This assignment illustrates how to perform Sampling and Reconstruction on MATLAB software. The results from MATLAB have also been explained

1. MULTIMEDIA UNIVERSITY OF KENYA
FACULTY OF ENGINEERING AND TECHNOLOGY
DEPARTMENT OF ELECTRICAL AND COMMUNICATION ENGINEERING
(ECE)
BSC. TELECOMMUNICATION AND INFORMATION ENGINEERING
NAME: MARTIN WACHIYE WAFULA
REG. NO: ENG-211-075/2012
UNIT CODE: ETI 2505
UNIT NAME: DIGITAL SIGNAL PROCESSING (DSP)
LECTURER: MR. JOSIAH MAKICHE
Date of Submission: 5th
October,2016
TITLE: MATLAB ASSIGNMENT 1 - SAMPLING AND
RECONSTRUCTION OF ANALOG SIGNALS

2. Objectives
1. To study the sampling principle and the effect of sampling on the frequency-domain
quantities
2. To study several approaches of reconstruction
Introduction
DSP provides alternative method for processing analog signals. In DSP applications, real world
analog signals are converted into discrete signals using sampling and quantization operations
called analog-to-digital conversion or ADC. These discrete signals are processed by the digital
signal processors, and the processed output signals are converted into analog signal using a
reconstruction operation called digital-to-analog conversion or DAC.
Fig1: Block diagram of a digital signal processing system
The relationship between an analog signal and its discrete time sampled version is necessary to
understand the operation of DSP system. This relationship in time domain is given by:
x(n) = xa (nTs)
where Ts is the sampling interval. This relation can also be shown in the frequency domain using
the Fourier analysis.
The continuous time Fourier transform is given by 



 dtetxFX Ftj
aa
2
)()( .
The inverse continuous time Fourier transform is given by 


 dFeFXtx Ftj
aa
2
)()( .
The discrete time Fourier transform is given by 




n
fnj
enxfX 2
)()( .
The inverse discrete time Fourier transform is given by 


 dFeFXtx Ftj
aa
2
)()( .
The relation between Xa(F) and X(f) is given by 



k
sas FkfXFfX ))(()( .

3. where Fs is a sampling frequency = 1/Ts. In other words, X(f) consists of infinite numbers of
copies of scaled Xa(F) separated by frequency interval f = 1. From the relation between discrete
time and analog frequencies
sF
F
f 
we get




k
sas
s
kFFXF
F
F
X )()(
in which )(
sF
F
X consists of infinite numbers of copies of scaled Xa(F) separated by interval F =
Fs.
Sampling Principle
To avoid aliasing, a band-limited signal xa(t) with bandwidth B can be reconstructed from
its sample values x(n) = xa(nTs) if the sampling frequency Fs = 1/Ts is greater than twice the
bandwidth B of xa(t).
BFs 2
Otherwise aliasing would result in x(n). The sampling rate of 2B for an analog band-limited
signal is called the Nyquist rate.
Reconstruction
From the sampling theorem and the above examples it is clear that if we sample band-
limited xa(t) above its Nyquist rate, then we can reconstruct xa(t) from its samples
x(n). Using an interpolation formula:
))((sinc)()( ss
n
a nTtFnxtx  


where
x
x
x

 )sin(
)(sinc  is an interpolating function derived from an ideal low pass
reconstruction filter. However, since an ideal low pass reconstruction filter cannot be
implemented, we usually estimated the ideal low pass filter by the following methods:
 Zero-order-hold (ZOH) interpolation: In this interpolation a given sample value is held
for the sample interval until the next sample is received.
( ) ( ), ( 1)a s sx t x n nT t n T   
which can be obtained by filtering the impulse train through an interpolating filter of the
form

4. s
0
1 0 t<T
( )
0 otherwise
h t

 

which is a rectangular pulse. The resulting signal is a piecewise-constant (staircase)
waveform which requires an appropriately designed analog post-filter for accurate
waveform reconstruction.
x(n)   xa(t)   xa(t)
 First-order-hold (FOH) interpolation: In this case the adjacent samples are joined by
straight lines. This can be obtained by filtering the impulse train through
s
1
1 0
( )
1 2
0 otherwise
s
s s
s
t
t T
T
h t t
T t T
T

  

    


