matlab assignment(complete)

7/24/2019 Matlab Assignment(COMPLETE)

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EMC 201 MEASUREMENT AND INSTRUMENTATION

SEM 1 2015/2016

NAME : MUHAMAD NURHAFIZ BIN RAHIM ( 111551)SCHOOL : MECHANICAL ENGINEERING

LECTURER : DR NORZALILAH MOHAMAD NOR


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EXPLANATION OF HOW THE DATA IS EXTRATED

F! "#$%&' *!+,-+. * 2000H

t = 1/fs= 1/2000

= 0.5ms

F! "#$%&' *!+,-+. * 500H

t = 1/fs

= 1/500

= 2ms

F! "#$%&' *!+,-+. * 1000H

t = 1/fs

= 1/1000

= 1ms

F! "#$%&' *!+,-+. * 000H

t = 1/fs

= 1/4000

= 0.25ms

Therefore, the image is overlayed with vertical lines with the interval of 0.5ms, 2ms, 1ms and

0.25ms for the sampling fre!encies of 2000"#, 500"#, 1000"# and 4000"# respectively. Then

the corresponding voltages are ta$en and shown in the ta%les respectively.


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(A) S#$%&' F!+,-+. * 2000 H

S#$%&+3 D#4# *! S#$%&' F!+,-+. * 2000 H

M#4 C3+

&'!rpose( i)To o%tain the fre!ency spectr!m!sing fft f!nction in *atla%

& ii)To determine the fre!ency

components and their corresponding amplit!des

Time, t+ms)

oltage,

V(V)

N

T'$+ ($") 7&4#+(7)

14 .5 -.4

15 -

1 -.5

1- 10

1 .5 2

1 -.5

20 .5 4.5

21 10 0.5

22 10.5

2 11 4.

24 11.5 2.525 12

N T'$+ ($") 7&4#+(7)

1 0 5.5

2 0.5 -

1 2

4 1.5 105 2

2.5 -.1

-

.5 2.5

4 2

10 4.5 4.5

11 5 4.5

12 5.5 1.5

1 4


4/16

close all &clear all the data

fid = fopen+3*62000"#.t7t3, 3r3) &open the file from the te7t

8 = fscanf+fid, 3&g &g3, 92 inf:) & read the data from the 2 col!mns of the te7t file;=83 & transfer the data from the te7t file

t=;+(,1) & 1 refer to the data at col!mn 1

v=;+(,2) &2 refer to the data at col!mn 2plot+t,v)

7la%el+3Time+ms)3)

yla%el+3


5/16


6/16

M#4 C3+

&'!rpose( i)To o%tain the fre!ency spectr!m !sing fft f!nction in *atla%& ii)To determine the fre!ency components and their corresponding amplit!des

close all &clear all the datafid = fopen+3*6500"#.t7t3, 3r3) &open the file from the te7t

8 = fscanf+fid, 3&g &g3, 92 inf:) & read the data from the 2 col!mns of the te7t file

;=83 & transfer the data from the te7t filet=;+(,1) & 1 refer to the data at col!mn 1

v=;+(,2) &2 refer to the data at col!mn 2

plot+t,v)

7la%el+3Time+ms)3)yla%el+3


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S#$%&+3 D#4# *! S#$%&' F!+,-+. * 1000H

N T'$+ ($") 7&4#+(7)

1 0 5.5

2 1 2

2 5

4

5 4 2

5 4.5

- 4

-

10

10 -.5

11 10 0.5

12 11 4.

1 12

oltage,

V(V)

Time, t+ms)


8/16

M#4 C3+

&'!rpose( i)To o%tain the fre!ency spectr!m !sing fft f!nction in *atla%& ii)To determine the fre!ency components and their corresponding amplit!des


8 = fscanf+fid, 3&g &g3, 92 inf:) & read the data from the 2 col!mns of the te7t file;=83 & transfer the data from the te7t file

t=;+(,1) & 1 refer to the data at col!mn 1


7la%el+3Time+ms)3)

yla%el+3


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(D) S#$%&' F!+,-+. * 000H

S#$%&+3 D#4# *! S#$%&' F!+,-+. * 000H

N

T'$+ ($") 7&4#+(7)

oltage,

V(

V)

Time, t+ms)

N T'$+ ($") 7&4#+(7)

1 0.00 5.50

2 0.25 -.00

0.50 -.00

4 0.-5 5.00

5 1.00 1.00

1.25 -.00

- 1.50 10.50

1.-5 10.50

2.00 5.00

10 2.25 2.00

11 2.50 .00

12 2.-5 -.20

1 .00 .00

14 .25 4.50

15 .50 2.50

1 .-5 0.00

1- 4.00 2.00

1 4.25 .50

1 4.50 4.5020 4.-5 5.00

21 5.00 4.50

22 5.25 .50

2 5.50 1.50

24 5.-5 0.50

25 .00 4.00


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2 .25 .00

2- .50 -.20

2 .-5 .50

2 -.00 .00

0 -.25 2.00

1 -.50 .00

2 -.-5 10.00

.00 10.00

4 .25 .00

5 .50 4.00

.-5 -.00

- .00 -.20

.25 .50

.50 5.00

40 .-5 .0041 10.00 0.50

42 10.25 1.50

4 10.50 .00

44 10.-5 4.0

45 11.00 4.0

4 11.25 4.00

4- 11.50 2.50

4 11.-5 0.50

4 12.00 2.50

M#4 C3+

&'!rpose( i) To o%tain the fre!ency spectr!m !sing fft f!nction in *atla%& ii) To determine the fre!ency components and their corresponding amplit!des


