Plotting fourier series partial sums

Hi. I am struggling to get my plot for my partial sums correct. I have an

A0 = 0.9

An even and odd = 0

Bn even and odd = -3./(pi*n).*(cos(pi*n*1.5./2.5))-1

I didn't expect my partial sums to look like it does (below).

Any ideas why it is coming out the way it is.

This is my code:

A = 3; % Amplitude (Dataset 5)
L = 5; % Period (Dataset 5)
dutycycle = 30;
sampling = 0.01; 
t = 0 : sampling : (L-sampling); % Define time vector

y(t<L*(dutycycle/100)) = A;
y(t>L*(dutycycle/100) & (t<L)) = 0;

x = 0:0.01:2*pi;
n = 25;
Fn = fourier(n,x);
plot(x,Fn)
title ('Partial Sums')
grid on
xlabel('X');
ylabel('Fn');

function Fn = fourier(n,x)
  x = x(:);
  a0 = 0.9;
  a = zeros(1,n);
  b = zeros(1,n);
  for i = 1:2:n
     b(i) = -3/(pi*i)*(cos(pi*i*1.5/2.5))-1;
  end
  Fn = a0/2 + sum(a.*cos((1:n).*x)+b.*sin((1:n).*x),2);
end

[Paul]

Double check the derivations of A0, A(n), and B(n). Fourier Series (eqn 5)

Why is the plot from 0 to 2*pi when L = 5?

Double check the expression for Fn. Here's one source: Fourier Series (eqn 2)

[AB29]

Pretty confident with my figures.

A0 = 0.9

An even and odd = 0

Bn even and odd = -3./(pi*n).*(cos(pi*n*1.5./2.5))-1

Think 0 to 2*pi was a mistake when carrying over from some previous work.

Very inexperienced on Matlab so i'm struggling to get to grips with this. any help is much appreicated

[Paul]

I got different answers than you for A0, A(n), and B(n). Did you compute those by hand or using Matlab? If the latter, you're likely to get more help if you post the code.

The expression for Fn is not the same as in the Wikipedia link I provide.

[Torsten]

@AB29

If your function has period L, the fourier series and its coefficients are computed differently.

https://en.wikipedia.org/wiki/Fourier_series

((eq.2) and (eq.5) with P = L).

By the way:

b(i) = -3/(pi*i)*(cos(pi*i*1.5/2.5))-1;

cannot be the sin Fourier coefficients of a function - they must converge to 0 as i -> Inf. Yours converge to -1.

[Paul]

The L in @Torsten's link is not the same as the L in this question (based on the comment in the code regarding what L is).

[Torsten]

@Paul

Sorry, I changed the link. In the Wikipedia link, I guess the P therein should be the OP's L now, shouldn't it ?

[AB29]

Hi Paul.

My figures were calculated by hand. I'm relatively new to fourier as a whole so very possible i made a mistake. This was my function:

[Torsten]

I get

syms n x
A_0 = 1/5 * int(3,x,0,1.5)

A_0 = 

A_n = 2/5 * int(3*cos(2*pi/5*n*x),x,0,1.5)

A_n = 

B_n = 2/5 * int(3*sin(2*pi/5*n*x),x,0,1.5)

B_n = 

x = -7:0.01:7;
n = 25;
Fn = fourier(n,x);
plot(x,Fn)
title ('Partial Sums')
grid on
xlabel('X');
ylabel('Fn');

function Fn = fourier(n,x)
  x = x(:);
  a0 = 0.9;
  a = 3./(pi*(1:n)).*sin(3*pi*(1:n)/5);
  b = 6./(pi*(1:n)).*(sin(3*pi*(1:n)/10)).^2;
  Fn = a0 + sum(a.*cos(2*pi/5*(1:n).*x)+b.*sin(2*pi/5*(1:n).*x),2);
end

[Paul]

@Torsten, yes, P in the wikipedia link, which is the same page that I linked to, is L in this problem.

Answer 1

[Hassaan]

To compute and plot the partial sums of a Fourier series based on the coefficients you've provided (A0 = 0.9, An = 0, and Bn given by a specific formula), Calculates these coefficients and then uses them to compute the Fourier series up to the nth term.

% Define the given parameters
A0 = 0.9; % DC offset
L = 5; % Period
dutycycle = 30; % Duty cycle in percent
sampling = 0.01; % Sampling rate
n = 25; % Number of terms in the partial sum

% Define the time vector
t = 0 : sampling : (L - sampling);

% Initialize the output signal
y = zeros(size(t));

% Define the signal based on the duty cycle
y(t < L * (dutycycle / 100)) = 3; % Amplitude is 3 for the duty cycle

% Define the x values for plotting the Fourier series
x = 0 : 0.01 : 2 * pi;

% Calculate the Fourier series up to the n-th term
Fn = fourierSeries(A0, n, x);

% Plot the result
plot(x, Fn)
title('Partial Sums of Fourier Series')
grid on

function Fn = fourierSeries(n, x, A0)
    Fn = A0 / 2; % Start with A0 term
    for k = 1:n
        An = 0; % Given An = 0
        Bn = -3 / (pi * k) * (cos(pi * k * 1.5 / 2.5)) - 1; % Calculate Bn based on the formula
        Fn = Fn + An * cos(k * x) + Bn * sin(k * x); % Sum up the series
    end
end

Note

Adjust the function as needed based on your actual series formula and make sure that the fourierSeries function is consistent with the series you want to compute.

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