Procedure
1. Sampling xa (t) at Fs = 50 samples/sec
ZOH Post-Filter

5. 1. %xa(t)= exp(-10|t|)
2. %CTFT is given by Xa(F) =
[(2*10)/(10^2 + (2*pi*F)^2)]
3. %sampling xa(t) at Fs = 50
samples/sec
4. clear all
5. tmin = -1;
6. tmax = 1;
7. %Analog signal
8. t = tmin: 0.001: tmax;
9. xa = exp(-10*abs(t));
10. %Sampling
rate(sample/second)
11. Fs = 50;
12. %Sample period
13. Ts = 1/Fs;
14. %Discrete time signal
15. n = tmin/Ts:tmax/Ts;
16. x = exp(-10*abs(n*Ts));
17. %Display signal in time
domain
18. figure(1)
19. subplot(211)
20. plot(t,xa)
21. title('Analog and
Discrete Time Signals')
22. xlabel('time(sec)')
23. ylabel('Analog Signal
x(t)')
24. subplot(212)
25. stem(n,x)
26. xlabel('n')
27. ylabel('Discrete time
signal x(n)')
28. %Computing FT
29. %Analog frequency (Hz)
30. F = -100:0.1:100;
31. W = (2*pi*F);
32. %DT Frequency
(circles/sample)
33. f = F/Fs;
34. w = 2*pi*f;
35. %Analog spectrum for
CTFT
36. XaF = 2.*(10./(10^2+
W.^2));
37. %DTFT
38. XF = x * exp(-1i*n'*w);
39. %Display spectra in
frequency domain
40. figure(2)
41. subplot(311)
42. plot(F,abs(XaF))
43. title('spectra of
signals')
44. xlabel('Freq(circle/sec
)')
45. ylabel('Original
Xa(F)')
46. subplot(312)
47. plot(F,abs(XF))
48. xlabel('Freq(circle/sec
)')
49. ylabel('X(F/Fs)')
50. subplot(313)
51. plot(f,abs(XF))
52. xlabel('Freq(circle/sec
)')
53. ylabel('X(f)')
54. %Display spectra in the
fundamental range
55. figure(3)
56. subplot(211)
57. plot(F,abs(XaF))
58. plot(F,abs(XF))
59. title('spectra in the
fundamental range')
60. xlabel('Freq(circle/sec
)')
61. ylabel('X(F/Fs)')
62. v = axis;
63. v(1:2) = [-Fs/2 Fs/2];
64. axis(v)
65. subplot(212)
66. plot(f,abs(XF))
67. xlabel('Freq(circle/sec
)')
68. ylabel('X(f)')
69. v = axis;
70. v(1:2) = [-1/2 1/2];
71. axis(v)
Experimental results

6. Explanation:
We have superimposed the discrete signal x(n) over x(t) to emphasize the sampling. The plot
shows that it is a scaled version (scaled by Fs = 50) of X(F). Clearly there is no aliasing.
Q2. Reconstruction of xa(t)
72. %Reconstruction of
xa(t)
73. t = tmin:0.001:tmax;
74. figure(4)
75. clf
76. subplot(211)
77. hold on
78. stem(n*Ts,x,'r')
79. for i= 1:size(x,2)
80. xsinc(i,:) =
x(i)*sinc(Fs*(t-(i+min(n)-
1)*Ts));
81. plot(t,xsinc(i,:))
82. end
83. title('Signal
reconstruction')
84. xlabel('time(second)')
85. ylabel('x(n)*Sinc(Fs*(t
-nTs))')
86. hold off
87. xar = sum(xsinc);
88. subplot(212)
89. plot(t,xar,'b-
',t,xa,'r:')
90.
91. legend('Reconstructed
signal','Origin signal')
92. ylabel('Reconstructed
signal xa(t)')
93.
94. xlabel('time (second)')
95. %reconstruction error
96. maxerror = max(abs(xa-
xar));
Experimental Results

7. Explanation:
The maximum error between the reconstructed and the actual analog signal is 0.0380. This is
because xa(t) is not strictly band-limited and we also have a finite number of samples. From the
reconstructed figure above, we note that visually the reconstruction is excellent.
Q3. For Fs = 10 samples/cycle, the results are as shown below:

8. Explanation:
Here Fs = 10 < 20. There is considerable amount of aliasing. From the figure above, this is
evident since shape of x(n) is different from that of Xa(F) and can be seen as adding overlapping
replicas of Xa(F).
Explanation:
The maximum error between the reconstructed and the actual analog signals is 0.1852, which is
significant and cannot be attributed to the nonband-limitedness of xa(t) alone. From Figure
above, the reconstructed signal differs from the actual one in many places over the interpolated
regions. This is the visual demonstration of aliasing in the time domain. (Proakis, 2012)
Reason why Fs = 50 samples/sec is better than Fs = 10 samples/sec:
- Since the bandwidth of xa(t) = 20Hz, the Nyquist rate Fo = 40 Hz according to the
Sampling theorem. To prevent aliasing, the analog signal should be sampled at a rate
higher than twice the highest frequency in the analog signal.
- Therefore, for Fs = 50 Hz > Fo there is no aliasing hence better compared Fs = 10
samples/sec < Fo . (Manolakis, 1996)
Q4. xa(t) = sin(20πt), 0≤ t ≤ 1. Samples at Ts = 0.01, 0.03, 0.05, 0.07 and 0.1
a) For each Ts, x(n) was plotted.
MATLAB Code
1. tmin = 0;
2. tmax = 1;
3. %Analog signal
4. t = tmin: 0.001: tmax;
5. xa = sin(20*pi*t);
6. %Sample period
7. Ts1 = .01;
8. %Discrete time signal
9. n1 = tmin/Ts1:tmax/Ts1;
10. x1 = sin(20*pi*n1*Ts1);
11. %Display signal in time
domain
12. figure(5)
13. subplot(211)
14. plot(t,xa)

