Sinusoidal currents and voltages occur frequently, and the electrical
engineer must thoroughly understand them. This requires an intimate familiarity
with the mathematical properties of sinusoids. Below is a diagram of a typical
sinusoidal voltage, given by the equation, v(t) = V_{m}cos(ωt +
φ) volts.

V_{m} is the magnitude of the voltage. ω is the angular
frequency in radians/second. φ is the phase angle in radians. Notice that
the wave will reach a peak at ωt = -φ; this point is noted in the
figure as t = -φ/ω. T is the period in seconds; in physics, this
would be known as the wave length. Period is related to frequency by the
equation,

f = 1/T.

f is the frequency in Hz or cycles/second. Naturally, the frequency f and the
angular frequency ω are related:

ω = 2πf

V_{peak-to-peak} is nothing more than what it says. Mathematically,
it's V_{max} - V_{min}, where V_{max} and
V_{min} are the maximum and minimum values of the sinusoid.

You may also have use for V_{avg}. This is the average value of the
wave. By inspection, it should be clear that the average value of the above
sinusoid is zero, because it spends as much time above zero as below zero. A DC
voltmeter connected to a sinusoidal source will read zero volts, the average
value.

Another useful parameter is the root-mean-square (rms) voltage. For all
sinusoids centered around zero, this is just the peak value divided by the
square root of 2. The rms value is also know as the *effective value*,
because a DC voltage equal to the rms value will cause the same average power
to be consumed in a resistor. An AC voltmeter connected to a sinusoidal source
will read the rms value. Because the rms value is the most useful for
sinusoidal voltages and currents, it is the default value. If a person says a
certain sinusoidal voltage is "120 volts," it is implied that this is the rms
value, not the peak value.

Preview of lessons to come: If we converted this voltage to a phasor, its
value would be V_{m}/φ. If you don't
understand what this means yet, don't panic; you soon will understand it.

Note that, in the expression cos(ωt + φ), ωt is radians/second times seconds, to give radians. Nevertheless, the angle φ is frequently given in degrees. It is therefore important to convert the units to the same type before attempting to compute the cosine. If you choose to convert φ to radians, you must also remember to configure your calculator to accept angles in radians, recalling that calculators default to assuming angles are in degrees. For this reason, most engineers convert the ωt term to degrees. This is easily done by multiplying the ωt term by 180/π, because there are π radians in 180°.

Electrical engineers prefer working with cosines, rather than sines. It's easy to convert a sine to a cosine:

sin(φ) = cos(φ - 90°)

Before going on, you should complete Tutorial 10 on
**sinusoids**.

Although the rms value of a sinusoid is the maximum value divided by the square root of 2, this is not true for other wave shapes. In general, one must calculate the root-mean-square. This is the square root of the mean of the square of the wave, as shown below.

In this equation, t_{0} is some arbitrary time, and time
t_{0} + T is the time one period later. Although the equation is
written for a voltage v, it works for current i also. The wave must be
periodic (repetitive), or it is not possible to find an rms value.

Before going on, you should complete Tutorial 10A on **root
mean square**.

Electrical engineers deal with many sinusoidal problems by converting sinusoidal currents and voltages to phasors. Phasors allow the engineer to replace difficult-to-solve time-domain differential equations with frequency-domain phasor equations, which are easier to solve. Phasors make use of complex numbers. Before studying phasors, we must therefore first understand how to manipulate complex numbers. You probably already learned how to do this in math courses, so the material below is just a review, with special emphasis on the way electrical engineers use complex numbers.

Although most mathematicians use *i* to represent the square root of
minus one, electrical engineers use *j*, to avoid confusion with their
usual variable name for current.

j = √-1

j × j = -1

1/j = -j

**M** = general form of a complex number, written in bold face,
or with an overline: M

Me^{jθ} = exponential form, where M is the magnitude and θ
is the phase angle, usually in radians

M/θ = polar form, where M is the magnitude and
θ is the phase angle, usually in degrees

a + jb = rectangular form, where a = Mcosθ, and b = Msinθ

(a , b) = rectangular form, as shown on many calculators

(a / b) = polar form, as shown on many
calculators

You must become adept at converting from one form to another. Most
scientific calculators have provisions for converting from rectangular to polar
coordinates, and *vice versa*. If your calculator cannot do that, then
you must use the sine, cosine, and tangent functions to do conversions. Take
care when using the tangent and arctangent functions. The complex number -5 +
j5 converts to 7.07/135°, not 7.07/-45°. The arctangent cannot distinguish between (-5)/5
and 5/(-5).

