# Matlab Code for Numerical Hessian Matrix

## Matlab code for numerical hessian matrix

In this post, I am sharing a Malab code calculating a numerical hessian matrix.

You can use NumHessian.m  with this syntax.

function hf = NumHessian(f,x0,varargin)

You can understand how to use simply by reading these two simple examples.

## Example 1

>> NumHessian(@cos,0)

ans =

-1.0000

## Example 2

function y=test_func(x,a)
y=ax(1)x(2)*x(3);

>> NumHessian(@test_func,[1 2 3]’,2)

ans =

0 6.0000 4.0000
5.9999 -0.0001 1.9998
3.9999 2.0000 0

I hope this Matlab code for numerical Hessian matrix helps your projects

This Matlab code is based on another Matlab function, NumJacob, which calculates a numerical Jacobian matrix. If you are interested in this, visit here.

If you want to know the theory on Hessian matrix, please read this Wiki.

I acknowledge that this code is originally made by Parviz Khavari, a visitor of this blog, and I modified to post here.

# Vertex angle from three points, Matlab code

I am sharing a Matlab code to get a vertex angle from three points.

function [ang] = get_vertex_ang_from_three_points(p0,p1,p2)

% get the vertex angle at p0 from three points p0,p1,p2
% p0, p1,p2 are two dim vector
%
%%% example
% p0=[0;0];p1=[1;0];p2=[0;1];
% get_vertex_ang_from_three_points(p0,p1,p2)
% answer = 1.57 (45 deg)
%%%

v1 = p1-p0;
v2 = p2-p0;
ang = acos(v1′v2/(norm(v1)norm(v2)));

It is very easy to use. you can get the vertex angle from three points, It is written in Matlab code

Good luck~~

# Lagrange Equation by MATLAB  with Examples

In this post, I will explain how to derive a dynamic equation with Lagrange Equation by MATLAB  with Examples. As an example, I will derive a dynamic model of a three-DOF arm manipulator (or triple pendulum). Of course you may get a dynamic model for a two-DOF arm manipulator by simply removing several lines. I am attaching demo codes for both two and three DOF arm manipulators.

If you know all theories and necessary skills and if you just want source code, you can just download from here.

In the attached file, “Symb_Development_3DOF.m” generates a dynamic model. “main_sim_three_dof_arm.m” runs a simulation. You may get a result like this.

## 1. Example system

Let’s suppose a three DOF arm manipulator shown in the below figure. I am assuming all of masses (M1, M2, M3) exist at the end of links for simplicity. Three actuators exist at each joint, and directly actuate the torques (u1,u2,u3). The manipulator kinematics is governed by three joint angles (q1, q2, q3).

three DOF arm manipulator

## 2. Theoretical Background

We will get a dynamic equation of this system by using Lagrangian mechanics. If you do not have a background knowledge of Lagrangian mechanics, please refer here.

The general dynamic equation is obtained by

$\frac{d}{dt}\left(&space;\frac{\partial&space;T}{\partial&space;\dot{q}_j}\right&space;)&space;-&space;\frac{\partial&space;T}{\partial&space;q_j}&space;+&space;\frac{\partial&space;V}{\partial&space;q_j}&space;=&space;P_j$

Where, T is the total kinetic energy, V is the total potential energy of the system. where  $P_j$ is the generalized force, t is time, $q_j$ is the generalized coordinates, $\dot{q}_j$ is the generalized velocity. For the example of three-DOF arm manipulator problem, $P_j$ is the torque at the j-th joint,  $q_j$  is the angle of the j-th joint, $\dot{q}_j$ is the angular velocity of the j-th joint.

