Determining optimal coefficients for Horwitz matrix or characteristic equation

How to find kp and kv so that H becomes Horwitz?

H =

[ 0, 0, 1, 0]

[ 0, 0, 0, 1]

[ -(2000*kp)/1477, -(1000*kp)/1477, -(2000*kv)/1477, -(1000*kv)/1477]

[ (1000*kp)/1477, -(1000*kp)/1477, (1000*kv)/1477, -(1000*kv)/1477]

Also, the coefficients of the characteristic equation are as follows in order from large to small

[ 1, (3000*kv)/1477, (3000000*kv^2)/2181529 + (3000*kp)/1477, (6000000*kp*kv)/2181529, (3000000*kp^2)/2181529]

[Jon]

When you say "becomes Horwitz" I assume you are asking what values of kp and kv will make the system stable. By the way it is Hurwitz, not Horwitz.

In general I would expect that there will be a range of kp and kv which will make the system stable, and that these values will depend upon each other. So for a given value of kp there will be a range of kv's for which the system is stable.

So what are you actually looking for here as an answer?

[mohammadreza]

Hello and thank you for your help

I am looking for a program that can help me determine the coefficients in such a way that my matrix is stable, although it should be said that the numerical values change over time.

The image and numerical values of the matrix were fixed for a time

Also, I apologize for the fact that I am not fluent in English and I had a problem writing it

[Sam Chak]

Hi @mohammadreza

Conventionally, the range of and may be determined from the Routh–Hurwitz Stability Criterion, which is a pretty tedious task to compute elements in the Table for high-order systems (), and then solving the equations simultaneously.

So far, there is no such function in the Control System Toolbox.

There are some 3rd-party programs written by expert MATLAB users, posted on File Exchange. You should search for the ones that construct the table symbolically.

If you are looking for the determination of the optimal coefficients for and , then it becomes an optimization problem, in which you most likely need a optimization algorithm to solve the problem.

Answer 1

[Sam Chak]

Hi @mohammadreza

Edit: It appears that and can make the matrix Hurwitz. However, the optimal coefficients are subjective, because they depend on the definition of the cost performance function.

If the quadratic cost function is used, perhaps these values and are optimal coefficients.

CASE 1: and

% CASE 1
kp = 0.25;
kv = kp;
A = [0, 0, 1, 0;
     0, 0, 0, 1;
   -(2000*kp)/1477, -(1000*kp)/1477, -(2000*kv)/1477, -(1000*kv)/1477;
    (1000*kp)/1477, -(1000*kp)/1477,  (1000*kv)/1477, -(1000*kv)/1477]

A = 4×4

0 0 1.0000 0 0 0 0 1.0000 -0.3385 -0.1693 -0.3385 -0.1693 0.1693 -0.1693 0.1693 -0.1693

eig(A)

ans = 
  -0.0012 + 0.5822i
  -0.0012 - 0.5822i
  -0.2527 + 0.4356i
  -0.2527 - 0.4356i

G = tf((3000000*kp^2)/2181529, [1, (3000*kv)/1477, (3000000*kv^2)/2181529 + (3000*kp)/1477, (6000000*kp*kv)/2181529, (3000000*kp^2)/2181529])

G =
 
                       0.08595
  --------------------------------------------------
  s^4 + 0.5078 s^3 + 0.5937 s^2 + 0.1719 s + 0.08595
 
Continuous-time transfer function.

step(G)

stepinfo(G)

ans = struct with fields:
         RiseTime: 2.5536
    TransientTime: 3.0953e+03
     SettlingTime: 3.0953e+03
      SettlingMin: 0.1465
      SettlingMax: 1.9260
        Overshoot: 92.5956
       Undershoot: 0
             Peak: 1.9260
         PeakTime: 8.3833

CASE 2: and

% CASE 2
kp = 0.6557;
kv = 1.3419;
G = tf((3000000*kp^2)/2181529, [1, (3000*kv)/1477, (3000000*kv^2)/2181529 + (3000*kp)/1477, (6000000*kp*kv)/2181529, (3000000*kp^2)/2181529])

G =
 
                     0.5912
  ---------------------------------------------
  s^4 + 2.726 s^3 + 3.808 s^2 + 2.42 s + 0.5912
 
Continuous-time transfer function.

step(G)

pole(G)

ans = 
  -0.8058 + 1.0032i
  -0.8058 - 1.0032i
  -0.5570 + 0.2164i
  -0.5570 - 0.2164i

stepinfo(G)

ans = struct with fields:
         RiseTime: 4.8988
    TransientTime: 9.2379
     SettlingTime: 9.2379
      SettlingMin: 0.9008
      SettlingMax: 1.0004
        Overshoot: 0.0379
       Undershoot: 0
             Peak: 1.0004
         PeakTime: 14.9637

[mohammadreza]

Hi and Thank you very much for your advice

I have a question ? How can the same question be done when the numerical values of H are variable?

Because H is obtained by multiplying several matrices, one of which is variable with time

In your opinion, by keeping one variable constant (giving it a value), it is possible to obtain another variable?

Also, I apologize for the fact that I am not fluent in English and I had a problem writing it

[Sam Chak]

Hi @mohammadreza

Don't apologize just because English is the common medium of communication used in this forum. It doesn't mean you are at fault. When people ask questions, try your best to clarify.

If the matrix is known with values, then you can compute the eigenvalues of . And if all eigenvalues have negative real parts, then the system is stable.

H = tril(-magic(4))

H = 4×4

-16 0 0 0 -5 -11 0 0 -9 -7 -6 0 -4 -14 -15 -1

eig(H)      % eigenvalues

ans = 4×1

-1 -6 -11 -16

If there are time elements in , then it is a linear time-varying (LTV) model whose dynamics vary with time. This type of system is characterized by the time-varying matrix . The full context of your problem is unspecified, so I'm not sure how to advise you to attack the problem.

Of course, you can also do the Integer Brute-force Search on and , and then narrow down the solutions. For a two-variable problem, I think it is still manageable without causing a combinatorial explosion.

[Sam Chak]

Hi @mohammadreza, Your latest update is a new problem, and the proposed method may not be applicable to the time-varying problem. Besides You should not reply in the "Answer" section.

If is a time-varying function (as shown in your supplied image), and without analyzing the function, then there is no guarantee that the system will be stable even though is designed to be Hurwitz as .

Answer 2

[mohammadreza]

Thank you very much for your help

R(t) is Time-varying matrix

My problem is finding kp and kv scalars

[mohammadreza]

H should always be Hurwitz