This article aims to explain some basic properties of sinusoidal waveforms, and three phase systems. It is important to have a thorough understanding of these concepts before diving into the more advanced topics of motor drive design, or electrics in general. I will assume that the reader is familiar with the use of complex numbers, ohm's law, basic circuit analysis and calculus.

## The square root of three

Only valid for symmetrical three phase systems, the square root of three is commonly seen as a relation between line and phase quantities.

V_L = \sqrt{3} V_f

The power in a three phase system is three times the power in a single phase system:

P = 3 \cdot V_f \cdot I_f
\label{eq:3-phase-power-phase-q}

However, when line quantities are considered, the equation is:

P = V_L \cdot I_L \cdot \sqrt{3}

The following derivations is an attempt to explain how this seemingly magic relation is obtained.

For a Y-connected system, the line voltage is the phasor sum of two phase voltages. I.e. the sum considering the phase difference of $$120^{\circ}$$.

V_L = V_f \enclose{phasorangle}{0^{\circ}} + V_f \enclose{phasorangle}{120^{\circ}} = V_f \enclose{phasorangle}{60^{\circ}}

For a phase voltage of one volts, the result in rectangular form is given as:

1 \enclose{phasorangle}{0^{\circ}} + 1 \enclose{phasorangle}{120^{\circ}} = 1 \enclose{phasorangle}{60^{\circ}} = \frac{1}{2} + j \frac{\sqrt{3}}{2}

The sum of the two phase voltage vectors may also be expressed as:

V_L = 2 \cdot V_f \cdot \sin(120^{\circ})

Furthermore the trigonometric function evaluates to:

\sin(120^{\circ}) = \frac{\sqrt{3}}{2}

Hence the resulting relation of phase to line voltage is given as:

V_L = 2 \cdot V_f \cdot \frac{\sqrt{3}}{2} = V_f \cdot \sqrt{3}

Still considering a Y-connected system, the line current equals the phase current:

I_{L} = I_{f}

Using the expressions for line quantities in the power equation \eqref{eq:3-phase-power-phase-q}, we obtain:

P = 3 \cdot V_f \cdot I_f = 3 \cdot \frac{V_L}{\sqrt{3}} \cdot I_L = \sqrt{3} \cdot V_L \cdot I_L

Because:

\frac{3}{\sqrt{3}} = \sqrt{3}

It is important to realize that for a $$\Delta$$-connected system a similar derivation to the above will result in exactly the same equation for power. The phase voltage will equal the line voltage, but the current will adhere to the square root of three relation.

Additionally the displacement power factor $$\cos(\phi)$$ is added to obtain the well known general power equation for balanced three phase systems:

P = V_L \cdot I_L \cdot \sqrt{3} \cdot \cos(\phi)

## RMS, and the square root of two

Another seemingly magic number is the square root of two, relating the RMS value of a sine wave to it's amplitude. Using voltage as a example, we have that:

\hat{V} = V_{RMS} \cdot \sqrt{2}

Where $$\hat{V}$$ is the amplitude voltage, i.e. peak value of the sine wave.

Again using voltage as a example, the RMS value is defined as:

V_{RMS} = \sqrt{\frac{1}{T} \int_{t_0}^{t} v(t)^2 \mathrm{d}t }

The general equation for a sine wave is given as:

v(t) = \hat{V} \cdot \sin(\omega t + \theta)

Where $$\hat{V}$$ is the amplitude of the voltage. Ignoring the phase shift $$\theta$$, we have the following expression for RMS voltage:

V_{RMS} = \sqrt{\frac{1}{T} \int_{t_0}^{t} \hat{V}^2 \cdot \sin^2(\omega t) \mathrm{d}t }

We also have that:

\sin^2(\omega t) = \frac{1 - \cos(2 \omega t)}{2}

The amplitude voltage squared $$\hat{V}^2$$ is a positive constant, and is moved it front of the expression. The exponent is canceled by the square root. The trigonometric identity is applied, and we obtain:

V_{RMS} = \hat{V}\sqrt{\frac{1}{T} \int_{t_0}^{t} \frac{1 - \cos(2 \omega t)}{2} \mathrm{d}t }

Evaluating the integral, we obtain:

V_{RMS} = \hat{V}\sqrt{\frac{1}{T} \left[ \frac{t}{2}  - \frac{\sin(2 \omega t)}{4 \omega} \right]_{t_0}^t }

By definition, the RMS value is taken over a whole number of complete cycles. Hence the $$\sin(2 \omega t)$$ term will cancel out.

V_{RMS} = \hat{V}\sqrt{\frac{1}{T} \left[ \frac{t}{2} \right]_{t_0}^t }

The period is given as:

T = t - t_0

Hence:

V_{RMS} = \hat{V}\sqrt{\frac{1}{t - t_0} \left[ \frac{t}{2} \right]_{t_0}^t } = \hat{V}\sqrt{\frac{1}{t - t_0} \left( \frac{t}{2} - \frac{t_0}{2} \right) } = \hat{V} \frac{1}{\sqrt{2}}

Q.E.D

## Average rectified value

The conceptual difference between RMS value, and average rectified value is something that can be intuitively difficult to comprehend. The average of a sine wave around it's symmetry line is always zero, because the positive and negative values are equal. The average rectified value however, is the average of the sine wave if both cycles are positive.

V_{ARV} = \frac{1}{T} \int_{t_0}^{t} |v(t)| \mathrm{d}t

For a sinusoidal waveform, the relation between amplitude, and average rectified value is given as:

\hat{V} = V_{ARV} \cdot \frac{\pi}{2}

The form factor is used to express the difference between the RMS, and the average rectified value:

k_f = \frac{V_{RMS}}{V_{ARV}}

Hence, for the sinusoidal example, the form factor is:

\frac{\pi}{2 \sqrt{2}} \approx 1.1107

I.e. for a regular line RMS voltage of $$230 \;V$$, the ARV is $$207.2 \;V$$.

### Intuition

So why do we use these two different numbers, and what do they represent?

The RMS is explained as the DC voltage that will dissipate the same amount of power as the AC voltage from which it is calculated. I.e. the use of RMS simplifies the calculation of power in AC circuits.

The explanation is that the power dissipated by a resistor is proportional to the square of the voltage. Double the voltage, and the power is increased by a factor of four:

P = \frac{V_{RMS}^2}{R} = \frac{\hat{V}^2}{2\cdot R}

The average on the other hand is useful when dealing with rectifiers, or other non-linear circuits. It is also used when calculating the magnetic flux produced by a coil.

\Phi = V_{ARV} \cdot t [Wb]

## The measurement of power in three phase systems

Three phase power measurement methods are often classified by the number of single phase watt meters that are used in the measurement.

The single watt meter method is only applicable to symmetrical loads, while the two and three watt meter methods are applicable to all kinds of loads. The difference between the two, and three watt meter method is that the latter supports unsymmetrical systems with neutral conductors, while the former requires a isolated neutral point.

### Blondel's theorem

Blondel's theorem, from electrical engineer André Blondel, states that for a system of $$N$$ conductors, the required number of watt meters is $$N-1$$.