[畢業(yè)設(shè)計(jì)精品]基于單片機(jī)的超聲波測距系統(tǒng)的研究與設(shè)計(jì) 外文翻譯

1、附 錄附錄a外文翻譯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。在電子電路的分析要考慮在后來的過程中，既dc和ac電壓源會(huì)適用于同一個(gè)網(wǎng)絡(luò)。這將是必要的了解或確定直流（或平均值）和交流電壓或電流元件在系統(tǒng)的各個(gè)部分。 例13.13確定的波形平均值 圖。 13.37。 圖。 13.37 例如13.13。 解決方案： 字母a.通過檢查，上述地區(qū)的軸以下面積等于多 一個(gè)周期，在一個(gè)零電壓的平均值結(jié)果。采用式。 （13.26）： 灣采用式。 （1

42、3.26）： 如圖所示。 13.38。 在現(xiàn)實(shí)生活中，波形圖。 13.37（二）僅僅是方波無花果。 13.37（1）為4直流電壓轉(zhuǎn)變，也就是說， 2版= v1的+4 v的例13.14查找下列波形的平均值在一個(gè)完整周期：字母a.圖。 13.39。 灣圖。 13.40。 圖。 13.38 定義為波形的平均值無花果。 13.37（b）項(xiàng)。圖。 13.39 例如13.14，第（1）。根據(jù)我們發(fā)現(xiàn)在前面的例子曲線地區(qū)的使用簡單的幾何公式。如果我們遇到一個(gè)正弦波或任何其他不尋常的形狀，但是，我們必須找到一些地區(qū)其他手段。我們可以獲取該地區(qū)良好的逼近試圖重現(xiàn)原始波形使用了若干小矩形或其他熟悉的形狀，面積，我

43、們已經(jīng)知道通過簡單的幾何公式。例如，積極的（或負(fù)區(qū)）的正弦波脈沖是凌晨2點(diǎn)。 逼近的兩個(gè)三角形（圖13.43）這個(gè)波形，我們得到 （使用面積_二分之一基地_為三角形的面積高）一個(gè)粗略的概念 實(shí)際面積： 保華阿更接近于可能有兩個(gè)類似三角形的矩形 （圖13.44）：這肯定接近實(shí)際面積。如果無限人數(shù) 形式使用，是凌晨2點(diǎn)確切的答案可以得到的。對于不規(guī)則波形，這種方法可能是特別有用的資料，例如 平均值是所期望的。該演算過程，給出了確切的解決辦法是凌晨2點(diǎn)被稱為整合。整合是這里唯一令 圖。 13.41 在直流米響應(yīng)波形 圖。 13.39。 圖。 13.42 在直流米響應(yīng)波形圖。 13.40。 圖。 13

44、.43 逼近的正脈沖形狀 正弦波形的兩個(gè)權(quán)三角形。圖。 13.44 一種形狀的更好的近似正脈沖波形的正弦。 識別方法給讀者，它是沒有必要精通它的使用繼續(xù)這個(gè)文本。這是一個(gè)有用的數(shù)學(xué)工具，但是， 并應(yīng)教訓(xùn)。在積極尋找該地區(qū)的脈沖正弦波使用一體化，我們已 是集成，0和p簽署的一體化的限制， 其中 我罪一是要集成的功能，達(dá)表示，我們結(jié)合方面答：整合，我們得到 因?yàn)槲覀冎涝谡妫ɑ蜇?fù)地區(qū)）的脈搏，我們可以很容易地確定正（或負(fù)平均值） 一個(gè)正弦波脈沖地區(qū)運(yùn)用方程。 （13.26）： 為了圖波形。 13.45，  （平均相同作為一個(gè)完整脈沖） 例13.15確定的正弦平均值 波形圖。 13.46

45、。 解決方案：通過檢查相當(dāng)明顯一個(gè)純正弦波形超過一個(gè)完整周期的平均價(jià)值為零。例13.16確定波形的平均價(jià)值圖。 13.47。 解答：峰值正弦函數(shù)的峰值16壓_為2 mv _ 18壓。正弦的波形峰值振幅因此，18毫伏/ 2 _ 9壓。倒數(shù)第2至9壓壓（或9壓從_16機(jī)動(dòng)車輛）在一個(gè)_7壓或dc平均水平的成果，正如由圖虛線。 13.47。 圖。 13.45 找到的一個(gè)平均值的一半正脈沖波形的正弦。圖。 13.46 例如13.15。 圖。 13.47 例如13.16。例13.17確定波形的平均價(jià)值 圖。 13.48。 解決方案： 例13.18為了圖波形。 13.49，確定是否平均價(jià)值是正面還是負(fù)面，