9. 15. title('Analog and
Discrete Time Signals')
16. xlabel('time(sec)')
17. ylabel('Analog Signal
x(t)')
18. subplot(212)
19. stem(n1,x)
20. xlabel('n1')
21. ylabel('Discrete time
signal x1(n)')
22. Ts2 = .03;
23. %Discrete time signal
24. n2 = tmin/Ts2:tmax/Ts2;
25. x2 = sin(20*pi*n2*Ts2);
26. %Display signal in time
domain
27. figure(6)
28. subplot(211)
29. plot(t,xa)
30. title('Analog and
Discrete Time Signals')
31. xlabel('time(sec)')
32. ylabel('Analog Signal
x(t)')
33. subplot(212)
34. stem(n2,x2)
35. xlabel('n2')
36. ylabel('Discrete time
signal x2(n)')
37. Ts3 = .05;
38. %Discrete time signal
39. n3 = tmin/Ts3:tmax/Ts3;
40. x3 = sin(20*pi*n3*Ts3);
41. %Display signal in time
domain
42. figure(7)
43. subplot(211)
44. plot(t,xa)
45. title('Analog and
Discrete Time Signals')
46. xlabel('time(sec)')
47. ylabel('Analog Signal
x(t)')
48. subplot(212)
49. stem(n3,x3)
50. xlabel('n3')
51. ylabel('Discrete time
signal x3(n)')
52. Ts4 = .07;
53. %Discrete time signal
54. n4 = tmin/Ts4:tmax/Ts4;
55. x4 = sin(20*pi*n4*Ts4);
56. %Display signal in time
domain
57. figure(8)
58. subplot(211)
59. plot(t,xa)
60. title('Analog and
Discrete Time Signals')
61. xlabel('time(sec)')
62. ylabel('Analog Signal
x(t)')
63. subplot(212)
64. stem(n4,x4)
65. xlabel('n4')
66. ylabel('Discrete time
signal x4(n)')
67. Ts5 = .1;
68. %Discrete time signal
69. n5 = tmin/Ts5:tmax/Ts5;
70. x5 = sin(20*pi*n5*Ts5);
71. %Display signal in time
domain
72. figure(9)
73. subplot(211)
74. plot(t,xa)
75. title('Analog and
Discrete Time Signals')
76. xlabel('time(sec)')
77. ylabel('Analog Signal
x(t)')
78. subplot(212)
79. stem(n5,x5)
80. xlabel('n5')
81. ylabel('Discrete time
signal x5(n)')

10. For Ts = 0.01, figure 5, Ts = 0.03 figure 6, Ts = 0.05 figure 7, Ts = 0.07 figure 8 and Ts = 0.1
figure 9 were plotted.
(b) Reconstruction of ya(t) code (Proakis, 2012)

11. (c) Results
The maximum error between the reconstructed and the actual analog signal is 0.0013. This is
because xa(t) is not strictly band-limited and there is a finite number of samples. From the
reconstructed figure above, we note that visually the reconstruction is excellent.
QUESTIONS
1. What is the MATLAB function that would be used to plot a staircase (ZOH) interpolation
of the analog signal?
stairs(n * Ts * frequency, x);
hold on
stem(n* Ts * frequency, x);
hold off
(Proakis, 2012)
2. What is the MATLAB function that would be used to plot a staircase (ZOH) interpolation
of the analog signal?
plot(n * Ts * frequency, x);
hold on
stem(n* Ts * frequency, x);
hold off
(Proakis, 2012)
3. From the experiment, describe why minimum sampling rate must be at least twice the
bandwidth of an analog signal?

12. To avoid aliasing
Explanation of the function of each part of DSP system above:
ADC - The output of the A /D converter is a digital signal that is appropriate as an input to the
digital processor. ADC coverts continuous time signal into discrete time signals through
sampling and quantization
The digital signal processor- may be a large programmable digital computer or a small
microprocessor programmed to perform the desired operations on the input signal. It may also be
a hard wired digital processor configured to perform a specified set of operations on the input
signal.
DAC – converts the digital output of the processor into analog signal. This is called
reconstruction.(Manolakis, 1996)
References
Manolakis, J. G. (1996). Digital Signal Processing - Principles, Algorithms and Applications (Third ed.).
New Jersey, USA: Prentice Hall International-Inc.
Proakis, V. K. (2012). DIGITAL SIGNAL PROCESSING USING MATLAB (3rd ed.). Stamford, USA: Cengage
Learning.