Following this question, I am trying to convert the state-space model from this tutorial back into its transfer-function form. I have tried the
R = 2.0; % Ohms
L = 0.5; % Henrys
Km = 0.1; % torque constant
Kb = 0.1; % back emf constant
Kf = 0.2; % Nms
J = 0.02; % kg.m^2/s^2
Td = 1; % models load disturbances
h1 = tf(Km, [L R]); % armature
h2 = tf(1, [J Kf]); % eqn of motion
dcm = ss(h2) * [h1 , Td]; % w = h2 * (h1 * Va + Td)
[b, a] = ss2tf(dcm.A, dcm.B, dcm.C, dcm.D);
dcm_tf = tf(b, a);
However, I get the error message:
IU must be specified for systems with more than one input.
According to this comment on Reddit, given the above system is Multiple-Input-Multiple-Output (MIMO), we need to specify the input index ni from the ss2tf()function. In this case the
[b, a] = ss2tf(dcm.A, dcm.B, dcm.C, dcm.D, 1);
returns the w / Va transfer function and the
[b, a] = ss2tf(dcm.A, dcm.B, dcm.C, dcm.D, 2);
returns the w / Td one.
Alternatively, one could simply use the tf(dcm) to convert the state-space model directly to transfer functions of both inputs.
Related
Find the error as a function of n, where the error is defined as the difference between two the voltage from the Fourier series (vF (t)) and the value from the ideal function (v(t)), normalized to the maximum magnitude (Vm ):
I am given this prompt where Vm = 1 V. Below this line is the code which I have written.
I am trying to write a function to solve this question: Plot the error versus time for n=3,n=5,n=10, and n=50. (10points). What does it look like I am doing incorrectly?
clc;
close all;
clear all;
% define the signal parameters
Vm = 1;
T = 1;
w0 = 2*pi/T;
% define the symbolic variables
syms n t;
% define the signal
v1 = Vm*sin(4*pi*t/T);
v2 = 2*Vm*sin(4*pi*t/T);
% evaluate the fourier series integral
an1 = 2/T*int(v1*cos(n*w0*t),0,T/2) + 2/T*int(v2*cos(n*w0*t),T/2,T);
bn1 = 2/T*int(v1*sin(n*w0*t),0,T/2) + 2/T*int(v2*sin(n*w0*t),T/2,T);
a0 = 1/T*int(v1,0,T/2) + 1/T*int(v2,T/2,T);
% obtain C by substituting n in c[n]
nmax = 100;
n = 1:nmax;
a = subs(an1);
b = subs(bn1);
% define the time vector
ts = 1e-2; % ts is sampling the
t = 0:ts:3*T-ts;
% directly plot the signal x(t)
t1 = 0:ts:T-ts;
v1 = Vm*sin(4*pi*t1/T).*(t1<=T/2);
v2 = 2*Vm*sin(4*pi*t1/T).*(t1>T/2).*(t1<T);
v = v1+v2;
x = repmat(v,1,3);
% Now fourier series reconstruction
N = [3];
for p = 1:length(N)
for i = 1:length(t)
for k = N(p)
x(k,i) = a(k)*cos(k*w0*t(i)) + b(k)*sin(k*w0*t(i));
end
% y(k,i) = a0+sum(x(:,i)); % Add DC term
end
end
z = a0 + sum(x);
figure(1);
plot(t,z);
%Percent error
function [per_error] = percent_error(measured, actual)
per_error = abs(( (measured - actual) ./ 1) * 100);
end
The purpose of the forum is helping with specific technical questions, not doing your homework.
I have always used R, so I am quite new to Matlab and running into some troubleshooting issues. I am running some code for a tensor factorization method (available here: https://github.com/caobokai/tBNE). To start I tried to run the demo code, which generates simulated data to run the method with, which results in the following error(s):
Error using feval
Undefined function or variable 'Sfun'.
Error in OptStiefelGBB (line 199)
[F, G] = feval(fun, X , varargin{:}); out.nfe = 1;
Error in tbne_demo>tBNE_fun (line 124)
S, #Sfun, opts, B, P, X, L, D, W, Y, alpha, beta);
Here is the block of code I am running:
clear
clc
addpath(genpath('./tensor_toolbox'));
addpath(genpath('./FOptM'));
rng(5489, 'twister');
m = 10;
n = 10;
k = 10; % rank for tensor
[X, Z, Y] = tBNE_data(m, n, k); % generate the tensor, guidance and label
[T, W] = tBNE_fun(X, Z, Y, k);
[~, y1] = max(Y, [], 2);
[~, y2] = max(T{3} * W, [], 2);
fprintf('accuracy %3.2e\n', sum(y1 == y2) / n);
function [X, Z, Y] = tBNE_data(m, n, k)
B = randn(m, k);
S = randn(n, k);
A = {B, B, S};
X = ktensor(A);
Z = randn(n, 4);
Y = zeros(n, 2);
l = ceil(n / 2);
Y(1 : l, 1) = 1;
Y(l + 1 : end, 2) = 1;
X = tensor(X);
end
function [T, W] = tBNE_fun(X, Z, Y, k)
% t-BNE computes brain network embedding based on constrained tensor factorization
%
% INPUT
% X: brain networks stacked in a 3-way tensor
% Z: side information
% Y: label information
% k: rank of CP factorization
%
% OUTPUT
% T is the factor tensor containing
% vertex factor matrix B = T{1} and
% subject factor matrix S = T{3}
% W is the weight matrix
%
% Example: see tBNE_demo.m
%
% Reference:
% Bokai Cao, Lifang He, Xiaokai Wei, Mengqi Xing, Philip S. Yu,
% Heide Klumpp and Alex D. Leow. t-BNE: Tensor-based Brain Network Embedding.
% In SDM 2017.
%
% Dependency:
% [1] Matlab tensor toolbox v 2.6
% Brett W. Bader, Tamara G. Kolda and others
% http://www.sandia.gov/~tgkolda/TensorToolbox
% [2] A feasible method for optimization with orthogonality constraints
% Zaiwen Wen and Wotao Yin
% http://www.math.ucla.edu/~wotaoyin/papers/feasible_method_matrix_manifold.html
%% set algorithm parameters
printitn = 10;
maxiter = 200;
fitchangetol = 1e-4;
alpha = 0.1; % weight for guidance
beta = 0.1; % weight for classification loss
gamma = 0.1; % weight for regularization
u = 1e-6;
umax = 1e6;
rho = 1.15;
opts.record = 0;
opts.mxitr = 20;
opts.xtol = 1e-5;
opts.gtol = 1e-5;
opts.ftol = 1e-8;
%% compute statistics
dim = size(X);
normX = norm(X);
numClass = size(Y, 2);
m = dim(1);
n = dim(3);
l = size(Y, 1);
D = [eye(l), zeros(l, n - l)];
L = diag(sum(Z * Z')) - Z * Z';
%% initialization
B = randn(m, k);
P = B;
S = randn(n, k);
S = orth(S);
W = randn(k, numClass);
U = zeros(m, k); % Lagrange multipliers
%% main loop
fit = 0;
for iter = 1 : maxiter
fitold = fit;
% update B
ete = (S' * S) .* (P' * P); % compute E'E
b = 2 * ete + u * eye(k);
c = 2 * mttkrp(X, {B, P, S}, 1) + u * P + U;
B = c / b;
% update P
ftf = (S' * S) .* (B' * B); % compute F'F
b = 2 * ftf + u * eye(k);
c = 2 * mttkrp(X, {B, P, S}, 2) + u * B - U;
P = c / b;
% update U
U = U + u * (P - B);
% update u
u = min(rho * u, umax);
% update S
tic;
[S, out] = OptStiefelGBB(...
S, #Sfun, opts, B, P, X, L, D, W, Y, alpha, beta);
tsolve = toc;
fprintf(...
['[S]: obj val %7.6e, cpu %f, #func eval %d, ', ...
'itr %d, |ST*S-I| %3.2e\n'], ...
out.fval, tsolve, out.nfe, out.itr, norm(S' * S - eye(k), 'fro'));
% update W
H = D * S;
W = (H' * H + gamma * eye(k)) \ H' * Y;
% compute the fit
T = ktensor({B, P, S});
normresidual = sqrt(normX ^ 2 + norm(T) ^ 2 - 2 * innerprod(X, T));
fit = 1 - (normresidual / normX);
fitchange = abs(fitold - fit);
if mod(iter, printitn) == 0
fprintf(' Iter %2d: fitdelta = %7.1e\n', iter, fitchange);
end
% check for convergence
if (iter > 1) && (fitchange < fitchangetol)
break;
end
end
%% clean up final results
T = arrange(T); % columns are normalized
fprintf('factorization error %3.2e\n', fit);
end
I know that there is little context here, but my suspicion is that I need to have Simulink, as Sfun is a Simulink related function(?). The script requires two toolboxes: tensor_toolbox, and FOptM.
Available at:
https://www.sandia.gov/~tgkolda/TensorToolbox/index-2.6.html
https://github.com/andland/FOptM
Thank you so much for your help,
Paul
Although SFun is an often used abbreviation for a Simulink S-Function, in this case the error has nothing to do with Simulink, and the name is a coincidence. (There is no Simulink related function specifically called Sfun, it is just a general term.)
Your error message has #Sfun in it, which is a way in MATLAB of creating a function handle to an (m-code) function called Sfun. I'd summize from the code you've shown that this is a cost function used in the optimization.
If you look at the code that your code is based on (tBNE_fun.m) you'll see that there is a function at the end of the file called Sfun. It is this that you are missing.
I'm currently learning Matlab's ODE-functions to solve simple vibration-problems.
For instance mx''+cx'+kx=F*sin(wt) can be solved using
function dx = fun(t,x)
m=0.02; % Mass - kg
k=25.0; % Stiffness - N/m
c=0.0125; % System damping - Ns/m
f=10; % Frequency
F=5;
dx= [x(2); (F*sin(2*pi*f*t)-c*x(2)-k*x(1))/m]
And then calling the ode45 function to get displacement and velocity
[t,x]=ode45(#fun,[0 10],[0.0;0.0])
My question, which I have not fully understood searching the web, is if it is possible to use ODE-function for a multiple degree of freedom system? For instance, if we have two masses, springs and dampers, which we excite att mass 1, we get the following equations:
m1*x1''+c1*x1'-c2*x2'+(k1+k2)*x1-k2*x2 = f1(t)
m2*x2''-c2*x1'+(c1+c2)*x2'-k2*x1+k2*x2 = 0
Here, the displacements x1 & x2 depend on each other, my question is how one should go about to solve these ODE's in Matlab?
There is no restriction that the inputs to the function solved by ODE45 be scalar. Just pass in an input matrix and expect out an output matrix. For example here is a function that solves the position of a 6 bar mechanism.
function zdot = cp_solve(t,z)
%% Constants
g = -9.81;
L1 = .2;
m0 = 0;
I0 = 0;
m1 = 1;
I1 = (1/3) * m1 * L1^2;
%% Inputs
q = z(1:6);
qdot = z(7:12);
%% Mass Matrix
M = zeros(6,6);
M(1,1) = m0;
M(2,2) = m0;
M(3,3) = I0;
M(4,4) = m1;
M(5,5) = m1;
M(6,6) = I1;
%% Constraint Matrix
Phiq = zeros(5,6);
Phiq(1,1) = 1;
Phiq(2,2) = 1;
Phiq(3,3) = 1;
Phiq(4,1) = 1;
Phiq(4,4) = -1;
Phiq(4,6) = (-L1/2)*sin(q(6));
Phiq(5,2) = 1;
Phiq(5,5) = -1;
Phiq(5,6) = (L1/2)*cos(q(6));
%% Generalized Forces
Q = zeros(6,1);
Q(5) = m1 * g;
%% Right Side Vector
rs = zeros(5,1);
rs(4) = (L1/2) * cos(q(6)) * qdot(6)^2;
rs(5) = (L1/2) * sin(q(6)) * qdot(6)^2;
%% Coefficient Matrix
C = [M Phiq'; Phiq zeros(5,5)];
R = [Q; rs];
%% Solution
Sol = C \ R;
zdot = [qdot; Sol(1:6)];
end
The inputs are the positions and velocities of the members. The outputs are the new positions and velocities.
You use it the same way you would any ODE45 problem. Setup the initial conditions, define a time and solve the problem.
%% Constants
L1 = .2;
C1 = L1/2;
theta1 = 30*pi/180;
theta_dot1 = 0;
tspan = 0:.001:2;
%% Initial Conditions
q = zeros(6,1);
q(6) = theta1;
q(4) = C1 * cos(q(6));
q(5) = C1 * sin(q(6));
qdot = zeros(6,1);
qdot(6) = theta_dot1;
z0 = [q; qdot];
%% Solve the problem
options = odeset('RelTol', 1.0e-9, 'AbsTol', 1.0e-6);
[Tout, Zout] = ode45(#cp_solve, tspan, z0, options);
In your case you have 2 equations and 2 unknowns. Set the problem up as a matrix problem and solve it simultaneously in your function. I would recommend the modal approach for your case.
I am trying to simulate the rotation dynamics of a system. I am testing my code to verify that it's working using simulation, but I never recovered the parameters I pass to the model. In other words, I can't re-estimate the parameters I chose for the model.
I am using MATLAB for that and specifically ode45. Here is my code:
% Load the input-output data
[torque outputs] = DataLogs2();
u = torque;
% using the simulation data
Ixx = 1.00;
Iyy = 2.00;
Izz = 3.00;
x0 = [0; 0; 0];
Ts = .02;
t = 0:Ts:Ts * ( length(u) - 1 );
[ T, x ] = ode45( #(t,x) rotationDyn( t, x, u(1+floor(t/Ts),:), Ixx, Iyy, Izz), t, x0 );
w = x';
N = length(w);
q = 1; % a counter for the A and B matrices
% The Algorithm
for k=1:1:N
w_telda = [ 0 -w(3, k) w(2,k); ...
w(3,k) 0 -w(1,k); ...
