Subscripted assignment dimension mismatch.

Hi everyone. I'm new to MATLAB. What i am trying to do with this code is to plot the sum of the n values vs t that has 0.005 time step but i couldnt figure it out and whatever i try i get an subscripted assignment dimension match in l,ne 26(surface(2,k)=n+surface(2,k);). Also i want to see all the results for each variable i created, anybody has suggestions for that too? Here is my code;

    clear
clc
d=0.4;
A=0.15;
B=1.6;
fo=0.0005;
e=((2*pi)*rand(1,50)); %epsilon
count=1;
for i=1:50;
    for t=0:0.005:300;
    f=(1.1^(i-1))*fo; %fi
    T=1/f; %ti
    k=2*pi/(dispersion(T,d)); %ki
    h=0.5*(1+((2*k*d)/sinh(2*k*d))); %ni
    ph=(1/(2*h))*(tanh(k*d))^2; %fi
    a=abs((0.2*f*(0.205*(A^2)*(B^-4)*(f^-5)*exp(-0.75*(B*(f^-4))*ph)))^0.5); %ai
    Ur(count)=2*a*(dispersion(T,d))^2/d^3; %Ursell Number
    n(count+1)=a*cos((2*pi*f*t)+e(i)); %n(t)
    count=count+1;
    end
end
for k=1:3000050;
    surface(2,k)=0;
    for i=1:50;
        surface(1,k)=t;
        surface(2,k)=n+surface(2,k);
    end
end
plot(surface(1,:),surface(2,:))

[madhan ravi]

What’s dispersion?

Answer 1

[Walter Roberson]

You assign to n(count+1) so n will be a vector. surface(2,k) is a scalar. When you add vector n to scalar you get a vector result, which cannot be stored back into a scalar location.

I notice that the maximum location stored at for n is exactly one larger than the upper limit of your for k loop. That suggests to me that you want to add n(k+1) instead of the entire n