自動檢測技術(shù)

1、自動檢測技術(shù)-許大宇-作業(yè)解 答第一章作業(yè)對某節(jié)流元件（孔板）開孔直徑d20進(jìn)行了 15次測量，測量數(shù)據(jù)如下（單位：mm）120.42120.43120.40120.42120.43120.39120.30120.40120.43120.41120.43120.42120.39120.40120.39試用格拉布斯準(zhǔn)則判斷上述數(shù)據(jù)是否含有粗大誤差，并寫出其測量結(jié)果。解：DiviV 2iviV. 2120.420.0160.0002560.0085717.34694E-05120.430.0260.0006760.0185710.000344898120.4-0.0041.6E-05-0.0114

2、30.000130612120.420.0160.0002560.0085717.34694E-05120.430.0260.0006760.0185710.000344898120.39-0.0140.000196-0.021430.000459184120.3-0.1040.010816120.4-0.0041.6E-05-0.011430.000130612120.430.0260.0006760.0185710.000344898120.410.0063.6E-05-0.001432.04082E-06120.430.0260.0006760.0185710.000344898120.

3、420.0160.0002560.0085717.34694E-05120.39-0.0140.000196-0.021430.000459184120.4-0.0041.6E-05-0.011430.000130612120.39-0.0140.000196-0.021430.000459184Gd = 120.4114V 2i15 -1 =0.032689v 0.08826, Gd = 120.4114V 2i15 -1 =0.032689v 0.08826, d 應(yīng) 新平均值：7支did = 4=115=120.404Gb = 2.7*0.0326891=0.08826i7=114 -1

4、=0.01612 n -12=0.0428均符合要求剔去算術(shù)平均值的標(biāo)準(zhǔn)差為:剔去b =咯=0.0043d = d + ZG = 120.4114 土 3 * 0.0043 = 120.4114 土 0.0129 20用手動平衡電橋測量電阻虬，已知R1=100Q, R2=1000Q, Rn=100Q，各橋臂電 阻的恒值系統(tǒng)誤差分別為 R1=0.1Q,AR2=0.5Q,ARn=01Q。求消除恒值系 統(tǒng)誤差后的虬值。解：RR=1 NR2r = RR=1B100=10 TOC o 1-5 h z x 0 R21000_ 8RdRdRAR = x AR + x AR + ARx dR1dR26RNR

5、AR1R AR1冷 AR2 + R ARNIB . 0.1 - y0 . 0.5+IB . 0.1 1000100021000= 0.015R = R 0 -AR = 10-0.015 = 9.985Q測量某電路的電流I=225mA,電壓U=12.6V,標(biāo)準(zhǔn)差分別為=0.5mA, % =0.1V,求所耗功率P=UI及其標(biāo)準(zhǔn)差。解：P = UI = 22.5mA - 12.6V = 283.5mW = 0.2835Wb =生% 2 + f 壟Tb 2 = Jl2b 2 + U2。2p 此8U) u 8IJ i * u I=(22.52 * 10-6 *0.12 + 12.62 * 10-6 *

6、0.52=0.00669P = 0.2835 土 Za p = 0.2835 土 0.020069, Z = 3第二章作業(yè)將一 R=120。，k=2.0的應(yīng)變片粘貼于軸向拉伸試件表面，應(yīng)變片軸線與試件軸線平行。試件材料為鋼，彈性模量E=2.1X 10iiN/m2。若施加應(yīng)力o=3000X105N/m2,求：（1）應(yīng)變片的阻值變化多少？（2）如將此應(yīng)變片粘貼于可產(chǎn)生較大彈性變形的試件，當(dāng)應(yīng)變從零增加到5000m時，應(yīng)變片阻值變化多少？（3）另一半導(dǎo)體應(yīng)變片的R=120 Q，k=100,也粘貼于上述鋼試件，當(dāng)試件應(yīng)力o=3000 X 105N/m2時，應(yīng)變片阻值變化多少？-解：b3000 x105

7、 N m2(1) & = = 1.43 x10-3E2.1x1011 N m2應(yīng)變片的阻值變化AR = K 0 Rs = 120 x 2 x 1.43 x 10-3 = 0.34( Q)(2) AR = K0Rs = 120 x2x5000 x 10-3 = 1.2(Q) AR = K0R E = 120 X100 X 3 = 17.1(Q)習(xí)題2-4已知：采用4等臂電橋（RjRjRjRjR） ,R=350Q, k=2.07,P=0.1W,檢流計內(nèi)阻Rp=350Q，靈敏度 KP=0.003mm/pA o（1）求這一系統(tǒng)（應(yīng)變片-電橋-檢流計）的靈敏度；（2）如以R=120 Q應(yīng)變片取代R=35

8、0Q的應(yīng)變片，上述系統(tǒng)的靈敏度能否提高？-解：八也 2 ）2 . （U 2 ）2P =x 4 = 120 x 4 = 0.1U = 5.96（V）電橋靈敏度：Su = U = 5.96檢測計將電橋輸出電壓轉(zhuǎn)換為電流傳遞函數(shù)為：S =-L = -L = 1_ i Aw R 100總靈敏度為：S = k - S - S - K =2.07x5.96x x0.003x106=241.976U I P100不提高。由以上靈敏度表達(dá)式可見，總靈敏度與應(yīng)變片電阻值無關(guān)。第三章作業(yè)某線性差動變壓器式傳感器用頻率1kHz，峰-峰值為6V的電源“1激勵，如圖（a） 所示。設(shè)銜鐵運(yùn)動d是頻率為100Hz的正弦運(yùn)動，位移幅值為 5mm。已知傳 感器的靈敏度為3V/mm，試畫出銜鐵位移d和輸出電壓uo的波形并說明理由。(a)uo(c)位移頻率為輸入激勵頻率的1/10,輸出電壓與位移同步。一臺變極極距非接觸式電容測微儀，其極板的極限半徑r=4mm，假設(shè)與被測工件的初始間隙d0=0.33mm,試求:i=j（1）若極板與工件間隙的變化量，則電i=jA.d H10R mL容的變化量是多少?l=Jw（2）若測量電路的靈敏度Ku=100mV/pF，則在l=JwAd 1叼時的輸出電壓是多少？ m解：（1）C0-孑=&85x*；：*X10-3）2 -1.347X10-1