## 3. Using Matlab symbolic toolbox

### First, let’s define the symbols.

I am using x to represent q, xd for $\dot{q}$, xdd for $\ddot{q}$  L is the length of each link. u is the torque of each joint. g is the gravity constant.

syms M1 M2 M3;
syms x1 x1d x1dd x2 x2d x2dd x3 x3d x3dd;
syms L1 L2 L3;
syms u1 u2 u3;
syms g

### Second, find the position and velocities of the point masses.

p1x = L1cos(x1);
p1y = L1
sin(x1);
p2x = p1x+L2cos(x1+x2);
p2y = p1y+L2
sin(x1+x2);
p3x = p2x+L3cos(x1+x2+x3);
p3y = p2y+L3
sin(x1+x2+x3);

v1x = -L1sin(x1)x1d;
v1y = +L1cos(x1)x1d;
v2x = v1x-L2sin(x1+x2)(x1d+x2d);
v2y = v1y+L2cos(x1+x2)(x1d+x2d);
v3x = v2x – L3sin(x1+x2+x3)(x1d+x2d+x3d);
v3y = v2y + L3cos(x1+x2+x3)(x1d+x2d+x3d);

### Third, define the kinetic energy, and the potential energy.

KE = 0.5M1( v1x^2 + v1y^2) + 0.5M2( v2x^2 + v2y^2) + 0.5M3( v3x^2 + v3y^2);
KE = simplify(KE);

PE = M1gp1y + M2gp2y + M3gp3y;
PE = simplify(PE);

Px1 = u1;
Px2 = u2;
Px3 = u3;

### Fifth,  solve the Lagrangian equation

We have to get the Lagrangian eqn. $\frac{d}{dt}\left(&space;\frac{\partial&space;T}{\partial&space;\dot{q}_j}\right&space;)&space;-&space;\frac{\partial&space;T}{\partial&space;q_j}&space;+&space;\frac{\partial&space;V}{\partial&space;q_j}&space;=&space;P_j$

Let’s obtain step by step.

$\left(&space;\frac{\partial&space;T}{\partial&space;\dot{q}}\right&space;)$ is obtained by

pKEpx1d = diff(KE,x1d);

To calculate, $\frac{d}{dt}\left(&space;\frac{\partial&space;T}{\partial&space;\dot{q}}\right&space;)$we need the chain rule,  that is$\frac{d}{dt}\left(&space;\frac{\partial&space;T}{\partial&space;\dot{q}_j}\right&space;)=&space;\sum^{3}_i&space;\left\{&space;\frac{\partial&space;}{\partial&space;{q}_i}\left(&space;\frac{\partial&space;T}{\partial&space;\dot{q}_j}\right&space;)\frac{dq_i}{dt}&space;+&space;\frac{\partial&space;}{\partial&space;\dot{q}_i}\left(&space;\frac{\partial&space;T}{\partial&space;{q}_j}\right&space;)\frac{d&space;\dot{q}_i}{dt}&space;\right\}$

ddtpKEpx1d = diff(pKEpx1d,x1)x1d+ …
diff(pKEpx1d,x1d)
x1dd+ …
diff(pKEpx1d,x2)x2d + …
diff(pKEpx1d,x2d)
x2dd + …
diff(pKEpx1d,x3)x3d + …
diff(pKEpx1d,x3d)
x3dd;

$\frac{\partial&space;T}{\partial&space;q_j},&space;\frac{\partial&space;V}{\partial&space;q_j}$ are easily obtained by

pKEpx1 = diff(KE,x1);
pPEpx1 = diff(PE,x1);

By summing all equations,

eqx1 = simplify( ddtpKEpx1d – pKEpx1 + pPEpx1 – Px1);

By repeating these procedures, we can get all governing equations.

### Sixth, rearrange the equations.

We love more simplified forms like

$\frac{d}{dt}&space;\begin{bmatrix}&space;x_1\\&space;\dot{x}_1\\&space;x_2\\&space;\dot{x}_2\\&space;x_3\\&space;\dot{x}_3&space;\end{bmatrix}=&space;f(x_1,\dot{x}_1,x_2,\dot{x}_2,x_3,\dot{x}_3)$

For this form, we need to rearrange the equations by

Sol = solve(eqx1,eqx2,eqx3,’x1dd,x2dd,x3dd’);
Sol.x1dd = simplify(Sol.x1dd);
Sol.x2dd = simplify(Sol.x2dd);
Sol.x3dd = simplify(Sol.x3dd);

