lab 5 long repot

8/13/2019 Lab 5 Long Repot

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Title: TMS320C6713 Fast Fourier Transform (FFT)

1.0 Objectives

At the end of this laboratory, student should able to:

i) Implement FFT using TMS320C6713 DSK.

ii) Analyze the frequency content of signal by implementing FFT on the targeted

DSK through a digital oscilloscope.

iii) Apply Matlab programming and compare the programming result with the

experimental result.

2.0 Equipment

i) TMS320C6713 DSKii) Signal generator

iii) Digital oscilloscope

iv) PC/workstation (installed with Code Composer Studio and Real Time DSP

Training System softwares)

3.0 Safety Guide

i) The DSK board should be handled carefully as some of the modules are very

fragile.

ii) Do not touch on unlikely components or parts (electrostatics energy from the

body might disturb the components attributes)

iii) Always wear the Lab Coat and shoes (fully-covered) when attending the lab

session.


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4.0 Theory

4.1 Fourier Transform

The Fourier Transform converts signals from a time domain to a frequency

domain. It is the basic method for many sound analysis and visualization algorithms.

Fourier transform can convert many type of signal that made up from sine and cosine

function into magnitudes and phases. The example of Fourier Transform graph is as

shown in the Figure 1 and Figure 2 below. [1]

Figure 1: Analog signal Figure 2: Fourier transform signal

From the Figure 1, we are taking 32 real valued points and the signal is then

undergoes Fourier transform and gets the signal shows in Figure 2. For the Fourier

transform, the plotting only takes consider the magnitude of the signal. The first 16 points

gives the result in Figure 2 while the other 16 points are the mirror image due to the

symmetry analog signal.

Notice spikes at entries 2, 5, and 7, which correlate to the periods of the

components of the input function. The complex value associated to the magnitude at 2 is

-1.3787+2.35648 i. The real and imaginary parts are of similar magnitude and are larger

than surrounding values. The similarity in magnitude is because the phase of the 2 t

component comes from two equal amplitude parts.

The component of the period 5 piece is 2.61789 1.00959 i, where the real and

imaginary parts have slightly more separation in magnitude. This is because the phase of

that part is "less" out of sync with the overall period of the signal. This combination of

magnitude and phase in each output entry gives the "strength" and relationship between

the various frequencies underlying the signal.


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4.2 Fast Fourier Transform

The fast Fourier transform (FFT) is a discrete Fourier transform algorithm that

reduces the number of computations needed for N points from 2N 2 to 2Nlog 2 N. If the

function that want to be transform in non harmonic frequency, the response of the FFT

will look like a sinc function. Spectrum leakage can happen in FFT but it can be reduce

apodization using apodization function. However, aliasing reduction is at the expense of

broadening the spectral response. [2]

FFTs were first discussed by Cooley and Tukey (1965), although Gauss had

actually described the critical factorization step as early as 1805 (Bergland 1969, Strang

1993). A discrete Fourier transform can be computed using an FFT by means of

the Danielson-Lanczos lemma if the number of points N is a power of two. If the numberof points N is not a power of two, a transform can be performed on sets of points

corresponding to the prime factors of N which is slightly degraded in speed.

The Fast Fourier Transform (FFT) is essentially the high speed implementation of

the DFT. The two approaches for implementing the FFT are decimation in time (DIT)

and decimation in frequency (DIF).

The DIT is:

= 2=

+ ( ) 2=

The DIF is:

= 22 1

=0

+ 22 1

=0


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The frequency resolution of the FFT and DFT can be calculated by using the

equation below:

= where f res is the frequency resolution of FFT

f s is the sampling frequency

N is the FFT sample size

Calculation of the frequency obtained at the oscilloscope by following the steps below:

