two variables one equation 3D plot

Dear,

I have two input vector matrices as variables alf and gam:

alf = [0, 0.03, 0.1, 0.5, 1, 1.5];

gam = [0, 0.03, 0.1, 0.5, 1, 1.5];

Then I have some complicated steps resulting in one value for each possible combination of the two variables alf and gam.

For example alf=0.03 and gam=1 gives one answer. And alf=0.5 and gam=0 gives an other answer etc.

How can I plot this as a 3D plot with a surface with alf and gam as X and Y axes and the outcome Z axis.

if I try:

[gamM, alfM]=meshgrid(gam,alf);

surf(gamM,alfM,log10(T))

it gives me the error saying:

Error using surf (line 71)

Z must be a matrix, not a scalar or vector.

kind regards Nickel

[Dyuman Joshi]

You might be storing T as row/column vector. Change it into a matrix.

Since you mentioned that you get a unique output for each gam and alf, then your T should have 36 elements. Convert it into a 6x6 matrix and try again.

[Nickel Blankevoort]

Dear Dyuman Joshi,

That is correct I get 36 individual elements for T.

But how do I convert that into a 6x6 matrix?

I run two For loops simultaneously for alf and gam, calculating T.

Kind regards Nickel

[Dyuman Joshi]

for i=1:numel(alf)
    for j=1:numel(gam)
        T(i,j)= alf(i)*gam(j) %this is just an exmample, use your existing formula
    end
end

[Nickel Blankevoort]

clc;clear

eL=0;                           
eR=0;                           
gammaL=-2;                      
gammaR=-2;                    
aL=-0.1;                       
aR=-0.1;                       

E=linspace(-0,0.0001,1);    
kL=acos((eL-E)/(2*gammaL
kR=acos((eR-E)/(2*gammaR));    
h=6.582119514*10^-16;


N = 3;
for    alf = [0, 0.03, 0.1, 0.5, 1, 1.5];
for    gam = [0, 0.03, 0.1, 0.5, 1, 1.5];
    
H = diag(gam.*ones(1,N-1),-1);

if N>2
H = diag(gam.*ones(1,N-1),-1)+diag(alf.*ones(1,N-2),-2);
else
H = diag(gam.*ones(1,N-1),-1);
end

H = H+H';

a=1;
b=N;
n=1;

    g=(((E(n))*eye(length(H))-H))^-1;
    gaa=g(a,a);
    gbb=g(b,b);
    gab=g(a,b);
    gba=g(b,a);
    vL=(2*gammaL*sin(kL(n)))/h;
    vR=(2*gammaR*sin(kR(n)))/h;
    gL=(exp(i*kL(n)))/(-gammaL);
    gR=(exp(i*kR(n)))/(-gammaR);
    sL=aL.^2*gL;
    sR=aR.^2*gR;
    d=1+sL*gaa+sR*gbb+sR*sL*(gaa*gbb-gab*gba);
    t=i*h*sqrt(vL)*aL*gL*(gba/d)*gR*aR*sqrt(vR);
    T(:,n)=[abs(t^2)];      %I would like to store the value of this T for every iteration
 
end
end

[Nickel Blankevoort]

Dear Dyuman,

thank you for the quick responses. I have included my code now for more clearity.

I try to store the values for T for every iteration.

Nickel

[Dyuman Joshi]

I can't see your code? Or did you mean, that you have included what I wrote in your code?

[Nickel Blankevoort]

Can you see my code now in the attachement?

Nickel

Answer 1

[Dyuman Joshi]

I have modified your code -

clc;clear
eL=0;                           
eR=0;                           
gammaL=-2;                      
gammaR=-2;                    
aL=-0.1;                       
aR=-0.1;                       
E=linspace(-0,0.0001,1);    
kL=acos((eL-E)/(2*gammaL));
kR=acos((eR-E)/(2*gammaR));    
h=6.582119514*10^-16;
N = 3;
alf = [0, 0.03, 0.1, 0.5, 1, 1.5];
gam = [0, 0.03, 0.1, 0.5, 1, 1.5];

for    i=1:6
for    j=1:6
    
H = diag(gam(i).*ones(1,N-1),-1);
if N>2
H = diag(gam(i).*ones(1,N-1),-1)+diag(alf(i).*ones(1,N-2),-2);
else
H = diag(gam(i).*ones(1,N-1),-1);
end
H = H+H';
a=1;
b=N;
n=1;
    g=(((E(n))*eye(length(H))-H))^-1;
    gaa=g(a,a);
    gbb=g(b,b);
    gab=g(a,b);
    gba=g(b,a);
    vL=(2*gammaL*sin(kL(n)))/h;
    vR=(2*gammaR*sin(kR(n)))/h;
    gL=(exp(i*kL(n)))/(-gammaL);
    gR=(exp(i*kR(n)))/(-gammaR);
    sL=aL.^2*gL;
    sR=aR.^2*gR;
    d=1+sL*gaa+sR*gbb+sR*sL*(gaa*gbb-gab*gba);
    t=i*h*sqrt(vL)*aL*gL*(gba/d)*gR*aR*sqrt(vR);
    T(i,j)=[abs(t^2)];      %I would like to store the value of this T for every iteration
end
end
[gamM, alfM]=meshgrid(alf,alf);
surf(gamM,alfM,log10(T))

[Nickel Blankevoort]

thanks a lot for the help, much appriciated!

[Nickel Blankevoort]

I didn't get the graphs I was expecting to get. Turned out you used i for the for loop too hahah, and in my equations I use i as imaginary... took me a couple of days to figure that out:)

But all happy now!

kind regards and big thanks again