Advanced methods for info. retrieval implemented in Matlab

Lab/Lab_1.m

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+%ADVANCED METHODS FOR INFORMATION REPRESENTATION
+%Lab_1
+
+clear all;
+close all;
+clc;
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%1)Given two vectors in R^2, e1=[e1x e1y]' and e2=[e2x e2y]' find the dual 
+%basis f1=[f1x f1y]' and f2=[f2x f2y]'.
+%The biorthogonal basis must satisfy the following conditions:
+%
+%1) <e1,f1>=[e1x e1y]'*[f1x f1y]=1
+%2) <e2,f1>=[e2x e2y]'*[f1x f1y]=0
+%3) <e2,f2>=[e2x e2y]'*[f2x f2y]=1
+%4) <e1,f2>=[e1x e1y]'*[f2x f2y]=0
+%
+%Solving the system composed by 1) and 2) i obtain f1:
+% e1x*f1x+e1y*f1y=1
+% e2x*f1x+e2y*f1y=0
+% A*f1=C
+%
+%Solving 3) and 4) i obtain f2:
+%
+% e2x*f2x+e2y*f2y=1
+% e1x*f2x+e1y*f2y=0
+% B*f2=C
+
+%Given base:
+e1=[1;0];
+e2=[1/2;1];
+
+%Coefficients of the systems:
+A=[e1(1) e1(2); e2(1) e2(2)];
+B=[e2(1) e2(2); e1(1) e1(2)];
+C=[1;0];
+
+%Biorthogonal base:
+f1=A\C;
+f2=B\C;
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%2)If N>=3 i have to evaluate more conditions, in case of N=3 i have nine
+%equations:
+%
+% e1x*f1x+e1y*f1y+e1z*f1z=1
+% e2x*f1x+e2y*f1y+e2z*f1z=0
+% e3x*f1x+e3y*f1y+e3z*f1z=0
+% A*f1=D
+% 
+% e2x*f2x+e2y*f2y+e2z*f2z=1
+% e1x*f2x+e1y*f2y+e1z*f2z=0
+% e3x*f2x+e3y*f2y+e3z*f2z=0
+% B*f2=D
+%
+% e3x*f3x+e3y*f3y+e3z*f3z=1
+% e1x*f3x+e1y*f3y+e1z*f3z=0
+% e2x*f3x+e2y*f3y+e2z*f3z=0
+% C*f2=D
+
+e1=[1;0;0];
+e2=[1/2;1;0];
+e3=[0;1/2;1];
+
+A=[e1(1) e1(2) e1(3); e2(1) e2(2) e2(3); e3(1) e3(2) e3(3)];
+B=[e2(1) e2(2) e2(3); e1(1) e1(2) e1(3); e3(1) e3(2) e3(3)];
+C=[e3(1) e3(2) e3(3); e1(1) e1(2) e1(3); e2(1) e2(2) e2(3)];
+D=[1;0;0];
+
+f1_1=A\D;
+f2_1=B\D;
+f3_1=C\D;
+
+%Note: the method is working properly, in fact it satisfies <ei,yj>=1 for
+%all i=j and zero otherwise! For higher N i have to iterate the same 
+%procedure.
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%3)The analysis consists of determining the coordinate coefficients a of 
+%the vector x given the basis y1,y2,...
+
+x=[1/4;1/2];
+y1=[1;0];
+y2=[0;1];
+G=[y1'*y1 y2'*y1; y1'*y2 y2'*y2];
+T=[x'*y1;x'*y2];
+a=G\T;
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%4)The synthesis consists of determining the vector x given the 
+%coefficients c1,c2,... and the basis y1,y1,...
+
+c=[1;1];
+y1=[1;0];
+y2=[1/2; 1];
+x=c(1)*y1+c(2)*y2;
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%5)Uniform quantizer with parameters Nlevels, and bounds +-ext
+
+t=0:0.01:1;
+s=sin(2*pi*2*t);
+Nlevels=8;
+ext=1;
+Yq=round(s/ext*(Nlevels/2))/(Nlevels/2)*ext;
+set=find(abs(Yq)>ext);
+Yq(set)=ext*sign(Yq(set));
+figure; plot(t,s); hold on; stem(t,Yq,'r'); title('Effect of quantization over a sinusoid');
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%6)100 random values from the normal distribution with mean=0 and var=10
+X1=random('norm',0,10,[100 2])';
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%7)100 random values from the uniform distribution with mean=0 and var=1
+X2=random('norm',0,1,[100 4])';
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%8)
+
+%Find the dual basis for the random vectors contained in X2:
+DUAL=zeros(size(X2));
+for i=1:length(X2)
+    DUAL(:,i)=biorb([X2(1,i) X2(2,i)],[X2(3,i) X2(4,i)]);
+end
+
+%Compute the analysis of the dual basis:
+for i=1:length(X2)
+    a=[DUAL(1,i);DUAL(2,i)];
+    b=[DUAL(3,i);DUAL(4,i)];
+    c(:,i)=analysis(X1(:,i),[a b]); 
+end
+
+%Quantize the dual basis coefficients with 16 levels and range 
+%abs(max(min(c),max(c))):
+ext=max(max(abs(c)));
+C=zeros(size(c));
+for i=1:length(c)
+   C(:,i)=quantizer(c(:,i),16,ext); 
+end
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%9)
+
+%Reconstructed vector:
+xrec=synth(C,X2);
+
+%Reconstruction error:
+error=zeros(1,length(xrec));
+for i=1:length(X2)
+    error(i)=norm((X1(:,i)-xrec(:,i)).^2,2);
+end
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%10)Repeat 9) for each x
+
+E=zeros(size(X2));
+for i=1:length(X2)
+    for j=1:length(X2)
+        E(i,j)=norm((X1(:,i)-xrec(:,j)),2)^2;
+    end
+end
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%11)
+
+figure; surf(E); shading('interp'); xlabel('x-xrec'); ylabel('x-xrec'); zlabel('error');
+title('Square norm of the errors');
+
+%Note: from the graph i can see that higher the error higher is the
+%distance between the signals x and the reconstructed ones, where
+%the error is low menas instead that the two signals are close.

