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: Defining the radian.
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:
$$\bbox[10px,border:1px solid grey]{\pi = 3.14159 26535 89793 23846 26433 . . .}$$
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.
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. (4) 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
$$f = {1 \over T} = {1 \over 12.57} = 79.58 Hz$$
Do you want to say or ask something?