### Seventh, substitute with y1,y2… variables.

Just for easier implementation of symbolic codes, let’s substitute x1, x1d, x2, … with y1,y2….

syms y1 y2 y3 y4 y5 y6
fx1=subs(Sol.x1dd,{x1,x1d,x2,x2d,x3,x3d},{y1,y2,y3,y4,y5,y6})
fx2=subs(Sol.x2dd,{x1,x1d,x2,x2d,x3,x3d},{y1,y2,y3,y4,y5,y6})
fx3=subs(Sol.x3dd,{x1,x1d,x2,x2d,x3,x3d},{y1,y2,y3,y4,y5,y6})

### Eighth, this is the result…. as you can see, it is almost impossible to solve by hand

fx1 =

(L2L3M2u1 – L2L3M2u2 + L2L3M3u1 – L2L3M3u2 – L1L3M2u2cos(y3) + L1L3M2u3cos(y3) – L1L3M3u2cos(y3) + L1L3M3u3cos(y3) – L2L3M3u1cos(y5)^2 + L2L3M3u2cos(y5)^2 + L1L2^2L3M2^2y2^2sin(y3) + L1L2^2L3M2^2y4^2sin(y3) – L1L2L3M2^2gcos(y1) + (L1^2L2L3M2^2y2^2sin(2y3))/2 + L1L2M2u3sin(y3)sin(y5) + L1L2M3u3sin(y3)sin(y5) + L1L3M3u2cos(y3)cos(y5)^2 – L1L3M3u3cos(y3)cos(y5)^2 + L1L2^2L3M2M3y2^2sin(y3) + L1L2^2L3M2M3y4^2sin(y3) – L1L2L3M1M2gcos(y1) – L1L2L3M1M3gcos(y1) – L1L2L3M2M3gcos(y1) + 2L1L2^2L3M2^2y2y4sin(y3) + (L1^2L2L3M2M3y2^2sin(2y3))/2 – L1L3M3u2cos(y5)sin(y3)sin(y5) + L1L3M3u3cos(y5)sin(y3)sin(y5) + L1L2L3M2^2gcos(y1)cos(y3)^2 – L1L2L3M2^2gcos(y3)sin(y1)sin(y3) + 2L1L2^2L3M2M3y2y4sin(y3) + L1L2L3M2M3gcos(y1)cos(y3)^2 + L1L2L3M1M3gcos(y1)cos(y5)^2 + L1L2L3^2M2M3y2^2cos(y5)sin(y3) + L1L2L3^2M2M3y4^2cos(y5)sin(y3) + L1L2L3^2M2M3y6^2cos(y5)sin(y3) + 2L1L2L3^2M2M3y2y4cos(y5)sin(y3) + 2L1L2L3^2M2M3y2y6cos(y5)sin(y3) + 2L1L2L3^2M2M3y4y6cos(y5)sin(y3) – L1L2L3M2M3gcos(y3)sin(y1)sin(y3))/(L1^2L2L3(M2^2 – M2^2cos(y3)^2 + M1M2 + M1M3 + M2M3 – M2M3cos(y3)^2 – M1M3*cos(y5)^2))

