Plotting Fourier Series (first 5, 15, 25 partial sums, (3 Plots))

Hi there. I am a very new inexperienced user of MAtlab and have been asked to plot the first 5, 15, 25 partial sums across 3 plots for my fourier series.

Fourier series values shared below - any help with simple anotated breakdown would be appreciated.

• a0 = 1.5
• aN odd and even = 0
• bN odd=
• bN even = 0

Answer 1

[Hassaan]

Assuming you get the formula for bN​ (odd), here's one of the approach you can use in MATLAB:

% Define the basic parameters
a0 = 1.5;
T = 2*pi; % Period of the Fourier series, change according to your function
t = linspace(0, 2*T, 1000); % Time variable from 0 to 2*T with 1000 points

% Function to calculate bN for odd N, replace this with the actual formula
function bN = calculate_bN(N)
    % Your formula for bN when N is odd goes here
    % For example, a common form might be: bN = (-1)^(someFunction(N))/N;
    bN = 1/N; % Placeholder, replace with actual formula
end

% Function to calculate the partial sum of the Fourier series
function F = fourierPartialSum(t, N)
    F = a0 / 2; % Start with a0/2
    for n = 1:N
        if mod(n,2) == 1 % If n is odd
            bN = calculate_bN(n);
            F = F + bN*sin(n*t); % Update the sum with the nth term
        end
        % Since aN and bN for even are 0, those terms are skipped
    end
end

% Calculate and plot the first 5, 15, and 25 partial sums
figure;
subplot(3,1,1);
plot(t, fourierPartialSum(t, 5));
title('First 5 Partial Sums');

subplot(3,1,2);
plot(t, fourierPartialSum(t, 15));
title('First 15 Partial Sums');

subplot(3,1,3);
plot(t, fourierPartialSum(t, 25));
title('First 25 Partial Sums');

• Formula for bN​: You need to provide the actual formula for bN​ when N is odd. Replace the placeholder function calculate_bN(N) with your specific formula.
• Time Domain and Period: Adjust the time variable t and the period T according to the specific function you're analyzing.
• Resolution: The number of points in the linspace function determines the resolution of the plot. Increase the number of points for a smoother curve.

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[AB29]

Hi thanks for the help. I have added in my values. However i am still getting an error code here.

My bn Value is

Any ideas what I may have done wrong.

Once again thank you so much for the help

[AB29]

For reference here are some of the other figures used.

[Torsten]

Please include your code fragments as plain ascii text, not as a picture.

Otherwise we cannot use it to reproduce problems you have or results you got.

[AB29]

Apologies. I’ll update my comments soon.

[AB29]

% Define the basic parameters
a0 = 1.5;
T = 2*pi; % Period of the Fourier series, change according to your function
t = 0 % Time variable from 0 to 2*T with 1000 points
% Function to calculate bN for odd N, replace this with the actual formula
A = 3;
Bn= (-A./(pi*n)).*(cos(pi*n)-(cos(0)))
    % Your formula for bN when N is odd goes here
    % For example, a common form might be: bN = (-1)^(someFunction(N))/N;
    Bn= (-A./(pi*n)).*(cos(pi*n)-(cos(0))) % Placeholder, replace with actual formula

% Function to calculate the partial sum of the Fourier series
function F = fourierPartialSum(t, N)
    F = a0 / 2; % Start with a0/2
    for n = 1:N
        if mod(n,2) == 1 % If n is odd
            bN = calculate_bN(n);
            F = F + bN*sin(n*t); % Update the sum with the nth term
        end
        % Since aN and bN for even are 0, those terms are skipped
    end
end
% Calculate and plot the first 5, 15, and 25 partial sums
subplot(3,1,1);

Function definitions in a script must appear at the end of the file.
Move all statements after the "fourierPartialSum" function definition to before the first local function definition.

plot(t, fourierPartialSum(t, 5));
title('First 5 Partial Sums');
subplot(3,1,2);
plot(t, fourierPartialSum(t, 15));
title('First 15 Partial Sums');
subplot(3,1,3);
plot(t, fourierPartialSum(t, 25));
title('First 25 Partial Sums');

[AB29]

A = 3; % Amplitude (Dataset 5)
L = 5; % Period (Dataset 5)
dutycycle = 50;
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;

[AB29]

Here is my code.

Answer 2

[Torsten]

You should be able to add the missing plots.

x = 0:0.01:2*pi;
n = 25;
Fn = fourier(n,x);
plot(x,Fn)

function Fn = fourier(n,x)
  x = x(:);
  a0 = 1.5;
  a = zeros(1,n);
  b = zeros(1,n);
  for i = 1:2:n
    b(i) = -3/(pi*i)*(cos(pi*i)-cos(0));
  end
  Fn = a0/2 + sum(a.*cos((1:n).*x)+b.*sin((1:n).*x),2);
end

[AB29]

Thank you so much for the help.

If i was to change my bn to this

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

Would I need to change my FN function.

This time i have values for bn when even too so I am assuming I modify 1:2:n to 1:1:n

[Torsten]

Just change the b values in the loop.

Your picture says

for i = 1:n
    b(i) = -3/(pi*i)*cos(pi*i*1.5/2.5)-1;
end

your code line says

for i = 1:n
    b(i) = -3/(pi*i)*(cos(pi*i*1.5/2.5)-1);
end

Decide yourself which one is correct.

[AB29]

Thanks. Can see my mistake with the brackets.

[AB29]

Hi once again. After looking at my code and the graph produced I still feel this does not follow the trend I expect to see. This is what I get. The peaks seem too high.

I am dealing with an A0 = 0.9

An even and odd = 0

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

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

However I expected to see something more along the lines of the original trend too?

Please let me know your thoughts any help is much appreciated.

[Torsten]

The code I gave you is for a 2*pi periodic function (P = 2*pi). Its Fourier expansion is described by (eq.2) and (eq.5) under

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

If you have a different period P, the Fourier series and its coefficients are computed differently (again see (eq.2) and (eq.5)).

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.