You must also be accomplished at doing math with complex numbers. Again, advanced scientific calculators can do math using complex numbers. If your calculator isn't that sophisticated, you'll have to make frequent use of these three important mathematical relationships:

(a + jb) + (c + jd) = a + c + j(b + d)

(a/λ)(b/φ) =
ab/(λ + φ)

(a/λ)/(b/φ) =
(a/b)/(λ - φ)

These relationships show that addition and subtraction are easy in rectangular coordinates, but multiplication and division are easy in polar coordinates. You should therefore convert complex numbers to rectangular form if you intend to add or subtract, and to polar form if you intend to multiply or divide.

Many calculators, even so-called "scientific" calculators, do not have adequate complex number features for electrical engineering use. If a calculator has "rectangular to polar conversion," such as the TI-82, this is helpful, but is far from ideal. When purchasing a calculator, it is wise to get one that has "complex number" capability, that is, a calculator able to directly handle mathematical operations such as (5 + j6)x(7/18°). The TI-83, -85, and -86 have this capability, as does the HPGX.

Below are some links that will help you learn how to use your calculator to do complex math:

Soon (in the next lesson), you will apply what you have learned about sinusoids and complex numbers to the solution of circuits in "the phasor domain." To do this, you must know how to convert currents and voltages in the time domain to their equivalents in the phasor domain. Although you will learn more about this in the next lesson, the basics are given below.

To convert a sinusoidal time-domain voltage or current to a phasor, drop the cosine and the ωt, and use only the magnitude and the phase angle. Example:

250 cos(65t + 73°) volts transforms to 250/73° volts.

To convert a phasor to a sinusoid, just reverse this process. Note that you must know the angular frequency to do the inverse transform. In the example below, assume the frequency is 50 Hz.

345/-81° amps transforms back to 345 cos(2π50 - 81°) amps = 345 cos(314t - 81°) amps.

Before going on to the homework, you should complete Tutorial 10B on
**complex numbers**.

1. A certain voltage is found to be 50sin(65t + 23°) volts. Find angular frequency in radians per second, frequency in Hz, period, maximum voltage, minimum voltage, peak-to-peak voltage, rms voltage, average voltage, voltage expressed as a phasor, the average power consumed by a 22 Ω resistor having this voltage, and the voltage at t = .02 s.

2. Convert the following complex numbers to polar form.

10 + j20, 6 + j89, 12 - j18, (-44) + j16.

3. Convert the following complex numbers to rectangular form.

5/11°, 110/-58°, 20/140°, 120/-175°

4. Evaluate these expressions. Give answers in both rectangular and polar form.

(510/64°)/(45 + j19), 18 + j22 - 19/23°

5. If the frequency is 60 Hz, what is the time function corresponding to a phasor of 90 + j70 volts?

6. A sinusoidal voltage has a maximum amplitude of 12 volts and a period of 17 ms. At t = 0, its value is 8.4 volts and is rising. Write an expression for this voltage as a function of time. Note: Do not ignore "and is rising."

7. A 440 V, 60 Hz sinusoidal source is connected to a 100 Ω load. What is the rms current, peak current, and average power consumed by the load?

8. Find the average and rms voltages of this periodic wave.

9. Find expressions for M and θ in symbolic form.

M/θ = (A/λ)(B/φ)/(C/α)

10. Find X and Y. Use your calculator (not Maple, Mathematica, or another math program). Note that X and Y are complex numbers; you must therefore solve a system of complex equations. Give your answer in rectangular form. X and Y must satisfy both equations. Also note that since X and Y are complex numbers, each has a real and imaginary part. Hint: In the solution for these equations, both the real and imaginary parts of X and Y are non-zero.

25X + j10Y = 5/26°

(13 - j9)X + 6Y
= 3 + j4

BONUS (No partial credit

**Find the Flaw in the Proof**

Given: t > 3

3t > 9

3t - t^{2} > 9 - t^{2}

t(3 - t) > (3 - t)(3 + t)

t > 3 + t

0 > 3