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Note for Electronic Instruments - EI By vtu rangers

  • Electronic Instruments - EI
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Text from page-3

lr/lod'yilp, - 1 M ea.++w et4ae.ant o"aqd-, E r r o-r The deviation of the first reading x I is d i and that of second reading x2 is d2 and so on lhe deviation fiom mean can be expressed 1'he algebraic sum as d, : x, - x, d, :az - x etc - of deviation is zero. Deviations : It is def,ned of an infinite number of data is defined as the square root of the sum of individual deviations squared divided by the number of readings. 3" Average Thus standard deviation is expressed as o= Practically the number of reading is finite, when the number of reading is small (n < 30) the denominator is (n - 1) and the standard deviation is represented by S. di +d?r+d]+......+d: S- The standard deviation is also called as mean square deviation. 5" Variance : The square ofstandard deviation is called variance. v = (o)' = [,8.l= n \r ) " Forsmallnumberof rea drng t 9.6 n Id' V=5"=n_l Problems on Statistical Analysis : For the given readings 1.34, 1.38 , 1.56, 1.47, 1.42, 1.43, 1.54, 1.48, 1.49, 1. Calculate. L."Arithmetic Mean 3. Standard deviation Sol rd' : 1.50. 2. Average deviation 4. Variance Given: Readings ; 1.34, 1.38, 1.56, L47, L43,1.54,1"48,1"49, 1.50 n: 1. 10 [ 10 readings] Arithmetic Mean : n Y" - i=, =a[x, + x2 + x, +,..... + xn] n n' ,(-J'^t x I 1.34 + 1.38 + 1.56 + 1.47 + 1.42 + I "43 + I.54 + 1.48 + 1.49 + 1.50 10 ii. Averagc deviation dr = Xr -x : = I .34 -1.461 = -0"121 d, -- \, - x = 1.38 - I 461= -0.081 Sfl,1star Exo,m Sr,z,nar.r 3

Text from page-4

I III Electrotnbllfi ,*ru,tnrueacf,o,fi.ox^v Sanq/EC/TC d: = X: - x = 1.56 -1.461= +0.099 do = X, - x = 1.47 - 1.461= +0.009 ds = Xs -i=1.42-1.461= +0.041 do = Xe -i =1.43 -1.461= +0.031 dt =x, - x = 1.54 -1.461= +0.079 - x = 1.48 -1.461= +0.019 ds = Xs -i=1.49 -I.461= -0.029 dr = Xs d,o = X,o -f = 1.50 -1.461= 0.039 -.-- f ldl ---..- - r---:-a: Averagedevlatlon ^ n 0.121 + 0.081 + 0.099+0.009 + 0.041 + 0.031 + 0.079 + 0.019 + 0.029 + 0.039 10 iii" Standard Deviation : Since the number of readings is < 30. Standard eviation is S = (o.ztz)' + (o"os /=-; ^lLo' V n-l r)' + (o.oee)'? + (o.ool ' + (o.o+t)' + (o.o: t ls = 01688l iv. Varia nce : V=Sj= (o.ooss)' = 0.004733 iy oro4d = 2. For the following readings calculate Arithmetic mean. ii. Deviation of each value iii. Algebraic sum of deviations i. 49.7 Sol : , 50.1, 50.3, 49.5, 49.7 Given : Readings : n: 5 12{ t!9Ji{'3 49"7,50.1, 50.3, 49.5,49.7 (i)e'ritirmeticMean:x = ,: xra x' I:::::-xo n5 = + 49'5 + 49'7 +q.ae (ii) Deviations from each vaiue d, = X, -i du = xz 4 = 49.7 - - * = SO.i - 49 "86 = -0^ 16 49.86 = +0.24 &ns*ar Fn<a,v.Su,naw

Text from page-5

l,4o*i,e/' 7 M ea.*ttr srvtp-vtt and, E r ror d: =X, -*=SO.:-49 86=+0"44 =X: -*=qg.S-49"86 =-0"36 ds =Xs -'x=49.1 -49.86=-0.16 do (iii)elgeUraic sum of deviation is 3. d,o,ur = -0'1 5 + 0'24 + 0'44 dtotur =+0'68-0'68 d,o,nt =0 - 0"36 - 0'16 The accuracy of five digital voltmeters are checked by using each of them to measure a standard 1.0000V from a calibration instrument. The voltmeter readings are as follows : V, = 1.001V, V, = 1.002, V. = 0"999V, V. = 0"998 and V, = 1'000 calculate the average measured voltage and average deviat;on. Sol : i. Average Measure Voltage _y+v,+%+vo+v, ;lr, _ A_ yav 5 = u", 1.001+ 1.002 + 0.999 + 0.998 + 1.000 (.- =i tr- 5 rvl ii. Average deviation , la,l+la,l+la,l+la.l+la'l _:T d, =V, -Vo, =1.001-1.000=0.001 d, =Y, - V"u = 1.002 - 1.000 = 0.002 d, =V, -V", =0'998-1'000=-0.001V - Vu, = 0.998 -1.000 = -0.002V d, = % - V", = 1.000 - 1.000 = 0V do = Vo a= 0.001+ 0.002 + 0.001 + 0.002 + 0 a-o'ooq=o.oo12 5 ld = 0.0012V1 Note: Probable error =0.67 45 o o = Standard deviation = Surrs*ar Exam Ex,arrnw di + d3 +...... + di n s

Text from page-6

III 1.7 SsywEClTC EldrowLo I yt ttr uwnenta.ttow Probability of Error : All observations include a large number of small random errors that may be both +ve and -ve. There is a greater probability of small and large errors. Probable'error of one reding rr = 0.6745o : (or) o"67a5s o forn>30 Sforn<30 Probable error of mean: I .- Vn-i t.or n <30 r,^ r_ = __6 forn >30 Vn 1.8 Limiting Errors : The manufactures specifu the accuracy of the instruments r,vithin a certain percentage of scale reading' R * 5%" 'llhis percentage inriicates the deviations tiom the national or : full lpecifled value of particular quantity" These c{eviations from the specified value are called Limiting Errors. lf The relative error is deflned as the ratio of the error to the specified magnitude of a quantity" Es Specified quantify- of a parameter & * Ea -+ Actual or Erroneous quantity then Error: Ea - Es RelativelimitingError p. = E, 'E" Percentage limiting Error o/oBr -E, =8, * 100 I 1.9 Problems 1. on Limiting Errors l For the following resistance yalues Rl 25Q * lYo' IL * 1.0f2, R. = 45f) + 47o Determine the following : Limiting Values limiting error i. Limiting value of resultant 1. 2. Fercent Sol : R, =25 ' res.istance. ti-r25t1.0f2 R,- =65r 100 J -100 x6-5= 65 :2.6t) R,=45=-J_'45=45:lge ' I00 6 Sqds*ar E-xaai ScannE r

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