While taking large range error is occur. How to solve this problem?

My code is as below:

f=1:1000:1000000000;
t1=1/0.4521;
t2=1/0.4908;
t3=1/0.61;
t4=1/1;
for i=1:1000:1000000000
    z1(i)=(3.68*10^(-7)*(t1)^(1/2)*(2*3.14*f(i))^0.5);
    z2(i)=(3.68*10^(-7)*(t2)^(1/2)*(2*3.14*f(i))^0.5);
    z3(i)=(3.68*10^(-7)*(t3)^(1/2)*(2*3.14*f(i))^0.5);
    z4(i)=(3.68*10^(-7)*(t4)^(1/2)*(2*3.14*f(i))^0.5);
 end
loglog(f,z1,'r',f,z2,'b',f,z3,'g',f,z4,'k');
xlabel('Frequency in Hz');
ylabel('Impedance in ohm');

Error is as below:

Error in ==> he_impe at 3
f=1:1000000000;

??? Attempted to access f(1e+006); index out of bounds because
numel(f)=1000000.

Error in ==> he_impe at 9
    z1(i)=(3.68*10^(-7)*(t1)^(1/2)*(2*3.14*f(i))^0.5);

[Jan]

"3.68*10^(-7)" is processed much more efficient by "3.68e-7". SQRT is much faster than "^(1/2)".

Answer 1

[Walter Roberson]

Yes, that makes sense. 1:1000:1E9 has 1E9/1000 elements in it, but you try to access those elements at indexes as high as 1E9.

I would suggest to you:

for i=1:length(f)

[Isha Deshpande]

Thanks for reply, my problem solved

Answer 2

[Jan]

You can replace the loop:

z1 = 3.68e-7 * sqrt(t1 * 2 * 3.14 * f);
z2 = 3.68e-7 * sqrt(t2 * 2 * 3.14 * f);
z3 = 3.68e-7 * sqrt(t3 * 2 * 3.14 * f);
z4 = 3.68e-7 * sqrt(t4 * 2 * 3.14 * f);

[Isha Deshpande]

Thanks for reply, my problem solved