group 3 sss lab protocol dfm mns 1

8/13/2019 Group 3 SSS Lab Protocol DFM MNS 1

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Fakultt fr Elektrotechnik und Informationstechnik

Professur fr Mess- und Sensortechnik

Smart Sensor Systems - Praktikum[245032-505]

Winter Semester 2013/2014

DIGITAL FREQUENCY MEASUREMENT

Faculty:

Mr.Bilel Kallel

Mr. Ammar Al-Hamry

Dipl.-Ing. Frank Ebert

Dr.-Ing. Thomas Keutel

Laboratory Protocol

By

GROUP 3on 16.01.2014

Micro & Nano Systems (MNS_1)

Name Matrikel-Nr. E-mail

Karthikeyan Manga 334707 [email protected]

LakshmiVaradharajan 353975 [email protected]

Binh Duong Nguyen 336644 [email protected]

LongqianXu 355643 [email protected]


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Lab Protocol_SSS_Group_3_MNS_1_WS2013 Page | 2

CONTENTS

1. Objectives 3

2. Components & Setup 3

3. Theory 4

4. Experimental Procedure 5

5. Experimental Tasks 5

5.1.Measurement in Time Domain 5

5.2.Measurement in Frequency Domain 8

6. Preparatory Tasks 33

7. Conclusion 34

8. References 35


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3. THEORY:

Frequency is usually represented as angular frequency in radians/second, or as in

seconds-1, also known as the unit hertz (Hz). You also can use beats per minute (BPM) and

revolutions per minute (RPM) to represent frequency. Angular frequency (rad/sec) and(Hz)

are related by the following formula: =2. Frequency is also spoken of in correlation to a

phase , which describes an offset of the wave from a specified reference point at the initial

time t0, and is usually given in degrees or radians. Taking the example of a sinusoidal wave, thewaveform function is

expressed in terms of time

as F(t) = Asin(t +),

with amplitude A,

angular frequency ,

and phase as constants.

Periodic analog

signals in real

applications are complex

and can rarely be

represented by a simple

sinusoid. Fourier

analysis is used to

decompose any complex

Figure 1: FFTANA Module


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waveform into a sum of simpler functions, either sines and cosines or complex exponentials.

The frequency components that make up such a signal are often the properties of interest, and

this analysis is known as frequency-domain or spectral analysis.

For a digital frequency acquisition of low-frequency signals, it is sufficient to use one counter,

or timebase. The rising edge of the input signal triggers the number of timebase ticks to be

counted. Because the timebase is of a known frequency, you can easily calculate the frequency

of the input signal.

When the frequency of the digital signal is very high or varies, it is better to use a two-counter

method described below. Note that the same hardware limitation applies to both two-counter

methods. That is, the frequency you are measuring cannot exceed the maximum input

frequency supported by the counter, even though it may exceed that of the internal timebase.

4. EXPERIMENTAL PROCEDURE:

To start the program FFTANA, you first have to call the program MATLAB in Windows.

After that you input FFTANA. To perform the wanted tasksof this experiment, below are

listed time signals in the form of data files, e. g. the file name (sin10_0) contains a sampled

sinusoidal harmonic with 10 time periods shown and recorded in the screen.

Click on the LOAD bottom in the interface of the program to select a signal file and run it.

The data are found in the path: C:\matlab4\toolbox\dfm\

1. sin2_0.mat 2. sin2_5.mat

3. sin10_0.mat 4. sin10_5.mat

5. sin369_0.mat 6. sin368_5.mat

Figure 2 : Digital Signal with Respect to Internal Timebase (One Counter for Low Frequency)

Figure 3 : Digital Signal Frequency Measured with Two Counters (High Frequency)


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7. sin922_0.mat 8. sin921_5.mat

9. rauschen.mat 10. rausch2.mat

11. saegz1_0.mat 12. gleichsp.mat

5. EXPERIMENTAL TASKS

5.1. Measurement in Time Domain

5.1.1 Compare the number of the samples per period which were determined during theexperiment with the results in the preparation task 3.5 for the measuring signals 1 8.

