sampling continuos signals

8/11/2019 Sampling Continuos Signals

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Laboratrio SDR

Analog to Digital Conversion Lab

UNICAMP IE309

1.Sampling and Aliasing

In signal processing,samplingis the reduction of a continuous signalto

a discrete signal. A common example is the conversion of a sound wave(a

continuous signal) to a sequence of samples (a discrete-time signal).

A sample refers to a value or set of values at a point in time and/or space.

A sampleris a subsystem or operation that extracts samples from a continuous

signal. A theoretical ideal samplerproduces samples equivalent to the

instantaneous value of the continuous signal at the desired points. In the Figure

1.1. we can see the presentation of a analog signal that has been sampled in

discrete time.

Figure 1.1.Signal sampling representation. The continuous signal is represented

with a green colored line while the discrete samples are indicated by the blue

vertical lines.

In this introduction we are going to see examples in sampling through Matlab

and Simulink, and finally we review a project to explore the aliasing in audio

tones signals.


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Experiment 1.1. Introduction in sampling in Matlab.

In this example we explore create the simulation of an analog signal (cosine

signal), and then create the sampled version, first a little over the Nyquist

Frequency and finally using oversampling to see the differences.

The example consists in

Create a cosine signal in a frequency of operation F0 = 60 Hz;

Sampling the original signal in 150 Hz, and the in 1 kHz.

Plot the analog signal, and the sampled versions.

Solution: The code that makes the experiment is show below, and the result is

showed in the Figure 1.2.

t = 0:1/2000:.02;x = cos(2*pi*60*t); % approx. to continuous-timet1 = 0:1/150:.02;n1 = 0:length(t1)-1;x150 = cos(2*pi*60/150*n1); % fs = 150 Hzt2 = 0:1/1000:.02;n2 = 0:length(t2)-1;x1000 = cos(2*pi*60/1000*n2); % fs = 1000 Hz

subplot(3,1,1)

plot(t,x)title('Analog signal')xlabel('time')ylabel('Amplitude')

subplot(3,1,2)stem(n1,x150,'filled')title('Sampled at 150 Hz')xlabel('n')ylabel('Amplitude')

subplot(3,1,3)

stem(n2,x1000,'filled')title('Samples at 1kHz')xlabel('n')ylabel('Amplitude')


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Figure 1.2. Simulation of the sampled version at rates (a) little above the

Nyquist rate, and (b) using the oversampling technique.

Exercise 1.1.- Sample a sinusoid (Fo = 2 kHz), sample the analog signal using

oversampling (Fs1 = 10 kHz), and undersampling (Fs2 = 3 kHz), and plot the

result.

Figure 1.3. Simulation of the sampled version at rates for a sinusoid Fo = 2 kHz

(a) oversampling (10 kHz) (b) undersampling technique (3 kHz).


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Task: In the above figure, the red signal (analog signal)is periodic with a Period

of T = 0.5 x 10, which means = 2 . Using the scope tool check theperiods and frequencies in the discrete time.

Experiment 1.2. In this experiment, we are going to explore the use of the

Simulink tool, which we are going to use along this practice together with the

Matlab. In this introduction, we are use the Sample and Hold. In this experiment

we are going summed up three sinusoid signals:

Fo1 = 500 Hz, A = 1 V.

Fo1 = 1500 Hz, A = 0.5 V.

Fo1 = 2500 Hz, A = 0.2 V.

The impulse generator feeds the Sampling and Hold Component. In another

words, this acts like an impulse train hitting the analog simulated signal. There

are three graphics; the first one is the Sinusoids Analog Signals. Which show us

the continuous three sinusoid that has been summed up. The plot of the Impulse

Samples, shows the impulses hitting in the analog signal. Finally, the Sample

and Hold Plot, figure out the sample and hold representation of the analog

signal.

In the figure 1.4. We can appreciate the experiment, and in the figure 1.5 we can

see the results from this example.

Figure 1.4. Simulation in Simulink: 3 sinusoids summed up, pass through the

sampled and hold


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Figure 1.5: Results from the Simulink model (a) Show the original simulation

(analog signal) (b) the signal been sampled at a frequency Fs = 8 KHz

Exercise 1.2.- Now to sample the signal, use an Zero-Order Hold from the

Discrete Library of the Simulink, setting the frequency sampling to 1/8000 (Fs =

8 kHz), and compare the result with the last example. There is any difference?

Experiment 1.3. In this example we are going to see the Spectral analysis in the

signal operating at = 2 , in this case we are going to do the samplingin = 5 , and review the spectrum of the signal. The codes that make thisexperiment generate the Figure 1.6.

% Espectral analisys

clc

clear allclose allFs = 5000; % sampling frequency% change sampling frequency to observe the effects of under

sampling% (Aliasing) and over samplingf = 2000; % signal frequency f Hz% change signal frequency and analyze the outputTmax = 10;t = 0:1/Fs:Tmax;s = 10*sin(2*pi*f*t) + rand(1,Tmax*Fs+1);Nfft = 2^8;[Pxx,freq] = pwelch(s,Nfft,[],[],Fs);%computing power spectral

densityplot(freq,Pxx), grid


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xlabel('frequency in hz');ylabel('power P')title('Signal PSD')

% frequency domain view% Compute the frequency response

w1=-Fs/2:Fs/1024:(Fs/2)-1/1024;S = fftshift(fft(s,1024));figureplot(w1,abs(S));gridxlabel('Frequency in hz');ylabel('Signal Amplitude A')title('Frequency Response')

Figure 1.6: FFT of the sampled signal, the = 5 , we are over samplingthe signal.

Exercise 1.3.- In this exercise change the frequency sampling to (a) =4 , and (b) = 3 , and see the results to this changes, answer if theresults correspond with the theory.

Exercise 1.4.- Now, using the example in the Experiment 1.2. put a BFFT to

explore the spectrum signal. The experiment should look like this one, in the

figure 1.7. Hint: use a buffer size, and FFT length of 262144.


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Figure 1.7: FFT analysis based on the experiment 1.2

Task: With the zoom scope analyze where the spectrum hits, and answer if this

correspond with the theory, change the Fs to do undersampling, and see what

happens with the spectrum.

Experiment 1.5. We are going to review the example with Aliasing, in this case

we plot the signal, in the right frequency sampling. The signal is cosine signal

= 60 , and then use a phase of /3. The signal is sampled at 400 Hz.

% Aliasing concept% Frequency in Fs = 400 Hzta = 0:1/4000:2/60; % analog time axis

xa1 = cos(2*pi*60*ta+pi/3);

tn = 0:1/400:2/60; % discrete-time axis as n*Tsxn1 = cos(2*pi*60*tn+pi/3);

plot(ta,xa1)hold onstem(tn,xn1,'filled', 'Color','red')axis([0 0.0333 -1.1 1.1])title('x(t) = cos(60t) + \pi/3 sampled in Fs = 400 Hz')

xlabel('n')ylabel('Amplitude')


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The result of this code is:

Figure 1.7: Cosine signal at an operation frequency in = 60 beensampled in = 400

Exercise 1.5- Based on the experiment 1.5. Now, you can create two another

signals this are the a cosine signal with a = 340 and a phase of =/3, and another cosine signal with a = 460 and a phase of =+/3. And then, plot the same discrete signal sampled at = 400 . Inanother words, the red signal is going to be the same xn1 of the above code

.Finally make the plot together with Hold Onto see the effect of the simulated

aliasing.

Task: Why the red discrete signal always beat with the analog simulated signal?

Experiment 1.6. In this experiment we are going to analyze the effect of the

aliasing with an audio signal. The experiment creates a signal that begins with a

frequency operation at = 200 , and this is going to be sampled at = 2 , as is obvious we are using oversampling. But in this case we aregoing to shift the frequency operation from 200 to reach2200 . With thisoperation if we maintain the frequency sampling, there is no more oversampling

to pass work under the Nyquist rate, and create the aliasing effect. Finally, we

create an audio signal to hear the effect of the aliasing. The code that makes this

project is presented below.


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clear all;close all;

%% Initialization

Fs = 2000; % Frequeny Smaplington_ds = 0.1; % Tone duration in seconds

N_samp = ton_ds*Fs; % Number of samplesn = [0:N_samp-1] / Fs; % n discre time vectorFs_cont = 10; % Fs Factor for continuous plot

(oversampling)t_const = [0:N_samp*Fs_cont-1] / (Fs*Fs_cont); % continuous time

vector

%% Figure configurationsset(gcf,'Position',[0 0 1023 768]);rect=get(gcf,'Position');

%% Input parameters to create the signals% We are going to create a sinusoidal signal with a begin frequency

% in Fo = 200 Hz, and this is going to shift until reach Fo = 2200 Hz% the amplitud is about A = 0.5; respectively.

% The idea is pass from low frequencies to the high frequency to see

% the spectrum and the effect of the aliasing. The sound of the signal% sampled properly and at low of the Nyquist point, to hear the effect

% of the aliasing.

f1 = 200; a1 = 0.5; % start parametersnum_plot_samples = 30; % Samples to plot the discrete signal

audio_out = []; % Empty array to fill the new created signal

frame_number = 0;forf_var = 0:10:2000

y = a1 * sin(2*pi*(f1+f_var)*n); % discrete signal

y_const = a1 * sin(2*pi*(f1+f_var)*t_const); % time constant sig

audio_out = [audio_out y];

stem(n(1:num_plot_samples),y(1:num_plot_samples)); hold on;

plot(t_const(1:num_plot_samples*Fs_cont),y_const(1:num_plot_samples*Fs

_cont),'red');set(gca,'ylim',[-1 1]);

title(strcat('y(n) = a1*sin(2*pi*n*',num2str(f1+f_var),') sampledat', num2str(Fs),' Hz'));

xlabel('n [s]');ylabel('y(n)');hold off;

frame_number = frame_number + 1;shift_frame(frame_number) = getframe(gcf,rect); %mantain the

signal moving to represnet

end;

wavwrite(audio_out,Fs,'sinus_aliasing.wav');


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The result is iterative, and if you run you can visually this effect. The final time

representation is showed in the figure 1.8.

Figure 1.8: Final representation of the signal been aliased because is under the

Nyquist sampling rate.

Finally, you can hear the effect of the aliasing phenomena, because is created the

audio signal of the tone been increased and decreased in the file, as the name

sinus_aliasing

Exercise 1.6- With the basis of the experiment 1.6. Now you have to build a

composed signal with frequencies= 200 , = 400 , and =400 . Setting this signals with amplitudes of = 0.5, 0.3 ! 0.2,respectively. Now create three plots, the first one is the time signal showing in

red the analog shifting signal, and in blue discrete samples showing the effect of

the discrete signal with aliasing. In the second plot, you could show the

spectrum sampled at frequency of = 20 , obviously is mean that is goingto be oversampling, and the spectrum is plotted in a right way. In the last plot,

show the spectrum of the aliasing. The signal could be similar to the shown inthe figure 1.9.


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Figure 1.9: Composed signal and their spectrum sampled at = 20 (middle), and showing the spectrum with the aliasing phenomena = 2 .

(last graphic)

Task: Create the audio signals sampled at = 2 and 20 , to hear thedifference, and realize the audio effect of the aliasing. Could you explain, what

is going on?

2.Band-Pass Sampling

If the signal is a bandpass signal with bandwidth B (Bandwidth is simply thedifference between the highest and lowest frequency present in the signal). In

order for a faithful reproduction and reconstruction of a bandpass analog signal

with bandwidthB. The signal should be sampled in a frequency sampling "#that is greater than or equal to twice the maximum bandwidth of the signal, it

means $ 2%.

For example, in the bandpass signal extends from &= 200 and '=300 , the bandwidth will be % = 100 . In this case the Nyquist rate isminimum around $ 600 , but we can use the bandpass, to sample in thedouble of the band, and now the $ 200 . As you can see is less even the


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highest frequency, and for that reason the bandpass sampling is called

undersampling.

In the figure 2.1, we can see the folding frequencies and aliasing zones, the zone

1 is a mirror image of the zone 2 (with frequency reversal). Similarly zone3 is a

mirror image of the zone 4, etc

Figure 2.1. Folding frequency and aliasing zones

Experiment 2.1: Consider an AM signal centered at carrier frequency 1MHz,

with two components offset by 10KHz 0.99 MHz and 1.01 MHz. So the AM

signal contains three frequency components at 0.99 MHz, 1 MHz and 1.01 MHz.

Our desire is to sample the AM signal. As with the usual sampling theorem

(baseband), we know that if we sample the signal at twice the maximumfrequency i.e Fs>=2*1.01MHz=2.02 MHz there should be no problem in

representing the analog signal in digital domain.

By the bandpass sampling theorem, we do not need to use a sampler running at

Fs>=2.02 MHz. Faster sampler implies more cost. By applying the bandpass

sampling theorem, we can use a slower sampler and reduce the cost of the

system. The bandwidth of the signal is 1.01MHz-0.99 MHz = 20 KHz. So, just

sampling at Fs>=2*20KHz=40KHz will convert the signal to digital domain

properly and we can also avoid using an expensive high rate sampler (if Fs >=2.02 MHz used according to baseband sampling theorem).

Lets set the sampling frequency to be Fs=120KHz (which is 3 times higher than

the minimum required sampling rate of 40KHz or oversampling rate =3). Now

we can easily find the position of the spectral components in the sampled output

by using the aliasing zone figure as given above. Since Fs=120 KHz, Fs/2 will

be 60KHz. So the zone 1 will be from 0 to 60 KHz, zone 2 -> 60-120KHz and

so on. The three spectral components at 0.99MHz, 1MHz and 1.01 MHz will fall

at zone 17 ( how? 0.99 MHz/60 KHz = 16.5 , 1MHz/60KHz = 16.67 and

1.01MHz/60KHz = 16.83 , all figures approximating to 17). By the aliasing zone


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figure, zone 16 contains a copy of zone 17, zone 15 contains a copy of zone 16,

zone 14 contains a copy of zone 15 and so on Finally zone 1 contains the copy

of zone 2 (Frequency reversal also exist at even zones). In effect, zone 1

contains a copy of zone 17. Since the original spectral components are at zone

17, which is an odd zone, zone 1 contains the copy of spectral components at

zone 17 without frequency reversal.

If we review the principle of the bandpass sampling through the figure 2.2. We

can see where the undersampling replica can beat.

Figure 2.2. Bandpass sampling: (a) original signal continuous spectrum (b)

sampled signal spectrum replications

Since there is no frequency reversal, in zone 1 and based on the figure 2.2, we

can compute where the discrete three components will be. For example for the

central frequency, we have to do successive rests from the ( = 1 ) 120 = **0 , and continuous do the rest to the zone 1, where the centralfrequency is located at 40 , by simply inspection the other components areat 30 and 50 .

Now, we are going to do the implementation based on the Simulink tool. First to

ensure the = 120 , dont cause aliasing, is needed to see the ratesallowed. Recall the formula from Lyons:

2 % $ - $2+ % + 1


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The index represents the zone where the replica spectrum will be. Theprogram to see the allowed band is quite simple:

fc = 1e6;

B = 20e3;m = input('Ingresa el ndice: ');a = (2*fc-B)/mb = (2*fc+B)/(m+1)

The design is shown in the figure 2.3. It consists in the sum of the three

components. After that, is multiplied with a oscillator represented by and

sinusoidal signal. There are three systems, the first one applies the Nyquiste rate

to sample the signal, the second one applies a bandpass filter to eliminate the

double sideband signal and then is oversampled to see of the spectrum is

presented. The third system is the system undersampled at a frequency =120 .

Figure 2.3.Simulink representation of the BandPass Sampling for the above

example.

The spectral result to check the frequencies response are shown in the figure 2.4.


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Figure 2.4.Spectrum response in the undersampling method to discretize an

analog bandpass signal.

Exercise 2.1. Based on the example in the exercise 2.1 compute the neededto undersampling the signal in & = 0. ), = 1 ), ' = 1.0 ),and make that the discrete spectrum beat at & = 0 , = 10 ,' = 20 .

3.Iterpolation and Reconstruction Filter

From the sampling theorem and the preceding examples, it is clear that if we

sample band-limited "# (analog signal) above its Nyquist rate, then we canreconstruct "#) from its samples . This reconstruction can be thought ofas a 2-step process:


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First the samples are converted into a weighted impulse train:

" 7-#8

9: 8= ; + 146" 7-# 3046" 7-# ;

Then the impulse is filtered through an ideal analog lowpass filter band-

limited to the 3


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Figure 3.2. Three sampled version (a) oversampling, (b) at the Nyquist rate and

(c) subsampling signal

Now, our objective is based on the sampled signals above create the signal in

continuous time. For that we will take the first example (the oversampling

signal), and interpolate it to create the analog signal. The code to create the

interpolate version is given below, and the figure 3.3 shoes the result from theinterpolation:

%plot x(n) for T_s = 0.01 sec% x_a(t) = cos(20*pi*t);clc; close all;

T_s1 = 0.01; n1 = [0:100];

x1 = cos(20*pi*n1*T_s1);Fs1 = 1/T_s1;t = 0:0.001:1;

nTs1 = n1*T_s1;xa = x1*sinc(Fs1*(ones(length(n1),1)*t-nTs1'*ones(1,length(t))));subplot(2,1,1); func = stem(n1,x1); axis([-5 105 -1.2 1.2]);set(func,'markersize',2); xlabel('n','FontSize',12);title(['x(n) = cos(20{\pi}nT_s) for T_s = 0.01

sec'],'FontSize',12);ylabel('x(n)','FontSize',14);

subplot(2,1,2)plot(t,xa)xlabel('t in sec','FontSize',12);title(['Sinc interpolation for T_s = 0.01 sec'],'FontSize',12);

ylabel('y_a(n)','FontSize',14);


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Figure 3.2. Interpolated versing of the oversampled signal 100 , thereconstructed signal is about = 10

Exercise 3.1: Based onthe above experiment, create the interpolated version for

the sampling frequencies at E= 20 and at = 10 .

Experiment 3.2: Now, in this experiment we are going to create a composed

analog signal and then sampling the composed signal from oversampling to

reach undersample the same signal. Then we are going to interpolate, to

reproduce the original analog signal. In a first moment the signal is

reconstructed well because we are working over the Nyquist rate, and finally

when the operation frequency of the analog signal exceeds the sampling

frequency, obviously the reconstructed signal cant reconstruct the original

sampled version.

First we need the interpolate function, based on the sinc function. In the next

code is showed the code that allow us, work the samples to get a interpolated

version.


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functionf = sinc_interpolate(s,fs_in,fs_out) % s = sampled

signal, fs_in = sampling frequency of input, fs_out = sampling

frequency of output

N = length(s);length_in_seconds = N / fs_in;

output_length_in_samples = ceil(length_in_seconds * fs_out);

Mat = [];forn = 1:N

t = [0:output_length_in_samples-

1]/output_length_in_samples*length_in_seconds; t = t * fs_in - (n-1);Mat = [Mat; s(n)*sinc(t)];

end;

f = sum(Mat);

We are going to create three sinusoidal signals, the initial frequency are 200 , = 400 and = 600 , the amplitudes are: = 0.5,= 0.3 and = 0.2 respectively. The code, that makes the project ispresented below, and the result is depicted in the figure 3.3.

clear all;

close all;

%% Initialization

Fs = 2000; % Frequeny Smapling

ton_ds = 0.1; % Tone duration in seconds

N_samp = ton_ds*Fs; % Number of samples

n = [0:N_samp-1] / Fs; % n discre time vector

Fs_cont = 10; % Fs Factor for continuous plot (oversampling)

t_const = [0:N_samp*Fs_cont-1] / (Fs*Fs_cont); % continuous time vector

%% Figure configurations

set(gcf,'Position',[0 0 1023 768]);

rect=get(gcf,'Position');

%% Input parameters to create the signals

% We are going to sum three sinusoisal signals, and we in frequencies% from 2200 Hz, 2400 Hz, and 2800 Hz. The amplitudes are 0.5, 0.3, and

0.2

% respectively. The idea is pass from low frequencies to the high

% frequency to see the spectrum and the effect od the aliasing.

% The sound of the signal sampled properly and at low of th Nyquist

% point, to hear the effect of the aliasing.

f1 = 200; a1 = 0.5;

f2 = 400; a2 = 0.3;

f3 = 600; a3 = 0.2;

num_plot_samples = 30; % Samples to plot

num_plot_samples1 = 50; % Samples to plot

audio_out = []; % Empty array to fill the new created signal

frame_number = 0;


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forf_var = 0:10:2000

y1 = a1 * sin(2*pi*(f1+f_var)*n); % discrete signal 1



y = y1 + y2 + y3; % discrete signal.

y_const1 = a1 * sin(2*pi*(f1+f_var)*t_const); % time constant sig 1



y_const = y_const1 + y_const2 + y_const3; % time signal

%audio_out = [audio_out y];

subplot(2,1,1);

stem(n(1:num_plot_samples),y(1:num_plot_samples));

hold on;

plot(t_const(1:num_plot_samples*Fs_cont),y_const(1:num_plot_samples*Fs_c

ont),'red');

set(gca,'ylim',[-1 1]);

hold off;

title(strcat('y(n) = a1*sin(2*pi*n*',num2str(f1+f_var),') +

a2*sin(2*pi*n*',num2str(f2+f_var),') +

a3*sin(2*pi*n*',num2str(f3+f_var),'), sampled at ',num2str(Fs),' Hz'));

xlabel('n [s]');

ylabel('y(n)');

interpolated_samples =

sinc_interpolate(y(1:num_plot_samples),Fs,Fs*Fs_cont);

subplot(2,1,2);

stem(n(1:num_plot_samples),y(1:num_plot_samples));

hold on;

plot(t_const(1:length(interpolated_samples)),interpolated_samples,'red')

;

set(gca,'ylim',[-1 1]);

hold off;

title('Reconstruction of y from its samples using Shannon''s

theorem');

xlabel('n [s]');

ylabel('y2(n)');

audio_out = [audio_out interpolated_samples];

frame_number = frame_number + 1;shift_frame(frame_number) = getframe(gcf,rect);

end;

wavwrite(audio_out,Fs,'audio_out1.wav');

%movie2avi(my_movie,'sinus_aliasing.avi','compression','indeo5','fps',10

);


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Figure 3.3. (a) Original signal, and the samples with aliasing effect, (b)

interpolated reconstruction of the signal.

Exercise 3.2: Based on the experiment 3.2, try to reconstruct the signal raising

the frequency sampling * , and interpolate the signal to get thereconstructed signal. In this case if we are using oversampling, it means that the

reconstructed interpolate signal have to be very similar to the original version.

Finally, create the audio signal for the original composed three tones and the

interpolated version and compare the results.

Experiment 3.3: This project consists in input a recorded audio signal, and

explore the downsample by a factor of 4, and after that interpolate the signal. In

this case we are going to interpolate by repeat the samples. The code that

implements this project is shown below.


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% The project consists in input a recorded audio signal, and

explore

% the downsample by a factor of 4.

% You have to record various audio signal

% with different frequency sampling

% Fs = (8000, 11025, 22050, 44100)

% In this first example you can run with a sample recorded% in the "audios" file

%% Audio read from previous recorded audio.

[x,fs]= wavread('audios\sonido');

% first shorten the sequence so that the length is multiple of 4

orig_sig=length(x);

N=floor(orig_sig/4)*4; % Signal multiple of 4

x = x(1:N); % Number of samples

disp([fs,N]); % display the frequency Sampling and the number of

samples.

%

%% Play the input signal

sound(x,fs); %orginal input signal

figure(1);

subplot(1,2,1)

bar(x(2000:2060),0.02);

axis tight;title('Waveform of Original Sound');

subplot(1,2,2)

psd(x,256,fs);

axis([0,fs/2,-80,0]);

xlabel('Hz');

ylabel('dB');

title('Spectrum of Original');

pause(2)

%% Downsample without prefiltering: take every 4th sample

y=x(1:4:N); %Take a sample ever 4 times

sound(y,fs/4); % hear the downsampled signal

figure(2);

subplot(1,2,1)

bar(y(2000/4:2060/4),0.02); %downsampled signal in time

axis tight;

title('Waveform Downsampled by 4');

subplot(1,2,2)

psd(y,256,fs/4); % spectral characteristic of the dow4

axis([0,fs/2,-65,0]);

xlabel('Hz');

ylabel('dB');

title('Spectrum of Down4');

pause(2);

%% Interpolation by repeating each sample 4 times

z=zeros(N,1); % Initialization vector

% interpolate by first copy the samples in x to every 4th

sample

% to z starting from 1st sample then starting at 2nd sample, and

so on

z(1:4:N) = y;z(2:4:N) = y;


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z(3:4:N) = y;

z(4:4:N) = y;

z1 = z';

sound(z1,fs);

figure(3);

subplot(1,2,1)bar(z1(2000:2060),0.02);

axis tight;

title('Waveform Upsampling by 4');

subplot(1,2,2)

psd(z1,256,fs);

axis([0,fs/2,-60,0]);

xlabel('Hz');

ylabel('dB');

title('Spectrum of Up4');

wavwrite(z1,fs,'audios_out\sonido_out');

%

%% Calculate the MSE

D = x - z1;

MSE1 = mean(D.^2);

MSE = 20*log10(MSE)

fprintf('\n Error between original and interpolated =

%g\n\n',MSE )

The result of this project is show in the figure 3.4. Where it shows the original

signal and their spectrum, the figure 3.5 shows the downsampled version by a

factor of 4. The figure 3.6 shows the interpolated version by repeat the samples

each four samples to reconstruct the original version.

Figure 3.4. (a)Part of the original signal in time (b) spectrum of the originalsignal


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Figure 3.5. (a)Part of the downsampled signal in time (b) spectrum of the

downsampled signal by 4

Figure 3.6. (a)Part of the upsampled signal in time (b) spectrum of the

upsampled signal in time


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Exercise 3.3: Based on the experiment 3.3, now we are going to use a

reconstruction filter. First create the file based on the experiment 3.3. Used the

same input, and then read it, and save in the variable , after that play theoriginal audio, then short the sequence so that the length is multiple of 4, like in

the experiment.

The next step is decimate the signal by a decimation filter. Use the following

command

y=decimate(x,4,30,'FIR');

With the decimated signal, create the audio signal through the sound command

to hear the effect of the decimation filter.

The next step is plot the decimation signal with the spectrum. The last step is

interpolate the signal by the use of the FIR filter, we use the interp command

from the signal processingtoolbox. The code is show as follows:

[z,h_intp]=interp(y,4); % Interpolate by a factor of 4

With this new interpolated version, create the plot of the signal and the

frequency response, and create the audio signal to hear the difference. Use the

minimum square error in dB, to research the error between the original version

and the reconstructed version. This version is better that the experiment 3.3?

Task: Create new audio signals in different frequency sampling, and N bits,

then modulate the signals to arise to a high frequency multiplying by an oscillate

signal and make the example of decimate and interpolate. The example to record

a new audio is

% r = audiorecorder(22050, 16, 1);

% record(r); % speak into microphone...% pause(r);% p = play(r); % listen% resume(r); % speak again% stop(r);% p = play(r); % listen to complete recording% mySpeech = getaudiodata(r, 'int16'); % get data as int16

array

Fs = 22050;y = wavrecord(5*Fs, Fs, 'int16');wavplay(y, Fs);wavwrite(y,Fs,16,'audios\sonido3.wav')


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4.PCM (Pulse Code Modulation)

PCM was first developed by Alec Reeves, a British Engineer in 1937. It is an

analog signal that's sampled at uniform intervals and converted to a digital

binary number. The technology was originally developed for telephone

communication system. However, today is also being used with other digital

audio formats such as audio CD's, DVD's, and Blu-Ray discs.

Figure 4.1. Pulse Code Modulation Signal representation.

Above is a basic graphic example of what occurs during the pulse code

modulation process. The blue line is an example of an audio signal, or sine

wave, and the jagged red line would be digital representation of that line. In this

example, the sampling points would be 0, 3, 6, 8, 10, 11, 12, 12, 12, 11, 10, 8, 6,

4, and 0 or in binary: 00000000 00000011 00000110 00001000 00001010

00001011 00001100 00001100 00001100 00001011 00001010 00001000

00000110 00000100 00000000.

Experiment 4.1: In the following we made the PCM representation, from a

sinusoidal signal, we made a uniform quantization and after that we did the

binary representation using just 2 and 3 bits. Finally the representation is showed

in the figure 4.2. The binary data can be saved in a excel sheet or it are showed

in the command pronto pf the Matlab


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% Conversin analogica digital en Matlab

A = input('Input the amplitude: '); % Amplitude

fo = input('Input the frequency: '); % Frrequency operation

p = input('Input the phase: '); % Phase

n = input('Input the number of samples by period: '); % N

b = input('Input the number of bits: ');

disp('Para archivo Excel')

file = input('\n\nWhich is the file name: ', 's');

hoja = input('Sheet where you want to work: ', 's');

desc =[{'N of sampes'},{'Time in sec.'},{'Voltage'},{'V. Binary'}]

T = 1/fo; % period

Tm = T/n; % sample time

switchb

case2 % case represent two bit to quantize

forx = 1:n % cicle / start in 1, increas 1 by 1, until

t = Tm*x;

y = (A/2)*sin(2*pi*fo*t+p) + (A/2);

ify > 0 && y (A/4) && y (A/2) && y (3*A/4) && y 0 && y (A/8) && y (2*A/8) && y (3*A/8) && y (A/2) && y


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fprintf('%g %f %f %s \n\n',x,t,y,f)

elseify > (5*A/8) && y (3*A/4) && y (7*A/8) && y


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Exercise 4.1: Based on the exampleshowed in the experiment 4.1, increment

the number of bits from 3 to 5 bits, and plot the binary representation of the

signal.

sampling continuos signals

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