Problem converting Matlab function to a Simulink function - matlab

I need to convert this function, which implements the gradient descent algorithm:
function [xopt, fopt, niteration, gnorm, dx] = grad_descent2(varargin)
if nargin == 0
x0 = [3 3]';
elseif nargin == 1
x0 = varargin{1};
end
tolerance = 1e-6;
maxiteration = 1000;
dxmin = 1e-6;
alpha = 0.01;
gnorm = inf;
x = x0;
niteration = 0;
dx = inf;
f = #(x1, x2) x1.^2 + 3*x2.^2;
figure(1); clf; fcontour(f, [-5 5 -5 5]); axis equal;hold on
f2 = #(x) f(x(1), x(2));
while and(gnorm >= tolerance, and(niteration <=maxiteration, dx >= dxmin))
g = grad(x);
gnorm = norm(g);
xnew = x - alpha*g;
plot([x(1) xnew(1)], [x(2) xnew(2)], 'ko-')
refresh
niteration = niteration + 1;
dx = norm(xnew - x);
x = xnew;
end
xopt = x;
fopt = f2(xopt);
niteration = niteration - 1;
end
function g = grad(X)
g = [2*X(1)
2*X(2)];
end
My X to minimize is the value to do of a model of a pmsm motor. At the moment, I write this script:
function[x1opt,x2opt] = grad_descent2(isdpred,isqerr,isdmiss,isqmiss)
x1=isdpred;
x2=isqerr;
x0=[isdmiss,isqmiss];
tolerance = 1e-6;
maxiteration = 1000;
dxmin = 1e-6;
alpha = 0.01;
gnorm = inf;
x = x0;
niteration = 0;
dx = inf;
f = x1.^2+x2.^2;
figure(1); clf; fcontour(f, [-5 5 -5 5]); axis equal;hold on
while and(gnorm >= tolerance, and(niteration <=maxiteration, dx >= dxmin))
g = grad(x);
gnorm = norm(g);
xnew = x - alpha*g;
niteration = niteration + 1;
dx = norm(xnew - x);
x = xnew;
end
x1opt=x(1);
x2opt=x(2);
niteration = niteration - 1;
end
function g = grad(X)
g = [2*X(1)
2*X(2)];
end
But Simulink reports this error:
This function does not fully set the dimensions of output port 2

Related

Steepest Descent using Armijo rule

I want to determine the Steepest descent of the Rosenbruck function using Armijo steplength where x = [-1.2, 1]' (the initial column vector).
The problem is, that the code has been running for a long time. I think there will be an infinite loop created here. But I could not understand where the problem was.
Could anyone help me?
n=input('enter the number of variables n ');
% Armijo stepsize rule parameters
x = [-1.2 1]';
s = 10;
m = 0;
sigma = .1;
beta = .5;
obj=func(x);
g=grad(x);
k_max = 10^5;
k=0; % k = # iterations
nf=1; % nf = # function eval.
x_new = zeros([],1) ; % empty vector which can be filled if length is not known ;
[X,Y]=meshgrid(-2:0.5:2);
fx = 100*(X.^2 - Y).^2 + (X-1).^2;
contour(X, Y, fx, 20)
while (norm(g)>10^(-3)) && (k<k_max)
d = -g./abs(g); % steepest descent direction
s = 1;
newobj = func(x + beta.^m*s*d);
m = m+1;
if obj > newobj - (sigma*beta.^m*s*g'*d)
t = beta^m *s;
x = x + t*d;
m_new = m;
newobj = func(x + t*d);
nf = nf+1;
else
m = m+1;
end
obj=newobj;
g=grad(x);
k = k + 1;
x_new = [x_new, x];
end
% Output x and k
x_new, k, nf
fprintf('Optimal Solution x = [%f, %f]\n', x(1), x(2))
plot(x_new)
function y = func(x)
y = 100*(x(1)^2 - x(2))^2 + (x(1)-1)^2;
end
function y = grad(x)
y(1) = 100*(2*(x(1)^2-x(2))*2*x(1)) + 2*(x(1)-1);
end