FFT sample size, N=1024 samples and sampling frequency, f s=8000 Hz are given

Frequency resolution, f res =

= = 80001024 = 7.8125 Hz

Sampling interval, T s =1

=1

8000 = 0.125 ms

Time duration between 2 peaks = division between 2-peak horizontal time-scale

= 8 divisions 20ms = 96 ms

Sample size between 2 peaks, n = =96

0.125 = 768 samples

Sample size for half peak,2 =

768

2 = 384 samples

Half FFT sample size,2 = 1024

2 = 512 samples

Peak FFT sample size, 512 384 = 128 samples

Peak frequency, f pk = f res 128 samples

= 7.8125 128 = 1000 Hz

4.3 Nyquist Sampling Theorem

Nyquist sampling theorem for digitalization on an analog signal is defined as the

minimum constrain on the sampling frequency of the analog signal. The sampling

frequency is refers to the number of samples that the system needs to process every

second of signal. For a perfect reconstruction, the Nyquist sampling criterion constrains

the minimum sampling rate to be greater than or equal to twice the maximum frequency

content in the signal. [3]


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In the equation form, Nyquist theorem can be written as:

2 where: f s is sampling frequency

f m is modulation or analog frequency

5.0 Procedures

5.1 Hardware Connectivity:

1) The connection was constructed as shown below in Figure 3.

2) The signal generator was set to sine wave 1000Hz, 2Vp-p. The signal was

monitored at the oscilloscope channel 1.

3) The TMS320C6713 DSK board was connected and the power was turned on.

Figure 3: Hardware connection diagram

5.2 Software Connectivity

Code Compose Studio

1) The DSK CCStudio icon on the workstation was double-clicked.

2) Debug Connect menu was used to establish a connection to DSK board.3)

The Fast Fourier Transform (FFT).pjt Code Compose was opened. It was locatedin the directory C:\CCStudio\real time dsp training system\Fast Fourier Transform

(FFT).

4) The file Fast Fourier Transform (FFT).pjt was chosen by choosing Project Open.


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Reviewing the Source Code:

1) In the Project View Window, Fast Fourier Transform (FFT).pjt was double-

clicked and the Source Folder was selected in the Project View Window.

2) fft.e file was double-clicked in the Project View to open the source code of the

program.

Build and Run the Program:

1) Project Rebuild all was chosen. All the files in the project were recompiled,reassembled, and relinked by the program. The process was continued only when

the build f rame at the bottom of the window displays message 0 Warning and 0

Error.

2) Fast Fourier Transform (FFT).out file was loaded by selecting File Load

Program. A file browser dialog was opened. Fast Fourier Transform (FFT).out

file was selected in the Debug directory in the file browser and the Open button

was hit to load the executable file. The compiled executable was reloaded every

time changes have been made to the program.

3) Debug Run option wa s selected under the Debug menu to run the program.

4) When the program is indeed running correctly, the program was stopped by

clicking Halt toolbar button on Debug Halt option.

Observation and Results:1) The FFT spectrum of fundamental frequency 1000Hz (at function generator) was

monitored at the digital oscilloscope channel 2. The oscilloscopes horizontal time

scale was set at 20ms (or until get 2 spectrums appeared at the monitor). The

run/stop button at the oscilloscope control panel was pressed and the FFT of the

output spectrum was calculated.

2) The accuracy of the result from channel 2 was compared with the theory from

DFT calculation.

3) The frequency at the signal generator was changed to 2 kHz and 4 kHz. Based on

all the result of this experiment, the fundamental frequency that aliasing

phenomenon occurred was determined. The detail was also explained and

discussed.


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6.0 Results

1) Fundamental frequency = 1000 Hz

Figure 4: Frequency set 1000Hz at signal generator

Figure 5: Waveform with 1000Hz frequency from signal generator

Figure 6: FFT spectrum at channel 2 in oscilloscope for 1000Hz


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Calculation of spectrum obtained for fundamental frequency = 1000 Hz

Time division = 20 ms

Amplitude division = 500 mV

FFT sample size, N = 1024 samples

Sampling frequency, f s = 8000 Hz


=8000

1024

= 7.8125 Hz


=

1

8000

= 0.125 ms

Time duration between 2 peaks = no of division time/division

= 4.8 divisions 20ms/div

= 96 ms

Sample size between 2 peaks, n =

=96

0.125

= 768 samples


768

2

= 384 samples

Half FFT sample size,2 =

1024

2

= 512 samples

Peak FFT sample size, 512 384 = 128 samplesPeak frequency, f pk = f res 128 samples

= 7.8125 128 = 1000 Hz


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2) Fundamental frequency = 2000 Hz