fx2 =

-(2L1^2L3M1u3 – 2L1^2L3M1u2 – 2L1^2L3M2u2 + 2L2^2L3M2u1 + 2L1^2L3M2u3 – L1^2L3M3u2 – 2L2^2L3M2u2 + L2^2L3M3u1 + L1^2L3M3u3 – L2^2L3M3u2 + L1L2^2M2u3cos(y3 – y5) + L1L2^2M3u3cos(y3 – y5) – L1^2L2M2u3cos(2y3 + y5) – L1^2L2M3u3cos(2y3 + y5) – L2^2L3M3u1cos(2y5) + L2^2L3M3u2cos(2y5) + L1^2L3M3u2cos(2y3 + 2y5) – L1^2L3M3u3cos(2y3 + 2y5) – L1L2^2M2u3cos(y3 + y5) – L1L2^2M3u3cos(y3 + y5) + 2L1^2L2M1u3cos(y5) + L1^2L2M2u3cos(y5) + L1^2L2M3u3cos(y5) + 2L1L2^3L3M2^2y2^2sin(y3) + 2L1^3L2L3M2^2y2^2sin(y3) + 2L1L2^3L3M2^2y4^2sin(y3) – L1L2L3M3u1cos(y3 + 2y5) + 2L1L2L3M3u2cos(y3 + 2y5) – L1L2L3M3u3cos(y3 + 2y5) + L1^2L2L3M2^2gcos(y1 + y3) – L1L2^2L3M2^2gcos(y1) – L1^2L2L3M2^2gcos(y1 – y3) + L1L2^2L3M2^2gcos(y1 + 2y3) + 2L1^2L2^2L3M2^2y2^2sin(2y3) + L1^2L2^2L3M2^2y4^2sin(2y3) + 2L1L2L3M2u1cos(y3) – 4L1L2L3M2u2cos(y3) + L1L2L3M3u1cos(y3) + 2L1L2L3M2u3cos(y3) – 2L1L2L3M3u2cos(y3) + L1L2L3M3u3cos(y3) + 2L1^3L2L3M1M2y2^2sin(y3) + L1^3L2L3M1M3y2^2sin(y3) + 2L1L2^3L3M2M3y2^2sin(y3) + 2L1^3L2L3M2M3y2^2sin(y3) + 2L1L2^3L3M2M3y4^2sin(y3) + 4L1L2^3L3M2^2y2y4sin(y3) – (L1^2L2L3M1M3gcos(y1 + y3 + 2y5))/2 – L1^3L2L3M1M3y2^2sin(y3 + 2y5) + L1L2^2L3^2M2M3y2^2sin(y3 + y5) + L1L2^2L3^2M2M3y4^2sin(y3 + y5) + L1L2^2L3^2M2M3y6^2sin(y3 + y5) + L1^2L2L3M1M2gcos(y1 + y3) + (L1^2L2L3M1M3gcos(y1 + y3))/2 + L1^2L2L3M2M3gcos(y1 + y3) – 2L1^2L2L3^2M1M3y2^2sin(y5) – L1^2L2L3^2M2M3y2^2sin(y5) – 2L1^2L2L3^2M1M3y4^2sin(y5) – L1^2L2L3^2M2M3y4^2sin(y5) – 2L1^2L2L3^2M1M3y6^2sin(y5) – L1^2L2L3^2M2M3y6^2sin(y5) – 2L1L2^2L3M1M2gcos(y1) – L1L2^2L3M1M3gcos(y1) – L1L2^2L3M2M3gcos(y1) + (L1^2L2L3M1M3gcos(y1 – y3 – 2y5))/2 + L1L2^2L3^2M2M3y2^2sin(y3 – y5) + L1^2L2L3^2M2M3y2^2sin(2y3 + y5) + L1L2^2L3^2M2M3y4^2sin(y3 – y5) + L1^2L2L3^2M2M3y4^2sin(2y3 + y5) + L1L2^2L3^2M2M3y6^2sin(y3 – y5) + L1^2L2L3^2M2M3y6^2sin(2y3 + y5) – L1^2L2L3M1M2gcos(y1 – y3) – (L1^2L2L3M1M3gcos(y1 – y3))/2 – L1^2L2L3M2M3gcos(y1 – y3) + L1L2^2L3M2M3gcos(y1 + 2y3) + (L1L2^2L3M1M3gcos(y1 – 2y5))/2 + (L1L2^2L3M1M3gcos(y1 + 2y5))/2 + 2L1^2L2^2L3M2M3y2^2sin(2y3) – L1^2L2^2L3M1M3y2^2sin(2y5) + L1^2L2^2L3M2M3y4^2sin(2y3) – L1^2L2^2L3M1M3y4^2sin(2y5) + 2L1^2L2^2L3M2^2y2y4sin(2y3) + 4L1L2^3L3M2M3y2y4sin(y3) + 2L1L2^2L3^2M2M3y2y4sin(y3 + y5) + 2L1L2^2L3^2M2M3y2y6sin(y3 + y5) + 2L1L2^2L3^2M2M3y4y6sin(y3 + y5) – 4L1^2L2L3^2M1M3y2y4sin(y5) – 2L1^2L2L3^2M2M3y2y4sin(y5) – 4L1^2L2L3^2M1M3y2y6sin(y5) – 2L1^2L2L3^2M2M3y2y6sin(y5) – 4L1^2L2L3^2M1M3y4y6sin(y5) – 2L1^2L2L3^2M2M3y4y6sin(y5) + 2L1L2^2L3^2M2M3y2y4sin(y3 – y5) + 2L1^2L2L3^2M2M3y2y4sin(2y3 + y5) + 2L1L2^2L3^2M2M3y2y6sin(y3 – y5) + 2L1^2L2L3^2M2M3y2y6sin(2y3 + y5) + 2L1L2^2L3^2M2M3y4y6sin(y3 – y5) + 2L1^2L2L3^2M2M3y4y6sin(2y3 + y5) + 2L1^2L2^2L3M2M3y2y4sin(2y3) – 2L1^2L2^2L3M1M3y2y4sin(2y5))/(L1^2L2^2L3(M2^2 – M2^2cos(2y3) + 2M1M2 + M1M3 + M2M3 – M2M3cos(2y3) – M1M3cos(2*y5)))

