AC and the Sine Wave
In Chapter 1, the sine wave was introduced by describing the development of an ac waveform with a single loop generator. The ac waveform was described in terms of its frequency or period, and another waveform parameter, the amplitude of the waveform, was mentioned briefly.
This chapter discusses in detail the amplitude descriptions of a sinusoidal waveform. Also, the time and frequency measurement of a waveform are reviewed, and an introduction to the trigonometric function will be presented.
This chapter discusses the amplitude descriptions of a sinusoid waveform and introduces the trigonometric functions that are essential to an analysis and understanding of ac circuits. At the end of this chapter you should be able to:
1. Explain the three ways to express the amplitude of a sinusoidal waveform and the relationship between them.
2. Understand the importance of the 0.707 constant and how it is derived.
3. Convert peak, peak-to-peak, and rms voltage and current values from one value to another.
4. Explain the sine, cosine, and tangent trigonometric functions.
5. Explain the relationship between the sine trigonometric function and an ac waveform.
6. Calculate the value of the sine of any angle between 0 degrees and 360 degrees.
REVIEW OF WAVEFORM FREQUENCY AND PERIOD RELATIONSHIPS
Any sinusoidal waveform can be described completely by identifying either time or frequency parameters and one of three possible amplitude specifications.
The frequency of a waveform is the number of cycles of the waveform which occur in one second of time. Common unit of measurement is hertz (Hz).
The period of a waveform, which sometimes is called its time, is the time required to complete one cycle of a waveform. It is measured in units of seconds, such as seconds, tenths of seconds, milliseconds, or microseconds.
If a waveform is to be properly described in terms of its period or frequency, it must be a repetitious waveform. A repetitious waveform is one in which each following cycle is identical to the previous cycle.
Frequency and Period Relationships
The frequency of a waveform in cycles per second is mathematically described in terms of the period, T, of the waveform as:
where f is the frequency of the waveform in hertz, and T is the time of one cycle of the waveform in seconds.
Frequency and Period Calculations
If the time of one cycle, or period, of a waveform is known, the frequency of the waveform can be calculated using equation 2–1. For example, Figure 2.1 shows one cycle of a waveform in which the cycle occurs in 0.025 milliseconds. Its frequency is calculated:
The frequency of the waveform is 40 kilohertz.
Or, for another example, a waveform has one cycle occurring every 10 milliseconds. Its frequency is calculated using equation 2–1.
The frequency of the waveform is 100 hertz.
equation 2–1 can be manipulated to describe the period of the waveform in terms of frequency.
The unit of the variables in the equation remain the same as in the frequency equation 2–1.
If the frequency of a waveform is known, the time of one cycle of the waveform can be calculated using equation 2–2.
For example, a waveform’s frequency is 120 hertz. The time of one cycle of that waveform is calculated:
The time of one cycle, or the period, is 8.33 milliseconds.
Or, for another example, a waveform has a frequency of 80 kilohertz. The period of one cycle is calculated:
The period of one cycle of the waveform is 12.5 microseconds.
Additional Frequency Calculation Examples
The following worked-out problems are additional examples of how to use equations 2–1 and 2–2 to calculate the frequency or period of a waveform.
For example, the frequency of a waveform is 8 hertz. Its period is calculated:
The period of the waveform is 125 milliseconds.
As another example, assume that the period of a waveform is 250 milliseconds. Therefore,
The frequency of the waveform is calculated to be 4 hertz, one-half of the frequency of the waveform with a period of 125 milliseconds. Since the period is twice as large the frequency is half as much. Thus the accuracy of the previous calculations is confirmed.
These two examples should show you vividly the frequency and period relationships of ac waveforms, and also, that one equation can be used to verify your calculations performed using the other.
WAVEFORM AMPLITUDE SPECIFICATIONS
In addition to frequency and period values, a third major specification of a waveform is the amplitude or height of the wave. There are three possible ways to express the amplitude of a sinusoidal waveformml: peak, peak-to-peak, and root-mean-square (rms).
Peak Amplitude Specifications
The peak amplitude of a sinusoidal waveform is the maximum positive or negative deviation of a waveform from its zero reference level. Recall from the discussion of the single-loop generator in Chapter 1 that this maximum voltage or current occurs as the loop of wire cut the magnetic flux at a 90-degree angle.
The sinusoidal waveform is a symmetrical waveform, so the positive peak value is the same as the negative peak value as shown in Figure 2.2. If the positive peak has a value of 10 volts, then the negative peak will also have a value of 10 volts. When measuring the peak value of a waveform, either positive or negative peaks can be used.
Peak-to-Peak Amplitude Specifications
The peak-to-peak amplitude is simply a measurement of the amplitude of a waveform taken from its positive peak to its negative peak as shown in Figure 2.3.
