Digital Control Systems

Links

• Textbooks
  • Introduction to Applied Digital Control, Greg Starr, University of New Mexico
    • Greg's ME481/ME581 web pages contain solutions to the book problems and other things.
  • Optimal Sampled-Data Control Systems, Chen & Francis
  • Digital Control System Analysis and Design, 3rd Ed., Philips
  • Control Systems and Control Engineering
  • Introduction to Signal Processing
  • Zoran Gajic's Book on Control Systems (Continuous & Discrete) and some other documents. Look for the documents entitled Chapter N.
• Lectures
  • Sampled Time Control
  • Zoran Gajic's notes on Observers. This is a very short set of lecture notes that explains the why's of observers, with several interesting derivations.
• Papers
  • David Luenberger's Introduction to Observers
  • F.L. Lewis Linear Quadratic Regulator Derivation and Examples
  • Jake Abbott's lecture on the Linear Quadratic Regulator

MATLAB/Octave

• Octave Control Systems Toolbox This is not the same thing that is on Octave Forge here.

Pendulum Specific Scripts

Single Pendulum

% Double Pendulum Parameters (Tentative: There are two pendulum setups, each with different parameters. I'm not sure which these go to.) % This script is for balancing only the long rod. % Run parameters %f = input('Control Frequency (Hz) = '); %Trun = input('Run Time (s) = '); %f=130; f=1000; Trun=30; kmax = round(f*Trun); T = 1/f; Maxpos = 0.25;  % Max carriage travel +- 0.25 m Maxangle = 0.175;  % Max rod angle -- 10 deg Maxvoltage = 20;  % Max motor voltage, V pstart = 0.005;  % Carriage position starting limit, m astart = 1*pi/180;  % Angle starting limit, rad g = 9.81;  % m/s^2 Gravitational constant % SYSTEM PARAMETERS % Measured Mechanical Parameters d1 = 0.323;  % m Length of pendulum 1 (long) d2 = 0.079;  % m Length of pendulum 2 (short) %mp1 = 0.0208;  % kg Mass of pendulum 1 mp1 = 0.0318; %mp2 = 0.0050;  % kg Mass of pendulum 2 mp2 = 0.0085; %m=.5; m = 0.3163;  % kg Mass of carriage rd = 0.0254/2;  % m Drive pulley radius md = 0.0375;  % kg Mass of drive pulley (cylinder) %mc1 = 0.0036;  % kg Mass of clamp 1* %mc2 = 0.0036;  % kg Mass of clamp 2* mc1 = 0.0085; mc2 = mc1; % *Clamp Dimensions % Rectangular 0.0254 x 0.01143 m % The pivot shaft is 0.00714 m from the end % Motor Parameters (Data Sheet) Im = 43e-7;  % kg m^2/rad Rotor moment of inertia R = 4.09;  % ohms Resistance kt = 0.0351;  % Nm/A Torque constant ke = 0.0351;  % Vs/rad Back emf constant % Derived Mechanical Parameters  % kg m^2/rad Moment of inertia, clamp 1 %Ic1 = mc1*(0.01143^2 + 0.0254^2)/12 + mc1*(0.0127-0.00714)^2; Ic1 = mc1*(0.0098^2 + 0.0379^2)/12; Ic2 = Ic1;  % kg m^2/rad Moment of inertia, clamp 2 Id = md*(rd^2)/2;  % kg m^2/rad Moment of inertia, drive pulley Imd = Im + Id;  % kg m^2/rad Moment of inertia, combined J1 = Ic1 + mp1*(d1^2)/3;  % Total moment of inertia, pendulum 1 (long) J2 = Ic2 + mp2*(d2^2)/3;  % Total moment of inertia, pendulum 2 (short) Jd = Im + Id;  % Total moment of inertia, motor drive Mc = m + mc1 + mc2;  % Total carriage mass % Friction Test Data % Carriage Slope = 19 deg; Terminal Velocity xdotss = 0.312 m/s; From % twincarriage.m; formula b = m g sin(theta)/xdotss % Pendulum 1 (long) Exponent a1 = 0.0756 1/s; From longfit.m % Pendulum 2 (short) Exponent a2 = 0.2922 1/s; From shortfit.m % formula b = 2 a J %alpha = 19; alpha = 12.2; %xdotss = 0.312; xdotss = 0.4852; %a1 = 0.0756; %a2 = 0.2922; a1 = 0.0185; a2 = 0.012;  % Ns/m Viscous friction of carriage system b = (Mc + mp1 + mp2)*g*sin(alpha*pi/180)/xdotss; b1 = 2*a1*J1;  % Nms/rad Viscous friction of pendulum 1 (rotational) b2 = 2*a2*J2;  % Nms/rad Viscous friction of pendulum 2 (rotational) scale = [rd*2*pi/4096 2*pi/4096 -0.05/250]; T = 1/f; % The data above comes from the fweb wiki. M=Mc; %mass of cart m=mp1; %mass of pendulum 1 b=b; %friction l=d1/2; %length of pendulum I=J1; %inertia of pendulum %q=(M+m)*(l+m*l^2)-(m*l)^2; %num=[m*l,0]; %numerator for transfer function %den=[q,b*(l+m*l^2),-m*g*l*(M+m),-b*m*g*l]; %denominator for transfer function %[A,B,C,D]=tf2ss(num,den) %A,B,C,D matricies for the state space model % x_vec is [x, x_dot, theta, theta_dot]' % See the web site: https://www.library.cmu.edu/ctms/ctms/examples/pend/invpen.htm A=[ 0 1 0 0; 0 ((-(I+m*l^2)*b)/(I*(M+m)+M*m*l^2)) ((m^2*g*l^2)/(I*(M+m)+M*m*l^2)) 0; 0 0 0 1; 0 ((-m*l*b)/(I*(M+m)+M*m*l^2)) ((m*g*l*(M+m))/(I*(M+m)+M*m*l^2)) 0]; B=[ 0; ((I+m*l^2)/(I*(M+m)+M*m*l^2)); 0; ((m*l)/(I*(M+m)+M*m*l^2))]; C=[ 1 0 0 0; 0 0 1 0]; D=[ 0; 0]; cont_sys = ss(A,B,C,D) rank_ctrb = rank(ctrb(A,B)) original_poles = eig(A) %poles for our system

