Hello All .I want a solution with a simple explanation

Since I'm not good at symbolic math in Matlab, I'll give an example of how to do this problem non-symbolically. The code below shows off Matlab's great ability to be smart about adding things up without using for loops.

a0=0.5; A=1; T=2; %mean value; peak2peak amplitude; period

t=(0:.001:2)*T; %time vector for two cycles

b=(2*A/pi)./(1:2:9999); %coefficients for sines (odd only)

w=2*pi*(1:2:9999)/T; %frequencies for sines (odd only)

f=a0-A/2+A*(mod(t,T)<T/2); %exact square wave function

f1=a0+b(1)*sin(w(1)*t); %Fourier series, n=1

f9=a0+b(1:5)*sin(w(1:5)'*t); %Fourier series, n=1:9 odd

f99=a0+b(1:50)*sin(w(1:50)'*t); %Fourier series, n=1:99 odd

f999=a0+b(1:500)*sin(w(1:500)'*t); %Fourier series, n=1:999 odd

f9999=a0+b(1:5000)*sin(w(1:5000)'*t); %Fourier series, n=1:9999 odd

plot(t,f,'-k',t,f1,'-r',t,f9,'-g',t,f99,'-b',t,f999,'-c',t,f9999,'-m');

legend('f(t)','f1','f9','f99','f999','f9999');

Try code above.

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Ahmed Ali 2021-11-1

@William Rose @Walter Roberson

Thank you so much. It's worked

Very grateful to you

William Rose 2021-11-1

@Ahmed Ali , you're welcome.

@Walter Roberson is really good at math and Matlab, so when he gives a hint, as he did in this case, I pay attention.

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Answer 2

Walter Roberson 2021-10-31

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1575298-hello-all-i-want-a-solution-with-a-simple-explanation#answer_821028

在 MATLAB Online 中打开

Fixing your syntax, but not attempting to fix the logic.

Question: why do you calculate ao, an, bn, if you are not going to use them?

Note: the values for an and bn are correct for those expressions and that integration interval.

syms t n

T = 2 ;

w0 = 2 * pi / T ;

n = [1, 10, 100, 1000, 10000];

ao = ( 1 / T ) * int ( sym(1) , t , 0,2 )

ao =

1

an = ( 2 / T ) * int ( sym(1) * cos ( n * w0 * t ) , t , 0,2 )

an =

bn = ( 2 / T ) * int ( sym(1) * sin ( n * w0 * t ) , t , 0,2 )

bn =

ft = (0.5 + 2/pi ) * sum ((1./n) .* sin(n*pi*t))

ft =

fplot(ft, [-10 10])

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Ahmed Ali 2021-11-1

@Walter Roberson These are all errors

Error using fcnchk (line 106)

If FUN is a MATLAB object, it must have an feval method.

Error in fplot (line 60)

fun = fcnchk(fun);

Error in Untitled (line 11)

fplot(ft, [-10 10])

>>

I am using the 2014b version

Walter Roberson 2021-11-1

在 MATLAB Online 中打开

syms t n

T = 2 ;

w0 = 2 * pi / T ;

n = [1, 10, 100, 1000, 10000];

ao = ( 1 / T ) * int ( sym(1) , t , 0,2 )

ao =

1

an = ( 2 / T ) * int ( sym(1) * cos ( n * w0 * t ) , t , 0,2 )

an =

bn = ( 2 / T ) * int ( sym(1) * sin ( n * w0 * t ) , t , 0,2 )

bn =

ft = (0.5 + 2/pi ) * sum ((1./n) .* sin(n*pi*t))

ft =

fth = matlabFunction(ft);

fplot(fth, [-10 10])

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Answer 3

Ahmed Ali 2021-11-17

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1575298-hello-all-i-want-a-solution-with-a-simple-explanation#answer_833949

clc

clear all

S=0;

H=0;

N=input('input N=');

for k=1:N

for I=1:40000

f=I/10000;

H(I)=(2/(pi*(2*k-1)))*sin((2*k-1)*pi*f);

end

S=S+H;

end

S=S+0.5;

plot(S)

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Ahmed Ali 2021-11-17

The typical solution to this question

William Rose 2021-11-17

@Ahmed Ali , your solution looks very nice.

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Answer 4

William Rose 2021-11-1

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1575298-hello-all-i-want-a-solution-with-a-simple-explanation#answer_821623

@Walter Roberson said he is correcting your syntax, but not your logic. @Walter Roberson sees that there are logical errors in your code, in addition to syntax errors. Therefore it will not compute the Fourrier series correctly, even when you solve the syntax errors.

I assume the figure you provided represents an infinite pulse train.

Since the function f(t) is odd, except for the mean value, the Fourier series will involve odd terms with sine only, and not any cosine components. In other words, when you solve the integral for the Fourier series coefficients, you will find that all the cosine coefficients are zero, and the sine coefficients are zero when n is even. Therefore only the odd sine coefficients, and the coefficient for the mean value, are non-zero. The Fourier series is therefore

where

(n odd)

where

is the mean value, A is the peak-to-peak amplitude, T is the period. In this case,

=0.5, A=1, and T=2. Your code should do the summation above, stopping after n=1, 9, 99, 999, 9999.