For the non-sinusoidal waveform shown in Figure 2.4, the peak-to-peak value of the voltage can be determined by adding the magnitude of the positive and the negative peak. In this example, the peak-to-peak amplitude is 18 volts plus 2 volts for a total of 20 volts, peak-to-peak.
For sinusoidal waveforms, if the positive peak value is 10 volts in magnitude, then the negative peak value of the same waveform is also 10 volts. Measuring from peak-to-peak, there is a total of 20 volts. Therefore, the value of the sinusoidal waveform in Figure 2.2 can be specified as either 10 volts peak or 20 volts peak-to-peak.
Summary of Peak and Peak-to-Peak Voltage Specifications
There is a mathematical relationship of peak and peak-to-peak amplitude specifications for all sine waves. Since the positive peak value is equal to the negative peak value, the peak-to-peak value is equal to two times the value of either peak voltage.
These voltage relationships can be expressed in these equations:
If the voltage applied to the ac circuits that will be studied in this text alternates in a sine wave fashion, then the current that flows will also vary according to a sine wave. Therefore, current relationships of peak and peak-to-peak amplitude specifications for all sine waves are the same as those used for voltage and can be expressed in these equations:
However, if a waveform is non-sinusoidal, these relationships may or may not hold true.
Root-Mean-Square Amplitude Specifications
The third specification for ac waveforms is called root-mean-square, abbreviated rms. This term allows the comparison of ac and dc circuit values. Root-mean-square values are the most common methods of specifying sinusoidal waveforms. In fact, almost all ac voltmeters and ammeters are calibrated so that they measure ac values in terms of rms amplitude.
RMS Relation to DC Heating Effect
An rms value is also known as the effective value and is defined in terms of the equivalent heating effect of direct current. The rms value of a sinusoidal voltage (or any time-varying voltage) is equivalent to the value of a dc voltage that causes an equal amount of heat (power dissipation) due to the circuit current flowing through a resistance. Since heating effect is independent of the direction of current flow, resistive power dissipation can be used as the basis for comparison of ac and dc values.
In other words, applying an ac voltage with a particular rms value to a resistive circuit will dissipate as much power in the resistors as a dc voltage would that has the same value.
The rms value of a sinusoidal voltage or current waveform is 70.7 percent or 0.707 of its peak amplitude value.
That is, as shown in Figure 2.5, a sinusoidal voltage with a peak amplitude of 1 volt has the same effect as a dc voltage of 0.707 volts as far as its ability to reproduce the same amount of heat in a resistance. Because the ac voltage of 1 volt peak or 0.707 volts rms is as effective as a dc voltage of 0.707 volts, the rms value of voltage is also referred to as the effective value.
Effective value and rms value are used interchangeably in electronics terminology. The rms value, however, is used more extensively and therefore it is the designation which will be used throughout this book.
Determining the 0.707 Constant
How is the 70.7 percent of peak-value constant derived? Essentially, the words root-mean-square tell how because they define the mathematical procedure used to determine the constant.
The word square comes form the square of instantaneous values of the ac waveform. Recall from dc circuit calculations that power dissipated in a circuit is equal to E, the voltage applied, times I, the current flowing in the circuit, as expressed in equation 2–9.
By Ohm’s law the applied voltage, E, is equal to the circuit current I times the circuit resistance. This is expressed in equation 2–10.
Substituting equation 2–10 for E in equation 2–9 provides equation 2–11, which allows the power dissipated in a circuit to be calculated if the circuit current and circuit resistance are given.
If the Ohm’s law equation 2–10 is rearranged then equation 2–12 results
Substituting equation 2–12 in equation 2–9 for I provides equation 2–13, which allows the power dissipated in a circuit to be calculated if the applied voltage E and total circuit resistance are given.
Note that in both equation 2–11 and 2–13 power is determined by the square of either the voltage or current values.
Power calculations in ac circuits are somewhat different. Since the applied voltage and the resulting circuit current are both sine waves, they are constantly changing.
Therefore, the power dissipated is constantly changing. For this reason a means of averaging the constantly changing values was derived.
The averaging is done as follows: The instantaneous values of voltage or current are sampled at regular equally-spaced points along the waveform as shown in Figure 2–6 — in this case every 15 degrees of the sine wave. Twelve of these samples would be made for the positive alternation. The voltage values are listed in table form as shown in the first column in Figure 2–7. The second column is the square of the instantaneous values in the first column. All of the squared values are summed together as shown in Figure 2–7 and an average or mean value calculated by dividing by the number of samples taken, in this case, twelve.
Figure 2.7 Instantaneous Voltage Values for the Voltage Sine Wave of Figure 2.6
Since the mean value is an average of squared values, the square root of the mean is the single value that is equivalent to a steady-state dc value. Thus, the result is the name root-mean-square since it is the square root of the mean of squared values. Mean is an average of the sum of the squares of instantaneous values of the voltage or current waveform. Root is the square root of the mean.