Lab/Lab_2.m

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+%ADVANCED METHODS FOR INFORMATION REPRESENTATION
+%Lab_2
+
+%NOTES: useful tools
+%sptool
+%spectrogram
+%fft,stft,cwt
+%sound(x,1000)
+%scale2freq
+%wavemenu
+
+clear all;
+close all;
+clc;
+
+%Load data
+load('cardio100.mat');
+w=hamming(2048)';
+
+if 1==0
+%Compute the fft of the signals
+k1=fftshift(fft(heart100,1000));
+k2=fftshift(fft(heart100.*w,1000));
+k3=fftshift(fft(heart100));
+k4=fftshift(fft(heart100.*w));
+
+figure;
+subplot(1,4,1); stem(abs(k1)); ylabel('|k1|'); xlabel('k'); title('fft 1000');
+subplot(1,4,2); stem(abs(k2)); ylabel('|k2|'); xlabel('k'); title('fft 1000 + win');
+subplot(1,4,3); stem(abs(k3)); ylabel('|k3|'); xlabel('k'); title('fft 2048');
+subplot(1,4,4); stem(abs(k4)); ylabel('|k4|'); xlabel('k'); title('fft 2048 + win');
+
+%Square the signals and compute the fft
+k1sq=fftshift(fft(heart100.^2,1000));
+k2sq=fftshift(fft((heart100.*w).^2,1000));
+k3sq=fftshift(fft(heart100.^2));
+k4sq=fftshift(fft((heart100.*w).^2));
+
+figure;
+subplot(1,4,1); stem(abs(k1sq)); ylabel('|k1sq|'); xlabel('k'); title('fft 1000');
+subplot(1,4,2); stem(abs(k2sq)); ylabel('|k2sq|'); xlabel('k'); title('fft 1000 + win');
+subplot(1,4,3); stem(abs(k3sq)); ylabel('|k3sq|'); xlabel('k'); title('fft 2048');
+subplot(1,4,4); stem(abs(k4sq)); ylabel('|k4sq|'); xlabel('k'); title('fft 2048 + win');
+
+%Recover the frequencies of the main harmonics
+N1=1000;
+N2=2048;
+fc=1000;
+
+m1=find(k1==max(k1));
+m2=find(k2==max(k2));
+m3=find(k3==max(k3));
+m4=find(k4==max(k4));
+
+f1=m1*fc/N1;
+f2=m2*fc/N1;
+f3=m3*fc/N2;
+f4=m4*fc/N2;
+
+m1sq=find(k1sq==max(k1sq));
+m2sq=find(k2sq==max(k2sq));
+m3sq=find(k3sq==max(k3sq));
+m4sq=find(k4sq==max(k4sq));
+
+f1sq=m1sq*fc/N1;
+f2sq=m2sq*fc/N1;
+f3sq=m3sq*fc/N2;
+f4sq=m4sq*fc/N2;
+
+%NOTES: The frequency of the heartbeat estimated roughly from the graph is
+%around 7.5 Hz (15 peaks in 2.048 s), here instead the frequency of the
+%main harmonic is around 500 Hz. In this case the fft is not useful to extract
+%information from our peaks: the main harmonic corresponds to the smaller
+%variations inside our signal (that have more energy).
+%In fact the Fourier transform of a peak is a constant and
+%the transform of a comb of delta is still a comb of delta, but here the
+%delta are not uniformally spaced in time and moreover they have not the
+%same amplitude, so the expected transform it's nor a constant nor a pure
+%comb of delta. The frequencial information of the peaks is spread over
+%the spectrum and so we are not able to detect their frequency.
+end
+
+%Some tests:
+% cfreq=centfrq('mexh',10,'plot');
+% a=scal2frq(1,'mexh',1);
+% b=scal2frq(1,'mexh',.001);
+% c=scal2frq(5,'mexh',.001);
+%t=0:0.01:1;
+%x=sin(2*pi*10*t);
+% save('x.mat','x');
+
+%load('x.mat');
+%c=cwt(x,1:32,'mexh','glb'); %contourf(c,8);
+%h=cwt(heart100,1:32,'mexh','glb');
+x=[1 1 1 1];
+
+d=dwt(x,'dmey');
+e=cwt(x,1:4,'dmey');
+figure; stem(d);
+figure; imagesc(e);