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=8000

1024

= 7.8125 Hz


=

1

8000

= 0.125 ms



= 64 ms


=64

0.125

= 512 samples


512

2

= 256 samples


1024

2

= 512 samples

Peak FFT sample size, 512 256 = 256 samples

Peak frequency, f pk = f res 256 samples

= 7.8125 256 = 2000 Hz


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3) Fundamental frequency = 4 kHz,





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=8000

1024

= 7.8125 Hz


=

1

8000

= 0.125 ms



= 400 ms


=400

0.125

= 3200 samples


3200

2

= 1600 samples


1024

2

= 512 samples

Peak FFT sample size, 512 1600 = 1088 samplesPeak frequency, f pk = f res 1088 samples

= 7.8125 1088 = 8500 Hz


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Observation:

Fundamental frequency, f m (Hz) Peak frequency, f pk (Hz)

1000 1000

2000 2000

4000 8500

Table 1: Result obtained

When the fundamental frequency, f m is at 1000 Hz and 2000 Hz, the calculation of the

output spectrum give back the value of the fundamental frequency. But when

fundamental frequency, f m is at 4000 Hz, the calculation of the output spectrum gives

result of 8500 Hz instead of the desired value 4000 Hz.

Therefore this means that the fundamental frequency that aliasing phenomenon occur at

the spectrum is at 4000 Hz.

7.0 Exercise

1) A continuous time signal is defined as = cos 2 1000 + 12

cos 2 2000

i) If the signal is sampled at 8000Hz, what is the suitable sample size for DFT that

can represent both signal components?

Solution:


Frequency resolution, f res = f 1 f 2

= 2000 1000

= 1000

Sample size, N =

,

, =

8000

1000

= 8


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ii) Plot the DFT of the signal by using Matlab for N = 16 and N = 32.

Figure 13: Matlab coding to plot DFT


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Figure 14: Graph obtained for N=16



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2) A discrete-time signal is defined as = cos4

.

i) Calculate he DFT for N = 8 and N = 16.

( ) = cos ( /4)

( ) = [ 4 + 42

]

= 21

=0

= 12 4 +12

4 21

=0

For N = 8,

= 12 4 +12

4 2 881

=0

= 12 2

8(1 )

7

=0

+12

2 8 (1+ )7

=0

= 0 + 1 ( )

For 0 , the function is maximum when 0 = 12

28

(1 )7

=0

0 = 12 2

8(0)

7

=0

0 = 12 17

=0

0 = 82 = 4 This is true if ( 1 ) = 0, thus k = 1.


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For 1 , the function is maximum when

1 = 12 2

8(1+ )

7

=0

1 = 12 2

8(0)

7

=0

1 = 12 17

=0

1 = 82 = 4This is true if ( 1 + ) = 0, thus k = -1. However, the value -1 is not in the range of

k, 0 k 7. Since the DFT is derived from the DFS that has a period of N = 8.Thus, the value of k = -1 also equal to k = 8-1 = 7.

Substitute k = 7 into 1 ,

1 = 12 2

8(1+7)

7

=0

1 = 12 27

=0

1 = 12 17

=0

1 = 82 = 4Thus, the results shows that the function is maximum at k=7.

For N = 16,

= 12 4 + 12 4 2

16

16

1

=0

= 12 2

8(1 2 )

15

=0

+12

2 8 (1+ 2 )15

=0

= 0 + 1 ( )


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28

(1 2 )15

=0

0 =12

2

8(0)

15

=0

0 = 12 115

=0

0 = 162 = 8This is true if ( 1 2 ) = 0, thus k = 2.


28

(1+2

)15

=0

1 = 12 2

8(0)

15

=0

1 = 12 115

=0

1 = 162 = 8This is true if ( 1 + 2 ) = 0, thus k = -2. However, the value of -2 is not in the range

of k, 0 k 15. Since the DFT is derived from the DFS that has a period of N =16. Thus, the value of k = -2 also equal to k = 16-2 = 14.

Substitute k = 14 into 1 , 1 = 12

28

(1+142

)15

=0

1 = 12 215

=0

1 = 12 115

=0

1 = 162 = 8Thus, the results shows that the function is maximum at to k=14.


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ii) Plot the DFT by using Matlab.