Power is dissipated on both the positive and negative alternation; it makes no difference which direction the current is flowing. However, squaring the instantaneous values eliminates any concern of direction or polarity.
equation 2–14 is an expression for the mean value of the voltage waveform of Figure 2.6.
The mean value is the sum of the squared instantaneous voltage values divided by the number of squared instantaneous voltage values.
Therefore, for the Ei2 values of Figure 2.7,
1,199.66 volts divided by 12 provides a mean value of 99.97 volts.
The square root of the mean, 99.97 volts, is approximately 10.0 volts:
Thus, the rms voltage value of the waveform of Figure 2.6 is approximately 10 volts. In the example, if a larger number of instantaneous points had been selected rather than 12, the rms value would have been exactly 10.0 volts.
The calculated rms voltage value can be proved to be 70.7 percent or 0.707 times the measured peak value of voltage of 14.14 volts.
Converting the 0.707 ratio of rms voltage to peak voltage to a percentage gives equation 2–17.
Also, it is of interest to note that if the column of instantaneous voltage values for the positive half sine wave shown in Figure 2.6 is averaged (by adding the values and dividing this sum by the number of values added), the average value obtained is approximately 0.637 of the peak value of the sine wave.
Average Voltage for the Half Sine Wave of
This is significant because voltmeters’ and various other instruments’ readings are derived from the response of the instrument to the average voltage of a half sine wave. The overall sine wave has an average voltage of 0 volts, but half a sine wave has an average voltage of 0.637 of its peak value.
It is also interesting to note that the average value 0.637 is equivalent to twice the reciprocal value of pi (recall pi =3.14). The above equation could also be written as
Relationship of an AC Waveform and a DC Waveform
Graphically, the relationship of a dc waveform to an ac waveform is as shown in Figure 2.8. The rms value, or 0.707 of the peak value, is located about three-quarters of the way up the ac waveform. And as you can see, the peak value of the ac waveform is considerably higher than the level of the dc waveform. This should not be surprising since an ac voltage is at its peak only momentarily and then drops back down.
The rms value of the wave can be determined if the peak voltage or current is known by rearranging the rms ratio equation 2–16. This can be rewritten as
It is interesting to note that the rms value of 0.707 is equivalent to the reciprocal of the square root of 2.
The above equation could also be written as:
It’s important to remember that rms or peak values can be used when referring to either voltage or current. For example, as shown in Figure 2.9, suppose a 100-volt peak sinusoidal voltage is applied across a resistor and one wants to know the value of dc voltage that could be applied across the resistor to create the same amount of power dissipation. The value of the ac voltage equals 0.707 of the peak value. Therefore, this can be calculated:
Thus the value of dc voltage that will create the same amount of power dissipation as an ac voltage with Epk of 100 volts is 70.7 volts dc.
If one wants to determine the peak value from the rms value, the ratio equation 2–16 can be rewritten:
This equation can be applied to either sinusoidal voltage or current waveforms. Therefore:
For example, using equation 2–19, the peak value of a waveform of 10 volts rms can be calculated:
The peak value of the waveform is 14.14 volts. Therefore, as shown in Figure 2.10 14.14 volts peak sinusoidal voltage is required to create the same amount of heat or power as that caused by 10 volts dc.
Standard Notation of Voltage RMS Values
Because ac voltage values are commonly specified as rms values, you will normally see voltages such as standard 60-hertz power voltages written as 115 VAC. The VAC notation is a simplified way of specifying rms voltage values.
Any ac voltage listed as VAC can be assumed to be an rms value, and where the type of specification is not provided, you can also assume rms values.
RMS, Peak, and Peak-to-Peak Conversion Examples
The following examples are typical conversion problems that you will encounter when working with ac circuits.
Peak and Peak-to-Peak from RMS
Calculate Epeak and Epeak-to-peak of 120 VAC. Use the conversion equations 2–19 and 2–3. The VAC designation indicates that the voltage is an rms value. From now on Epk will designate peak voltage instead of Epeak and Epp will designate peak-to-peak voltage instead of
This is summarized in Figure 2.11.
Peak and RMS from Peak-to-Peak
If a peak-to-peak value is given, the conversion will be to peak and rms values using equations 2–4 and 2–7. For example, Epk and Erms of a 60-volt peak-to-peak sine wave is:
This is summarized in Figure 2.12. Remember then that ac voltage amplitudes can be specified in one of three ways: peak, peak-to-peak, or rms, and that the same specifications can be used for current amplitudes.
THE SINE WAVE AND SINE TRIGONOMETRIC FUNCTION
The term sinusoidal has been used to describe a waveform produced by an ac generator. The term sinusoidal comes from a trigonometric function called the sine function.