Undefined function or variable 'x'

I am trying to evaluate two matrixes which I defined outside of the function MetNewtonSist using subs and I get the error Undefined function or variable 'x' whenever I try to run the code.
[edit] I added the code for the GaussPivTot function which determines the solution of a liniear system.
syms x y
f1 = x^2 + y^2 -4;
f2 = (x^2)/8 - y;
J = jacobian( [ f1, f2 ], [x, y]);
F = [f1; f2];
subs(J, {x,y}, {1, 1})
eps = 10^(-6);
[ x_aprox,y_aprox, N ] = MetNewtonSist( F, J, 1, 1, eps )
function [x_aprox, y_aprox, N] = MetNewtonSist(F, J, x0, y0, eps)
k = 1;
x_v(1) = x0;
y_v(1) = y0;
while true
k = k + 1;
z = GaussPivTot(subs(J, {x, y}, {x_v(k-1), y_v(k-1)}),-subs(F,{x, y}, {x_v(k-1), y_v(k-1)}));
x_v(k) = z(1) + x_v(k-1);
y_v(k) = z(1) + y_v(k-1);
if norm(z)/norm([x_v(k-1), y_v(k-1)]) < eps
return
end
end
N = k;
x_aprox = x_v(k);
y_aprox = y_v(k);
end
function [x] = GaussPivTot(A,b)
n = length(b);
A = [A,b];
index = 1:n;
for k = 1:n-1
max = 0;
for i = k:n
for j = k:n
if A(i,j) > max
max = A(i,j);
p = i;
m = j;
end
end
end
if A(p,m) == 0
disp('Sist. incomp. sau comp. nedet.')
return;
end
if p ~= k
aux_line = A(p,:);
A(p,:) = A(k, :);
A(k,:) = aux_line;
end
if m ~= k
aux_col = A(:,m);
A(:,m) = A(:,k);
A(:,k) = aux_col;
aux_index = index(m);
index(m) = index(k);
index(k) = aux_index;
end
for l = k+1:n
M(l,k) = A(l,k)/A(k,k);
aux_line = A(l,:);
A(l,:) = aux_line - M(l,k)*A(k,:);
end
end
if A(n,n) == 0
disp('Sist. incomp. sau comp. nedet.')
return;
end
y = SubsDesc(A, A(:,n+1));
for i = 1:n
x(index(i)) = y(i);
end
end
By default, eps is defined as 2.2204e-16 in MATLAB. So do not overwrite it with your variable and name it any word else.
epsilon = 1e-6;
Coming to your actual issue, pass x and y as input arguments to the MetNewtonSist function. i.e. define MetNewtonSist as:
function [x_aprox, y_aprox, N] = MetNewtonSist(F, J, x0, y0, epsilon, x, y)
%added x and y and renamed eps to epsilon
and then call it with:
[x_aprox, y_aprox, N] = MetNewtonSist(F, J, 1, 1, epsilon, x, y);

