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1、附 录附录A外文翻译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
2、 various parts of the system.EXAMPLE 13.13 Determine the average value of the waveforms of Fig. 13.37.FIG. 13.37Example 13.13.Solutions:a. By inspection, the area above the axis equals the area below over one cycle, resulting in an average value of zero volts.b. Using Eq.(13.26):as shown in Fig. 13.
3、38.In reality, the waveform of Fig. 13.37(b) is simply the square wave of Fig. 13.37(a) with a dc shift of 4 V; that is v2 =v1 + 4 VEXAMPLE 13.14 Find the average values of the following waveforms over one full cycle:a. Fig. 13.39.b. Fig. 13.40.Solutions: We found the areas under the curves in the p
4、receding 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 rectangle
5、s 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. 13.43), we obtain(using area 1/2 base height for the area of a triangle)
6、a rough idea of the actual area:A closer approximation might be a rectangle with two similar triangles(Fig. 13.44):which is certainly close to the actual area. If an infinite number of forms were used, an exact answer of 2Am could be obtained. For irregular waveforms, this method can be especially u
7、seful if data such as the average value are desired. The procedure of calculus that gives the exact solution 2Am 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 i
8、s a useful mathematical tool, however,and should be learned. Finding the area under the positive pulse of a sine wave using integration, we havewhere is the sign of integration, 0 and p are the limits of integration, Am sin a is the function to be integrated, and da indicates that we are integrating
9、 with respect to a. Integrating, we obtainSince 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. (13.26):For the waveform of Fig. 13.45,EXAMPLE 13.15 Determine the average valu
10、e of the sinusoidal waveform of Fig. 13.46.Solution: By inspection it is fairly obvious thatthe average value of a pure sinusoidal waveform over one full cycle iszero.EXAMPLE 13.16 Determine the average value of the waveform of Fig. 13.47.Solution: The peak-to-peak value of the sinusoidal function i
11、s16 mV +2 mV =18 mV. The peak amplitude of the sinusoidal waveform is, therefore, 18 mV/2 =9 mV. Counting down 9 mV from 2 mV(or 9 mV up from -16 mV) results in an average or dc level of -7 mV,as noted by the dashed line of Fig. 13.47.EXAMPLE 13.17 Determine the average value of the waveform of Fig.
12、 13.48.Solution:EXAMPLE 13.18 For the waveform of Fig. 13.49, determine whether the average value is positive or negative, and determine its approximate value.Solution: From the appearance of the waveform, the average value is positive and in the vicinity of 2 mV. Occasionally, judgments of this typ
13、e will have to be made.InstrumentationThe dc level or average value of any waveform can be found using a digital multimeter (DMM) or an oscilloscope. For purely dc circuits,simply set the DMM on dc, and read the voltage or current levels.Oscilloscopes are limited to voltage levels using the sequence
14、 of steps listed below:1. First choose GND from the DC-GND-AC option list associated with each vertical channel. The GND option blocks any signal to which the oscilloscope probe may be connected from entering the oscilloscope and responds with just a horizontal line. Set the resulting line in the mi
15、ddle of the vertical axis on the horizontal axis, as shown in Fig. 13.50(a).2. Apply the oscilloscope probe to the voltage to be measured (if not already connected), and switch to the DC option. If a dc voltage is present, the horizontal line will shift up or down, as demonstrated in Fig. 13.50(b).
16、Multiplying the shift by the vertical sensitivity will result in the dc voltage. An upward shift is a positive voltage (higher potential at the red or positive lead of the oscilloscope), while a downward shift is a negative voltage (lower potential at the red or positive lead of the oscilloscope). In general,1. Using the GND option, reset the horizontal line t
