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

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 $x(t)$ $x(t)$ with $h(t)$ $h(t)$ , you flip shift $h(t)$ $h(t)$ into $h(t-t_{0})$ $h(t-t_{0})$ on the $t_{0}$ $t_{0}$ axis, then you multiply it by $x(t_{0})$ $x(t_{0})$ to get $x(t_{0})h(t-t_{0})$ $x(t_{0})h(t-t_{0})$ , then you integrate with respect to $t_{0}$ $t_{0}$ , so that the convolution is: $x(t)*h(t)=\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}$ $x(t)*h(t)=\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}$ . The first animation shows this happening with sampled waveforms: $x_{s}(t)=x(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }x(nT)\delta (t-nT)$ $x_{s}(t)=x(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }x(nT)\delta (t-nT)$ and $h_{s}(t)=h(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }h(nT)\delta (t-nT)$ $h_{s}(t)=h(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }h(nT)\delta (t-nT)$ .
  - Discrete convolution animations
- Notice that this is really the same as Polynomial Multiplication.