The comparison is tabulated as shown below:

5.1.2 Determine the frequency of the measured signals 1 8 with using the cycle duration in timedomain. Note the following:

- The determination of the cycle duration is based on the measuring arrangement shown in

Figure 1. That means that the comparator switches when crossing the zero of the harmonic

signal. So the time measurement shall always start with the first positive value of the

sampled signal after passing the zero and end before the first positive value after the zerocrossing. The frequency divider is fixed to 1:1.

- Determine the frequency with the help of the duration of just one cycle and also by using

more cycles (10), (signals 4-8).

The results are tabulated as shown below:

SignalSamples/cycle

- Calc.Samples/cycle

- Exp.

1 1024 1024

2 819.2 819

3 204.8 204

4 195.047619 195

5 5.550135501 5.5

5 5.550135501 5.55

6 5.557666214 5.6

6 5.557666214 5.55

7 2.221258134 2.2

7 2.221258134 2.2

8 2.222463375 2.3

8 2.222463375 2.25

Table 1


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5.1.2.1 Calculate the frequencies of the measured signals using number of cycles and the samplingfrequency. Give the values of the measurement deviations (absolute and relative error) of

each frequency that you determined in this experiment.

The results are tabulated as follows :

Signal SamplingRate fs (Hz)

No. ofcycles

Samples/cycle- Calc.

Frequency -Calc. (Hz)

Samples/cycle- Exp.

Frequency -Exp. (Hz)

1 2.00E+07 2 1024 19531.25 1024 19531.25

2 2.00E+07 2.5 819.2 24414.0625 819 24420.02442

3 2.00E+07 10 204.8 97656.25 204 98039.21569

4 2.00E+07 10.5 195.047619 102539.0625 195 102564.1026

5 4.00E+06 369 5.550135501 720703.125 5.5 727272.7273

5 4.00E+06 369 5.550135501 720703.125 5.55 720720.7207

6 4.00E+06 368.5 5.557666214 719726.5625 5.6 714285.71436 4.00E+06 368.5 5.557666214 719726.5625 5.55 720720.7207

7 4.00E+06 922 2.221258134 1800781.25 2.2 1818181.818

7 4.00E+06 922 2.221258134 1800781.25 2.2 1818181.818

8 4.00E+06 921.5 2.222463375 1799804.688 2.3 1739130.435

8 4.00E+06 921.5 2.222463375 1799804.688 2.25 1777777.778

Table 2


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SignalSampling

Rate fs (Hz)No.ofcycles


Frequency -Calc. (Hz)

Samples/cycle- Exp.

Frequency -Exp. (Hz)

SamplingCount (No.of Signals)

Relative errorAbsoluterror (Hz

1 2.00E+07 2 1024 19531.25 1024 19531.25 0.000976563 19.0734862 2.00E+07 2.5 819.2 24414.0625 819 24420.02442 0.001221001 29.816879

3 2.00E+07 10 204.8 97656.25 204 98039.21569 0.004901961 480.58439

4 2.00E+07 10.5 195.047619 102539.0625 195 102564.1026 0.005128205 525.96975

5 4.00E+06 369 5.550135501 720703.125 5.5 727272.7273 55 (10) 0.018181818 13223.14

5 4.00E+06 369 5.550135501 720703.125 5.55 720720.7207 111 (20) 0.009009009 6492.9794

6 4.00E+06 368.5 5.557666214 719726.5625 5.6 714285.7143 56 (10) 0.017857143 12755.102

6 4.00E+06 368.5 5.557666214 719726.5625 5.55 720720.7207 111 (20) 0.009009009 6492.9794

7 4.00E+06 922 2.221258134 1800781.25 2.2 1818181.818 22 (10) 0.045454545 82644.62

7 4.00E+06 922 2.221258134 1800781.25 2.2 1818181.818 44 (20) 0.022727273 41322.314

8 4.00E+06 921.5 2.222463375 1799804.688 2.3 1739130.435 23 (10) 0.043478261 75614.366

8 4.00E+06 921.5 2.222463375 1799804.688 2.25 1777777.778 45 (20) 0.022222222 39506.172

Table 3


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5.1.1 Discuss the results! Under which condition can you actually determine a frequency from thesampled signal?

