Project - Position control of mass spring damper system

 Transfer Function:

 The transfer function defines the relation between the output and the input of a dynamic system,

 written in complex form (s variable). For a dynamic system with an input u(t) and an output y(t), the

 transfer function H(s) is the ratio between the complex representation (s variable) of the

 output Y(s) and input U(s).

 

 

 Formula:

 The simplest representation of a system is through Ordinary Differential Equation (ODE) When dealing

 with ordinary differential equations, the dependent variables are function of a positive real

 variable t (often time). By applying Laplace’s transform we switch from a function of time to a function

 of a complex variable s (frequency) and the differential equation becomes an algebraic

 equation.

 

 

 Transfer function  for a mass spring damper system:

             

  Image: Translational mass with spring and damper

 

The ordinary differential equation describing the dynamics of the system is:

   ...........(1)

 

 

 where:

 m [kg] – mass

 k [N/m] – spring constant (stiffness)

 c [Ns/m] – damping coefficient

 F [N] – external force acting on the body (input)

 x [m] – displacement of the body (output)

 

 The input of the system is the external force F(t) and the output is the displacement x(t). First we’ll

 apply the Laplace transform to each of the terms of the equation (1):

 

 

   

         

 The initial conditions of the mass position and speed are:

                  x(0) = 0

                  dx(0) / dt = 0

 Replacing the Laplace transforms and initial conditions in the equation (1) gives:

             F(s) = ms^2X(s) + csX(s) + kX(s)

               F(s) = X(s)(ms^2 + cs + k)

 

               X(s) / F(s) = 1 / ms^2 + cs + k

 

 We have now found the transfer function of the translational mass system with spring and

 damper:

               

 Given data for a system:

 m = 3.6; %kg

 k = 400; %N/m

 c = 100; %Ns/m

 

 Script:

 % design parameters

 Jm=0.01;

 Bm=0.1;

 Km=0.01;

 Kt=0.01;

 Ra=1;

 La=0.5;

 

 %PD design

 %A = [-B/J kt/J; -km/La -Ra/La]

 %B = [0; 1/La]

 

 A = [-Bm/Jm Kt/Jm ; -Km/La -Ra/La];

 B = [0; 1/La];

 

 % pole placement

 fn=1;

 wn=2*pi*fn;

 zeta=0.5;

 p1=-zeta*wn+wn*(zeta^2-1);

 p2=-zeta*wn-wn*(zeta^2-1);

 P=[p1;p2];

 K = place(A,B,P);

 

 

 

 Simulink Model:

  

 

 DC motor model :-

 

 

 In this simulink model we have provided the pulse generator as a reference signal as per the 

 question criterion.

 

 Blocks used:

 1. Pulse Generator block

 2. To workspace block

 3. Gain block

 4. Sum block

 5. Scope block

 6. Inport 

 7. outport

 8. Clock

 9. Integrator block

 10. Transfer function block

 

 On running the simulation for 27.5 sec. 

 we get,

 

1] K(1) = Kp = 12.4056,  K(2) = Kd= -2.8584 and Ki = 1 (value of Ki taken manually).

 

 

 

 

 2]  K(1) = Kp = 12.4056,  K(2) = Kd= -2.8584 and Ki = 10

 

 

 

 3] K(1) = Kp = 12.4056,  K(2) = Kd= -2.8584 and Ki = 100

 

 

 4] K(1) = Kp = 12.4056,  K(2) = Kd= -2.8584 and Ki = 500

 

 

 4] K(1) = Kp = 20  K(2) = Kd= -2.8584 and Ki = 100

 

 

 5] K(1) = Kp = 40  K(2) = Kd= -2.8584 and Ki = 100

   

 From above all the results we can observe that if we keep on increasing the Ki values with increasing

 Ki we get response of more overshoot and undershoot. also if we change the Kd gain then system

 suffers more condition of undershoot and overshoot.  Respone we are getting is not bounded at all.

 Lets observe the behaviour with different reference speed signal.

 

 1] K(1) = Kp = 12.4056,  K(2) = Kd= -2.8584 and Ki = 100

 

 

 

2]  K(1) = Kp = 12.4056,  K(2) = Kd= -2.8584 and Ki = 10

 

 

 3] K(1) = Kp = 12.4056,  K(2) = Kd= -2.8584 and Ki = 1

 We are getting  same response when we are changing the reference signal.

 

 PID Controller:

 

 

 PID controllers are found in a wide range of applications for industrial process control.

 Approximately 95% of the closed-loop operations of the industrial automation sector

 use PID controllers. PID stands for Proportional-Integral-Derivative. These three

 controllers are combined in such a way that it produces a control signal. As a feedback

 controller, it delivers the control output at desired levels. Before microprocessors were

 invented, PID control was implemented by the analog electronic components. But today

 all PID controllers are processed by the microprocessors. Programmable logic controllers

 also have the inbuilt PID controller instructions. Due to the flexibility and reliability of the

 PID controllers, these are traditionally used in process control applications.

 

  Simulink model without PID Block:

 

 1] For a resference of 230:

 

 

  2] For a reference of 100

 

 

  3] For a reference of 50

 

 

 4] For a reference of 350

 

 

 we have used the same script . 

 For plotting the command we have used 

 

 using pid controller block 

 

 Tunned response we are getting is shown below,

 

 

 When reference signal is 230

 

 

 when reference signal  is 100

 

 

 This is how we are maintaining stability of a dynamic system using pid controller. 

 

 Parameters: