CODEWITHSUNDEEP
loopsmatlabiterationsymbolic-mathfunction-handle
Jonathan Frutschy
Jonathan Frutschy

Reputation: 191

Replicate Summation Operator in MATLAB

I'm trying to model the following time dependent function in MATLAB:

enter image description here

where gamma,beta, and delta are constants. I'm not sure what the best way to approach the summation is. I could make u(t) and y(t) function handles, but symsum can only be used for symbolics. Is my only option to use a for loop to manually add up y(t-i) and u(t-m)? What I have done so far to generate y(t) is shown below:

N = 100;
y_out = zeros(N,1);
u = zeros(N,1);

for t = 1:1:N
u(t) = cos(beta*t);
end

y_out(1) = NARMA(beta,delta,gamma,true);
for t=2:N
    y_out(t) = NARMA(beta,delta,gamma);
end


function y_out = NARMA(beta,delta,gamma,first_call)
    % Set first_call to true only the first time you call the function
    persistent y
    if nargin>3 && first_call
        y = 0;
    end

    for i=1:1:m
    y = y + gamma*u + delta;
    end

    y_out = y;

end

Upvotes: 2

Views: 82

Answers (1)

rahnema1
rahnema1

Reputation: 15867

Here is a possible solution that doesn't require loop (not tested):

t = 1:N;
u = @(t) cos(beta * t);
y = cumsum(gamma * u(t - m) .* u(t) + delta);
y(m+1:end) = y(m+1:end) - y(1:end-m);

EDIT:

It turned out that the above solution is wrong. I think this is correct:

u = @(t) cos(beta*t);
y = zeros(N+m, 1);
for t = 1:N
    y(t+m) = sum(y(t:t+m-1)) + gamma * u(t-m) * u(t) + delta;
end
y = y (m+1:end);

EDIT2:

With further simplifications the summation can be reduced to a simple for loop:

u = @(t) cos(beta*t);
y = zeros(N+m, 1);
e = gamma * u((1:N)-m) .* u(1:N) + delta;
de = diff(e);
y(1+m) = e(1);
for t = 2:N
    y(t+m) = 2 * y(t+m-1) - y(t-1) + de(t-1);
end
y = y (m+1:end);

Upvotes: 2

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