fx3 =

-(L1L3^2M3^2u2 – L1L2^2M3^2u3 – L1L2^2M2^2u3 – L1L3^2M3^2u3 – L2L3^2M3^2u1cos(y3) + L2L3^2M3^2u2cos(y3) – 2L1L2^2M1M2u3 – 2L1L2^2M1M3u3 + 2L1L3^2M1M3u2 – 2L1L2^2M2M3u3 – 2L1L3^2M1M3u3 + 2L1L3^2M2M3u2 – 2L1L3^2M2M3u3 + L1L2^2M2^2u3cos(2y3) + L1L2^2M3^2u3cos(2y3) + L1L3^2M3^2u2sin(2y3)sin(2y5) – L1L3^2M3^2u3sin(2y3)sin(2y5) + L1L2L3M3^2u2cos(y5) – 2L1L2L3M3^2u3cos(y5) – 2L2L3^2M2M3u1cos(y3) + 2L2L3^2M2M3u2cos(y3) – 2L2^2L3M3^2u1sin(y3)sin(y5) + 2L2^2L3M3^2u2sin(y3)sin(y5) + L2L3^2M3^2u1cos(2y5)cos(y3) – L2L3^2M3^2u2cos(2y5)cos(y3) + 2L1L2^2M2M3u3cos(2y3) – L2L3^2M3^2u1sin(2y5)sin(y3) + L2L3^2M3^2u2sin(2y5)sin(y3) – L1L3^2M3^2u2cos(2y3)cos(2y5) + L1L3^2M3^2u3cos(2y3)cos(2y5) – L1L2^2L3^2M2M3^2y2^2sin(2y3) – L1L2^2L3^2M2^2M3y2^2sin(2y3) + 2L1L2^2L3^2M1M3^2y2^2sin(2y5) – L1L2^2L3^2M2M3^2y4^2sin(2y3) – L1L2^2L3^2M2^2M3y4^2sin(2y3) + 2L1L2^2L3^2M1M3^2y4^2sin(2y5) + L1L2^2L3^2M1M3^2y6^2sin(2y5) + 2L1L2L3M1M3u2cos(y5) – 4L1L2L3M1M3u3cos(y5) + L1L2L3M2M3u2cos(y5) – 2L1L2L3M2M3u3cos(y5) – 2L2^2L3M2M3u1sin(y3)sin(y5) + 2L2^2L3M2M3u2sin(y3)sin(y5) + 2L1L2L3^3M1M3^2y2^2sin(y5) + 2L1L2^3L3M1M3^2y2^2sin(y5) + L1L2L3^3M2M3^2y2^2sin(y5) + 2L1L2L3^3M1M3^2y4^2sin(y5) + 2L1L2^3L3M1M3^2y4^2sin(y5) + L1L2L3^3M2M3^2y4^2sin(y5) + 2L1L2L3^3M1M3^2y6^2sin(y5) + L1L2L3^3M2M3^2y6^2sin(y5) – L1L2L3M3^2u2cos(2y3)cos(y5) + 2L1L2L3M3^2u3cos(2y3)cos(y5) + L1L2L3M3^2u2sin(2y3)sin(y5) – 2L1L2L3M3^2u3sin(2y3)sin(y5) – L1^2L2L3^2M1M3^2y2^2sin(y3) – 2L1^2L2L3^2M2M3^2y2^2sin(y3) – 2L1^2L2L3^2M2^2M3y2^2sin(y3) – 2L1L2^2L3^2M2M3^2y2y4sin(2y3) – 