Plot equally spaced markers along a spiral

I want to move a red star marker along the spiral trajectory with an equal distance of 5 units between the red star points on its circumference like in the below image.
vertspacing = 10;
horzspacing = 10;
thetamax = 10*pi;
% Calculation of (x,y) - underlying archimedean spiral.
b = vertspacing/2/pi;
theta = 0:0.01:thetamax;
x = b*theta.*cos(theta)+50;
y = b*theta.*sin(theta)+50;
% Calculation of equidistant (xi,yi) points on spiral.
smax = 0.5*b*thetamax.*thetamax;
s = 0:horzspacing:smax;
thetai = sqrt(2*s/b);
xi = b*thetai.*cos(thetai);
yi = b*thetai.*sin(thetai);
plot(x,y,'b-');
hold on
I want to get a figure that looks like the following:
This is my code for the circle trajectory:
% Initialization steps.
format long g;
format compact;
fontSize = 20;
r1 = 50;
r2 = 35;
r3= 20;
xc = 50;
yc = 50;
% Since arclength = radius * (angle in radians),
% (angle in radians) = arclength / radius = 5 / radius.
deltaAngle1 = 5 / r1;
deltaAngle2 = 5 / r2;
deltaAngle3 = 5 / r3;
theta1 = 0 : deltaAngle1 : (2 * pi);
theta2 = 0 : deltaAngle2 : (2 * pi);
theta3 = 0 : deltaAngle3 : (2 * pi);
x1 = r1*cos(theta1) + xc;
y1 = r1*sin(theta1) + yc;
x2 = r2*cos(theta2) + xc;
y2 = r2*sin(theta2) + yc;
x3 = r3*cos(theta3) + xc;
y3 = r3*sin(theta3) + yc;
plot(x1,y1,'color',[1 0.5 0])
hold on
plot(x2,y2,'color',[1 0.5 0])
hold on
plot(x3,y3,'color',[1 0.5 0])
hold on
% Connecting Line:
plot([70 100], [50 50],'color',[1 0.5 0])
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0, 1, 1]);
drawnow;
axis square;
for i = 1 : length(theta1)
plot(x1(i),y1(i),'r*')
pause(0.1)
end
for i = 1 : length(theta2)
plot(x2(i),y2(i),'r*')
pause(0.1)
end
for i = 1 : length(theta3)
plot(x3(i),y3(i),'r*')
pause(0.1)
end
I can't think of a way to compute distance along a spiral, so I'm approximating it with circles, in hopes that it will still be useful.
My solution relies on the InterX function from FEX, to find the intersection of circles with the spiral. I am providing an animation so it is easier to understand.
The code (tested on R2017a):
function [x,y,xi,yi] = q44916610(doPlot)
%% Input handling:
if nargin < 1 || isempty(doPlot)
doPlot = false;
end
%% Initialization:
origin = [50,50];
vertspacing = 10;
thetamax = 5*(2*pi);
%% Calculation of (x,y) - underlying archimedean spiral.
b = vertspacing/(2*pi);
theta = 0:0.01:thetamax;
x = b*theta.*cos(theta) + origin(1);
y = b*theta.*sin(theta) + origin(2);
