part (d): MSD controller design

The design process of the feedback controller for our 3-mass 4-spring system is shown in the Matlab script below. The resulting K matrix is:

2890

118

-3727 ]
These values can be interpreted as multipliers on each of the state variables as they are fed back into the system via the input. Recall that these values modify the effective system matrix via (A – BK), or through being multiplied by the input matrix B. This multiplication results in:

1.0e+04 *

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0.9340 0.1445 0.0862 0.0059 -1.6056 -0.1863

0 0 0 0 0 0

0 0 0 0 0 0

% MSDcntlDesign.m

%

% MSD Control Design for HW 11, using the 3 mass, 4 spring continuing

%

% J Cardell, April 2017

% Define constants, calculated separately from problem statement, for a

% second order system with PO = 3% and ts = 3s, see pages 240-241

zeta = 0.7448; % damping ratio

wn = 1.790; % natural frequency rad/s

%% Generic Second Order System

% Define the TRANSFER FUNCTION, standard form for a 2nd order system (see

% page 238)

num = wn^2; % numerator

dem = [1 2*zeta*wn wn^2]; % denominator

% the roots of the denominator = eigenvalues

eigs = roots(dem);

% define the transfer functoin for this 2nd order system

TF2nd = tf(num, dem)

% Plot the step response of this generic 2nd order system, and use the

% Matlab figure window options to lable the performance characteristics -

% right click within the resulting figure window and select the desired

% characteristic.

step(TF2nd)

set(gca, 'FontSize', 18)

grid

%% Augemented for 6th order system

% define 4 additional eigenvalues, to be at least 10x further to the left

% in the LHP (left-half plane), of the complex plane. See page 247. Be sure

% to avoid creating repeated eigenvalues.

for i = 3:6

eigs(i) = real(eigs(1)*10 - (i-3));

end

% Define the characteristic polynomial that has these 6 eigenvalues as its

charPoly6 = poly(eigs);

% Note, the final term from 'poly(eigs)' is the constant in the polynomial

% expression. If the coefficients of lambda (or s) are 'ci' then the

% polynomial is c1*s^6 + c2*s^5 ... c6*s + c7. Recognize that if c7 = 0

% then this is a homogeneous equation. If c7 <> 0 then this is a

% non-homogeneous equation and c7 is in fact the forcing function - i.e.,

% it is the input. Therefore, this terms is the numerator of the transfer

% function. The denominator of a transfer function is always the

% characteristic equation, or characteristic polynomial, of the system.

%

% ALSO, see page 249 and generic characteristic polynomials defined in the

TFsix = tf([charPoly6(7)], charPoly6)

% Plot the step response of this system as above. Also plot the step

% response of the two transfer functions, together, in order to allow

% visual comparison.

figure

set(gca, 'FontSize', 18)

grid

figure

set(gca, 'FontSize', 18)

grid

%% Control Law (K matrix) definition for specific MSD system of our

% From previous homework we have the A and B matrices. Use these with the

% 'place' command in order to find the specific values of the feedback

% matrix K that willforce our 3-mass 4-spring system to have the

% eigenvalues defined above

% Constants / Parameters

k1 = 10; % spring constant, N/m

k2 = 20; k3 = 30; k4 = 40;

c1 = 0.4; % damping coefficient, Ns/m

c2 = 0.8; c3 = 1.2; c4 = 1.6;

m1 = 1; % mass, kg

m2 = 2; m3 = 3;

A = [0 1 0 0 0 0;

-(k1+k2)/m1 -(c1+c2)/m1 k2/m1 c2/m1 0 0

0 0 0 1 0 0

k2/m2 c2/m2 -(k2+k3)/m2 -(c2+c3)/m2 k3/m2 c3/m2

0 0 0 0 0 1

0 0 k3/m3 c3/m3 -(k3+k4)/m3 -(c3+c4)/m3 ];

B = [ 0; 0; 0; 1/m2; 0; 0];

C = [ 1 0 0 0 0 0];

D = [0];

K = place(A, B, eigs)

sysCntl = ss((A-B*K), B, (C-D*K), D);

figure

% Normalize responses.

setoptions(hMSD6,'Normalize','on');

set(gca, 'FontSize', 18)

grid

figure

setoptions(hFinal,'Normalize','on');