-w(2,k) w(1,k) 0 ];
if k == N % to handle the problem of the last iteration
w_dash(:,k) = (-w(:,k))/Ts;
else
w_dash(:,k) = (w(:,k+1)-w(:,k))/Ts;
end
a = kron( w_dash(:,k)', eye(3) ) + kron( w(:,k)', w_telda );
A(q:q+2,:) = a; % a 3N*9 matrix
B(q:q+2,:) = u(k,:)'; % a 3N*1 matrix % u(:,k)
q = q + 3;
end
% Forcing J to be diagonal. This is the case when we consider our quadcopter as two thin uniform
% rods crossed at the origin with a point mass (motor) at the end of each.
A_new = [A(:, 1) A(:, 5) A(:, 9)];
vec_J_diag = A_new\B;
J_diag = diag([vec_J_diag(1), vec_J_diag(2), vec_J_diag(3)])
eigenvalues_J_diag = eig(J_diag)
error = norm(A_new*vec_J_diag - B)
where my dynamic model is defined as:
function [dw, y] = rotationDyn(t, w, tau, Ixx, Iyy, Izz, varargin)
% The output equation
y = [w(1); w(2); w(3)];
% State equation
% dw = (I^-1)*( tau - cross(w, I*w) );
dw = [Ixx^-1 * tau(1) - ((Izz-Iyy)/Ixx)*w(2)*w(3);
Iyy^-1 * tau(2) - ((Ixx-Izz)/Iyy)*w(1)*w(3);
Izz^-1 * tau(3) - ((Iyy-Ixx)/Izz)*w(1)*w(2)];
end
Practically, what this code should do, is to calculate the eigenvalues of the inertia matrix, J, i.e. to recover Ixx, Iyy, and Izz that I passed to the model at the very begining (1, 2 and 3), but all what I get is wrong results.
Is the problem with using ode45?
Well the problem wasn't in the ode45 instruction, the problem is that in system identification one can create an n-1 samples-signal from an n samples-signal, thus the loop has to end at N-1 in the above code.
I am trying to implement a Kalman Filter for estimating the state 'x' (displacement and velocity) of an oscillator. The code is below and should be simple to follow.
clear; clc; close all;
% io = csvread('sim.csv');
% u = io(:, 1);
% y = io(:, 2);
% clear io;
% Estimation of state of a single degree-of-freedom oscillator using
% the Kalman filter
% x[n + 1] = A x[n] + B u[n] + w[n]
% y[n] = C x[n] + D u[n] + v[n]
% Here, x[n] is 2 x 1, u[n] is 1 x 1
% A is 2 x 2, B is 2 x 1, C is 1 x 2, D is 1 x 1
%% Code begins here
N = 1000;
u = randn(N, 1); % Synthetic input
y = randn(N, 1); % Synthetic output
%% Definitions
dt = 0.005; % Time step in seconds
T = 1.50; % Oscillator period
zeta = 0.05; % Damping ratio
sv0 = max(abs(u)) * dt;
sd0 = sv0 * dt;
Q = [sd0 ^ 2 0.0; 0.0 sv0 ^ 2]; % Prediction error covariance matrix
smeas = 0.001 * max(abs(u));
R = smeas ^ 2; % Measurement error (co)variance scalar
wn = 2. * pi / Ts;
c = 2.0 * zeta * wn;
k = wn ^ 2;
A = [0. 1.; -k -c];
Ad = expm(A * dt);
Bd = A \ (Ad - eye(2));
Bd = Bd(:, 2);
C = [-k -c];
D = -1.0;
%% State-space model and Kalman filter
sys = ss(Ad, Bd, C, D, dt, 'inputname', 'u', 'outputname', 'y');
[kest,L,P] = kalman(sys, Q, R, []);
Here is my problem. I get the error: 'In the "kalman(SYS,QN,RN,NN,...)" command, QN must be a real square matrix with at most 1 rows.'.
I thought that QN = Q = const and should be 2 x 2, but it is asking for a scalar. Perhaps I don't understand the difference between Q and QN in MATLAB's 'kalman' help description. Any insights?
Thanks.
MATLAB is assuming the process noise is only one stochastic variable and not two like Q is representing.
So you have to add the G and H matrices to your sys like so:
G = eye(2);
H = [0,0];
sys = ss(Ad, [Bd, G], C, [D, H], dt, 'inputname', 'u', 'outputname', 'y');
Just as a reminder, using MATLAB's syntax:
x*=Ax+Bu+Gw
y=Cx+Du+Hw+v