# The Pythagorean theorem: engineering applications

August 15, 2017 (8/15/17) was Pythagorean Theorem Day (see I missed Pythagorean Theorem Day). I was not even aware that there was such a day until this year, but it brought to my attention some really interesting engineering applications of which I was not previously aware. Let’s take a look at a few interesting ones.

**Antenna analysis ^{1}**

A probe designed for elliptically polarized field measurements is employed for a system with *n* identical dipoles which all intersect at the center of an *xy* plane in a Cartesian coordinate system. The angles between those dipoles are all identically equal to π/*n*. The first dipole crosses the *y* axis at an angle γ (**Figure 1**).

**Figure 1 **An *n* dipole system diagram

In this example, the probe is illuminated by a monochromatic plane wave with angular frequency Ω which propagates in the *z* plane direction. Its electrical components are as follows:

In which the amplitudes are *A* and *B.*

Next, the author makes the assumption that the sizes of the system are small as compared with the wavelength, so the mutual coupling influences between the antennas are negligible. The author next makes the assumption that all of the antennas are loaded with square-law detectors, so with these conclusions we can describe their directional patterns to be sinusoidal in nature.

Now, the square of the output voltage of the *i*th antenna (ν) is as follows:

Where **1_{i}** is the vector of the

*i*th dipole.

Now, we can calculate the output voltage from *n* dipoles by the use of the Pythagorean Theorem as follows from Reference 1:

If, in a circle are inscribed *n* secants which intersect themselves in the center of that circle, the angle between two adjacent secants is π/*n*, and the first of them crosses the axis of symmetry of the circle at angle γ.

Then the sum of the directional sines and cosines of the secants is then constant and equals *n*/2.

We can write this result as follows:

The formulas in Equation 3 are valid for *n* __>__ 2. The notation used in the formulas is identical with that of Figure 1. When *n* = 2, the formulas are in the form of the Pythagorean Theorem.

By taking the sum of Equation 2 and using Equation 3 for *n* detectors, we get the square of the output voltage, *V*, of the system. For elliptical polarization, this is as follows:

Now, for circular polarization, *A* = *B* and

And for linear polarization, such as, *B* = 0, and we get

Here we can see that for the formulas in Equation 3, for *n* being even, it is a simple sum of *n*/2 “standard” Pythagorean Theorems.

The author is puzzled regarding the result when *n* is odd, so how can it be shown in this case, the last part of the sum equals ½, or its angle is a multiple of π/4 ?

Can anyone solve this? The author offers a premium glass of wine for anyone who does. (I have not yet seen any comments or solutions for this by IEEE audience members.)

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