for loops and matrices in matlab - matlab

Question

I am trying to link two equations, where I use a for loop calculate the value of k from the range of frequencies (eg. 1-5 Hz) then use each of the k values and substitute the values of k into a 6x6 matrix. Can anyone help show me how to create a matrix for each value of k in Matlab?
1st Equation
for f = 1:5; % Range of Frequencies (Hz)
f;
w = 2.*pi.*f; % Angular Frequency (Hz)
p = 8050;% Density of Mild Steel(kg/m^3)
v = 0.30; % Poissons Ratio of Mild Steel
R = 0.02; % Radius of Pipe (m)
E = 210*10^9; % Youngs Modulus of Mild Steel (pa)
a = (w.^2).*p;
b = (p.*(1-(v.^2)).*(R.^2).*(w.^2)-E);
c = (p.*(R.^2).*(w.^2)-E).*E;
**k(f) = sqrt((a.*b)/c); % k = Wave Number**
end
2nd Equation (6x6 Matrix)
k =
L1=0.1;
L2=0.6;
L3=0.6;
D= [0,0,exp(-k*L1),exp(-k*L2),0,0; exp(-k*L1),1,exp(-k*L1),exp(-k*L2),0,0; -k*exp(-k*L1),k,k*exp(-k*L1),-k*exp(-k*L2),0,0;0,0,exp(-k*(L1+L2)),k,-exp(-k*(L1+L2)),-exp(-k*L3);0,0,-k*exp(-k*(L1+L2)),1,k*exp(-k*(L1+L2)),k*exp(-k*L3);0,0,exp(-k*(L1+L2)),1,0,0]

Answer 1

(Answer modified following the comment)
You can try definig the output array D as a 3 dimensions array.
D=NaN(6,6,length(k))
for i=1:length(k)
D(:,:,i)= [0,0,exp(-k(i)*L1),exp(-k(i)*L2),0,0; exp(-k(i)*L1),1,exp(-k(i)*L1),exp(-k(i)*L2),0,0; -k(i)*exp(-k(i)*L1),k(i),k(i)*exp(-k(i)*L1),-k(i)*exp(-k(i)*L2),0,0;0,0,exp(-k(i)*(L1+L2)),k(i),-exp(-k(i)*(L1+L2)),-exp(-k(i)*L3);0,0,-k(i)*exp(-k(i)*(L1+L2)),1,k(i)*exp(-k(i)*(L1+L2)),k(i)*exp(-k(i)*L3);0,0,exp(-k(i)*(L1+L2)),1,0,0]
end
The D array will be:
D(:,:,1) =
0 0 0.9999 0.9993 0 0
0.9999 1.0000 0.9999 0.9993 0 0
-0.0012 0.0012 0.0012 -0.0012 0 0
0 0 0.9991 0.0012 -0.9991 -0.9993
0 0 -0.0012 1.0000 0.0012 0.0012
0 0 0.9991 1.0000 0 0
D(:,:,2) =
0 0 0.9998 0.9985 0 0
0.9998 1.0000 0.9998 0.9985 0 0
-0.0025 0.0025 0.0025 -0.0025 0 0
0 0 0.9983 0.0025 -0.9983 -0.9985
0 0 -0.0025 1.0000 0.0025 0.0025
0 0 0.9983 1.0000 0 0
D(:,:,3) =
0 0 0.9996 0.9978 0 0
0.9996 1.0000 0.9996 0.9978 0 0
-0.0037 0.0037 0.0037 -0.0037 0 0
0 0 0.9974 0.0037 -0.9974 -0.9978
0 0 -0.0037 1.0000 0.0037 0.0037
0 0 0.9974 1.0000 0 0
D(:,:,4) =
0 0 0.9995 0.9971 0 0
0.9995 1.0000 0.9995 0.9971 0 0
-0.0049 0.0049 0.0049 -0.0049 0 0
0 0 0.9966 0.0049 -0.9966 -0.9971
0 0 -0.0049 1.0000 0.0049 0.0049
0 0 0.9966 1.0000 0 0
D(:,:,5) =
0 0 0.9994 0.9963 0 0
0.9994 1.0000 0.9994 0.9963 0 0
-0.0061 0.0062 0.0061 -0.0061 0 0
0 0 0.9957 0.0062 -0.9957 -0.9963
0 0 -0.0061 1.0000 0.0061 0.0061
0 0 0.9957 1.0000 0 0
Hope this helps.

Answer 2

I will try to answer your question...
Firstly, you don't need the initial for loop. You can calculate all values of wavelength without loops.
f = 1:5; % Range of Frequencies (Hz)
w = 2.*pi.*f; % Angular Frequency (Hz)
p = 8050;% Density of Mild Steel(kg/m^3)
v = 0.30; % Poissons Ratio of Mild Steel
R = 0.02; % Radius of Pipe (m)
E = 210*10^9; % Youngs Modulus of Mild Steel (pa)
a = (w.^2).*p;
b = (p.*(1-(v.^2)).*(R.^2).*(w.^2)-E);
c = (p.*(R.^2).*(w.^2)-E).*E;
k = sqrt((a.*b)./c); % k = Wave Number
L1=0.1;
L2=0.6;
L3=0.6;
Then you can use a for loop to calculate the D matrix and a cell array to store the results:
results = cell(1, length(k));
for i = 1:length(k)
results{i} = [
0,0,exp(-k(i)*L1),exp(-k(i)*L2),0,0;...
exp(-k(i)*L1),1,exp(-k(i)*L1),exp(-k(i)*L2),0,0;...
-k(i)*exp(-k(i)*L1),k(i),k(i)*exp(-k(i)*L1),-k(i)*exp(-k(i)*L2),0,0;...
0,0,exp(-k(i)*(L1+L2)),k(i),-exp(-k(i)*(L1+L2)),-exp(-k(i)*L3);...
0,0,-k(i)*exp(-k(i)*(L1+L2)),1,k(i)*exp(-k(i)*(L1+L2)),k(i)*exp(-k(i)*L3);...
0,0,exp(-k(i)*(L1+L2)),1,0,0
]
end
P.S. Without trying to get on to you, I think that what you just want is to someone solve an excercise for you. If this is not the case, ignore my p.s.