Frequency from a sampled signal can be obtained under the following conditions

Sampling frequency must satisfy Nyquist Theorem

No. of samples must be relatively higher per cycle.

ADC should have reasonable resolution to measure the desired amplitude range.

5.2 Evaluation in frequency domainA frequency determination in frequency domain is useful, if the original shape of the sampled

signal cannot be represented in time domain anymore, e. g. when the number of sampled values

in one cycle of a harmonic signal is less than 10. Especially when having signals that just exist

for a short time (like the ones that result from measuring methods based on the Doppler Effect)

the frequency domain is advantageous. Then a linear (flat) interpolation within the main lobe

can be useful for such frequency spectra. For such interpolation, the centroid method is used.

5.2.1 Transform the measuring signals 1 8 into frequency domain and discuss the representation.The transformation always happens with a rectangular window (idx 1 2048). Use the ZOOM

function and if necessary print some examples for further explanation (details about the zeros).

The corresponding signals in time domain were transformed to their frequency domain

counterparts using the module FFTANA, as shown below :

T

Figure 4: Signal 2; Frequency = 19.53125 KHz


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Figure 5 : Signal 2.5 (a); Dominant peak Frequency (1) = 19.53215 KHz

Figure 6 : Signal 2.5 (b); Dominant peak Frequency (2) = 29.29688 KHz; Average frequency (KHz)= (19.53215 + 29.29688)

/ 2 = 24.414515 KHz


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Figure 9 : Signal 10.5 (b); Dominant Frequency (2) = 107.4219 KHz; Average frequency (KHz)= (107.4219 + 97.65625) / 2 =

102.539075 KHz


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Figure 10 : Signal 369; Frequency = 720.7031 KHz

Figure 11 : Signal 368.5 (a); Dominant Frequency (1) = 718.7500 KHz


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Figure 12 : Signal 368.5 (b); Dominant Frequency (2) = 720.7031 KHz

Figure 13 : Signal 922; Frequency = 1.800781 MHz


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Figure 14 : Signal 921.5 (a); Dominant Frequency (1) = 1.800781 MHz

Figure 15 : Signal 921.5 (b); Dominant Frequency (2) = 1.798828 MHz


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5.2.2 Load measuring signal 3 and find out the time interval for exactly one cycle (idx j ... idx k). Openthe submenu OPTION and set the idx-values. By pushing OK you get back and see the time

interval with detail showing the single cycle. Transform this signal and discuss the result. Give

reasons why the spectral line with the maximum amplitude can lie on another frequency than

that determined easily in time domain. Repeat this task by setting other time intervals you can

choose on your own.

Figure 16 : Signal 10 in Time Domain

The spectral line with the maximum amplitude can lie on another frequency than that determined easily

in time domain accounting to spectral width error due to

Quantization error (Discretization error)

Very few samples

Low Sampling Frequency


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Figure 17 : Signal 10 in Frequency Domain; Frequency = 78.12500 KHz

5.2.3 Load the measuring signal 12 (do not cut off the offset). Discuss the representation in timedomain. Give the number of bits the A-D converter has to be used when the measuring range for

the sampling of the signal is 2 V.

The measuring signal represented in the time domain as shown in Figure 19 is a DC

signal.

Quantization Voltage level, Vq= (0.664 0.656) V = 8 mV

Levels of Amplitude required to measure 2 V = (2 / 0.008) = 250 ~ 256

Thus, No. of Bits required by the ADC to measure a 2 V range, n, is determined from:

2n= 256 = > n = 8


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Figure 18 : Signal 12; DC Signal in time domain

5.2.4 Show how the spectra of box-car pulses (rectangular) look like. The duration of the pulses haveto be set according to the preparation task 3.4 (OPTION). Compare the roots with the calculated

values.