8 = fscanf+fid, 3&g &g3, 92 inf:) & read the data from the 2 col!mns of the te7t file

;=83 & transfer the data from the te7t filet=;+(,1) & 1 refer to the data at col!mn 1


7la%el+3Time+ms)3)yla%el+3


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amplit!de=a%s+y)A2/@ &calc!late the act!al amplit!de

f=+1(@)/@ &normali#e the scale

fre!ency=fAf6sampling &calc!late the act!al fre!encyfig!re, plot+fre!ency,amplit!de) &plot a graph of act!al amplit!de vers!s act!al fre!ency

7la%el+3?re!ency+"#)3)

yla%el+38mplit!de3)title+3?re!ency >pectr!m for >ampling ?re!ency 4000"#3)

Figure $:Graph "f Voltage(V) Versus Time(s) Figure %: Frequency

Spectrum

EXTRACTED FRE8UENC9 SPECTRA

(A) F! S#$%&' F!+,-+. * 2000 H

(B) F! S#$%&' F!+,-+. * 500 H

F!+,-+.(H) A$%&'4-3+ (7)

-1.4 1.14

142. 2.51

214. 5.14-

F!+,-+.

(H)

A$%&'4-3+(7)

0 0.0

10 0.254

240 0.11

20 0.055--400 .-

40 0.15

50 2.

40 0.1

-20 1.20-

00 0.11-

0 0.25

0 0.451040 0.44


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25.- 4.10

(C) F! S#$%&' F!+,-+. * 1000 H


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(D) F! S#$%&' F!+,-+. * 000H

N F!+,-+.(H) A$%&'4-3+(7)

1 1. 0.142

2 1.2 0.55

244. 0.4042

4 2.52 0.004

5 40.15 .

4.- 0.42

- 5-1.41 2.4

5.04 0.22-

-4.- 1.04

10 1. 0.1-1

11 -. 0.42

12 -.5 0.1114

1 101.1 0.20

14 1142.2 0.0-

15 1224.45 0.141

1 10.0 0.002

1- 1-.-1 0.15-1 14.4 0.01

1 1550.- 0.0-41

20 12. 0.110

21 1-14.2 0.0-

22 1-5. 0.10-

2 1--.4 0.10

24 15.12 0.101

25 2040.-5 0.052

F!+,-+.(H) A$%&'4-3+(7)

-.2 0.14

15. 0.5

20. 0.-

0-.- 0.554

4. .-

41.5 2.0

5.5 2.2


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COMPARISONS BETWEEN ACTUAL AND MEASURED DATA

(A) F! S#$%&' F!+,-+. * 2000H

@o.

?re!ency, f +"#) 'ercentage ofBifference

8mplit!de, +


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'ercentage of 'ercentage of

Bifference +&)

8ct!al *eas!red 8ct!al *eas!red

1st 22. 214. .5- .- 5.14- 2.1

2nd 4. 25.- 40. 2. 4.10 41.5

rd 45.2 0.

(C) F! S#$%&' F!+,-+. * 1000 H

@o.

?re!ency, f +"#) 'ercentage ofBifference

8mplit!de, +


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fre!encies arenDt. Therefore, the first and second amplit!de o%tained contain small percentage

error of 4.& and 2.0-& whereas the third signal has large percentage error of 14.11&.

Therefore, aliasing occ!rs for the 45.2"# signal fre!ency.

F! "#$%&' *!+,-+. * 000 H, the @y!ist fre!ency is 2000"#. of the signal

fre!encies detected are less than this @y!ist fre!ency. "ence, the percentage error for thefre!ency and the amplit!de o%tained are small which range from 1& to 2-& and 1& to 25&

respectively. Therefore, no aliasing occ!r.

@y!ist fre!ency state that signals with fre!ency lower thanf&yq =fs'2are acc!rately

sampled. >ignals with fre!encies greater than or e!al tof&yq are not acc!rately sampled and

appears as lower fre!encies in the discrete sample.

Ef the fre!ency !sed is lower than the @y!ist rate. Et will e7hi%it aliasing. 8liasing is the

presence of !nwanted components in the reconstr!cted signal and it occ!rs whenever the @y!ist

fre!ency falls %elow the signal fre!ency.

Therefore to get more acc!rate res!lt, it is necessarily to sample at twice the ma7im!m

fre!ency of the signal. The sampling theorem state that a signal can %e e7actly reprod!ces if itis sample at a fre!ency, ? where ? is greater than twice the ma7im!m fre!ency in the signal.

;y comparing fig!re 1, , 5 and - with the trace from the oscilloscope, it is fo!nd that

sampling fre!ency of 2000"# and 4000 "# can plot a !ite similar graph to the original tracefrom the oscilloscope whereas sampling fre!encies of 500"# and 1000 "# acts otherwise. This

is %eca!se aliasing occ!r in %oth 500"# and 1000 "# sampling fre!encies.

CONCLUSION:

?rom the a%ove calc!lation !sing *atla%, it is concl!ded that aliasing will occ!r if the

highest signal fre!ency is greater than the @y!ist fre!ency.

?or this trace from an oscilloscope +

matlab assignment(complete)

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