How to find the maximum value of constant, so that my function doesnt go over a certain value?

Hello!

I am working on a unit step function, and what I want to do is to find the maximum value of a constant that keeps my function Y value as close as 1.3 as possible.

This is the function of the code:

function H = transferH(K)
syms s;
% Declaring transfer function G(s)
G = 1/(s*(s+4));
H = (G*K)/(1+G*K);
end
function y = create_y_step(K)
syms s t
H = transferH(K);
y = ilaplace(H*(1/s)); %Unit step
end 

Let's say im creating the codeline f = create_y_step(K);

This gives me the output:

1 - exp(-2*t)*(cosh(t*(4 - K)^(1/2)) + (2*sinh(t*(4 - K)^(1/2)))/(4 - K)^(1/2))

What I essentially want to do is finding the appropriate/max K value, so that my funtion above, y = create_y_step_(K) is equal to 1.3.

This is the code im trying to work with:

y = create_y_step(K);
y1 = y;
syms K;
y2 = diff(y1,K) == 1.3;
x = solve(y2,K);
X = max(x);
disp(X);


function H = transferH(K)
syms s;
% Declaring transfer function G(s)
G = 1/(s*(s+4));
H = (G*K)/(1+G*K);
end
function y = create_y_step(K)
syms s t
H = transferH(K);
y = ilaplace(H*(1/s)); %Unit step
end 

However the output just becomes

"warning unable to find explicit solution".

I have tried done some manually testing with wolfram and the K value should be around 30-31, but im not getting it correct in matlab sadly.

Would really appreciate any help from here! All answers greatly appreciated!!

[Can Atalay]

I would evaluate the output of the symbolic variable (that represents the time response) for various K values over time=0:0.01:10 (or however long you'd like the step function to reach 1.3) using two for loops within each other and record the discrete time response in a double. Then I would do logicals on the double to pick the best one (like the ones that are closest to 1.3 without going over 1.3). I tried some things below but the response starts at 1.3 and reapproaches 1.3 so I think my symbolic variable that represents the time response is not correct.

syms K;
y = create_y_step(K);
y1 = y;

y2 = diff(y1,K) == 1.3;
x = solve(y2,K);
X = max(x);
disp(X);


K=-500:1:500;
t=0:1:100;
[row1,col1]=size(K);
[row2,col2]=size(t);
all_values=zeros(col1,col2);
for ii1=1:col1
    K_now=K(ii1);
    for ii2=1:col2
        t_now=t(ii2);
        all_values(ii1,ii2) = abs(-exp(-2*t_now)*(sinh(t_now*(4 - K_now)^(1/2))/(4 - K_now)^(3/2) + (t_now*cosh(t_now*(4 - K_now)^(1/2)))/(K_now - 4) - (t_now*sinh(t_now*(4 - K_now)^(1/2)))/(2*(4 - K_now)^(1/2))) - 13/10);
        % incorrect expression above. delete the abs(), look for the proper
        % time response
    end
end

good_ones = [];
indices_good_ones = [];
for ii3=1:col1
    if all_values(ii3,:)<=1.3
        good_ones = [good_ones;  all_values(ii3,:)];
        indices_good_ones=[indices_good_ones; ii3];
    end
end



function H = transferH(K)
syms s;
% Declaring transfer function G(s)
G = 1/(s*(s+4));
H = (G*K)/(1+G*K);
end
function y = create_y_step(K)
syms s t
H = transferH(K);
y = ilaplace(H*(1/s)); %Unit step
end

[MathNemesis]

Thank you so much for your answer! I tried using the code you sent me but Im having a bit of hard time finding the best K value, after running the code, in which variable would the best K value be present? I removed the abs part and all the values are negative in "good ones" and "indicies" run from 0-100. I assume you want me to use the function logicals to retrieve the best K value?

Greatly appreciating your help!

[Paul]

As you showed, y is a function of time, t. So what do you mean by "keeps my function Y value as close as 1.3 as possible"? As close as 1.3 as possible at which time? What is the "Y value"? Do you mean you want the steady state value of y(t) to be 1.3? Maybe you want the peak value to be 1.3?

Answer 1

[Paul]

Based on your statement that the answer should be around 30-31, I'm guessing that you want to find the value of K such that the peak value of the step response is 1.3, You can do this in the way that I think you're attempting, except you should start by solving for the time(s), t, when the derivative of y is equal to 0, because that's the time(s) when the function y is at a max or min (or inflection point). The next step I'll leave to you.

However, there is a much simpler approach for a second order system (unless you're required to use the approach you're trying). Just use the closed form expression for the peak value and in a few steps you can solve for K. Here's one reference that might be helpful.