Z Transforms

• Relationship between the Laplace and Z transforms
• Convolution and Z Transforms
  • Z Transforms and Convolution
  • Here is an animation of convolution with continuous signals. Look at it so you understand what is happening with the mechanics of convolution. To convolve $x(t)$ with $h(t)$, you flip shift $h(t)$ into $h(t-t_0)$ on the $t_0$ axis, then you multiply it by $x(t_0)$ to get $x(t_0)h(t-t_0)$, then you integrate with respect to $t_0$, so that the convolution is: $x(t)*h(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t_0)h(t-t_0)d{t}_0$. The first animation shows this happening with sampled waveforms: $x_s(t)=x(t)\sum _{n=0}^{\mathrm{\infty }}\delta (t-nT)=\sum _{n=0}^{\mathrm{\infty }}x(nT)\delta (t-nT)$ and $h_s(t)=h(t)\sum _{n=0}^{\mathrm{\infty }}\delta (t-nT)=\sum _{n=0}^{\mathrm{\infty }}h(nT)\delta (t-nT)$.
    • Discrete convolution animations
      • Here is an animation of discrete convolution to illustrate the above equations.
      • This one lets you try several functions.
      • This one shows what happens if your signals happen prior to time zero.
  • Notice that this is really the same as Polynomial Multiplication.

Discretization

• This is how the c2d zero order hold works in MATLAB/octave. It uses the "exact" solution to the discretization problem.
• Also check Chapter 7.3 of Greg Starr's book.

Scilab/Xcos

• Xcos Beginners Guide
• Scilab/Scicos Book