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6th-Semester-Spring-2024/SysCon/Labs/Lab7/SysConLab7.mlx Normal file
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6th-Semester-Spring-2024/SysCon/routh_hurwitz.m/license.txt Normal file
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Copyright (c) 2019, Varun Bhatia
All rights reserved.
Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are met:
* Redistributions of source code must retain the above copyright notice, this
list of conditions and the following disclaimer.
* Redistributions in binary form must reproduce the above copyright notice,
this list of conditions and the following disclaimer in the documentation
and/or other materials provided with the distribution
* Neither the name of nor the names of its
contributors may be used to endorse or promote products derived from this
software without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
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SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
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%% Routh-Hurwitz Stability Criterion Table Generator V1.1
% Author: Varun Bhatia [bhatiav16@berkeley.edu]
% Date: July 28, 2019
% This function goes through the process of setting up a Routh-Hurwitz
% table to determine information regarding the in/stability of a control
% system given a closed/open-loop transfer function. More specifically, this
% will solve for and output how many closed/open-loop poles are in the
% right-half plane, the left-half plane, and on the jw-axis. Additionally,
% this function will take into account and generate the Routh_Hurwitz table
% given two special cases:
% 1)First element of a row is 0.
% 2)Entire Row is 0.
%Recall that this function operates under the assumption that you are
%inputting a CLOSED LOOP TRANSFER FUNCTION. Later developments will
%incorporate inputs of open loop forward transfer functions with unity
%and/or nonunity feedback.
%Any usage of MATLAB functions is accredited to MathWorks.
%UPDATE 1.1: This program now takes into account UNITY feedback.
%%
function []= routh_hurwitz()
prompt_n = 'Enter highest power in numerator: ';
num_hp = input(prompt_n);
num_hp_added = num_hp + 1;
num = zeros(1, num_hp_added);
for i = 1:num_hp+1
num(i) = input('Enter coefficients, starting with the highest power: ');
end
prompt_d = 'Enter highest power in denominator: ';
den_hp = input(prompt_d);
den_hp_added = den_hp + 1;
den = zeros(1, den_hp_added);
for i = 1:den_hp+1
den(i) = input('Enter coefficients, starting with the highest power: ');
end
prompt_TF = 'Are you entering an open loop or closed loop TF? If closed loop type "C", if open loop type "O" with quotes around letter: ';
TF_type = input(prompt_TF);
if TF_type == "O"
H = tf(num,den);
K = 1;
G = feedback(H,K)
[~,d] = tfdata(G, 'v');
den = d; %to acquire the new denominator coefficients from post-feedback
elseif TF_type == "C"
G = tf(num,den)
end
if den_hp == 0
cols = 1;
elseif den_hp == 1 | den_hp == 2
cols = 2;
elseif den_hp == 3 | den_hp == 4
cols = 3;
elseif den_hp == 5 | den_hp == 6
cols = 4;
elseif den_hp == 7 | den_hp == 8
cols = 5;
elseif den_hp == 9 | den_hp == 10
cols = 6;
elseif den_hp == 11 | den_hp == 12
cols = 7;
elseif den_hp == 13 | den_hp == 14
cols = 8;
end
%the matrix with proper number of rows and columns
RH_Table = zeros(den_hp+1, cols);
%set up the first element in row 1 & row 2 since MATLAB doesn't do 0 indexing
if cols == 1
RH_Table(1,1) = den(1);
elseif cols > 1
RH_Table(1,1) = den(1);
RH_Table(2,1) = den(2);
end
if cols == 1
fprintf('\n *Only one element in the RH Table, thus, you have a stable system.* \n');
elseif cols > 1
first_count = 3; %since added first element (element 1) above so now skip one and start at 3rd element for first row
second_count = 4; %since added second element (element 2) above so now skip one and start at 4th element for second row
%filling in the first row of the RH table
for i = 2:cols %start at 2 because filled in first element above
if first_count <= den_hp_added
RH_Table(1,i) = den(first_count);
first_count = first_count + 2;
else
RH_Table(1,i) = 0;
end
end
%filling in the second row of the RH table
for i = 2:cols %start at 2 because filled in first element above
if second_count <= den_hp_added
RH_Table(2,i) = den(second_count);
second_count = second_count + 2;
else
RH_Table(2,i) = 0;
end
end
%now fill in remaining elements of RH Table with determinants
X = zeros(2,2);
for i = 1:(den_hp-1)
for j = 1:(cols-1)
X(1,1) = RH_Table(i,1);
X(2,1) = RH_Table(i+1,1);
X(1,2) = RH_Table(i,j+1);
X(2,2) = RH_Table(i+1,j+1);
RH_Table(i+2,j) = -det(X)/RH_Table(i+1,1);
end
end
RH_Table % to show full RH table
%now analyze the first column of the table for sign changes to
%determine how many poles in the right half-plane.
sign_changes = 0;
for i = 1:(den_hp)
if (RH_Table(i,1) * RH_Table(i+1,1) < 0)
sign_changes = sign_changes + 1;
end
end
if sign_changes > 0
fprintf(['\n *You have %d right-half poles and %d left-half poles.',...
' Thus, you have an unstable system.* \n'],sign_changes,den_hp-sign_changes)
else
fprintf(['\n *You have %d right-half poles and %d left-half poles.',...
