banana

%%Problem 2

% b)We wish to find the error in reconstruction. To do that we firstly need
% to find the second norm of f(t). By applying Parseval's theorem we can
% compute this as the area under the FT of f(t) squared.
normF = 2 * integral(@(x) (8000 - 4*x).^2, 0, 2000);
%Know we need to find the norm of the difference in a similar matter
part1 = 2 * integral (@(x) ( (8000-4*x)-(1600) ).^2, 1600, 1800);
part2 =  2 * integral (@(x) (8000-4*x).^2, 1800, 2000);
normDiff = part1 + part2;
%now we can express the error as shown below
relativeError = normDiff/normF;

% c) We limit f(t) within frequnecy components between -1800 and 1800
% (apply low-pass). We recalcualte the error for this new signal in a
% similar way as before.
normDiffC  = temp2;
relativeErrorC = normDiffC / normF;

%d)To avoid the effects of aliasing it would be better to filter the signal
%f(t) with a lowpass (with cut-off frequncy ws=1800). The effects of this
%in the reconstructed signal are shown above, and as we noticed, are better
%then suffering the aliasing effect.
%% Problem 3

f = @(t) (cos(30*pi.*t)+ sin(40*pi.*t) - 2*cos(20*pi.*t));
%According to the Nyquist criteria we need 8 samples per period to properly
%sample this signal, but since there is a sin component in it, we need to
%chose any rate higehr then this for proper sampling (say 16 samples per
%period) T=0.2s, we chose N=16.

t=0:0.0125:15*0.0125;
x=f(t);
f=0.2/16*fft(x); %We multiply by T/N
w=0:10*pi:150*pi;
subplot(1,2,1);
stem(w,real(f));
hold
stem(w,imag(f),'r');
title ('Imaginary and Real part of FT Y');
xlabel ('w, rad/s');
ylabel('Y, Volts');
subplot(1,2,2);
stem(w,abs(f));
title ('Abs part FT Y');
xlabel ('w, rad/s');
ylabel('Y, Volts');
%We can see from the graphs that we have properly reconstructed the FT of
%the signal from its DFT
%% Problem 4
%Our signla is trumcated between -5 and 5 (outside is zero). We assume this
%signal repeats periodically outside this bounds.
N=71 %We chose this N to achieve a point-wise errer less than 0.05.
t1=0:10/N:5;
t2=t1(2:length(t1));
t2=fliplr(t2);
t=[t1,t2];
x=exp(-abs(t));
y=10./N*fft(x,N);
k=0:round(N/2)-1;
f1=2./(1+(k*pi./5).^2);
f2=f1(2:length(f1));
f2=fliplr(f2);
f=[f1,f2];
err=norm(y-f)./N^0.5;
%% Problem 5
N=41 %We use this sampling rate
t1=0:10/N:5;
t2=t1(2:length(t1));
t2=fliplr(t2);
t=[t1,t2];
x=sinc(t);
y=10./N*fft(x,N);
k=0:round(N/2)-1;
f1=rectangularPulse((-pi+0.01),(pi+0.01),(k*pi/5));
%We have to define a wider pulse, becasue of the matlab implementation
%This will not effect the result however
f2=f1(2:length(f1));
f2=fliplr(f2);
f=[f1,f2];
err=norm(y-f)./N^0.5;
%We can' reconstruct the signla's FT from its DFT no matter how many
%samples we take, because the signal is not periodic in time domain
%% Problem 6
N=200001;%We need to take roughly this many samples, to have an erro less then 0.01
t1=0:(3*pi)/N:3*pi/2;
t2=t1(2:length(t1));
t2=fliplr(t2);
t=[t1,t2];
x=cos(t);
helper = @(w) sin(3*pi.*w./2)./w;
F = @(k) helper(2.*k./3 -1) + helper(2.*k./3 +1);
y=10./N*fft(x,N);
k=0:round(N/2)-1;
f1=F(k);
f2=f1(2:length(f1));
f2=fliplr(f2);
f=[f1,f2];
err=norm(y-f)./N^0.5;
%You can replace N=7 in this code to simualte a samplin with rate T=pi/2