1、单片机 外文翻译 外文文献 英文文献 基于单片机的超声波测距系统的研究与设计单片机 外文翻译 外文文献 英文文献 基于单片机的超声波测距系统的研究与设计附 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 det
2、ermine the dc (or average value) and ac components of the voltage or current in various parts of the system. EXAMPLE 13.13 Determine the average value of the waveforms of Fig. 13.37. FIG. 13.37 Example 13.13. Solutions: a. By inspection, the area above the axis equals the area below over one cycle,
3、resulting in an average value of zero volts. b. Using Eq.(13.26): as shown in Fig. 13.38. 26 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 V EXAMPLE 13.14 Find the average values of the following waveforms over one ful
4、l cycle: a. Fig. 13.39. b. Fig. 13.40. 27 Solutions: 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
5、 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 wavefo
6、rm by two triangles (Fig. 13.43), we obtain(using area 1/2 base height for the area of a triangle) a rough idea of the actual area: A closer approximation might be a rectangle with two similar triangles(Fig. 13.44): 28 which is certainly close to the actual area. If an infinite number of forms were
7、used, an exact answer of 2Am 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 2Am is known as integration. Integration is presented here only to make the method reco
8、gnizable 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 where ? is the sign of integration, 0 and p are the
9、 limits of integration, Am sin a is the function to be integrated, and da indicates that we are integrating with respect to a. Integrating, we obtain 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 sin
10、e wave pulse by applying Eq. (13.26): For the waveform of Fig. 13.45, 29 EXAMPLE 13.15 Determine the average value of the sinusoidal waveform of Fig. 13.46. Solution: By inspection it is fairly obvious that the average value of a pure sinusoidal waveform over one full cycle is zero. EXAMPLE 13.16 De
11、termine the average value of the waveform of Fig. 13.47. Solution: The peak-to-peak value of the sinusoidal function is16 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 leve
12、l 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. 13.48. Solution: 30 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 t
13、he appearance of the waveform, the average value is positive and in the vicinity of 2 mV. Occasionally, judgments of this type will have to be made. Instrumentation The dc level or average value of any waveform can be found using a digital multimeter (DMM) or an oscilloscope. For purely dc circuits,
14、simply set the DMM on dc, and read the voltage or current levels.Oscilloscopes are limited to voltage levels using the sequence 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
15、probe may be connected from entering the oscilloscope and responds with just a horizontal line. Set the resulting line in the middle 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 s
16、witch 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). Multiplying the shift by the vertical sensitivity will result in the dc voltage. An upward shift is a positive voltage (higher 31 potential at the red or positive lead of
17、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 to the middle of the screen. 2. Switch to AC (all dc components of the signal to which the probe is connec
18、ted will be blocked from entering the oscilloscope only the alternating, or changing, components will be displayed). Note the location of some definitive point on the waveform, such as the bottom of the half-wave rectified waveform of Fig. 13.51(a); that is, note its position on the vertical scale.
19、For the future, whenever you use the AC option, keep in mind that the computer will distribute the waveform above and below the horizontal axis such that the average value is zero; that is, the area above the axis will equal the area below. 3. Then switch to DC (to permit both the dc and the ac comp
20、onents of the waveform to enter the oscilloscope), and note the shift in the chosen level of part 2, as shown in Fig. 13.51(b). Equation (13.29) can then be used to determine the dc or average value of the waveform. For the waveform of Fig. 13.51(b), the average value is about The procedure outlined
21、 above can be applied to any alternating waveform such as the one in Fig. 13.49. In some cases the average value may require moving the starting position of the waveform under the AC option to a different region of the screen or choosing a higher voltage scale. DMMs can read the average or dc level
22、of any waveform by simply choosing the appropriate scale. 32 13.7 EFFECTIVE (rms) VALUES This section will begin to relate dc and ac quantities with respect to the power delivered to a load. It will help us determine the amplitude of a sinusoidal ac current required to deliver the same power as a pa
23、rticular dc current. The question frequently arises, How is it possible for a sinusoidal ac quantity to deliver a net power if, over a full cycle, the net current in any one direction is zero (average value 0)? It would almost appear that the power delivered during the positive portion of the sinuso
24、idal waveform is withdrawn during the negative portion, and since the two are equal in magnitude, the net power delivered is zero. However, understand that irrespective of direction, current of any magnitude through a resistor will deliver power to that resistor. In other words, during the positive
25、or negative portions of a sinusoidal ac current, power is being delivered at each instant of time to the resistor. The power delivered at each instant will, of course, vary with the magnitude of the sinusoidal ac current, but there will be a net flow during either the positive or the negative pulses
26、 with a net flow over the full cycle. The net power flow will equal twice that delivered by either the positive or the negative regions of sinusoidal quantity. A fixed relationship between ac and dc voltages and currents can be derived from the experimental setup shown in Fig. 13.52. A resistor in a
27、 water bath is connected by switches to a dc and an ac supply. If switch 1 is closed, a dc current I, determined by the resistance R and battery voltage E, will be established through the resistor R. The temperature reached by the water is determined by the dc power dissipated in the form of heat by
28、 the resistor. If switch 2 is closed and switch 1 left open, the ac current through the resistor will have a peak value of Im. The temperature reached by the water is now determined by the ac power dissipated in the form of heat by the resistor. The ac input is varied until the temperature is the sa
29、me as that reached with the dc input. When this is accomplished, the average electrical power delivered to the resistor R by the ac source is the same as that delivered by the dc source. The power delivered by the ac supply at any instant of time is 33 The average power delivered by the ac source is
30、 just the first term, since the average value of a cosine wave is zero even though the wave may have twice the frequency of the original input current waveform. Equating the average power delivered by the ac generator to that delivered by the dc source, which, in words, states that the equivalent dc
31、 value of a sinusoidal current or voltage is 1/2 or 0.707 of its maximum value. The equivalent dc value is called the effective value of the sinusoidal quantity. In summary, As a simple numerical example, it would require an ac current with a peak value of 2 (10) 14.14 A to deliver the same power to
32、 the resistor in Fig. 13.52 as a dc current of 10 A. The effective value of any quantity plotted as a function of time can be found by using the following equation derived from the experiment just described: 34 which, in words, states that to find the effective value, the function i(t) must first be squared. After i(t) is squared, the area under the curve isfound by integration. It is then divided by T, the length of the cycle or the period of the waveform, to obtain the average or
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