Lab/Lab_3.m

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+%ADVANCED METHODS FOR INFORMATION REPRESENTATION
+%Lab_3
+
+clear all;
+close all;
+clc;
+
+%Load data
+load('cardio100.mat');
+
+g0=1/(4*sqrt(2))*[(1+sqrt(3)), (3+sqrt(3)), (3-sqrt(3)), (1-sqrt(3))];
+h0=1/(4*sqrt(2))*[(1-sqrt(3)), (3-sqrt(3)), (3+sqrt(3)), (1+sqrt(3))];
+g1=1/(4*sqrt(2))*[0, -(1-sqrt(3)), (3-sqrt(3)), -(3+sqrt(3)), (1+sqrt(3))];
+h1=1/(4*sqrt(2))*[(1+sqrt(3)), -(3+sqrt(3)), (3-sqrt(3)), -(1-sqrt(3)), 0];
+
+figure;
+subplot(2,2,1); stem(g0); title('g_0(n)');
+subplot(2,2,2); stem(h0); title('h_0(n)');
+subplot(2,2,3); stem(g1); title('g_1(n)');
+subplot(2,2,4); stem(h1); title('h_1(n)');

Lab/analysis.m

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+function [c] = analysis(x, Y)
+%This function determine the coordinate coefficients c of the vector x given
+%the basis y1,y2,... stored in Y=[y1 y2 ...] where:
+%c=G^-1*T
+%G=[<y1,y1> <y2,y1> ... <yn,y1>; <y1,y2> <y2,y2> ... <yn,y2>; ...; <y1,yn> <y2,yn> ... <yn,yn>]
+%T=[<x,y1>; <x,y2>; ...; <x,yn>]
+G=zeros(size(Y));
+T=zeros(size(Y(:,1)));
+for i=1:length(Y)
+    for j=1:length(Y)
+        G(i,j)=Y(:,j)'*Y(:,i); 
+    end
+        T(i,:)=x'*Y(:,i); 
+end
+    c=G\T;
+end

Lab/biorb.m

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+function [F] = biorb(e1, e2)
+%This function returns a biorthogonal basis from the given vectors e1, e2.
+%The result satisfy the conditions on the scalar products <ei,fj>=1 for all
+%i=j and zero otherwise.
+A=[e1(1) e1(2); e2(1) e2(2)];
+B=[e2(1) e2(2); e1(1) e1(2)];
+C=[1;0];
+F=[A\C; B\C];
+end

Lab/quantizer.m

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+function [Yq] = quantizer(Y,Nlevels,ext)
+%This function returns the quantized version of the vector Y over Nlevels
+%and in the range +ext/-ext.
+Yq=round(Y/ext*(Nlevels/2))/(Nlevels/2)*ext;
+set=find(abs(Yq)>ext);
+Yq(set)=ext*sign(Yq(set));
+end

Lab/synth.m

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+function [xrec] = synth(C,basis)
+%This function returns the synthesis of a vector given a base and the
+%coefficients.
+xrec=zeros(size(C));
+    for i=1:length(basis(1,:))
+        xrec(:,i)=C(1,i)*[basis(1,i) basis(2,i)]+C(2,i)*[basis(3,i) basis(4,i)];
+    end
+end