{array}{rcl} \int_{- π}^{π} {sin}^{2} n t d t & = & \frac{1}{2} \int_{- π}^{π} (1 - cos 2 n t) d t = π \end{array}$
When $n = m = 0$ , $\int_{- π}^{π} sin n t sin m t d t = 0.$

We summarise these results in the following Key Point:

Key Point 3

For integers $n, m$

\begin{array}{rcl} \int_{- π}^{π} sin n t cos m t d t & = & 0 \\ \int_{- π}^{π} cos n t cos m t d t & = & \{\begin{matrix} 0 & n \neq m \\ π & n = m \neq 0 \\ 2 π & n = m = 0 \end{matrix} \\ \int_{- π}^{π} sin n t sin m t d t & = & \{\begin{matrix} 0 & n \neq m, n = m = 0 \\ π & n = m \end{matrix} \end{array}

All these results hold for any integration range of width $2 π$ .

2 Functions of period 2 π

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2.1 Orthogonality properties of sinusoids

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2 Functions of period $2 π$