46、并確定其近似值。解決方案：從波形的出現(xiàn)，平均價(jià)值是積極的，在2壓附近。有時(shí)候，這個(gè)判決類型將要作出。儀表直流水平或任何波形的平均值，可以發(fā)現(xiàn)使用一數(shù)字萬用表（dmm）或示波器。對于純粹的直流電路，簡單地設(shè)置的直流數(shù)字多用表，并讀取電壓或電流的水平。示波器僅限于電壓等級使用的一系列步驟列舉如下： 1。首先選擇從直流接地，接地，交流相關(guān)的選項(xiàng)列表每個(gè)垂直通道。接地模塊選擇的任何信號其中示波器探頭可能會(huì)連接進(jìn)入示波器，只需水平線響應(yīng)。設(shè)置結(jié)果在垂直軸中間的水平線軸，如圖所示。 13.50（1）。 圖。 13.49 例如13.18。（二）垂直靈敏度= 50 mv / div時(shí)。移= 2.5股利。（一）

47、 圖。 13.50 使用示波器測量直流電壓：（一）設(shè)置接地的條件; （二）的垂直轉(zhuǎn)移時(shí)轉(zhuǎn)移到區(qū)選擇一個(gè)直流電壓產(chǎn)生的。2。應(yīng)聘示波器探頭的電壓進(jìn)行測量（如尚未連接），并切換到dc的選擇。如果直流電壓 是目前，水平線將轉(zhuǎn)向上漲或下跌，因?yàn)轱@示了圖。 13.50（b）項(xiàng)。乘以垂直移位敏感性將導(dǎo)致直流電壓。一個(gè)上移是一個(gè)正電壓（高潛力在紅色或積極的主導(dǎo)作用示波器），而向下轉(zhuǎn)移是負(fù)電壓（較低的潛能示波器的紅色或積極牽頭）。一般來說， 伏_（垂直轉(zhuǎn)移組。）_（垂直靈敏度在v /分區(qū)。）（13.29） 為了圖波形。 13.50（b）項(xiàng)，伏_（2.5分區(qū)。）（50毫伏/分區(qū)。）_ 125壓示波器也可用于測量

48、dc或平均水平波形的任何使用以下順序： 1。使用接地選項(xiàng)，重置水平線中間在屏幕上。2。切換到交流（信號的所有組成部分的直流探頭連接將被阻止進(jìn)入示波器只有交替，或改變，部分將顯示）。請注意一些明確的點(diǎn)的位置上的波形，例如作為下半波整流無花果波形。13.51（1），即說明其對縱坐標(biāo)位置。對于今后，只要您使用ac選項(xiàng)時(shí)，請記住，計(jì)算機(jī)將派發(fā)上述波形及以下水平軸這樣的平均值為零，也就是說，面積上述軸將等于下面的區(qū)域。3。然后切換到dc（允許同時(shí)直流和交流組件對進(jìn)入波形示波器），并注意轉(zhuǎn)移第2部分選擇的水平，如圖所示。 13.51（b）項(xiàng)。方程（13.29），然后可以用于確定dc或平均值波形。為了圖波形

49、。 13.51（二），平均價(jià)值約 變風(fēng)量_伏_（0.9分區(qū)。）（5伏/分區(qū)。）_ 4.5 v 圖。 13.51確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。上述的程序可以適用于任何交流如波形圖之一。 13.49。在某些情況下的平均值可能需要?jiǎng)幼h下波形的起始位置交流選擇屏幕的一個(gè)不同地區(qū)或選擇更高電壓規(guī)模。數(shù)字萬用表可以讀取任何波形或dc平均水平只需選擇適當(dāng)?shù)囊?guī)模。13.7有效（均方根）值本節(jié)將開始與有關(guān)直流和交流的數(shù)量權(quán)力交付給負(fù)載。這將幫助我們確定的振幅一個(gè)正弦交流電流必須將它作為一種同樣的權(quán)力特別是直流電流。經(jīng)常發(fā)生的問題，怎么可能正弦交流的數(shù)量，以提供凈功率的是，經(jīng)過一個(gè)完整周期，凈在任何一個(gè)方向的電流為零（平均價(jià)值0）？它幾乎看來，動(dòng)力，積極發(fā)表波形的正弦部分撤回在負(fù)部分，而且由于兩個(gè)幅度相等，凈功率傳送是零。然而，理解，不論 方向，電流通過電阻器將提供任何規(guī)模權(quán)力的電阻。換句話說，在正面還是負(fù)面部分的正弦交流電流，功率，正在每個(gè)交付時(shí)間瞬間的電阻。交付的權(quán)力在每個(gè)瞬間當(dāng)然會(huì)，而改變正弦交流電流的大小，但將在凈流量無論是正面還是負(fù)面一觸即發(fā)