I am new to matlab. All i wanted was to integrate f(t)*cos(pi*n*t) over t through quadgk inside a loop where n varies from 1 to 10. But I guess the problem is that quadgk needs a function where only t is a variable and not n. how to do it???
t=linspace(0,12,1000);
T=2;
Y=g(t)';
rep=(t(end)-t([1]))/T;
f=0;
N=5;
hold on;
h=zeros(N,1);
clr=lines(N);
for n=1:2:N;
line(t,Y,'linewidth',5)
grid on;hold on;
A=(1/T)*quadgk(@g,0,T);
P=Y.*(cos(n*pi*t))';
Q=Y.*(sin(n*pi*t))';
Ax=(2/T)*quadgk(@s,0,T);
Bx=(2/T)*trapz(t,Q)/rep;
f=f+(Ax*cos(pi*n*t)+Bx*sin(pi*n*t));
final=A+f;
h(n)=plot(t,final,'linewidth',2,'Color',clr(n,:));
end
hold off;
legend(h, num2str((1:N)','harmonic-%d'))
function f=g(x)
z=2*floor(x/2);
x1=x-z;
y1=(1).*(0<=x1 & x1<=0.8);
y2=(-1).*(0.8<x1 & x1<2);
f=y1+y2;

 Réponse acceptée

Matz Johansson Bergström
Matz Johansson Bergström le 24 Août 2014
Modifié(e) : Matz Johansson Bergström le 24 Août 2014
To solve this problem you can use a anonymous function handle to a function with x and n as arguments.
f = @(x) myfun(x, n);
So you define your function with 'function myfun(x,n)' and pass the handle f to quadgk like
quadgk(f,0,T)
Please note that I don't use @ here.

6 commentaires

function f=g(x)
z=2*floor(x/2);
x1=x-z;
y1=(1).*(0<=x1 & x1<=0.8);
y2=(-1).*(0.8<x1 & x1<2);
f=y1+y2;
t=linspace(0,12,1000);
T=2;
Y=g(t)';
f=0;
N=50;
hold on;
h=zeros(N,1);
clr=lines(N);
for n=1:25:N;
line(t,Y,'linewidth',5)
grid on;hold on;
A=(1/T)*quadgk(@g,0,T);
P=@(t)g(t).*(cos(n*pi*t));
Q=@(t)g(t).*(sin(n*pi*t));
Ax=(2/T)*quadgk(P,0,T);
Bx=(2/T)*quadgk(Q,0,T);
f=f+(Ax*cos(pi*n*t)+Bx*sin(pi*n*t));
final=A+f;
h(n)=plot(t,final,'linewidth',2,'Color',clr(n,:));
end
hold off;
legend(h, num2str((1:N)','harmonic-%d'))
Thank You Sir. It worked and I think the plots are correct.
Please check once and correct me if needed. Thank You once again. :)
Matz Johansson Bergström
Matz Johansson Bergström le 24 Août 2014
This is not the final code, right?
Nabhdeep Bansal
Nabhdeep Bansal le 24 Août 2014
This is the final code, Sir. What happened?
Matz Johansson Bergström
Matz Johansson Bergström le 24 Août 2014
Your code works, but I thought you wanted to pass n as an extra argument to the function, which is nicer, but if you are happy it's ok with me ;-)
Nabhdeep Bansal
Nabhdeep Bansal le 24 Août 2014
What you are saying is true Sir. The output curves are incorrect. How to pass n as an extra argument?
Matz Johansson Bergström
Matz Johansson Bergström le 24 Août 2014
You want to calculate the quadrature of f(t)*cos(pi*n*t), how do you define f?

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