Right-Triangle Side and Angle Relationships
As you may know, trigonometry is the study of triangles and their relationships. The basic triangle studied in trigonometry is a right triangle, which is a triangle that has a 90-degree angle as one of its three angles. A 90-degree triangle has a unique set of relationships from which the rules for trigonometry are derived.
Figure 2.13 shows a right triangle and identifies the 90-degree angle. In order to study the relationships of a right triangle one of the remaining two angles will be labeled with the Greek letter theta,θ.
To help distinguish the sides of a right triangle from one another, a name is given to each side. The sides of the triangle are named with respect to the angle theta. The side of the triangle across from or opposite to the angle theta is called the opposite side. The longest side of a right triangle is called the hypotenuse. The remaining side is called the adjacent side because it lies beside or adjacent to the angle. These names are commonly abbreviated to their first initials: O, H, and A.
As long as the angle theta remains unchanged, the sides of the right triangle will retain the same relationship to one another. Figure 2.14 shows an example to illustrate this point. The sides of that right triangle are 6, 8, and 10 inches long. The 10-inch side can be compared to the 6-inch side by dividing the length of one side by the length of the other:
The result is the decimal fraction 0.6. Therefore, since a ratio is defined as the value obtained by dividing one number by another, the ratio of the 6-inch side to the 10-inch side is 0.6.
Keeping theta constant, what will happen to the ratio if the lengths of the sides of the triangle are doubled? Figure 2.15 shows a triangle which has sides which are 12, 16, and 20 inches long — double the length of the sides of the triangle of Figure 2.14. The same two sides, the opposite side and the hypotenuse, are compared by dividing 12 by 20:
Therefore, the ratio is 0.6, the same ratio as the first triangle. The ratio of the length of the sides remained constant because the angle theta remained constant. The ratios of the sides of right triangles will remain constant no matter what the lengths of their sides as long as the angle theta is not changed. However, if the angle theta is changed, then the ratio of the length of the sides will also change.
Basic Trigonometric Functions
In trigonometry, these ratios have specific names. The three most commonly-used ratios in the study of right triangles are called sine, cosine, and tangent. The sine of the angle theta is equal to the ratio formed by the length of the opposite side divided by the length of the hypotenuse:
The cosine of the angle theta is equal to the ratio formed by length of the adjacent side divided by the length of the hypotenuse:
The tangent of the angle theta is equal to the ratio formed by length of the opposite side divided by the length of the adjacent side:
Remember that the sine, cosine, and tangent represent a ratio of sides of triangles. Specifically, these ratios compare the lengths of the sides of a triangle.
The names sine, cosine, and tangent are often abbreviated sin for sine, cos for cosine, and tan for tangent.
The cosine and tangent functions will be used in later chapters. In this chapter the sine function will be discussed.
The Sine Function Related to an AC Waveform
In the previous two examples, the lengths of two sides of two triangles were compared. In both examples the ratio of the opposite side to the hypotenuse was determined. In fact, what was determined was the conditions for the angle theta. For both triangles theta is 37 degrees. The sine of theta for both triangles is 0.60. It was stated that no matter how large the triangle might become or how long the sides, as long as the angle theta does not change, the ratio of the sides will not change and the sine of 37 degrees will always be 0.60181502 (rounded off to two places this is 0.60).
Every angle theta has its own specific sine value. This is true for all angles between 0 and 360 degrees. The sine for each angle between 0 and 360 degrees could be calculated by dividing the length of the opposite side of the triangle by the length of the hypotenuse. Fortunately, this has already been done and published as a set of trigonometric tables. Figure 2.16 shows a part of a trigonometric table.
Moreover, the values of these common trigonometric functions are provided as functions available on scientific calculators. Therefore, such calculators can be used to find the sine, cosine, or tangent of an angle instead of a trigonometric table. Specific examples are included at the end of this chapter.
Rotating AC Generator
You may be wondering why all of this emphasis is being placed on the sine function. It is for one important reason: the sine of an angle theta is an exact mathematical statement describing the voltage produced by an ac generator. Therefore, the table which tabulates the sine of theta for any angle between 0 and 360 degrees enables ac voltages to be expressed at any point in a cycle. To explain these statements, the relationship of the rotation of an ac generator, its sinusoidal output, the resulting vector right triangles, and the sine function will be discussed.
To demonstrate, a circle with a rotating arrow as shown in Figure 2.17 will be used to represent an ac generator producing one cycle of an ac waveform. The arrow in Figure 2.17 has been rotated 30 degrees, and a line perpendicular to the horizontal axis has been drawn downward from the tip of the arrow. As you can see, a right triangle has been formed. The arrow is the hypotenuse. The perpendicular line is the opposite side. The horizontal axis is the adjacent side. When the generator is rotated to different angles, the length of the hypotenuse (arrow) will never change. Only the opposite and adjacent sides will change length.
