2 Periodic functions
A function $f\left(t\right)$ is periodic if the function values repeat at regular intervals of the independent variable $t$ . The regular interval is referred to as the period . See Figure 1.
Figure 1
If $P$ denotes the period we have
$\phantom{\rule{2em}{0ex}}f\left(t+P\right)=f\left(t\right)$
for any value of $t$ . The most obvious examples of periodic functions are the trigonometric functions $sint$ and $cost$ , both of which have period $2\pi $ (using radian measure as we shall do throughout this Workbook) (Figure 2). This follows since
$\phantom{\rule{2em}{0ex}}sin\left(t+2\pi \right)=sint\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}cos\left(t+2\pi \right)=cost$
Figure 2
The amplitude of these sinusoidal functions is the maximum displacement from $y=0$ and is clearly 1. (Note that we use the term sinusoidal to include cosine as well as sine functions.)
More generally we can consider a sinusoid
$\phantom{\rule{2em}{0ex}}y=Asinnt$
which has maximum value, or amplitude, $A$ and where $n$ is usually a positive integer.
For example
$\phantom{\rule{2em}{0ex}}y=sin2t$
is a sinusoid of amplitude 1 and period $\frac{2\pi}{2}=\pi $ (Figure 3). The fact that the period is $\pi $ follows because
$\phantom{\rule{2em}{0ex}}sin2\left(t+\pi \right)=sin\left(2t+2\pi \right)=sin2t$
for any value of $t$ .
Figure 3
We see that $y=sin2t$ has half the period of $sint$ , $\pi $ as opposed to $2\pi $ (Figure 4). This can alternatively be phrased by stating that $sin2t$ oscillates twice as rapidly (or has twice the frequency) of $sint$ .
Figure 4
In general $\phantom{\rule{1em}{0ex}}y=Asinnt$ has amplitude $A$ , period $\frac{2\pi}{n}$ and completes $n$ oscillations when $t$ changes by $2\pi $ . Formally, we define the frequency of a sinusoid as the reciprocal of the period:
$\phantom{\rule{2em}{0ex}}\text{frequency}\phantom{\rule{1em}{0ex}}=\phantom{\rule{1em}{0ex}}\frac{1}{\text{period}}$
and the angular frequency , often denoted the Greek Letter $\omega $ (omega) as
$$\begin{array}{rcll}\text{angularfrequency}=2\pi \times \phantom{\rule{1em}{0ex}}\text{frequency}=\frac{2\pi}{\text{period}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& & & \text{}\end{array}$$Thus $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}y=Asinnt$ has frequency $\frac{n}{2\pi}$ and angular frequency $n$ .
Task!
State the amplitude, period, frequency and angular frequency of
 $y=5cos4t$
 $y=6sin\frac{2t}{3}.$
amplitude 5, period $\frac{2\pi}{4}=\frac{\pi}{2}$ , frequency $\frac{2}{\pi}$ , angular frequency 4
amplitude 6, period $3\pi $ , frequency $\frac{1}{3\pi}$ , angular frequency $\frac{2}{3}$
2.1 Harmonics
In representing a nonsinusoidal function of period $2\pi $ by a Fourier series we shall see shortly that only certain sinusoids will be required:

${A}_{1}cost$
(and
${B}_{1}sint$
)
These also have period $2\pi $ and together are referred to as the first harmonic (or
fundamental harmonic ).

${A}_{2}cos2t$
(and
${B}_{2}sin2t$
)
These have half the period, and double the frequency, of the first harmonic and are referred to as the second harmonic .

${A}_{3}cos3t$
(and
${B}_{3}sin3t$
)
These have period $\frac{2\pi}{3}$ and constitute the third harmonic .
In general the Fourier series of a function of period $2\pi $ will require harmonics of the type
$\phantom{\rule{2em}{0ex}}{A}_{n}cosnt\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left(\text{and}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}_{n}sinnt\right)\phantom{\rule{2em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}n=1,2,3,\dots $
2.2 Nonsinusoidal periodic functions
The following are examples of nonsinusoidal periodic functions (they are often called “waves”).
Square wave
Figure 5
Analytically we can describe this function as follows:
$\phantom{\rule{2em}{0ex}}f\left(t\right)=\left\{\begin{array}{ccc}\hfill 1& \hfill & \hfill \pi <t<0\\ \hfill +1& \hfill & \hfill 0<t<\pi \end{array}\right.\phantom{\rule{2em}{0ex}}$ (which gives the definition over one period)
$\phantom{\rule{2em}{0ex}}f\left(t+2\pi \right)=f\left(t\right)$ (which tells us that the function has period $2\pi $ )
Sawtooth wave
Figure 6
In this case we can describe the function as follows:
$\phantom{\rule{2em}{0ex}}f\left(t\right)=2t\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}0<t<2\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}f\left(t+2\right)=f\left(t\right)$
Here the period is 2, the frequency is $\frac{1}{2}$ and the angular frequency is $\frac{2\pi}{2}=\pi $ . Triangular wave
Figure 7
Here we can conveniently define the function using $\pi <t<\pi $ as the “basic period”:
$\phantom{\rule{2em}{0ex}}f\left(t\right)=\left\{\begin{array}{ccc}\hfill t& \hfill & \hfill \pi <t<0\\ \hfill t& \hfill & \hfill 0<t<\pi \end{array}\right.$
or, more concisely,
$\phantom{\rule{2em}{0ex}}f\left(t\right)=\leftt\right\phantom{\rule{2em}{0ex}}\pi <t<\pi $
together with the usual statement on periodicity
$\phantom{\rule{2em}{0ex}}f\left(t+2\pi \right)=f\left(t\right).$
Task!
Write down an analytic definition for the following periodic function:
$$\begin{array}{rcll}f\left(t\right)=\left\{\begin{array}{ccc}\hfill 2t& \hfill & \hfill 0<t<3\\ \hfill 1& \hfill & \hfill 3<t<5\end{array}\right.\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}f\left(t+5\right)=f\left(t\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& & & \text{}\end{array}$$
Task!
Sketch the graphs of the following periodic functions showing all relevant values:
 $f\left(t\right)=\left\{\begin{array}{ccc}\hfill {t}^{2}\u22152& \hfill & \hfill 0<t<4\\ \hfill 8& \hfill & \hfill 4<t<6\\ \hfill 0& \hfill & \hfill 6<t<8\end{array}\right.\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}f\left(t+8\right)=f\left(t\right)$
 $f\left(t\right)=2t{t}^{2}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}0<t<2\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\left(t+2\right)=f\left(t\right)$
Figure 9