《毕业论文外文翻译-直流值相等.docx》由会员分享，可在线阅读，更多相关《毕业论文外文翻译-直流值相等.docx（29页珍藏版）》请在taowenge.com淘文阁网|工程机械CAD图纸|机械工程制图|CAD装配图下载|SolidWorks_CaTia_CAD_UG_PROE_设计图分享下载上搜索。
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). I
17、n 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 connected will be blocked from entering the oscilloscope only the alternating, or changing, components will be displayed).Note the location
18、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. For the future, whenever you use the AC option, keep in mind that the computer will distribute the waveform above and below the horizon
19、tal 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 components 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
20、(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 aboutThe procedure outlined 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 st
21、arting 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 of any waveform by simply choosing the appropriate scale.13.7 EFFECTIVE (rms) VALUESThis section will begin to relate dc and ac quantities
22、 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 particular dc current. The question frequently arises, How is it possible for a sinusoidal ac quantity to deliver a net power if, over a full cyc
23、le, 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 sinusoidal waveform is withdrawn during the negative portion, and since the two are equal in magnitude, the net power delivered is zero. However, und
24、erstand that irrespective of direction, current of any magnitude through a resistor will deliver power to that resistor. In other words, during the positive or negative portions of a sinusoidal ac current, power is being delivered at eachinstant of time to the resistor. The power delivered at each i
25、nstant 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 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 sinus
26、oidal 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 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
27、 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 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 re
28、ached 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 same 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
29、 same as that delivered by the dc source. The power delivered by the ac supply at any instant of time isThe average power delivered by the ac source is 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 curren
30、t waveform. Equating the average power delivered by the ac generator to that delivered by the dc source,which, in words, states thatthe equivalent dc 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
31、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 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
32、equation derived from the experiment just described: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 wavefor
33、m, to obtain the average or mean value of thesquared waveform. The final step is to take the square root of the meanvalue. This procedure gives us another designation for the effectivevalue, the root-mean-square (rms) value. In fact, since the rms term isthe most commonly used in the educational and
34、 industrial communities,it will used throughout this text.EXAMPLE 13.19 Find the rms values of the sinusoidal waveform in each part of Fig. 13.53.Solution: For part (a), Irms 0.707(12 103 A) 8.484 mA.For part (b), again Irms 8.484 mA. Note that frequency did notchange the effective value in (b) abov
35、e compared to (a). For part (c),Vrms 0.707(169.73 V) 120 V, the same as available from a home outlet.EXAMPLE 13.20 The 120-V dc source of Fig. 13.54(a) delivers 3.6 W to the load. Determine the peak value of the applied voltage (Em) and the current (Im) if the ac source Fig. 13.54(b) is to deliver t
36、he same power to the load.Solution:EXAMPLE 13.21 Find the effective or rms value of the waveform of Fig. 13.55.Solution:EXAMPLE 13.22 Calculate the rms value of the voltage of Fig. 13.57.Solution:EXAMPLE 13.23 Determine the average and rms values of the square wave of Fig. 13.59.Solution: By inspect
37、ion, the average value is zero.The waveforms appearing in these examples are the same as thoseused in the examples on the average value. It might prove interesting tocompare the rms and average values of these waveforms.The rms values of sinusoidal quantities such as voltage or currentwill be repres
38、ented by E and I. These symbols are the same as thoseused for dc voltages and currents. To avoid confusion, the peak valueof a waveform will always have a subscript m associated with it: Imsin qt. Caution: When finding the rms value of the positive pulse of asine wave, note that the squared area is
39、not simply (2Am)2 4A2m; itmust be found by a completely new integration. This will always bethe case for any waveform that is not rectangular.A unique situation arises if a waveform has both a dc and an ac componentthat may be due to a source such as the one in Fig. 13.61. Thecombination appears fre
40、quently in the analysis of electronic networkswhere both dc and ac levels are present in the same system.The question arises, What is the rms value of the voltage vT? Onemight be tempted to simply assume that it is the sum of the rms valuesof each component of the waveform; that is, VT rms 0.7071(1.
