Question

ми М. -5 -4 -3 -2 -1 0 1 2 3 4 5 1

f) Calculate the coefficients of the trigonometric form of the Fourier series
numerically in MATLAB and graphically represent the one-sided spectrum
(width and phase) frequency for n up to 10 compared to the analytics results.
g) From the coefficients of the trigonometric form of the Fourier series ,
calculate the coefficients of the exposure series and present the two-sided spectrum (width and phase) frequency.
h) Find the average and active value of the signal from the Fourier expansion.
i) Check if the sum of the harmonics actually creates the original signal. To  verify this you need to add the fixed component and all the harmonious (for n = 1 to n = ). Since we can't do that in practice, try adding to MATLAB the first 5 harmonics in the constant component, then the constant component and the first 10 harmonics. Present graphically in the same chart the two last sums and the original signal .

Answer 1

Matlab code for Fourier Series clear all close all All time values X=linspace(-2,2,1001); Loop for creating the function for i=1:length(x) if x(i)>=-2 && X(i)<-1 zz(i)=0; elseif x(i)>=-1 && x(i)<0 zz (i)=-4.*x(i); elseif x(i)>=0 && X(i) <1 zz(i)=4. *X(i); else zz(i)=0; end end figure (1) %Plotting the function hold on plot (x,zz, 'linewidth', 3) xlabel('x') ylabel('f(x)') title('Plotting of Actual data') lgnd{1}='Actual data'; al=X (1); bl=X(end); l= (bl-al)/2; Fourier series of the function for finding a and b coefficients for j=1:20 ssl=zz.*cos(j*pi*X/1); fall a values of the Fourier series aa (j)=(1/1)*trapz (X, ss1); ss2=zz.*sin(j*pi*x/1); fall b values of the Fourier series bb (j)=(1/1)*trapz(x, ss2); end %a0 value of Fourier series aa0=(1/1)*trapz (X, 22); X=linspace(-5,5,6001); s=aa 0/2; fall an and bn terms fprintf('Printing few terms for Fourier series \n') for i=1:10 fprintf('\tThe value of a&d=%f and b&d=ff. \n\n',i,aa(i),i,bb(i)) end %Fourier series of the function for i=1:5

s=s+bb(i)*sin(i*pi*X/1)+aa (i)*cos(i*pi*X/1); plot(X,s) lgnd{i+1}=sprintf('Fourier %d terms.',i); end legend (lgnd, 'location', 'best') box on %%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%% Printing few terms for Fourier series The value of al=0.925326 and bl=-0.008000. The value of a2=-0.810580 and b2=-0.000000. The value of a3=-1.028933 and b3=0.008000. The value of a4=-0.000000 and b4=-0.000000. The value of a5=0.444403 and b5=-0.008000. The value of a6=-0.0900 74 and b6=0.000000. The value of a7=-0.396814 and b7=0.008000. The value of a8=-0.000000 and b8=0.000000. The value of a9=0.262847 and b9=-0.008000. The value of a10=-0.032433 and b10=0.000000.

Plotting of Actual data Actual data Fourier 1 terms. Fourier 2 terms. Fourier 3 terms. Fourier 4 terms. Fourier 5 terms. -5 -4 -3 -2 -1 0 1 2 3 4 5 Published with MATLAB® R2018a

%Matlab code for Fourier Series
clear all
close all
%All time values
X=linspace(-2,2,1001);
%Loop for creating the function
for i=1:length(X)
    if X(i)>=-2 && X(i)<-1
        zz(i)=0;
    elseif X(i)>=-1 && X(i)<0
        zz(i)=-4.*X(i);
    elseif X(i)>=0 && X(i)<1
        zz(i)=4.*X(i);
    else
        zz(i)=0;
    end
end
figure(1)
%Plotting the function
hold on
plot(X,zz,'linewidth',3)
xlabel('x')
ylabel('f(x)')
title('Plotting of Actual data')
lgnd{1}='Actual data';

a1=X(1); b1=X(end);
l=(b1-a1)/2;
%Fourier series of the function for finding a and b coefficients
for j=1:20
    ss1=zz.*cos(j*pi*X/l);
    %all a values of the Fourier series
    aa(j)=(1/l)*trapz(X,ss1);
    ss2=zz.*sin(j*pi*X/l);
    %all b values of the Fourier series
    bb(j)=(1/l)*trapz(X,ss2);
end

%a0 value of Fourier series
aa0=(1/l)*trapz(X,zz);
X=linspace(-5,5,6001);
s=aa0/2;

%all an and bn terms
fprintf('Printing few terms for Fourier series\n')
for i=1:10
    fprintf('\tThe value of a%d=%f and b%d=%f. \n\n',i,aa(i),i,bb(i))
end
  
%Fourier series of the function
for i=1:5
    s=s+bb(i)*sin(i*pi*X/l)+aa(i)*cos(i*pi*X/l);
    plot(X,s)
    lgnd{i+1}=sprintf('Fourier %d terms.',i);  
end
legend(lgnd,'location','best')
box on
%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%