Recall from equation 2–21 that the sine theta is equal to the length of the opposite side divided by the length of the hypotenuse:
In Figure 2.17 the length of the hypotenuse never changes, it is a constant. By setting the constant length of the hypotenuse equal to 1 or unity, it remains constant at 1 as the generator is rotated. This simplifies the equation for sine theta as shown in equation 2–24 because now the opposite side will be divided by 1:
So in this instance the sine of theta is equal to the length of the opposite side itself:
Therefore, examining the length of the opposite side as theta varies will help demonstrate how the sine function varies.
During the following discussion the length of the opposite side of the resulting right triangle at different degree points around the circle will be compared to the unity value of the hypotenuse. The resulting values of these ratios will be the value of the sine function for different angles. The values will be plotted on the graph shown in Figure 2.18.
Let’s begin when the rotating arrow in Figure 2.17 is at 0 degrees. In this case, the length of opposite side of the right triangle is zero units long. Therefore,
The original position of the arrow in Figure 2.17 is 30 degrees. If you were to measure the opposite side you would find that it would be one-half the length of the hypotenuse, or 0.5 units long. Therefore,
This point is plotted at 30 degrees on the graph shown in Figure 2.18. Remember this point indicates the voltage amplitude when the hypotenuse has been rotated 30 degrees from the 0 degrees reference.
In Figure 2.19, the generator has been rotated to 60 degrees. The length of the opposite side is 86.6 percent of the length of the hypotenuse, or 0.866 units long. Therefore,
This point is plotted at 60 degrees on the graph shown in Figure 2.18.
In Figure 2.20, the rotating hypotenuse is at 90 degrees. Notice that at this point the length of the opposite side must be equal to the length of the arrow (hypotenuse). Therefore,
As shown plotted at 90 degrees on the graph of Figure 2.18, the amplitude of this point is a maximum.
In Figures 2.21 and 2.22 with the arrow at 120 degrees and 150 degrees, respectively, right triangles are produced with angles of theta equal to the 60-degree and 30-degree angles. The opposite sides will be respectively, 0.866 and 0.5 units.
These are plotted at their respective points on Figure 2.18.
When the hypotenuse reaches 180 degrees as shown in Figure 2.23, the length of the opposite side is again zero units long, and,
This is plotted at 180 degrees on Figure 2.18. Note that all the values plotted are positive values. In Figure 2.24 all of the points are connected with a curve indicating the continuously changing values as the generator rotates.
Note that from 0 to 90 degrees of rotation the amplitude of the sine wave goes from 0 to 1, and from 90 to 180 degrees that the amplitude comes down from 1 to 0 following a similarly-shaped curve. Therefore, the amplitude values of a sine wave repeat themselves during the second 90 degrees of rotation in reverse order from the amplitude variations during the first 90 degrees of rotation.
Thus far, all of the points have been in the positive region of the graph because the perpendicular line has extended in an upward, or positive, direction. If the hypotenuse is rotated to the bottom half of the circle as shown in Figure 2.25, right triangles are produced which duplicate the right triangles in the top part of the circle. However, the opposite side of the right triangle is now extending in a downward direction which gives the sine of the angles negative values. The points are plotted on Figure 2.18 and connected by a curve in Figure 2.24. Note that the negative portion of the curve is an exact replica of the positive portion; both sides are symmetrical.
In this discussion and demonstration, you should have become aware of the fact that the sine wave is an exact mathematical statement of the output of an ac generator. Also, the sine wave is merely a plot of the sines of angles between 0 and 360 degrees. Therefore the voltage or current output of an ac generator at any angle of rotation, theta, depends directly on the sine of the angle theta. This dependence is a very important point, and, in a later chapter, more explicit use will be made of the sine function in relation to the ac sine wave.
USING A 0 TO 90 DEGREES TRIGONOMETRIC TABLE
Many tables of trigonometric functions list only the function values from 0 to 90 degrees. If you have a good understanding of how the angles of the trigonometric functions are related, this will cause no problems. To help you understand the trigonometric relationship and the relationship to the rotation of an ac generator the rotation will be divided into four equal parts. Each part is called a quadrant.
Figure 2.26 shows a graph that is divided into four quadrants with each quadrant labeled with a Roman numeral I, II, III, or IV. Notice that the Roman numerals increase in a counter-clockwise direction around the graph.
Therefore, as shown in Figure 2.27, as the generator goes one-quarter turn from 0 to 90 degrees in quadrant I, the sine wave follows a curve through a set of amplitudes that increase from 0 to 1. Through the next quarter-turn from 90 to 180 degrees in quadrant II, the sine wave follows a curve through the same set of values, but now in reverse order and decreasing from 1 to 0. During the next quarter-turn from 180 to 270 degrees in quadrant III, and the next quarter-turn from 270 to 360 degrees in quadrant IV, the same curves through the same set of amplitude values are gone through from 0 to 1 and 1 to 0 as were gone through in quadrant I and II. In quadrants III and IV, however, the amplitude of the sine wave is negative.
