How to solve Routh Hurwitz with constant K

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justin nabbs
justin nabbs am 26 Apr. 2015
Kommentiert: ami raa am 8 Mai 2022
I am attempting to solve for system stab1ility for the equation: s^4+19*s^3+111*s^2+189*s+K*s+5*K=0
The constant K is throwing me off. I have an idea how to solve this with one variable "s" but need help on how to insert the K as a constant in matlab. thank you in advanced
  3 Kommentare
justin nabbs
justin nabbs am 27 Apr. 2015
I normally use the characteristic equation 1+G(s)*H(s)=0 to solve for the equation s^4+19*s^3+111*s^2+189*s+K*s+5*K=0
I then put the constants in the Routh Hurwitz formula to solve. Normally the constants i'm working with are just numbers. I don't know how to define K in matlab so that I can put K in the Routh Hurwitz formula
justin nabbs
justin nabbs am 27 Apr. 2015
Bearbeitet: Geoff Hayes am 9 Apr. 2018
Solving for stability using Routh Hurwitz gives you the b1,b2 etc. But how do i enter the constant K when i'm entering the coefficients of a characteristic equation through Routh Hurwitz because it gives me a K is undefined error. My equation is s^4+19*s^3+111*s^2+189*s+ K *s+5*K=0 and I used the following syntax:
%%routh hurwitz criteria
clear
clc
%%firstly it is required to get first two row of routh matrix
e=input('enter the coefficients of characteristic equation: ');
disp('-------------------------------------------------------------------------')
l=length(e);
m=mod(l,2);
if m==0
a=rand(1,(l/2));
b=rand(1,(l/2));
for i=1:(l/2)
a(i)=e((2*i)-1);
b(i)=e(2*i);
end
else
e1=[e 0];
a=rand(1,((l+1)/2));
b=[rand(1,((l-1)/2)),0];
for i=1:((l+1)/2)
a(i)=e1((2*i)-1);
b(i)=e1(2*i);
end
end
%%now we genrate the remaining rows of routh matrix
l1=length(a);
c=zeros(l,l1);
c(1,:)=a;
c(2,:)=b;
for m=3:l
for n=1:l1-1
c(m,n)=-(1/c(m-1,1))*det([c((m-2),1) c((m-2),(n+1));c((m-1),1) c((m-1),(n+1))]);
end
end
disp('the routh matrix:')
disp(c)
%%now we check the stablity of system
if c(:,1)>0
disp('System is Stable')
else
disp('System is Unstable');
end

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Akzeptierte Antwort

Julius
Julius am 10 Dez. 2018
Bearbeitet: Julius am 1 Nov. 2020
Hi, maybe a bit late, but anyway here is my solution using Matlab and Routh criterion for evaluation of K for stability (root locus does it perfectly in a graphical way by showing critical value of K if locus crosses jw axis or whatever)
syms Kp s
G = (5*s + 2)/(s*(s^2 + 3*s + 2)) % plant TF
Gc = (Kp*(s + 3))/(s + 5) % controller TF
chareq = 1+G*Gc==0
cheq = expand(simplify(chareq))
% haven't figure out how to extract char equation from symbolic, but you can simply copy coefs
% or adapt to you existing char.eq.
%% Update in 2020 - have figured out how to extract coefficients out of char eq
[n, d] = numden(cheq)
cheq = n == 0
cheq = collect(n,s) == 0
R = coeffs(n,s)
%% Now coefficients can be accessed from the vector R and put into Rouht Table
% RT = [R(1,5) R(1,3) R(1,1);
% R(1,4) R(1,2) 0]
% Routh table first two rows from coefs of char.eq. (from cheq)
RT = [1 17+5*Kp 6*Kp;
8 10+17*Kp 0];
% the rest of the table
b1 = (RT(2,1)*RT(1,2)-RT(1,1)*RT(2,2))/RT(2,1);
b2 = (RT(2,1)*RT(1,3)-RT(1,1)*RT(2,3))/RT(2,1);
b3 = 0;
c1 = (b1*RT(2,2)-RT(2,1)*b2)/b1;
c2 = (b1*RT(2,3)-RT(2,1)*b3)/b1;
c3 = 0;
d1 = (c1*b2-b1*c2)/c1;
d2 = (c1*b3-b1*c3)/c1;
d3 = 0;
% full Routh table
RT = [1 17+5*Kp 6*Kp;
8 10+17*Kp 0;
simplify(b1) b2 b3;
simplify(c1) c2 c3;
simplify(d1) d2 d3]
% coeficient Kp values for stability to satisfy condition when b1=0, c1=0 and d1=0
K1 = vpasolve(b1, Kp)
K2 = vpasolve(c1, Kp)
K3 = vpasolve(d1, Kp)
Haven't checked for limitations and what if there is row of zeros or zero in 1st column.
  10 Kommentare
fatima mouffok
fatima mouffok am 11 Mär. 2022
chareq = 1+G*Gc==0 remove ==0
chreq = 1+G*Gc like that it's gonna work!!
ami raa
ami raa am 8 Mai 2022
thank you so much

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J. Carlos Aguado
J. Carlos Aguado am 9 Apr. 2018
I assume that the origin of this K is a proportional controller. Therefore, the simplest way (according to MatLab implementations, not to mathematical simplicity) to study its effect on stability is to use the root locus of the plant. MatLab will draw it for you and give you the gain value that marks the frontier with instability

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