6 Engineering Example 3
6.1 Inverse square law decay of electromagnetic power
Introduction
Engineers are concerned with using and intercepting many kinds of wave forms including electromagnetic, elastic and acoustic waves. In many situations the intensity of these signals decreases with the square of the distance. This is known as the inverse square law . The power received from a beacon antenna is expected to conform to the inverse square law with distance.
Problem in words
Check whether the data in the table below confirms that the measured power obeys this behaviour with distance.
Power received, $W$  0.393  0.092  0.042  0.021  0.013  0.008 
Distance from antenna, $m$  1  2  3  4  5  6 
Mathematical statement of problem
Represent power by $P$ and distance by $r$ . To show that the data fit the function $\phantom{\rule{1em}{0ex}}P=\frac{A}{{r}^{2}}$ where $A$ is a constant, plot log( $P$ ) against log( $r$ ) (or plot the ‘raw’ data on loglog axes) and check
 how close the resulting graph is to that of a straight line
 how close the slope is to 2.
Mathematical analysis
The values corresponding to log( $P$ ) and log( $r$ ) are
log( $P$ )  0.428  1.041  1.399  1.653  1.851  2.012 
log( $r$ )  0  0.301  0.499  0.602  0.694  0.778 
These are plotted in Figure 11 and it is clear that they lie close to a straight line.
Figure 11
The slope of a line through the first and third points can be found from
$\phantom{\rule{2em}{0ex}}\frac{1.399\left(0.428\right)}{0.4990}=2.035$
The negative value means that the line slopes downwards for increasing $r$ . It would have been possible to use any pair of points to obtain a suitable line but note that the last point is least ‘in line’ with the others. Taking logarithms of the equation $P=\frac{A}{{r}^{n}}$ gives $log\left(P\right)=log\left(A\right)nlog\left(r\right)$
The inverse square law corresponds to $n=2$ . In this case the data yield $n=2.035\approx 2$ . Where $log\left(r\right)=0$ , $log\left(P\right)=log\left(A\right)$ . This means that the intercept of the line with the $log\left(P\right)$ axis gives the value of $log\left(A\right)=0.428$ . So $A=100.428=0.393$ .
Interpretation
If the power decreases with distance according to the inverse square law, then the slope of the line should be $2$ . The calculated value of $n=2.035$ is sufficiently close to confirm the inverse square law. The values of $A$ and $n$ calculated from the data imply that $P$ varies with r according to
$\phantom{\rule{2em}{0ex}}P=\frac{0.4}{{r}^{2}}$
The slope of the line on a loglog plot is a little larger than $$ 2. Moreover the points at 5 m and 6 m range fall below the line so there may be additional attenuation of the power with distance compared with predictions of the inverse square law.
Exercises

Sketch the graphs of
 $y={\text{e}}^{t}$
 $y={\text{e}}^{t}+3$
 $y={\text{e}}^{t}$
 $y={\text{e}}^{t}1$

The figure below shows the graphs of
$y={\text{e}}^{t}$
,
$y=2{\text{e}}^{t}$
and
$y={\text{e}}^{2t}$
.
State in words how the graphs of $y=2{\text{e}}^{t}$ and $y={\text{e}}^{2t}$ relate to the graph of $y={\text{e}}^{t}$ .

The figures below show graphs of
$y={\text{e}}^{t}$
,
$y=4{\text{e}}^{t}$
and
$y=43{\text{e}}^{t}$
.
Use the above graphs to help you to sketch graphs of
 $y=5{\text{e}}^{t}$
 $y=52{\text{e}}^{t}$


The graph a) in the figure below has an equation of the form
$y=A+{\text{e}}^{kt}$ , where $A$ and $k$ are constants. What is the value of $A$ ?
 The graph (b) below has an equation of the form $y=A{\text{e}}^{kt}$ where $A$ and $k$ are constants. What is the value of $A$ ?
 Write down a possible form of the equation of the exponential graph (c) giving numerical values to as many constants as possible.
 Write down a possible form of the equation of the exponential graph (d) giving numerical values to as many constants as possible.

The graph a) in the figure below has an equation of the form

 $y=2{\text{e}}^{t}$ is the same shape as $y={\text{e}}^{t}$ but with all $y$ values doubled.
 $y={\text{e}}^{2t}$ is much steeper than $y={\text{e}}^{t}$ for $t>0$ and much flatter for $t<0$ . Both pass through $\left(0,1\right)$ . Note that $y={\text{e}}^{2t}={\left({\text{e}}^{t}\right)}^{2}$ so each value of $y={\text{e}}^{2t}$ is the square of the corresponding value of $y={\text{e}}^{t}$ .


 2
 5
 $y=64{\text{e}}^{kt}$
 $y=1+2{\text{e}}^{kt}$