Because of this repetition, most trigonometric function tables, as shown in Figure 2.28, only give the values of the function between 0 and 90 degrees. However, the generator rotates through 360 degrees; therefore, one must be able to find the sine of any angle from 0 to 360 degrees. That can be done by using the table given for quadrant I to also determine the sine of angles of quadrant II, III, and IV, as well.
To find the sine of an angle less than 90 degrees as shown in Figure 2.29 value is read directly from the table. For example, suppose you want to find the sine of 30 degrees. You would look in the table, such as the one shown in Figure 2.28, under the angle column for 30 degrees. Then you would look to the right of the angle in the sine column to locate the value of the sine of this angle. You can see in the table that it is 0.5. Thus, the sine of 30 degrees is 0.5.
The sine of quadrant II angles, that is, angles between 90 and 180 degrees, can be determined using a 0 to 90 degrees trig table by first converting the angle in the second quadrant to what is called a first–quadrant equivalent angle. For angles in the second quadrant, thetaII, the first-quadrant equivalent, thetaI, is determined by subtracting thetaII from 180 degrees:
For example, suppose that you want to determine the sine of the second quadrant angle thetaII = 120°, as shown in Figure 2.30. first–quadrant equivalent angle of 120–degrees is
Now, using the 0 to 90 degrees table of Figure 2.28, you would look up the sine of 60 degrees. This is 0.8660 and which, therefore, also equals the sine of 120 degrees.
Angles in quadrant III and quadrant IV also must be converted to quadrant I equivalents to be able to use a 0 to 90 degrees table to find the sine values of these angles. Use
equation 2–27 to determine first–quadrant equivalents for angles in quadrant III.
Use equation 2–28 for angles in quadrant IV.
The equations and where to use them are summarized in Figure 2.31 enable you to find the sine of any angle between 0 and 360 degrees using a 0 to 90 degrees trigonometric table. Remember, however, that the sine has positive values in quadrants I and II, but that it has negative values in quadrants III and IV. This also is summarized in Figure 2.31.
Most modern scientific calculators take into account that the sine has positive values for angles between 0 and 180 degrees and negative values for angles between 180 and 360 degrees. If you are using such a calculator, you need only input the angle for which you wish to determine the value of the sine and press the SIN (sine) key. The calculator will display the value of the sine of the angle with the appropriate positive or negative value.
Additional Trigonometric Calculations
You have seen how a sine wave is generated and how evaluating the amplitude of the sine wave corresponds to measuring the opposite side of a right triangle with hypotenuse equal to 1, as the hypotenuse rotates from 0 to 360 degrees.
Since you will encounter two other common trigonometric functions, cosine and tangent, it will be worthwhile to review them briefly also.
Cosine
Recall from equation 2–22 that the cosine of an angle theta of a right triangle is equal to the ratio of the adjacent side divided by the hypotenuse.
The cosine function is generated in the same way as the sine function except that now the amplitude of the cosine waveform corresponds to measuring the adjacent side of a right triangle with hypotenuse equal to 1.
In Figure 2.32a shown the same hypotenuse that was rotated for the sine wave. Its amplitude is kept constant at 1 (equal to unity). The hypotenuse is positioned at a rotation of 15 degrees in solid lines and at 75 degrees in dotted lines. Note that at 15 degrees the adjacent side is almost 1 so that the ratio of the adjacent side to the hypotenuse is 0.966, as plotted on the cosine curve of Figure 2.32b 15 degrees. If the hypotenuse were rotated back to 0 degrees, the adjacent side equals the hypotenuse and the cosine is equal to 1.
At a rotation of 75 degrees, as shown in Figure 2.32a, the adjacent side is reduced to 0.259 of the hypotenuse and therefore, the cosine is 0.259. This is plotted at 75 degrees in Figure 2.32b. Therefore, as the hypotenuse rotates through the first quadrant from 0 to 90 degrees, the cosine decreases from 1 to 0.
You can go through the exercise of rotating the hypotenuse through quadrants II, III and IV, calculate the equivalent first–quadrant angle, and verify that the cosine varies from 0 to −1 in quadrant II, from −1 to 0 in quadrant III, and from 0 to 1 in quadrant IV. It is particularly important to recognize that the adjacent side is a negative value for quadrants II and III.
Tangent
The tangent of the angle theta of a right triangle from equation 2–23 is equal to the ratio of the side opposite to the side adjacent.
