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;

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Matz Johansson Bergström
Matz Johansson Bergström 2014년 8월 24일
편집: Matz Johansson Bergström 2014년 8월 24일

0 개 추천

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.

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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. :)
This is not the final code, right?
Nabhdeep Bansal
Nabhdeep Bansal 2014년 8월 24일
This is the final code, Sir. What happened?
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 2014년 8월 24일
What you are saying is true Sir. The output curves are incorrect. How to pass n as an extra argument?
You want to calculate the quadrature of f(t)*cos(pi*n*t), how do you define f?

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