2L1L2^2L3^2M2^2M3y2y4sin(2y3) + 4L1L2^2L3^2M1M3^2y2y4sin(2y5) + 2L1L2^2L3^2M1M3^2y2y6sin(2y5) + 2L1L2^2L3^2M1M3^2y4y6sin(2y5) + 2L1L2^3L3M1M2M3y2^2sin(y5) + 2L1L2^3L3M1M2M3y4^2sin(y5) – L1L2L3M2M3u2cos(2y3)cos(y5) + 2L1L2L3M2M3u3cos(2y3)cos(y5) + 4L1L2L3^3M1M3^2y2y4sin(y5) + 4L1L2^3L3M1M3^2y2y4sin(y5) + 2L1L2L3^3M2M3^2y2y4sin(y5) + 4L1L2L3^3M1M3^2y2y6sin(y5) + 2L1L2L3^3M2M3^2y2y6sin(y5) + 4L1L2L3^3M1M3^2y4y6sin(y5) + 2L1L2L3^3M2M3^2y4y6sin(y5) + L1L2L3M2M3u2sin(2y3)sin(y5) – 2L1L2L3M2M3u3sin(2y3)sin(y5) – L1L2L3^3M2M3^2y2^2cos(2y3)sin(y5) – L1L2L3^3M2M3^2y2^2sin(2y3)cos(y5) – L1L2L3^3M2M3^2y4^2cos(2y3)sin(y5) – L1L2L3^3M2M3^2y4^2sin(2y3)cos(y5) – L1L2L3^3M2M3^2y6^2cos(2y3)sin(y5) – L1L2L3^3M2M3^2y6^2sin(2y3)cos(y5) + 2L1^2L2^2L3M1M3^2y2^2cos(y3)sin(y5) – 2L1^2L2L3^2M1M2M3y2^2sin(y3) + L1L2L3^2M1M3^2gsin(y1)sin(y3) + 2L1L2L3^2M2M3^2gsin(y1)sin(y3) + 2L1L2L3^2M2^2M3gsin(y1)sin(y3) + L1^2L2L3^2M1M3^2y2^2cos(2y5)sin(y3) + L1^2L2L3^2M1M3^2y2^2sin(2y5)cos(y3) – L1L2L3^2M1M3^2gcos(2y5)sin(y1)sin(y3) – L1L2L3^2M1M3^2gsin(2y5)cos(y3)sin(y1) + 4L1L2^3L3M1M2M3y2y4sin(y5) + 2L1^2L2^2L3M1M2M3y2^2cos(y3)sin(y5) – 2L1L2L3^3M2M3^2y2y4cos(2y3)sin(y5) – 2L1L2L3^3M2M3^2y2y4sin(2y3)cos(y5) – 2L1L2L3^3M2M3^2y2y6cos(2y3)sin(y5) – 2L1L2L3^3M2M3^2y2y6sin(2y3)cos(y5) – 2L1L2L3^3M2M3^2y4y6cos(2y3)sin(y5) – 2L1L2L3^3M2M3^2y4y6sin(2y3)cos(y5) – 2L1L2^2L3M1M3^2gcos(y3)sin(y1)sin(y5) + 2L1L2L3^2M1M2M3gsin(y1)sin(y3) – 2L1L2^2L3M1M2M3gcos(y3)sin(y1)sin(y5))/(L1L2^2L3^2M3(M2^2 – M2^2cos(2y3) + 2M1M2 + M1M3 + M2M3 – M2M3cos(2y3) – M1M3cos(2y5)))