For N=8 and N=16



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3) A continuous time signal is defined as = cos 2 1000 + 12

sin 2 1250

i) If the signal is sampled at 8000Hz, what is the suitable sample size for DFT that

can represent both signal components?

Solution:


Frequency resolution, f res = f 1 f 2

= 1250 1000

= 250

Sample size, N = ,

,

=8000

250

= 32

ii) Plot the DFT of the signal by using Matlab



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8.0 Discussions:

1) Discrete Fourier Transform (DFT) decomposes the sequence of values into

components of different frequencies, forming the peaks at the frequency

spectrum displayed at the oscilloscope. DFT can be implemented by using

Digital Signal Processing Processor in hardware experiment or Matlab software

in computer calculation.

2) DFT can be computed efficiently in practice by using FFT (Fast Fourier

Transform) algorithm. Both of these methods are almost same but FFT has

better performance as compared to DFT in its simplicity. Calculation by using

FFT is far more time efficient than DFT therefore FFT is more suitable to be use

when doing experiment in laboratory.

3) In doing experiment which involves hardware, firstly we have to make sure that

the connections between hardware are correct. Connections also have to be

made thoroughly to avoid measurement error and signal leakage.

4) In this experiment, the TMS320C6713 DSK hardware operates according to the

coding in the software. When the coding run, it set the TMS320C6713 DSK to

operates at sampling frequency, f s=8000Hz and number of sample N=1024.


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5) When the program runs, waveform is obtained on the digital oscilloscope. From

the waveform, peak frequency was calculated as shown in the result. By

comparing the calculation result (peak frequency) and the fundamental

frequency, we can say that at 1000 Hz and 2000 Hz, no aliasing occur. But at

4000Hz, aliasing occur.

6) According to the Nyquist Theorem, f s 2f m where, f s is the sampling frequency

and f m is the maximum frequency of the signal. Aliasing or the overlapping of

the signals will occur if Nyquist rate is not fulfilled.

Using Nyquist Theorem, f s 2f m

8 kHz 2f m

f m4 kHz

Theoretically for sampling frequency f s = 8000 Hz, aliasing will not occur forthe modulation frequency f m = 4000 Hz. But in our experiment, aliasing start to

occur at modulation frequency of 4000 Hz.

7) Although in our experiment we had set our frequency to not exceed 4 kHz and

expecting aliasing should not be occurred, aliasing do occur at 4 kHz. This is

because of the fluctuation of the frequency cause the unstable performance of

the signal. Besides, the value of frequency gives out by the function generator

might not stable as the frequency increase. There might be some tolerance in the

measured value.

8) Aliasing causes different signals to become indistinguishable in sampling. It is

also known as distortion that happens as the signal reconstructed from samples

becomes different from the original continuous signal.

9) From the experiment, it is known that when the sampling frequency increases,

the sample size between two peaks will decrease. This is known by according to

Ts=1/f s.

10) Using Matlab software enables us to do DFT and obtain answer in easier way. It

gives accurate result since the computational done by the computer.

11) For the signal that has more than one frequency, we need to determine the

suitable number of sample N. This is to avoid spectrum leakage to occur in our

DFT graph.


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9.0 Conclusion

After completing the lab session, students were able to implement FFT using

TMS320C6713 DSK. The method of obtaining the output waveform can be demonstrate

and explained. Besides that, students are also able to analyze the frequency content by

implementing FFT on the targeted DSK through a digital oscilloscope. It can be conclude

that the sampling frequency should be at least twice the modulation frequency to avoid

aliasing occur as stated in the Nyquist sampling theorem. Lastly, Matlab program was

applied to obtain result for any DFT problem.

10.0 Reference

[1] L, Christ. Introduction , The Fast Fourier Transform , pp. 1-2, 2010

[2] W. Eric, Fast Fourier Transform , (MathWorld A Wolfram Web Resources),[online], http://mathworld.wolfram.com/FastFourierTransform.html (Accessed:

2 September 2013).

[3] A. K. Shoab, Sampling Rate Consideration , Digital Design of Signal

Processing Systems: A Practical Approach , United Kingdom: Wiley, 2011, pp.

7-45
http://mathworld.wolfram.com/FastFourierTransform.htmlhttp://mathworld.wolfram.com/FastFourierTransform.html

lab 5 long repot

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