How can I plot this state space like the graph I attached by using tf() and step() command? Thank you! I2/E0=1/(s^3+s^2+3*s+1) NOTE:- Matlabsolutions.com provide latest MatLab Homework Help, MatLab Assignment Help , Finance Assignment Help for students, engineers and researchers in Multiple Branches like ECE, EEE, CSE, Mechanical, Civil with 100% output.Matlab Code for B.E, B.Tech,M.E,M.Tech, Ph.D. Scholars with 100% privacy guaranteed. Get MATLAB projects with source code for your learning and research. Try these codes below please; clc; clear; close all; numerator = 1; denominator = [1,1,3,1]; sys = tf(numerator,denominator); yyaxis left SEE COMPLETE ANSWER CLICK THE LINK https://www.matlabsolutions.com/resources/how-to-plot-transfer-functions-in-matlab-.php
I have:
and I want to minimize it on the unit square without using differentials.
I use the following code,
close all; clear; clc; options = optimoptions(@fminunc,'Display','iter','Algorithm','quasi-newton'); xy_guess = [0,0]; [xy_opt,fval] = fminunc(@quadratic,xy_guess,options) function f = quadratic(in) x = in(1); y = in(2); f = -5.*x - 5.*y + 10.*x.^2 + 2.*x.*y f = 1/200.*(-1000.*x - 1000.*y + 400.*x.*y + 1200.*y.^2 + 5.*cos(30.*x) + 4.*cos(80.*x.^2) + 5.*cos(30.*y) + 4.*cos(80.*y^2))
But declaring f twice does not work. How should I declare this double-function as an input for fminunc ?
NOTE:-
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You dont seem to have a double function.
So here is a code to solve your problem. You have local minimas so fmincon and brute force approach gives you different results, since brute force approach is a global optimization.
f = @(x) [x(1) x(2)]*[10 2;2 6]*[x(1);x(2)]-[5 5]*[x(1);x(2)]+(cos(30*x(1))+cos(30*x(2)))/40+(cos(80*x(1)^2)+cos(80*x(2)^2))/50; options = optimoptions(@fmincon); %options = optimoptions(@fmincon,'Display','iter','OptimalityTolerance',1e-12); lb = [0 0];% lower bounds ub = [1 1];% upper bounds x0 = [.5 .5]; % initial guess [xSol,fval,exitflag,output] = fmincon(f,x0,[],[],[],[],lb,ub,[],options); % solve optimization
Feasible point with lower objective function value found, but optimality criteria not satisfied. See output.bestfeasible.. Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
% brute force searching the entire space for min function value x1 = 0:0.001:1; x2 = 0:0.001:1; [X1,X2] = meshgrid(x1,x2);
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