Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: how do the integration?
Replies: 2   Last Post: May 18, 2013 7:57 PM

 Messages: [ Previous | Next ]
 ghasem Posts: 118 Registered: 4/13/13
Re: how do the integration?
Posted: May 18, 2013 7:57 PM

==========
sorry,f1 is:
f1(z,theta) = I(z)*sin(theta)*exp(jk*z*cos(theta))
==========
> Then you say that z is a vector
> z= linspace(0,0.7,100);
> then later say that
> I(z) = is a given vector same length with
> z => length(I(z)) = length(z)
> is I(z) a function of z?
> If so, what is the definition of this function I(z) ?

==========
yes,I(z) was a function from z (but no explicity).you suppose I have a vector from z and
I(z) that are same size.
in fact I note that I(zi) & k(zi) are correspond with zi , where i=1,2,3,...,length(z);

I have to plot f2 versus theta in polar diagram.
in fact my f2 is as following (I decompose it):

f2(theta) = integral[f1(z1,theta),dz,0,z1]+integral[f1(z2,theta),dz,z1,z2]+...
integral[f1(z3,theta),dz,z2,z3]+...;
where:
f1(z1,theta)=I(z1)*sin(theta)*exp(jk(z1)*z1*cos(theta))
f1(z2,theta)= I(z2)*sin(theta)*exp(jk(z2)*z2*cos(theta))
...
where:
zi = i*delta_z = i*[0.7/length(z)] i=1,2,3,...,lenght(z)
I(zi) = ith element in vector of value I(z) ; i=1,2,3,...,lenght(z)
k(zi) = ith element in vector of value k ; i=1,2,3,...,lenght(z)
I have all of them.
now,I want to calculate and next plot f2(theta) versus theta.
how integrate from f2(theta) relation and next with theta=linspace(0,2*pi,360) plot it?
for example you suppose we have:
m=100;
z=linspace(0,0.7,m);
real_k = linspace(10,100,m);
imag_k = linspace(1,5,m);
k = real(k)+1j*imag(k);
I = linspace(12,120,m); % I= I(z)
theta = linspace(0,2*pi,360);
NOW,how integrate from f2 relation?

Date Subject Author
5/18/13 ghasem
5/18/13 Nasser Abbasi
5/18/13 ghasem