Frequency, F = 1 / T ; T = Time Period

T = N . Ts ; N = Total no. of Samples

Ts = Sampling time interval

N = 2048 / n ; n = no. of samples per cycle

When n = 10 and Ts= (1 / 20 MHz) = 50 ns

F = 1 / (( 2048/10) . (50 x 10-9)) = 97656.25 Hz = 97.65625 KHz

This value can be verified from the first root on the frequency domain, i.e. 97.65625 KHz


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Figure 19 : Box Car Pulse in time domain

Figure 20 : Box Car Pulse in Frequency Domain


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Figure 21 : Box Car Pulse in time domain (Scaled)

Figure 22 : Box Car Pulse in Frequency Domain


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F = 1 / (( 2048/100) . (50 x 10-9)) = 976562.5 Hz = 976.5625 KHz

This value can be verified from the first root on the frequency domain which is

comparable to the calculated value, i.e. 986.3281 KHz

5.2.5 Evaluate the spectra of box- car pulses according to task 5.2.4for the following pulse durations:six-, two- and one-fold of the sampling cycle duration! Document what you find out.




When N = 6 and Ts= (1 / 20 MHz) = 50 ns

F = 1 / (6 . (50 x 10-9

)) = 3333333.3 Hz = 3.333333 MHz

This value can be verified from the first root on the frequency domain which is comparable to .e.

3.330078 MHz.


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Figure 23 : Spectra of Box Car Pulse (N = 6)


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Figure 24: Spectra of Box Car Pulse (N = 2)





F = 1 / (2 . (50 x 10-9

)) = 10 MHz

This value can be verified from the first root on the frequency domain, i.e. 10.0000 MHz.


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Figure 25 : Spectra of Box Car Pulse (N = 1)





F = 1 / (1 . (50 x 10-9

)) = 20.0000 MHz

This signal cannot be transformed by FFT as the sampling frequency does not satisfy theNyquist Theorem or criterion.


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5.2.6 Load measuring signal 6, make a window of 200 samples (from OPTIONS, e. g. idx 100 to idx300), show the spectrum with the ZOOM function. Calculate the minima of the signal in the

spectrum and decide whether an interpolation within the main lobe of the spectrum is sensible.

Calculate the measuring frequency more exactly by using 6 spectral lines in your calculation

(centroid method). Compare the result with the frequency line that has the maximum amplitude

in the spectrum and with the evaluation in time domain.

Figure 26 : Signal 6


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Figure 27 : Frequency Spectrum of Signal 6



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The frequency for such a signal having a wide amplitude spectrum width can be computed using the

Centroid formula:

F = Sum (fi.Ai) / Sum of (Ai) ; fi = Frequency of dominating frequency

Ai = Amplitude of the spectral line

i = Index

F = 4138.3575 / 5.751301 = 719.5515 KHz

The frequency spectrum with the maximum amplitude is 718.7500 KHz.

fi (KHz) Ai fi.Ai

714.8438 0.907796 648.9323

716.7969 0.968744 694.3927

718.7500 1 717.7500

720.7031 0.999795 720.5554

722.6562 0.968139 699.6317

724.6094 0.906827 657.0954

Sum 5.751301 4138.3575


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The value of frequency of the signal found out from the Centroid formula is comparable to the

frequency with the maximum amplitude.

5.2.7 The measuring signal 9 is a digitalized noise signal (white noise). Determine the spectrum thatis belonging to it and interpret the result.


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Figure 33 : White Noise in Time Domain

Figure 34 : Frequency Spectra of White Noise

As observed from the frequency spectrum of white noise, it consists of superposition of

various signals from a wide range of frequency. FFT decomposes the components of the


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white noise and helps to resolve the frequencies of the individual signals contained in the

white noise.

5.2.8 Measuring signal 10 was measured as an output signal of an operational amplifier, while theinput quantity was white noise generated by a noise generator. Determine the spectrum of the

output signal, interpret it and compare it with the result from task 5.2.7. How would the figure

change if the signal was not sampled just 2048 times, but with a measuring time (measuring

window) that is ten times longer, so that the signal is sampled 20,480 times now?