41、5 V) 6 V 1.06 V 6 V 7.06 V. However, the rms value is actuallydetermined bywhich for the above example is直流值相等。在电子电路的分析要考虑在后来的过程中，既DC和AC电压源会适用于同一个网络。这将是必要的了解或确定直流（或平均值）和交流电压或电流元件在系统的各个部分。 例13.13确定的波形平均值 图。 13.37。 图。 13.37 例如13.13。 解决方案： 字母a.通过检查，上述地区的轴以下面积等于多 一个周期，在一个零电压的平均值结果。采用式。 （13.26）： 湾采用式。 （1
42、3.26）： 如图所示。 13.38。 在现实生活中，波形图。 13.37（二）仅仅是方波无花果。 13.37（1）为4直流电压转变，也就是说， 2版= V1的+4 V的例13.14查找下列波形的平均值在一个完整周期：字母a.图。 13.39。 湾图。 13.40。 图。 13.38 定义为波形的平均值无花果。 13.37（b）项。图。 13.39 例如13.14，第（1）。根据我们发现在前面的例子曲线地区的使用简单的几何公式。如果我们遇到一个正弦波或任何其他不寻常的形状，但是，我们必须找到一些地区其他手段。我们可以获取该地区良好的逼近试图重现原始波形使用了若干小矩形或其他熟悉的形状，面积，我
43、们已经知道通过简单的几何公式。例如，积极的（或负区）的正弦波脉冲是凌晨2点。 逼近的两个三角形（图13.43）这个波形，我们得到 （使用面积_二分之一基地_为三角形的面积高）一个粗略的概念 实际面积： 保华阿更接近于可能有两个类似三角形的矩形 （图13.44）：这肯定接近实际面积。如果无限人数 形式使用，是凌晨2点确切的答案可以得到的。对于不规则波形，这种方法可能是特别有用的资料，例如 平均值是所期望的。该演算过程，给出了确切的解决办法是凌晨2点被称为整合。整合是这里唯一令 图。 13.41 在直流米响应波形 图。 13.39。 图。 13.42 在直流米响应波形图。 13.40。 图。 13
44、.43 逼近的正脉冲形状 正弦波形的两个权三角形。图。 13.44 一种形状的更好的近似正脉冲波形的正弦。 识别方法给读者，它是没有必要精通它的使用继续这个文本。这是一个有用的数学工具，但是， 并应教训。在积极寻找该地区的脉冲正弦波使用一体化，我们已 是集成，0和p签署的一体化的限制， 其中 我罪一是要集成的功能，达表示，我们结合方面答：整合，我们得到 因为我们知道在正面（或负地区）的脉搏，我们可以很容易地确定正（或负平均值） 一个正弦波脉冲地区运用方程。 （13.26）： 为了图波形。 13.45， （平均相同作为一个完整脉冲） 例13.15确定的正弦平均值 波形图。 13.46。 解决方案
45、：通过检查相当明显一个纯正弦波形超过一个完整周期的平均价值为零。例13.16确定波形的平均价值图。 13.47。 解答：峰值正弦函数的峰值16压_为2 mV _ 18压。正弦的波形峰值振幅因此，18毫伏/ 2 _ 9压。倒数第2至9压压（或9压从_16机动车辆）在一个_7压或DC平均水平的成果，正如由图虚线。 13.47。 图。 13.45 找到的一个平均值的一半正脉冲波形的正弦。图。 13.46 例如13.15。 图。 13.47 例如13.16。例13.17确定波形的平均价值 图。 13.48。 解决方案： 例13.18为了图波形。 13.49，确定是否平均价值是正面还是负面，并确定其近似
46、值。解决方案：从波形的出现，平均价值是积极的，在2压附近。有时候，这个判决类型将要作出。仪表直流水平或任何波形的平均值，可以发现使用一数字万用表（DMM）或示波器。对于纯粹的直流电路，简单地设置的直流数字多用表，并读取电压或电流的水平。示波器仅限于电压等级使用的一系列步骤列举如下： 1。首先选择从直流接地，接地，交流相关的选项列表每个垂直通道。接地模块选择的任何信号其中示波器探头可能会连接进入示波器，只需水平线响应。设置结果在垂直轴中间的水平线轴，如图所示。 13.50（1）。 图。 13.49 例如13.18。（二）垂直灵敏度= 50 mV / div时。移= 2.5股利。（一） 图。 13
47、.50 使用示波器测量直流电压：（一）设置接地的条件; （二）的垂直转移时转移到区选择一个直流电压产生的。2。应聘示波器探头的电压进行测量（如尚未连接），并切换到DC的选择。如果直流电压 是目前，水平线将转向上涨或下跌，因为显示了图。 13.50（b）项。乘以垂直移位敏感性将导致直流电压。一个上移是一个正电压（高潜力在红色或积极的主导作用示波器），而向下转移是负电压（较低的潜能示波器的红色或积极牵头）。一般来说， 伏_（垂直转移组。）_（垂直灵敏度在V /分区。）（13.29） 为了图波形。 13.50（b）项，伏_（2.5分区。）（50毫伏/分区。）_ 125压示波器也可用于测量DC或平均水
48、平波形的任何使用以下顺序： 1。使用接地选项，重置水平线中间在屏幕上。2。切换到交流（信号的所有组成部分的直流探头连接将被阻止进入示波器只有交替，或改变，部分将显示）。请注意一些明确的点的位置上的波形，例如作为下半波整流无花果波形。13.51（1），即说明其对纵坐标位置。对于今后，只要您使用AC选项时，请记住，计算机将派发上述波形及以下水平轴这样的平均值为零，也就是说，面积上述轴将等于下面的区域。3。然后切换到DC（允许同时直流和交流组件对进入波形示波器），并注意转移第2部分选择的水平，如图所示。 13.51（b）项。方程（13.29），然后可以用于确定DC或平均值波形。为了图波形。 13.51（二），平均价值约 变风量_伏_（0.9分区。）（5伏/分区。）_ 4.5 V 图。 13.51确定非正弦波形的平均值使用示波器：（1）垂直渠道的交流方式;（二）在垂直通道DC模式。上述的程序可以适用于任何交流如波形图之一。 13.49。在某些情况下的平均值可能需要动议下波形的起始位置交流选择屏幕的一个不同地区或选择更高电压规模。数字万用表可以读取任何波形或DC平均水平只需选择适当的规模。13.7有效（均方根）值本节将开始与有关直流和交