Question

1. Use the Routh-Hurwitz test to determine if the system described by the following transfer function is stable. If the system is unstable, how many poles are outside the LHP? Use Matlab to check your answers. C() 10-8) R(s) s2 +7s +28 2. Repeat problem 1) above for the system with transfer function C (s) R(5s +Bs+ 40 s2 +2s+4 3. Use the Routh-Hurwitz test to determine if the system described by the following characteristic equation is stable. If the system is unstable, how many poles are outside the LHP? Use Matlab to check your answers. s +2s3 +12s+27s 85 4. Repeat problem 3) above for the system with characteristic equation s6+12s5 +25s +280s3 +120s2 +220s + 800 5. Repeat problem 3) above for the system with characteristic equation + 7.5s5 + 63.5s'+ 210s+709s2+ 802.5s+2465

Answer 1

ols-3) CtS) RLS) here 6 rlem is Slable. and RHS Pole is Zevo. S3 Sa R. H. Pole=0 Pole on Tmeiny axis Synlem is magina Stekle 90 egn: S1Ho33 o o s 8s Sgflem is insteble.

S 1 25 0 Syf tem is unstable and No of R-Hs pole 63วั £02.5 S 3s Syple is tab No. of RHs Pole o

To Verify this answer in Matlab, Matlab Code is given Below :

clear ;
close all;
clc
% Taking coefficients vector and organizing the first two rows
coeffVector = input('input vector of your system coefficients: i.e. [an an-1 an-2 ... a0] = ');
ceoffLength = length(coeffVector);
rhTableColumn = round(ceoffLength/2);
% Initialize Routh-Hurwitz table with empty zero array
rhTable = zeros(ceoffLength,rhTableColumn);
% Compute first row of the table
rhTable(1,:) = coeffVector(1,1:2:ceoffLength);
% Check if length of coefficients vector is even or odd
if (rem(ceoffLength,2) ~= 0)
% if odd, second row of table will be
rhTable(2,1:rhTableColumn - 1) = coeffVector(1,2:2:ceoffLength);
else
% if even, second row of table will be
rhTable(2,:) = coeffVector(1,2:2:ceoffLength);
end
%% Calculate Routh-Hurwitz table's rows
% Set epss as a small value
epss = 0.01;
% Calculate other elements of the table
for i = 3:ceoffLength

% special case: row of all zeros
if rhTable(i-1,:) == 0
order = (ceoffLength - i);
cnt1 = 0;
cnt2 = 1;
for j = 1:rhTableColumn - 1
rhTable(i-1,j) = (order - cnt1) * rhTable(i-2,cnt2);
cnt2 = cnt2 + 1;
cnt1 = cnt1 + 2;
end
end
  
for j = 1:rhTableColumn - 1
% first element of upper row
firstElemUpperRow = rhTable(i-1,1);
  
% compute each element of the table
rhTable(i,j) = ((rhTable(i-1,1) * rhTable(i-2,j+1)) - ....
(rhTable(i-2,1) * rhTable(i-1,j+1))) / firstElemUpperRow;
end
  
  
% special case: zero in the first column
if rhTable(i,1) == 0
rhTable(i,1) = epss;
end
end
%% Compute number of right hand side poles(unstable poles)
% Initialize unstable poles with zero
unstablePoles = 0;
% Check change in signs
for i = 1:ceoffLength - 1
if sign(rhTable(i,1)) * sign(rhTable(i+1,1)) == -1
unstablePoles = unstablePoles + 1;
end
end
% Print calculated data on screen
fprintf(' Routh-Hurwitz Table: ')
rhTable
% Print the stability result on screen
if unstablePoles == 0
fprintf(' it is a stable system! ')
else
fprintf(' it is an unstable system! ')
end
fprintf(' Number of right hand side poles =%2.0f ',unstablePoles)