serie fourier using matlab

Question

nour yousfi 2017-10-15

0
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https://ww2.mathworks.cn/matlabcentral/answers/361396-serie-fourier-using-matlab

编辑： nour yousfi 2017-10-16

hi; I want a fourier transform of a piecewise function using matlab . The function is defined as follows:

0.6+t/0.3*pi when 0 <t < 0.12*pi

1 when 0.12*pi <t < 1.08*pi

4.6-t/0.3*pi when 1.08*pi <t < 1.2*pi

0.6 when 1.2*pi <t <2*pi

any help will be appreciated

I tried to do that but I didn't found a good result

Matlab file: clear all syms t n T1=2*pi;

I1 = int((0.6+(t/0.3*pi)), t,0,0.12*pi); I2 = int(1, t,0.12*pi,1.08*pi); I3 = int((4.6-(t/0.3*pi)), t,1.08*pi,1.2*pi); I4 = int(0.6, t,1.2*pi,2*pi);

a0 = (1/T1)*(I1 + I2 + I3+I4); fourierSum = 0;

II1 = int((0.6+(t./0.3*pi))*cos((2*n*pi*t)/T1), t,0,0.12*pi); II2 = int((1)*cos((2*n*pi*t)/T1), t,0.12*pi,1.08*pi); II3 = int((4.6-(t./0.3*pi))*cos((2*n*pi*t)/T1), t,1.08*pi,1.2*pi); II4 = int((0.6)*cos((2*n*pi*t)/T1), t,1.2*pi,2*pi); III1 = int((0.6+(t./0.3*pi))*sin((2*n*pi*t)/T1), t,0,0.12*pi); III2 = int(sin((2*n*pi*t)/T1), t,0.12*pi,1.08*pi); III3 = int((4.6-(t./0.3*pi))*sin((2*n*pi*t)/T1), t,1.08*pi,1.2*pi); III4 = int((0.6)*sin((2*n*pi*t)/T1), t,1.2*pi,2*pi);

for k = 1:40 an = subs(II1 + II2 + II3+II4,n,k); bn = subs(III1 +III2+III3+III4,n,k); term = an*cos((2*k*pi*t)/T1); termm = bn*sin((2*k*pi*t)/T1); fourierSum = fourierSum+term+termm; fourierSum=(2/T1)*fourierSum; end

ezplot(vpa(a0+fourierSum,6), [0 10])