%% Calculation of equidistant (xi,yi) points on spiral.
DST = 5; cRes = 360;
numPts = ceil(vertspacing*thetamax); % Preallocation
[xi,yi] = deal(NaN(numPts,1));
if doPlot && isHG2() % Plots are only enabled if the MATLAB version is new enough.
figure(); plot(x,y,'b-'); hold on; axis equal; grid on; grid minor;
hAx = gca; hAx.XLim = [-5 105]; hAx.YLim = [-5 105];
hP = plot(xi,yi,'r*');
else
hP = struct('XData',xi,'YData',yi);
end
hP.XData(1) = origin(1); hP.YData(1) = origin(2);
for ind = 2:numPts
P = InterX([x;y], makeCircle([hP.XData(ind-1),hP.YData(ind-1)],DST/2,cRes));
[~,I] = max(abs(P(1,:)-origin(1)+1i*(P(2,:)-origin(2))));
if doPlot, pause(0.1); end
hP.XData(ind) = P(1,I); hP.YData(ind) = P(2,I);
if doPlot, pause(0.1); delete(hAx.Children(1)); end
end
xi = hP.XData(~isnan(hP.XData)); yi = hP.YData(~isnan(hP.YData));
%% Nested function(s):
function [XY] = makeCircle(cnt, R, nPts)
P = (cnt(1)+1i*cnt(2))+R*exp(linspace(0,1,nPts)*pi*2i);
if doPlot, plot(P,'Color',lines(1)); end
XY = [real(P); imag(P)];
end
end
%% Local function(s):
function tf = isHG2()
try
tf = ~verLessThan('MATLAB', '8.4');
catch
tf = false;
end
end
function P = InterX(L1,varargin)
% DOCUMENTATION REMOVED. For a full version go to:
% https://www.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections
narginchk(1,2);
if nargin == 1
L2 = L1; hF = #lt; %...Avoid the inclusion of common points
else
L2 = varargin{1}; hF = #le;
end
%...Preliminary stuff
x1 = L1(1,:)'; x2 = L2(1,:);
y1 = L1(2,:)'; y2 = L2(2,:);
dx1 = diff(x1); dy1 = diff(y1);
dx2 = diff(x2); dy2 = diff(y2);
%...Determine 'signed distances'
S1 = dx1.*y1(1:end-1) - dy1.*x1(1:end-1);
S2 = dx2.*y2(1:end-1) - dy2.*x2(1:end-1);
C1 = feval(hF,D(bsxfun(#times,dx1,y2)-bsxfun(#times,dy1,x2),S1),0);
C2 = feval(hF,D((bsxfun(#times,y1,dx2)-bsxfun(#times,x1,dy2))',S2'),0)';
%...Obtain the segments where an intersection is expected
[i,j] = find(C1 & C2);
if isempty(i), P = zeros(2,0); return; end
%...Transpose and prepare for output
i=i'; dx2=dx2'; dy2=dy2'; S2 = S2';
L = dy2(j).*dx1(i) - dy1(i).*dx2(j);
i = i(L~=0); j=j(L~=0); L=L(L~=0); %...Avoid divisions by 0
%...Solve system of eqs to get the common points
P = unique([dx2(j).*S1(i) - dx1(i).*S2(j), ...
dy2(j).*S1(i) - dy1(i).*S2(j)]./[L L],'rows')';
function u = D(x,y)
u = bsxfun(#minus,x(:,1:end-1),y).*bsxfun(#minus,x(:,2:end),y);
end
end
Result:
Note that in the animation above, the diameter of the circle (and hence the distance between the red points) is 10 and not 5.