The hypotenuse equal to unity is again rotated as shown in Figure 2.33a the opposite side of the formed right triangle must be divided by the adjacent side in order to form the tangent ratio. The ratio does not correspond to the measurement of just one side as it did for the sine and cosine functions. For this reason the tangent waveform has a much different shape than the sine and cosine waveform.
Recall that as the sine function is formed it varies from 0 to 1 as the rotation goes from 0 to 90 degrees. In other words the opposite side is increasing from 0 to 1. From Figure 2.32 the cosine, the adjacent side decreases from 1 to 0 through this same rotation. Therefore, the tangent has the numerator (the opposite side) increasing from 0 to 1 and the denominator decreasing from 1 to 0,
as shown in equation 2–29, for theta from 0 to 90 degrees.
Let’s look at Figure 2.33a.When the rotation is at 0 degrees, the adjacent side is 1 and the opposite side is 0. Therefore, the tangent is 0 as plotted on Figure 2.33b. 15 degrees rotation the opposite side has taken on a small value and the adjacent side is still near 1 (0.966 as shown on Figure 2.32b). The tangent is 0.268 and is plotted on Figure 2.33b at 15 degrees. At 45 degrees both the opposite side and the adjacent side are equal to the same value so that the tangent is 1 as plotted on Figure 2.33b. As the rotation increases above 45 degrees the opposite side is larger than the adjacent side and the tangent is increasing and greater than 1 (specifically 1.73 at 60 degrees). At 90 degrees the opposite side is 1 and the adjacent side is 0. Division by 0 results in the tangent having a value of infinity at 90 degrees and again at 270 degrees as shown in Figure 2.33b. The opposite side is positive and the adjacent side is positive so that the tangent is positive in the first quadrant.
As the rotation continues, the tangent changes value from positive to negative as the rotation goes beyond 90 degrees because the adjacent side becomes negative. Therefore, it decreases from infinity to smaller negative values as shown in Figure 2.33b. At 180 degrees the tangent is again 0. In the third quadrant, the opposite side is negative and the adjacent side is negative, therefore, the tangent is positive and increases, from 0 to infinity in the same fashion as it did for the first quadrant. In like fashion, as shown in Figure 2.33b, the rotation through the fourth quadrant repeats the same tangent values as for the second quadrant. Each time the equivalent first quadrant angle is 45 degrees in any of the quadrants, the tangent is equal to 1. These points are plotted at 45, 135, 225, and 315 degrees in Figure 2.33b.
Finding Sides When Theta and Hypotenuse Are Given
One more important extension of right triangle mathematics and trigonometry that is important concerns calculating or evaluating the sides of a right triangle when the angle theta is given.
Look at Figure 2.34a. What is known about the sides in relationship to the angle theta? First of all, the ratio of the lengths of the sides is known by the equations 2–21, 2–22, and 2–23. (Abbreviations are inserted for convenience, H for hypotenuse, O for opposite, and A for adjacent.)
Secondly, when the hypotenuse and the angle are given as shown in Figure 2.34b, the opposite side and adjacent side can be determined easily. Rearranging equation 2–21 into equation 2–30.
allows the opposite side to be calculated if the hypotenuse and angle theta are given.
Rearranging equation 2–22 into equation 2–31,
allows the adjacent side to be calculated if the hypotenuse and angle theta are given.
Find the Hypotenuse and Angle When Sides Are Given
Thirdly, if the opposite and adjacent sides are given, as shown in Figure 2.34c, the hypotenuse and angle theta can be calculated easily.
From the given opposite and adjacent sides the tangent of theta can be determined by using equation 2–23. Knowing that theta is an angle that has a tangent equal to the calculated ratio, the angle can be determined from the trigonometric tables or by using a calculator that has the trigonometric functions such as the TI-34 II shown in Figure 2.35.
When the angle theta is known the hypotenuse can be determined by using equation 2–30 or 2–31, rearranged.
First Example
Here’s an example of how to determine unknowns. Figure 2.36a shows a right triangle with an angle, theta, of 39 degrees and a hypotenuse of 100.
Using equation 2–30,
The opposite side equals 62.9.
You can use a trigonometric table or a calculator. If a trigonometric table is used, you enter the table in the degrees column and scan across to the sine column and read the sine of 39 degrees and use it in the equation.
When a calculator is used, the result can be obtained by following these keystrokes:
Using equation 2–31,
The adjacent side equals 77.7.
The calculator keystrokes are:
Second Example
In Figure 2.36b shows a right triangle that has the opposite side equal to 90 and the adjacent side equal to 60. Using equation 2–23
Locating 1.5 in the tangent column of a trigonometric function table and scanning across to the degree column you find that the angle is 56.31 degrees.
If a calculator is used, the following keystrokes are followed to find the angle:
The hypotenuse can be determined by using equation 2–32,
This is easily calculated using a calculator with the following keystrokes:
SUMMARY
In this chapter the relationship of period to frequency was reviewed and you were shown how to specify the amplitude of a sine wave in terms of its peak, peak-to-peak, and rms values. The definitions of the various trigonometric functions were presented and related to right triangles and a quadrant graph. You were shown how to determine the value of the sine of any angle between 0 and 360 degrees and how to use it, the cosine, and the tangent to solve right triangle problems.
In the next chapter, you will learn about an instrument that provides an electronic picture of an ac sine wave, and you will learn how to use it to study further the characteristics of sine waves.
1. Given the frequency of waveform, determine the period, or time of one cycle. f = 16 kHz, T = ______
2. the time of one cycle, determine the frequency of the waveform. T = 265 μs, f = ______
3. Given the rms amplitude of a sine waveform, determine the peak and peak–to–peak values.
Epk = 1.414 × Erms = 1.414 × 8.3 V = 11.7 V
Epp = 2 × Epk = 2 × 11.7 V = 23.4 V
4. Given the peak amplitude of a sine waveform, determine the rms and peak–to–peak values.
Erms = 0.707 × Epk = 0.707 × 8 mV = 5.66 mV
Epp = 2 × Epk = 2 × 8 mV = 16 mV
5. Given the peak–to–peak amplitude of a sine waveform, determine the peak and rms values. Ipp =16 mA
Ipk =0.5 ×Ipp =0.5 ×16 mA = 8 mA
Irms =0.707 ×Ipk =0.707 ×8 mA = 5.66 mA
a. Given the following right triangle, determine the sine of the angle and with a calculator or using tables find the angle.
Calculator Solution for Angle:
Enter the ratio into the calculator so it is displayed. Press inv and sin keys or sin−1 key and the calculator will display the angle in decimal degrees.
Enter the table of natural trigonometric functions by locating the ratio in the sine column. Read across to the angle in decimal degrees in the degree column.
b. Given the following right triangle, determine the cosine of the angle and with a calculator or using tables find the angle.
Calculator Solution for Angle:
Enter the ratio into the calculator so it is displayed. Press inv and cos keys or the cos−1 key and the calculator will display the angle in decimal degrees.
Enter the table of natural trigonometric functions by locating the ratio in the cosine column. Read across to the angle in decimal degrees in the degree column.
7. Given the following right triangle, determine the sine, cosine, and tangent of the angle and the angle itself using a calculator or table as in problem 6.
There may be as much as 0.2 to 0.3 of a degree variation in the angle calculated from the sine, cosine or tangent depending on the rounding off of the sine, cosine, or tangent ratio as it is calculated.
8. Given the following right triangle, determine the remaining sides.
Calculator Solution for Sine, Cosine, or Tangent:
Enter the angle in the calculator so it is displayed. Press the sine, cosine, or tangent key for the respective function required and the value will read out directly on the display.
Table Solution for Sine, Cosine, or Tangent:
Enter the table of natural trigonometric functions by locating the angle in the degrees column. Scan across to the respective sine, cosine, or tangent column and read the value in whole numbers and/or decimals.
9. If a hypotenuse of length equal to unity (length = 1) is rotated from the zero-degree position by the following angles, identify the quadrant in which the hypotenuse is located.
a. 278° is greater than 270° and less than 360° — fourth quadrant
b. 153° is greater than 90° and less than 180° — second quadrant
c. 94° is greater than 90° and less than 180° — second quadrant
d. 10° is greater than 0° and less than 90° — first quadrant
e. 229° is greater than 180° and less than 270° — third quadrant
100° is greater than 90° and less than 180° — second quadrant
g. 191° is greater than 180° and less than 270° — third quadrant
h. 89° is greater than 0° and less than 90° — first quadrant
1. Given the following frequencies, calculate the period of the waveform.
2. Given the following times of one cycle of a waveform, calculate the frequency of each waveform.
3. Given the following peak-to-peak amplitudes, calculate the peak and rms amplitude values.
4. Given the following rms amplitudes, calculate the peak and peak-to-peak amplitude values.
5. Given the following right triangles, calculate the sine, cosine, and tangent of the angles.
6. Find the angles of the following ratios:
7. Determine the value of the hypotenuse H of the following right triangles with opposite side 0 and adjacent side A and the angle θ.
8. Determine the missing sides for the following right triangles with the hypotenuse and angle θ given.
9. In which quadrant are the following angles?
10. Find the quadrant I equivalent angle of the following angles.
| θ | |
| a. | 46° |
| b. | 100° |
| c. | 205° |
| d. | 420° |
1. Given the following periods or frequencies, determine the frequency or time of one cycle of the waveform.
5. Using the data given, determine if the opposite side of a right triangle is greater than, equal to, or less than the adjacent side when the angle theta is:
