Matlab Signal Representation Examples

8/2/2019 Matlab Signal Representation Examples

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Montadar Abas Taher Matlab Examples for Signal Representation Prof. Mahamod Ismail

P54852 Department of Electrical, Electronic & Systems EngineeringFaculty of Engineering & Built Environment

National University of Malaysia%% Example 1 [Time domain for analogue signals]

clear all;close all;clc;

Fs=8000; % Sampling Frequency

fm=10; % the signal's frequency in Hz.t1=[0:0.1*Fs]'/Fs; % Sampling time for 0.1 second

t2=[0:Fs]'/Fs; % Time vector for all the signal

x=sin(2*pi*fm*t1); % representation of 0.1 second of signal x

% so, there are one cycle in 0.1 second %because

fm=10 Hz

x2=sin(2*pi*fm*t2); % representation of 1 second of signal x

% Hence, there are 10 cycles in each second according

% to its frequency fm=10 Hz.

subplot(211); plot(t1,x);grid;

subplot(212);,plot(t2,x2);grid;

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National University of Malaysia%% Example 2 [Time domain for analoge signals]



fm=100; % the signal's frequency in Hz.

Fc=2000; % Carrier frequency is 2 kHz.t1=[0:0.1*Fs]'/Fs; % Sampling time for 0.1 second

t2=[0:Fs]'/Fs; % Time vector for all the signal

x=sin(2*pi*fm*t1); % representation of 0.1 second of signal x

% so, there are 10 cycles in 0.1 second %because fm=100 Hz

y=x.*sin(2*pi*Fc*t1); % Modulated signal Fc=2 kHz.

x2=sin(2*pi*fm*t2); % representation of 1 second of signal x

% Hence, there are 100 cycles in each second %according

% to its frequency fm=100 Hz.

y2=x2.*sin(2*pi*Fc*t2); % Modulated signal Fc=2 kHz.

subplot(221); plot(t1,x,'r');grid;

subplot(222);,plot(t1,y,'r');grid;subplot(223); plot(t2,x2,'k');grid;

subplot(224);,plot(t2,y2,'k');grid;

%% Example 3 [Time/Frequency domain for analoge signals]



T=1/Fs; % sampling time

L=1000; % length of signal

%% Time Domain Plot

t=[0:L-1]*T; % time vector

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National University of Malaysiax=0.7*sin(2*pi*150*t)+sin(2*pi*250*t); % Original Signal

y=x+2*randn(size(t)); % Noisy Signal

subplot(221); plot(Fs*t(1:50),x(1:50),'r');grid; xlabel('time in ms')

subplot(222);,plot(Fs*t(1:50),y(1:50),'r');grid; xlabel('time in ms')

%% Frequency Domain PlotNFFT=L; % Length of FFT

X=fft(y,NFFT)/L; % Fourier Transform of signal x

Y=fft(y,NFFT)/L; % Fourier Transform of signal y

f=Fs/2*linspace(0,1,NFFT/2+1); % Frequency index

subplot(223);plot(f,2*abs(X(1:NFFT/2+1)));grid; xlabel('Frequency in Hz')

subplot(224);plot(f,2*abs(Y(1:NFFT/2+1)));grid; xlabel('Frequency in Hz')

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National University of Malaysia%% Example 4 [Time/Frequency domain for analoge signals]



Fm=3.4e3; % message frequency

t=(0:0.01*Fs)'/Fs; % time indexingx=0.35*sin(2*pi*Fm*t); % Original Signal

L=length(t);

X=fft(x,L)/L;

f=Fs/2*linspace(0,1,L/2+1); % Frequency index

subplot(211); plot(t,x,'r');grid;

subplot(212);plot(f,2*abs(X(1:L/2+1)));grid; xlabel('Frequency in Hz')

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National University of Malaysia%% Example 5 [Time/Frequency domain for analoge signals]



Fm=1750; % message frequency

t=(0:0.01*Fs)'/Fs; % time indexingN=length(t);

x=0.8*square(2*pi*Fm*t); % Original Signal

X=fft(x,N)/N;

f=Fs/2*linspace(0,1,N/2+1); % Frequency index

subplot(211); plot(t,x,'r');grid;

subplot(212);plot(f,2*abs(X(1:N/2+1)));grid; xlabel('Frequency in Hz')

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National University of Malaysia%% Example 6 [Time domain for Digital signals]


Ts=1/1000; % Sampling Period is 1 ms

N=2000; % Number of bits are 2000 bits

data=randi([0 1],N,1);t=(0:1:N-1)*Ts; % Time vector

subplot(211); stem(t(1:40),data(1:40),'^r','filled');grid;

subplot(212); stairs(t(1:40),data(1:40),'k');grid;

%% Example 7 [Time domain for Digital signals]


Ts=0.1; % Sampling Period is 0.1 s

N=120; % Number of bits are 120 bitst=(0:1:N-1)*Ts; % Time vector

data=randi([0 1],N,1);

subplot(211); stem(t,data);grid;

subplot(212); stairs(t,data);grid;

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National University of Malaysia

%% Example 8 [Time domain for Digital signals]


Fs=8000; % Sampling rate

N=512; % Number of bits are 512 symbols

M=16; % Alphabit size

data=randi([0 M-1],N,1); %Digital Symbols

t=(0:0.01*Fs)'/Fs; % Time vector

stem(t(1:10),data(1:10));grid;

%

% In this example, the plot was for only 10 ms. We saw the sampling

% frequency Fs=8000 Hz, this means Ts=1/Fs=0.125 ms. Hence; in each 1.25 m

% there are 10 symbols.

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%% Example 9 [Frequency domain for Digital signals]


% A sinusoidal signal of frequency = 1 kHz, sampled at Fs=8 kHz.

% This signal must be plotted in time domain and frequency domain both in% one-sided and two sided for 1000 points.

Fs=8e3; % sampling frequency

N=1000; % number of data points

Fm=1e3; % signal frequency

t=[0:1:N-1]'/Fs; % time index

x=sin(2*pi*Fm*t);

figure(1);

plot(t(1:N/10),x(1:N/10));grid

X=abs(fft(x))/N;

P=X.*X;

f=[0:1:N-1]*Fs/N;figure(2);

subplot(211);plot(f,X);grid;

xlabel('Frequency in Hz');

subplot(212);plot(f,P);grid;


%% converting to one side

X(2:N)=2*X(2:N);

P=X.*X;

f=[0:1:N/2]*Fs/N;

figure(3);

subplot(211);plot(f,X(1:N/2+1));grid;


subplot(212);plot(f,P(1:N/2+1));grid;


% For those working with OFDM system, if you want to find the

% Peak to Average Power Ratio of the signal:

L=length(X);

p=max(P);

m=sum(abs(X.^2))/L;

paprvalue=10*log10(p/m);

msgbox(num2str(paprvalue),'PAPR')

figure(4);

a=real(X);b=imag(X);

subplot(411);stem(a);

subplot(412);stem(b);

c=abs(X);

d=angle(X);

subplot(413);stem(c);

subplot(414);stem(d);

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Matlab Signal Representation Examples