' Thus, you have a stable system.* \n'],sign_changes,den_hp-sign_changes)
end
end
%Special Case #1: Check if there exists a full row of zeros. If so, the
%following code will be utilized.
%first check if any array contains a row of zeros
row_of_zeros = 0;
tol = 1.e-6;
for i = 1:(den_hp + 1)
if abs(RH_Table(i,:) - 0) < tol
row_of_zeros = 1;
row = i;
break; %break because you have identified a row of zeros
end
end
%If the check shows that you do indeed have a row of zeros, the following
%code will run
if (row_of_zeros == 1)
row_above = RH_Table(row-1,:);
aux_poly = poly2sym(row_above);
diff_poly = diff(aux_poly);
diff_poly_coeffs = flip(coeffs(diff_poly));
for j = 1:cols
if numel(diff_poly_coeffs) < cols
diff_poly_coeffs = [diff_poly_coeffs, 0]; %add zero to the end of coefficients to make sure dimensions match when adding back to table
else
break;
end
end
RH_Table(row,:) = diff_poly_coeffs; %replacing row in RH Table with the coefficients of row above, differentiated
X = zeros(2,2);
for i = row-1:(den_hp-1)
for j = 1:(cols-1)
X(1,1) = RH_Table(i,1);
X(2,1) = RH_Table(i+1,1);
X(1,2) = RH_Table(i,j+1);
X(2,2) = RH_Table(i+1,j+1);
RH_Table(i+2,j) = -det(X)/RH_Table(i+1,1);
end
end
%Split up the sign changes process into two:
%1) Check sign changes until two rows above row of zeros normally and
%store these sign changes.
%2) Check sign changes from one row above row of zeros down to bottom.
%By symmetry, the right half poles must equal the left half poles, and
%anything left over are jw-axis poles. Additionally, the row above the
%row of zeros is required to be an even polynomial.
%1)
first_sign_changes = 0;
second_sign_changes = 0;
for k = 1:(row-2)
if (RH_Table(k,1) * RH_Table(k+1,1) < 0)
first_sign_changes = first_sign_changes + 1;
end
end
%2)
for k = row-1:den_hp
if (RH_Table(k,1) * RH_Table(k+1,1) < 0)
second_sign_changes = second_sign_changes + 1;
end
end
total_sign_changes = first_sign_changes + second_sign_changes;
left_half_poles =((row-2)-first_sign_changes + second_sign_changes);
jw_axis_poles = den_hp - (row-2) - (2*second_sign_changes);
RH_Table
if (total_sign_changes) > 0
fprintf(['\n *Upon changes made to the RH Table, you now have %d right-half',...
' poles and %d left-half poles. Additionally, you have %d jw-axis poles. Thus, you have an unstable',...
' system.* \n'],total_sign_changes,left_half_poles,jw_axis_poles);
elseif (total_sign_changes == 0)
fprintf(['\n *Upon changes made to the RH Table, you now have %d',...
' right-half poles and %d left-half poles. Additionally, you have %d jw-axis poles.',...
' Thus, you have a marginally stable system.* \n'],total_sign_changes,left_half_poles, jw_axis_poles);
end
end
%Special Case #2: Check if zero exists in first column and if so, form reciprocal
%polynomial and redo process above to get determinants. Have a separate
%method for this portion, however, for the purposes of File Exchange, I've
%attached that portion below.
for i = 1:(den_hp+1)
if RH_Table(i,1) == 0
den = flip(den);
if cols == 1
RH_Table(1,1) = den(1);
elseif cols > 1
RH_Table(1,1) = den(1);
RH_Table(2,1) = den(2);
end
if cols == 1
fprintf('\n *Only one element in the RH Table, thus, you have a stable system.* \n');
elseif cols > 1
first_count = 3;
second_count = 4;
for k = 2:cols
if first_count <= den_hp_added
RH_Table(1,k) = den(first_count);
first_count = first_count + 2;
else
RH_Table(1,k) = 0;
end
end
for l = 2:cols
if second_count <= den_hp_added
RH_Table(2,l) = den(second_count);
second_count = second_count + 2;
else
RH_Table(2,l) = 0;
end
end
X = zeros(2,2);
for m = 1:(den_hp-1)
for j = 1:(cols-1)
X(1,1) = RH_Table(m,1);
X(2,1) = RH_Table(m+1,1);
X(1,2) = RH_Table(m,j+1);
X(2,2) = RH_Table(m+1,j+1);
RH_Table(m+2,j) = -det(X)/RH_Table(m+1,1);
end
end
sign_changes = 0;
for k = 1:(den_hp)
if (RH_Table(k,1) * RH_Table(k+1,1) < 0) %if the element * next element is negative, must be a sign change
sign_changes = sign_changes + 1;
end
end
RH_Table
if sign_changes > 0
fprintf(['\n *Upon changes made to the RH Table, you now have %d right-half',...
' poles and %d left-half poles. Thus, you have an unstable',...
' system.* \n'],sign_changes,den_hp-sign_changes)
else
fprintf(['\n *Upon changes made to the RH Table, you now have %d',...
' right-half poles and %d left-half poles.',...
' Thus, you have a stable system.* \n'],sign_changes,den_hp-sign_changes)
end
end
end
end %end special case #2%
end