Average Value

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Even though the concept of the average value is an important one in most technical fields, its true meaning is often misunderstood. In Fig. 1(a), for example, the average height of the sand may be required to determine the volume of sand available. The average height of the sand is that height obtained if the distance from one end to the other is maintained while the sand is leveled off, as shown in Fig. 1(b).
Fig. 1: Defining average value.
The area under the mound of Fig. 1(a) will then equal the area under the rectangular shape of Fig. 1(b) as determined by $$A = b \times h$$ In Fig. 1 the distance was measured from one end to the other. In Fig. 2(a) the distance extends beyond the end of the original pile of Fig. 1. The situation could be one where a landscaper would like to know the average height of the sand if spread out over a distance such as defined in Fig. 2(a). The result of an increased distance is as shown in Fig. 2(b). The average height has decreased compared to Fig. 1. Quite obviously, therefore, the longer the distance, the lower is the average value.
Fig. 2: Effect of distance (length) on average value.
If the distance parameter includes a depression, as shown in Fig. 3(a), some of the sand will be used to fill the depression, resulting in an even lower average value for the landscaper, as shown in Fig. 3(b).
For a sinusoidal waveform, the depression would have the same shape as the mound of sand (over one full cycle), resulting in an average value at ground level (or zero volts for a sinusoidal voltage over one full period).
Fig. 3: Effect of depressions (negative excursions) on average value.
After traveling a considerable distance by car, some drivers like to calculate their average speed for the entire trip. This is usually done by dividing the miles traveled by the hours required to drive that distance. For example, if a person traveled $225 mi$ in $5 h$, the average speed was $225 mi/5 h$, or $45 mi/h$. This same distance may have been traveled at various speeds for various intervals of time, as shown in Fig. 4.
Fig. 4: Plotting speed versus time for an automobile excursion.
By finding the total area under the curve for the 5 h and then dividing the area by 5 h (the total time for the trip), we obtain the same result of $45 mi/h$; that is, $$\bbox[10px,border:1px solid grey]{\text{Average speed} = {\text{Area under curve} \over \text{length of curve}}} \tag{1}$$ $$\text{Average speed} = {A1 + A2 \over 5h }$$ $$\text{Average speed} = {60mi/h (2h) + 50mi/h(2.5h) \over 5h }$$ $$\text{Average speed} = {225mi \over 5h } = 45mi/h$$ Equation (1) can be extended to include any variable quantity, such as current or voltage, if we let G denote the average value, as follows: $$\bbox[10px,border:1px solid grey]{\text{(G) Average speed} = {\text{Algebraic sum of areas} \over \text{length of curve}}} \tag{2}$$ The algebraic sum of the areas must be determined, since some area contributions will be from below the horizontal axis. Areas above the axis will be assigned a positive sign, and those below, a negative sign. A positive average value will then be above the axis, and a negative value, below.
The average value of any current or voltage is the value indicated on a dc meter. In other words, over a complete cycle, the average value is the equivalent dc value. In the analysis of electronic circuits to be considered in a later course, both dc and ac sources of voltage will be applied to the same network. It will then be necessary to know or determine the dc (or average value) and ac components of the voltage or current in various parts of the system.
Example 1: Find the average values of the following waveforms.
a. Fig. 5.
b. Fig. 6.
Fig. 5
Fig. 6
Solution:
a. Given Fig. 5: $$G = {+(3V)(4ms) - (1V)(4ms) \over 8ms} \\ = {12V-4V \over 8ms} = 1V$$ b. Given Fig. 6: $$G = {-(10V)(2ms) - (4V)(2ms) - (2V)(2ms)\over 10ms} \\ = {-20V+8V-4V \over 8ms} = -1.6V$$
We found the areas under the curves in the preceding example by using a simple geometric formula. If we should encounter a sine wave or any other unusual shape, however, we must find the area by some other means. We can obtain a good approximation of the area by attempting to reproduce the original wave shape using a number of small rectangles or other familiar shapes, the area of which we already know through simple geometric formulas.
For example, the area of the positive (or negative) pulse of a sine wave is 2Am. Approximating this waveform by two triangles (Fig. 7), we obtain (using $area = 1/2 (\text{base}) \times (\text{height})$ for the area of a triangle) a rough idea of the actual area: $$\text{Area shaded} = 2({1\over 2} b h) = 2({1\over 2} {\pi \over 2} A_m)\\ = {\pi \over 2} A_m$$
Fig. 7: Approximating the shape of the positive pulse of a sinusoidal waveform with two right triangles.
Fig. 8: A better approximation for the shape of the positive pulse of a sinusoidal waveform.
A closer approximation might be a rectangle with two similar triangles (Fig. 8): $$\text{Area} = A_m {\pi \over 3} + 2({1\over 2} b h) \\ = A_m {\pi \over 3} + A_m {\pi \over 3} = {2 \pi \over 3} A_m\\ = 2.094A_m$$ which is certainly close to the actual area. If an infinite number of forms were used, an exact answer of $2A_m$ could be obtained. For irregular waveforms, this method can be especially useful if data such as the average value are desired.
The procedure of calculus that gives the exact solution $2A_m$ is known as integration. Integration is presented here only to make the method recognizable to the reader; it is not necessary to be proficient in its use to continue with this text. It is a useful mathematical tool, however, and should be learned. Finding the area under the positive pulse of a sine wave using integration, we have $$Area = \int_0^{\pi} A_m \sin \alpha \, d\alpha$$ where $\int$ is the sign of integration, 0 and $\pi$ are the limits of integration, $A_m sin \alpha$ is the function to be integrated, and $d\alpha$ indicates that we are integrating with respect to $\alpha$. Integrating, we obtain $$Area = \int_0^{\pi} A_m \sin \alpha \, d\alpha = A_m [-\cos \alpha ]_0^{\pi}\\ = -A_m [\cos \pi - cos(0)]=-A_m [-1 - 1]\\ =-A_m [-2] = 2A_m\\$$ Since we know the area under the positive (or negative) pulse, we can easily determine the average value of the positive (or negative) region of a sine wave pulse by applying Eq.(2) $$G = {2A_m \over \pi} = 0.637A_m$$
Example 2: Determine the average value of the sinusoidal waveform of Fig. 9.
Fig. 9: Effect of distance (length) on average value.
Solution: By inspection it is fairly obvious that
the average value of a pure sinusoidal waveform over one full cycle is zero
.
Eq. (2): $$G = {2A_m - 2A_m\over 2\pi} = 0V$$

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