## 4. Now it is ready!!! let’s run a simulation!

copy the result and paste the symbols in a Matlab function. Then, run an ODE with the function.

A sample code is here.

In the attached file, “Symb_Development_3DOF.m” generates a dynamic model. “main_sim_three_dof_arm.m” runs a simulation. Then, you can get this result.

So far, I have explained how to derive a Lagrange Equation by MATLAB  with Examples.  I hope that this post helps your project and save your time. Please leave a message if you have any question.

Update on 02/21/2016

I updated some code and posting about typo. “simple” -> “simplify” There is no function “simple” Now all programs are running well.

-Mok-

Reference

[1] http://en.wikipedia.org/wiki/Lagrangian_mechanics

[3] http://www.mathworks.com/matlabcentral/fileexchange/23037-lagrange-s-equations

# MATLAB code for Numerical calculation of Jacobian matrix for a complex number

## In this post, I share a MATLAB code for numerical calculation of Jacobian matrix for a complex number.

This code was developed by Sithan Kanna, based on my code. If you are just interested in Jacobian of a real number, please see this post.

http://youngmok.com/numerical-jacobian-matrix-matlab/

I really thank to Sithan Kanna for sharing this nice code. If you have a question on this code please send an email to sithankanna a-t gmail d-o-t com

## This is a sample result of the code

>> h=@(x)[x(1) ; conj(x(1))]; % nonlinear equation

conjug = 0;
real = 0;
x_test = [0.5*1j];
H = NumJacob(h, x_test, conjug, real)

H =

1.0000
-0.0000

Have a nice day, and thank you for visiting my blog.

-Mok-

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I am Youngmok Yun, and writing about robotics theories and my research.

My main site is http://youngmok.com, and Korean ver. is  http://yunyoungmok.tistory.com.

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# Maxon EPOS2 Driver C++ Class using USB-CAN Gateway for Multiple Motor Actuation

I am sharing a C++ Class for multiple Maxon motor actuation by EPOS2 via USB-Can gateway.

The source code is written for two motors, but it is possible to modify easily to actuate multiple motors.

If you want to see a simpler version for one motor actuation, see this article.

-Mok-

# UDP Server C++ Class with a listening thread

I am sharing a code for “UDP Server C++ Class with a listening thread”. In the class “UDPThread”, a thread is running to receive a UDP packet.

Just compile a demo program with this command. Then you will understand.

This is the result.

I hope this code helps your project.

-Mok-

# Minimal c++ class for Maxon EPOS2

I am sharing the minimal C++ class for Maxon EPOS2. It includes a basic “initialization”, “Move”, “Read position”, “Close device” functions. The program consists of a class called “cmaxonmotor”.  You can download the class and demo program here

You can see a simple demo program here.

#include <stdio.h>
#include <iostream>
#include "cmaxonmotor.h"

using namespace std;

int main(int argc, char *argv[])
{
CMaxonMotor motor("USB0",1);
motor.initializeDevice(); // initialize EPOS2

long TargetPosition = -200000;
int CurrentPosition = 0;

motor.Move(TargetPosition); // move to the target position

cout << "Press <Enter> to stop and quit..." << endl;
getchar();
motor.GetCurrentPosition(CurrentPosition); // get the current position
cout << "Current Position: " << CurrentPosition << endl;

motor.closeDevice(); // close EPOS2

return 0;
}


You can control one motor just via USB, and also able to control multiple motors via USB-CAN gateway. For this version, see this article.

I hope this helps your project.

-Mok-