EDIT: Data seems to be unsorted while plotting

Click here to see the figure. When I run and plot X and solution, it seems like the data is not sorted. The previous error regarding subscripted dimension mismatch is gone now. Below is my main code:
clear all
y0 = 20000; % initial conditions
iter = 0;
years = 5;
solution = [];
X = [];
for p = 1:1:years
iter = iter+1;
display(iter)
ll = 273;
ul = 273 + 91;
wl = (ul-ll).*rand(1,1) + ll;
yearlength = wl+0.1;
finaltime = p*yearlength;
t = 0:finaltime;
mugen;
Kgen;
d1gen;
n = 2;
deltat = 1;
tspan = 0:deltat:finaltime;
options = odeset('RelTol',1e-10,'AbsTol',1e-10);
sol = ode23s(#(t,y)para_1d(t,y,n,mug,Kg,d1g),tspan,y0,options);
X = [X sol.x];
Y = (sol.y)';
ny = length(Y);
Ytrans = Y';
solution = [solution Ytrans(1,:)] ;
%display(Ytrans(1,end))
clearvars -except solution solution1 iter X
y0 = solution(end,end);
display(y0)
end
plot(x, solution)
This is the function where my ode is:
function dy = para_1d(t, y,n, mug, Kg, d1g)
count = ceil(t)+1;
dy(1,1) = (mug(count).*(y(1).^n)/(Kg(count).^n+y(1).^n)) - d1g(count).*y(1);
and the parameter files are:
x1 = [0 91.25 91.26 182.5 182.51 273.75 273.76 wl];
clear x
for n = 1:p;
x(n,:) = (n-1)*(wl+0.1) + x1;
end
counter = 0;
for j=1:p
for i=1:8
counter = counter+1;
xnew(counter) = x(j,i);
end
end
y1 = [500 500 1500 1500 500 500 0 0];
clear y
for n = 1:p;
y(n,:) = y1;
end
counter = 0;
for j = 1:p
for i = 1:8
counter = counter+1;
ynew(counter) = y(j,i);
end
end
w = y(1,:);
for l = 2:p
w = [w y(l,:)];
end
v = x(1,:);
for q = 2:p
v = [v x(q,:)];
end
mug = pchip(v,w,t);
The parameter K is:
x1 = [0 91.25 91.26 182.5 182.51 273.75 273.76 wl];
clear x
for n = 1:p;
x(n,:) = (n-1)*(wl + 0.1) + x1;
end
counter = 0;
for j = 1:p
for i = 1:8
counter = counter + 1;
xnew(counter) = x(j,i);
end
end
%y1 = [8000 8000 27000 27000 8000 8000 6000 6000 ];
%y1 = [8000 8000 12000 12000 8000 8000 6000 6000 ];
y1 = [6000 6000 8000 8000 6000 6000 6000 6000 ];
clear y
for n = 1:p;
y(n,:) = y1;
end
counter = 0;
for j = 1:p
for i = 1:8
counter = counter+1;
ynew(counter) = y(j,i);
end
end
w = y(1,:);
for l = 2:p
w = [w y(l,:)];
end
v = x(1,:);
for q = 2:p
v = [v x(q,:)];
end
Kg = pchip(v,w,t);
and the last parameter is:
x1 = [0 91.25 91.26 182.5 182.51 273.75 273.76 wl];
clear x
for n=1:p;
x(n,:) = (n-1)*(wl + 0.1) + x1;
end
counter=0;
for j=1:p
for i=1:8
counter=counter+1;
xnew(counter) = x(j,i);
end
end
y1 = [ 0.02272 0.02272 0.04 0.04 0.02272 0.02272 0.005263 0.005263 ];
clear y
for n=1:p;
y(n,:) = y1;
end
counter=0;
for j=1:p
for i=1:8
counter=counter+1;
ynew(counter) = y(j,i);
end
end
w=y(1,:);
for l=2:p
w=[w y(l,:)];
end
v=x(1,:);
for q=2:p
v=[v x(q,:)];
end
d1g = pchip(v,w,t);
You can simply look at the main code and suggest me where I am doing mistake. The parameter codes and ode file is only for those who want to run the code for a clearer view. Thanks a lot for your time!

How do I plot the output of a system with an impulse response in matlab?

I am very new to matlab and need to plot y1[n] = x[n] + y1[n − 1] where x[n] = [1,2,4] and an impulse response, h[n] = [1,1,1,1,1] and am not sure if I have went about it the right way
My code so far is
x = [1,2,4];
h = [1,1,1,1,1];
y = [];
for n=1:length(x)
if (n==1)
y(n) = x(n);
else
y(n) = (x(n)*h(n)) + (y(n-1)*h(n));
end
end
stem(y);
Please note that I cannot use the conv() function
I don't really know why it got so complicated,
x = [1,2,4];
h = [1,1,1,1,1];
y = [];
lh = length(h);
lx = length(x);
t = -lh - lx : lh + lx;
x(end + 1 : end + lh + 1) = 0;
h(end + 1 : end + lx + 1) = 0;
x = padarray(x,[0 max(t)],'pre');
h = padarray(h,[0 max(t)],'pre');
xinv = x(end:-1:1);
for n = 1 : length(t)
xinv = circshift(xinv,[0 1]);
y(n) = sum(xinv .* h);
end
y = circshift(y,[0 find(t == 0)]);
subplot(311)
stem(t,x);
xlim([-10 10])
subplot(312)
stem(t,h);
xlim([-10 10])
subplot(313)
stem(t,y);
xlim([-10 10])
It works fine but I believe it can be coded in a more simpler way.
Can you use fft?
lx = numel(x);
lh = numel(h);
m = max(lx, lh);
y = ifft(fft([h zeros(1,max(lx-lh,0)+m)]) .* fft([x zeros(1,max(lh-lx,0)+m)]));
y = y(1:lx+lh-1);