Fourier Series Signal in Cosine Form and Exponential Form

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Black Santa
Black Santa 2021-10-4
回答： Aashray 2025-6-13
I'm trying to make this graph in its Fourier Series Signal in Cosine and Exponential Form. Can you please correct my codes.
t= [-3 -3 -2 -2 -1 -1 0 0 1 1 2 2 3 3]; % values of t
xt= [1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1]; % values of x
subplot(3,1,1) % to plot 3 graphs in a single page
plot(t,xt,'LineWidth',1.5)% plotting the original signal
title('Graph of the Original Signal')
xlabel('Time, (t)') % label the x-axis
ylabel('Amplitude, x(t)') % label the y-axis
grid on % putting a grid
subplot (3,1,2) % to plot 3 graphs in a single page
t = (-3:0.001:3); % exponential form
f = 0*t; % creates a zero valued function
N=5; % number of harmonics
for k=-N:1:N
if(k==0) % skip the zeroth term
continue;
end;
C_k = ((1)/(pi*1i*k))*(1-exp(-pi*1i*k)); % computes the k-th Fourier coefficient of the exponential form
f_k = C_k*exp(2*pi*1i*k*t); % k-th term of the series
f = f + f_k; % adds the k-th term to f
end
plot(t, f, 'LineWidth', 1.5);
title('Graph of the Fourier Series Signal in Exponential Form')
ylabel('Amplitude, x(t)') % label the y-axis
xlabel('Time, (t)') % label the x-axis
grid on % putting a grid
subplot (3,1,3)% to plot 3 graphs in a single page
a0=1/2; w0=pi; % cosine form
t=-3:0.001:3
sum=0;
for m=1:3:1
if mod(m+4-1,4)==0
Qm=0;
else Qm=-pi;
end
sum=sum+2/(m*pi)*cos(m*w0*t+Qm);
end
xt=a0+sum;
plot(t,xt,'LineWidth',1.5),hold on % plotting the fourier series signal
title('Graph of the Fourier Series Signal in Cosine Form')
ylabel('Amplitude, x(t)') % label the y-axis
xlabel('Time, (t)') % label the x-axis
grid on % putting a grid

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回答（1 个）

Aashray
Aashray 2025-6-13
Ran in:
Hello @Black Santa!
I went through the code that you have shared, and suggest the following changes:
  1. Use proper expression for a symmetric square wave:Ck = (1/(2i*pi*k)) * (exp(-1i*pi*k) - 1)
  2. Only plot real part using real(f_exp)
  3. Use sine terms with odd harmonics only (because square wave is odd function): bn = 4/(n*pi); xt_trig = xt_trig + bn * sin(n * w0 * t);
  4. Since the square wave is symmetric around zero: a0 = 0.
This correct code might help you understand the changes better:
% Original Signal Definition
t_discrete = [-3 -3 -2 -2 -1 -1 0 0 1 1 2 2 3 3];
xt_discrete = [1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1];
subplot(3,1,1)
plot(t_discrete, xt_discrete, 'LineWidth', 1.5)
title('Graph of the Original Signal')
xlabel('Time, (t)')
ylabel('Amplitude, x(t)')
grid on
%% Exponential Fourier Series Approximation
subplot(3,1,2)
t = -3:0.001:3; % fine time vector
T = 2; % assume period = 2
w0 = 2*pi/T; % fundamental frequency
f_exp = zeros(size(t));
N = 20; % number of harmonics
for k = -N:N
if k == 0
continue; % skipping DC term, assume zero average
end
% Fourier coefficients for square wave from -1 to 1, amplitude ±1
Ck = (1/(2i*pi*k)) * (exp(-1i*pi*k) - 1);
f_exp = f_exp + Ck * exp(1i * k * w0 * t);
end
plot(t, real(f_exp), 'LineWidth', 1.5)
title('Fourier Series Signal in Exponential Form')
xlabel('Time, (t)')
ylabel('Amplitude, x(t)')
grid on
%% Trigonometric (Cosine/Sine) Fourier Series Approximation
subplot(3,1,3)
t = -3:0.001:3;
xt_trig = zeros(size(t));
a0 = 0; % DC term for symmetric square wave is zero
N = 20; % number of sine harmonics
for n = 1:N
if mod(n, 2) == 1 % odd harmonics only
bn = 4/(n*pi); % coefficient for sine terms of square wave
xt_trig = xt_trig + bn * sin(n * w0 * t);
end
end
xt_trig = a0 + xt_trig;
plot(t, xt_trig, 'LineWidth', 1.5)
title('Fourier Series Signal in Sine (Trigonometric) Form')
xlabel('Time, (t)')
ylabel('Amplitude, x(t)')
grid on
●
A side note: The second subplot (exponential form) shows peaks where valleys are expected, due to phase shifts in the exponential representation and the absence of DC term.

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