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Note for Electronic Instruments - EI By vtu rangers

  • Electronic Instruments - EI
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Text from page-2

I Elprh,e,u-oIn*&.wrweatf,a.{uotu IItr SetwEC/TC A constant deviation of the operation of the instrument is known as 'Ihe systematic errors are mainly resulting due to the short comings of systematic errors. the instrument and the characteristics of the instrument, such as wom parts, ageing efrect, Systematic Errors : environmental effects etc. There are 3 types of systematic errors L lnstrumentai Errors 2" Envirorunental Errors 3. Observed Errors. Random Errors : The random errors are accidental, small and independent" These errors cannot be predicted and cannot be determined in the ordinary process of taking the measurement. These errors are generally small. I-lence these errors are of real concern only u'hen high degree of accuracy is required" 'fhe only way to reduce these errors is by increasing the number of obsen'ations and using the statistioai method to obtain the best approximation of the reading' t4.4 Absolute and Relative Errors : Absolute Error : When the error is specified in terms of an absolute quantity and not as a percentage, then it is cailed an absolute error^ * 0.5V as an absolute elror. Relative Error : When the en"or is expressed as a percentage or as a fraction of the toral quantity to be measured, then it is called relative error' Eg : 10 + 0.5V indicates Eg:100f) t /. 5% then + 5%o \ or + { } ]it tt . r.lative error [ 20, Relative error is also called fractional error. t,{"5 Statistical Analysis ' : The objective of the statistical method is to achieve consistencl'of the measured value and not their accuracy. To make statistical analysis meaningful a large number of measurements is usually required. 1. Arithmetic mean : The most probable value of measured variable is the arithmetic mean of the number of readings taken. The best approximation of the quantity is possible when the number of readings of the quantity is very large' The arithmetic mean of n measurements of the variable x is given by the expression. Xl + X: t -. -"-, Where i *""""* Xn lxn -+ fuithmeticmean xr toxn -+ n X, i" tonth readingtaken -> Total number of readings 2. Deviation from the mean : This is the deviation of a given reading from arithmetic mean of the group ofvalues. ' &nsfar Facal,,.Sr,z.nnw

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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

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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

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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

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