Figure 35 : Signal 10 in Time Domain


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Figure 36 : Frequency Spectra of Signal 10

When compared with the frequency spectrum from the previous task, it is observed that

the signal components of low frequencies have been filtered. Hence this implies that

the operational amplifier acts as a low pass filter.

And the frequency spectrum does not produce more resolved results upon increasing

the sampling range. Hence no change would be observed in the frequency spectrum

upon sampling 10 times more.


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6. Preparatory Tasks

6.1What are the cycle durations for the given sampling frequencies? Cycle duration = 1/fs

Cycle duration for the the given frequencies are calculated as follows

Frequency Cycle Duration

20 MHz 1/(20 x 106) = 50 ns

10 MHz 1/(10 x 106) = 100 ns

40 MHz 1/(40 x 106) = 250 ns

2 MHz 1/(2 x 106) = 500 ns

6.2. Compare the characteristic values in time and frequency domain. Name the interactions

between (grid points) sampling rate, width of the measuring window (time domain), the number of

(usable) spectral lines, frequency resolution and the maximum frequency (frequency domain).

The relationship between sampling rate, measuring window in time domain is

as follows :

f = 1/ NTA

f = Spectral Width

NTA = Measuring Time Window

Sampling frequency, fs= 1/TA

6.3. For which frequency range is it physically meaningful to describe the calculated amplitudedensity spectrum?

According to the Nyquist criterion, fs2 fm. Hence the frequency of the

measuring signal should satisfy this criterion.

6.4.For a periodic rectangular pulse (a periodic box-car pulse) amounts to the ratio of pulse durationto period duration = window width T=1/f0, a) 1: 10 and b) 1: 100. At which frequencies

(reference frequency f0) the calculation gives roots in the spectrum?





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F = 1 / (( 2048/10) . (50 x 10-9)) = 97656.25 Hz = 97.65625 KHz


F = 1 / (( 2048/100) . (50 x 10-9)) = 976562.5 Hz = 976.5625 KHz

6.5.Calculate the expected number of samples per cycle for the signals 1-8 we want to analyze.Sampled signal 1-4: fsampling=20 MHz; sampled signal 5-8: fsampling=4 MHz

Number of samples per cycle = No. of samples / No. of cycles

Number of samples = 2048

Expected number of samples for signals (1-8) (from experimental procedure)

CONCLUSION:

From the practical experiment carried out, the concepts and operating conditions of Digital

Frequency Measurement were studied. The governing parameters of such a measurement were

also studied. FFT and counter methods are widely employed to analyse signals in various types ofspectroscopies.

SignalNo. of

SamplesNo. ofcycles


1 2048 2 1024

2 2048 2.5 819.2

3 2048 10 204.8

4 2048 10.5 195.047619

5 2048 369 5.550135501

6 4.00E+06 368.5 5.557666214

7 4.00E+06 922 2.221258134

8 4.00E+06 921.5 2.222463375


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REFERENCES:

1. Lab Manual (2014), Digital Frequency Measurement, retrieved on 10.01.2014 from

https://www.tu-chemnitz.de/etit/messtech/material/secure/psss/practical_dfm.pdf

2. National Instruments (2013), Frequency Measurements: How-To Guide, retrieved on10.01.2014 fromhttp://www.ni.com/white-paper/7111/en/
https://www.tu-chemnitz.de/etit/messtech/material/secure/psss/practical_dfm.pdfhttps://www.tu-chemnitz.de/etit/messtech/material/secure/psss/practical_dfm.pdfhttp://www.ni.com/white-paper/7111/en/http://www.ni.com/white-paper/7111/en/http://www.ni.com/white-paper/7111/en/http://www.ni.com/white-paper/7111/en/https://www.tu-chemnitz.de/etit/messtech/material/secure/psss/practical_dfm.pdf

group 3 sss lab protocol dfm mns 1

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