# The Sine Wave

The terms defined in the previous section can be applied to any type of periodic waveform, whether smooth or discontinuous. The sinusoidal waveform is of particular importance, however, since it lends itself readily to the mathematics and the physical phenomena associated with electric circuits. Consider the power of the following statement:
The sinusoidal waveform is the only alternating waveform whose shape is unaffected by the response characteristics of R, L, and C elements. Fig. 1: The sine wave is the only alternating waveform whose shape is not altered by the response characteristics of a pure resistor, inductor, or capacitor.
In other words, if the voltage across (or current through) a resistor, coil, or capacitor is sinusoidal in nature, the resulting current (or voltage, respectively) for each will also have sinusoidal characteristics, as shown in Fig. 1. If a square wave or a triangular wave were applied, such would not be the case.  Fig. 2: Sine wave and cosine wave with the horizontal axis in degrees.
The unit of measurement for the horizontal axis of Fig. 2 is the degree. A second unit of measurement frequently used is the radian (rad). It is defined by a quadrant of a circle such as in Fig. 3 where the distance subtended on the circumference equals the radius of the circle.  Fig. 3.1: Arch length formula.

### Arc Length Formula

In general, if the length of the arc is $s$ units and radius is $r$ units as shown in the Fig. 3.1, then the ratio of $s$ to $r$ is $$\bbox[10px,border:1px solid grey]{\theta = {s \over r}} \tag{i}$$ and $$\bbox[10px,border:1px solid grey]{s = \theta r} \tag{ii}$$ if we rotate the radius around the circle for $360^\circ$, then the arc length will be the circumference of the circle, which is $$\bbox[10px,border:1px solid grey]{s= C = 2\pi r} \tag{iii}$$ comparing eq.(ii) and eq.(iii), we can say $$\bbox[10px,border:1px solid grey]{\theta = 2\pi = 360^\circ}$$ Fig. 4: There are $2\pi$ radians in one full circle of $360 \circ$
Therefore, there are $2\pi$ rad around a $360^\circ$ circle, as shown in Fig. 4, and $$\bbox[10px,border:1px solid grey]{2 \pi \, rad = 360^\circ} \tag{1}$$ with $$\bbox[10px,border:1px solid grey]{1 \, rad = {360^\circ \over 2 \pi} ={180^\circ \over \pi}= 57.3^\circ}$$ A number of electrical formulas contain a multiplier of $\pi$. For this reason, it is sometimes preferable to measure angles in radians rather than in degrees.
The quantity $\pi$ is the ratio of the circumference of a circle to its diameter.
$\pi$ has been determined to an extended number of places primarily in an attempt to see if a repetitive sequence of numbers appears. It does not. A sampling of the effort appears below: $$\pi = 3.14159 26535 89793 23846 26433 . . .$$ Although the approximation $\pi = 3.14$ is often applied, all the calculations in this text will use the $\pi$ function as provided on all scientific calculators. For $180^\circ$ and $360^\circ$, the two units of measurement are related as shown in Fig. 4. The conversion equations between the two are the following: $$\bbox[10px,border:1px solid grey]{Radians = {\pi \over 180^\circ} \,(Degrees)} \tag{2}$$ $$\bbox[10px,border:1px solid grey]{Degrees = { 180^\circ \over \pi} \,(Radians)} \tag{3}$$ Applying these equations, we find
$90^\circ:$ $$Radians = {\pi \over 180^\circ} \,(90^\circ) = {\pi \over 2} rad$$ $30^\circ:$ $$Radians = {\pi \over 180^\circ} \,(30^\circ) = {\pi \over 6} rad$$ ${\pi \over 3} rad:$ $$Degrees = {180^\circ \over \pi} \,({\pi \over 3} rad ) = 60^\circ degrees$$ Using the radian as the unit of measurement for the abscissa, we would obtain a sine wave, as shown in Fig. 5. Fig. 5: Plotting a sine wave versus radians.
It is of particular interest that the sinusoidal waveform can be derived from the length of the vertical projection of a radius vector rotating in a uniform circular motion about a fixed point. Starting as shown in Fig. 6(a) and plotting the amplitude (above and below zero) on the coordinates drawn to the right [Figs. 6(b) through (i)], we will trace a complete sinusoidal waveform after the radius vector has completed a $360^\circ$ rotation about the center.   Fig. 6: Generating a sinusoidal waveform through the vertical projection of a rotating vector.
The velocity with which the radius vector rotates about the center, called the angular velocity, can be determined from the following equation: $$\bbox[10px,border:1px solid grey]{\text{Angular Velocity} = {\text{distance (degrees or radians)} \over \text{time}} } \tag{4}$$ Substituting into Eq. (13.8) and assigning the Greek letter omega ($\omega$) to the angular velocity, we have $$\bbox[10px,border:1px solid grey]{ \omega = {\alpha \over t}} \tag{5}$$ and $$\bbox[10px,border:1px solid grey]{ \alpha = \omega t} \tag{6}$$ Since $\omega$ is typically provided in radians per second, the angle $\alpha$ obtained using Eq. (4) is usually in radians.
In Fig. 6, the time required to complete one revolution is equal to the period (T) of the sinusoidal waveform of Fig. 6(i). The radians subtended in this time interval are $2\pi$. Substituting, we have $$\bbox[10px,border:1px solid grey]{\omega= {2\pi \over T}} \, (rad/s) \tag{7}$$ In words, this equation states that the smaller the period of the sinusoidal waveform of Fig. 6(i), or the smaller the time interval before one complete cycle is generated, the greater must be the angular velocity of the rotating radius vector. Certainly this statement agrees with what we have learned thus far. We can now go one step further and apply the fact that the frequency of the generated waveform is inversely related to the period of the waveform; that is, $f =1/T$. Thus, $$\bbox[10px,border:1px solid grey]{\omega= 2\pi f} \, (rad/s) \tag{8}$$ This equation states that the higher the frequency of the generated sinusoidal waveform, the higher must be the angular velocity. Equations (7) and (8) are verified somewhat by Fig. 7, where for the same radius vector, $\omega = 100 \,rad/s$ and $500 \,rad/s$. Fig. 7: Demonstrating the effect of $\omega$ on the frequency and period.
Example 1: Determine the angular velocity of a sine wave having a frequency of 60 Hz.
Solution: $$\omega = 2\pi f = (2\pi)(60 Hz) = 377 \text{rad/s}$$ (a recurring value due to 60-Hz predominance)
Example 2: Determine the frequency and period of the sine wave of Fig. 7(b).
Solution: Since $\omega = 2\pi /T$, $$T = {2 \pi \over \omega} = {2 \pi \over 500}= 12.57ms$$ and $$f = {1 \over T} = {1 \over 12.57} = 79.58 Hz$$