Hello All .I want a solution with a simple explanation

[Sulaymon Eshkabilov]

This looks like a HW assignment. Could you show what you have do so far?

[Ahmed Ali]

@Sulaymon Eshkabilov

syms t n 
T = 2 ; 
w0 = 2 * pi / T ; 
n = 1:10:100:1000:10000
ao = ( 1 / T ) * int ( 1 , t , 0,2 ) 
an = ( 2 / T ) * int ( 1 * cos ( n * w0 * t ) , t , 0,2 ) 
bn = ( 2 / T ) * int ( 1 * sin ( n * w0 * t ) , t , 0,2 )
ft = (0.5 + 2/pi )* sum ((1/n)*sin(n*pi*t))
plot (T,ft)

[Ahmed Ali]

@Sulaymon Eshkabilov The most I've done

[Sulaymon Eshkabilov]

@Ahmed Ali

Good attempts!

Always, post your work along with your question that helps the audience to see where and how they can help you with your question.

[Ahmed Ali]

@Sulaymon Eshkabilov

Thanks. I will do it next time

Answer 1

[William Rose]

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.

[William Rose]

@Ahmed Ali ,

Here's something interesting: To see the Gibbs phenomenon in more detail, use a time step that is shorter than the highest frequency in the summation.

a0=0.5; A=1; T=2;                     %mean value; peak2peak amplitude; period
t=(0:1e-6:6e-3)*T;                    %time vector for .006 cycle
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');

[Ahmed Ali]

@William Rose @Walter Roberson

Thank you so much. It's worked

Very grateful to you

[William Rose]

@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.

Answer 2

[Walter Roberson]

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])

[Ahmed Ali]

@Walter Roberson

Hello . thanks for your time. There is a problem with the last line that the program tells me. I made several changes, but it didn't solve the problem. Can you take a look at it?

Big thanks to you

[Walter Roberson]

What error did you get? What matlab version are you using? The output I show above is from running that exact matlab code in R2021b

[Ahmed Ali]

@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]

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])

Answer 3

[Ahmed Ali]

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)

[Ahmed Ali]

The typical solution to this question

[William Rose]

@Ahmed Ali , your solution looks very nice.

Answer 4

[William Rose]

@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.