Category Archives: Control

three DOF arm manipulator

Lagrange Equation by MATLAB with Examples

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

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( \frac{\partial T}{\partial \dot{q}_j}\right ) - \frac{\partial T}{\partial q_j} + \frac{\partial V}{\partial q_j} = 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
p2x = p1x+L2cos(x1+x2);
p2y = p1y+L2
p3x = p2x+L3cos(x1+x2+x3);
p3y = p2y+L3

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);

Fourth, define the generalized forces, here torques.

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

Fifth,  solve the Lagrangian equation

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

Let’s obtain step by step.

\left( \frac{\partial T}{\partial \dot{q}}\right ) is obtained by 

pKEpx1d = diff(KE,x1d);

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

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

\frac{\partial T}{\partial q_j}, \frac{\partial V}{\partial 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} \begin{bmatrix} x_1\\ \dot{x}_1\\ x_2\\ \dot{x}_2\\ x_3\\ \dot{x}_3 \end{bmatrix}= 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

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.









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. 

I hope this article makes your life easier.


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.

g++ demo_udp_server.cpp UDPThread.cpp -lpthread -o demo_udp_server

This is the result.

Screenshot from 2015-03-04 15:44:40



I hope this code helps your project.



Minimal c++ class for Maxon EPOS2

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;
    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.




Sliding Mode Control (SMC), Robust control algorithm for nonlinear system

In this article, I am going to write about the Sliding Mode Control (SMC) algorithm.

Sliding Mode Control algorithm is a robust controller for nonlinear systems.

Robust control means that even though the system model has a certain error, if the controller can control the system, we say that the controller is robust.

SMC has two fundamental ideas.

1. To attract the system states to the surface.

2. To make the state slide on the surface toward the origin.

To explain the above two ideas more easily, let’s assume a typical control problem.

\dot x = f_1 (x,\dot x)   — (1)

\ddot x = f_2 (x, \dot x) + u

y=Cx — (2) if it is nonlinear, you need to know Lie Derivatives.

To achieve the first idea, we need to define a surface like the below. (Let’s assume that y is a scalar and differentiable.) 

s(x)=a_0 e + a_1 \dot{e}  where  e=y-y_d — (3)

Here we need to select a_0 and a_1 to make y->0 as t-> inf,s=0. You will see the reason after more several lines.

Let’s take the derivative of s(x) w.r.t. x. Then we can get the below equation.

\dot{s}(x)=a_0 \dot e + a_1 \ddot e  — (4)

In addition, if we add the additional term -\eta \text{sign}(s) we can obtain the below equation,

\dot{s}(x)=a_0 \dot e + a_1 \ddot e=-\eta \text{sign}(s) — (5)

Through a simple Lyapunov theorem, we can easily prove s-> 0 at t->inf .

if s=0, e-> 0 and \dot(e)-> 0, because we selected a_0 and a_1 to be like this.

Let’s look at (5),

a_0 \frac{d}{dt}(y-y_d)+a_1 \frac{d^2}{dt^2}(y-y_d)=-\eta \text{sign}(s)

=> C(a_0 \dot x +a_1 \ddot x)=-\eta \text{sign}(s)

=>C(a_0 f_1 +a_1 (f_2+u)))=-\eta \text{sign}(s)

=>u=((-\eta \text{sign} (s))/C - a_0 f_1)/a_1 -f_2

With this control input, we can control a nonlinear system with SMC.

it is difficult to explain quickly within short explanations…., if you need more explanation or questions, please leave me a reply.


Phase plane analysis and Matlab code toolbox

Phase plain analysis is a useful visualization tool to understand the characteristics of systems including not only linear system but also nonlinear system. For example, we can determine stability of the system from this phase plane analysis.

The attachment file <here> is Matlab toolbox to draw phase plain. The attached file includes a simple demo and the below is the result. You can draw phase plane, magnify where you are interest recursively. You can see how to use the Matlab code in the following Youtube video.



How to draw?




we can find the below equation


From \frac{dx_2}{dx_1}, we can find the direction of the phase change at the point of (x_1, x_2).



  1. It is an exact method. We can see the change of system’s state including transient response.
  2. Simple graphical method. It is very intuitive and easy to understand its characteristics.


  1. Limited to the 2nd order system. It is expandable, but hard to visualize. 


1. Lecture of Prof. Fernadez in Mech. Eng, The Univ. of Texas at Austin.

Extended Luenberger Observer for nonlinear system control


Herman (who commented for this posting) told me that there are two different versions of ELO. One is just to linearize a nonlinear function (this posting will handle it), and the other is using Lie-algebraic approach (refer to M. Zeitz 1987 “The extended Luenberger Observer”).

Thank you Herman.


This article is to explain the use of Luenberger observer for nonlinear system control. In other words, it is Extended Luenberger Observer, (ELO, just like Kalman Filter (KF), and Extended KF).

The basic idea is to linearize nonlinear system around the interesting point. The below is the description of ELO and how to select gain values for Extended Luenberger Observer. The below description assumes that you already know about Luenberger Observer for linear system. If you don’t know visit here.



I wish this can help your understanding about Luenberger observer, if you have any question, please leave me a comment below.


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

My main site is, and Korean ver. is