Chapter 23.1 Introduction
Nyquist is a stability benchmark frequnency type. The next one is Hurwitz stability benchmark. It’s algebraic type. We test equation M(s)=0, where polynomian M(s) is a G(s) denominator.
23-1
Fig. 23-1
G(s) as a fraction.
Conclusion
Automatics should have a knowledge about  the equations type M(s)=0.
23-2
Fig. 23-2
An example of the equation 5 degree.

Chapter 23.2 Quadratic equations, higher degree equations and complex numbers
23-3a
Fig. 23-3

23-4a

Fig. 23-4
The discriminator isn’t  negative –> there are real roots  x1=3 i x2=1. It’s easy to write product form of this binomial. It’s more
convenient form than general because the roots are visible here.
Let’s solve the next quadratic equation.
23-5a
Fig. 23-5
Houston. There is a problem! Negative discriminator. The secondary school student says “no solutions”.  But if we assume that there is a strange number–>Imaginary number as above–>equation have solutions!
Complex number consists of 2 separate parts. Real and Imaginary part. You used this number when you were 6 years old.  Real numbers are the complex numbers too, but their imaginary parts are null. When you buy 3 kg potatoes or (3 +j*0) kg potatoes you buy the same potatoes weight.
23-6a
Fig. 23-6
Geometric interpretation of the real and complex numbers as a quadratic equation solution.

Product polynomian form
23-7a
Fig. 23-7
4
degree polynomian as a product.

Chapter 23.3 The real roots of the M(s)=0 only. What’s about stability now?
Call Desktop/PID/09_kryterium_Hurwitza/01_stabilny.zcos
23-8a
Fig. 23-8
The “lower” ang “higher” objects G(s) are the same. But the “higher” is product type and the roots s1=-3 i s2=-1  are visible. Both roots are negative. What’s the conclusion?
Click “start”
23-9
Fig. 23-9
The x(t) Dirac type input pulse unbalanced the system. But it return to its y(t)=0 steady state without oscillations.
The most important conclusion:

G(s) is stable, when all M(s)=0 equation roots are negative.

The statement may be generalized for all M(s) degree  n.
Call Desktop/PID/09_kryterium_Hurwitza/02_niestabilny.zcos
23-10
Fig. 23-10
One root s2=+0.075 is positive.
Click “start”
23-11
Fig. 23-11
The system is instable. The positive root s2=+0.075 is a reason.
We tested 2 objects with the real number roots–>Fig. 23-8 and 23-10
But what with the complex number roots? Other words- the discriminator is negative.

Chapter 23.4 The complex roots of the M(s)=0. What’s about stability now?
Call Desktop/PID/09_kryterium_Hurwitza/03_stabilny_oscylacyjny.zcos
23-12a
Fig. 23-12
Negative discriminator–>There are complex type roots of the  M(s)=0. How to calculate the roots? see  Fig. 23-5.
The real 2 root parts are -1, i.e. are negative. I suspect that the system is stable. Let’s check by the Dirac hammer pulse x(t)
Click “start”
23-13
Fig. 23-13
The system returned to its steady state y(t)=0.–>System is stable. But there are decaying oscillations now, on the contrary to the Fig 23-9.
We can generalize the blue statement under Fig 23-9 now.

G(s) is stable, when all M(s)=0 equation real parts of the roots are negative.

The statement is valid for for all M(s) degree  n.
And what about positive real parts of the roots? Please note that we treat the roots as a complex numbers now.  Do you conjecture?
Call Desktop/PID/09_kryterium_Hurwitza/04_niestabilny_oscylacyjny.zcos

23-14a

Fig. 23-14
The real 2 root parts are +0.05, i.e. are positive.
Click “start”
23-15
Fig. 23-15
The system is instable. The pair of 2 complex number roots with positive real part is a reason of the instability.

Chapter 23.5 Hurwitz benchmark – The  roots M(s) values aren’t necessary to know G(s) stability.
The denominator M(s) of the  open loop transfer function G(s) is given in the product form. So the roots are known –>Fig.23-16a.. The roots aren’t known when closed loop Gz(s) is calculated–>Fig.23-16c.   There isn’t a big problem here. It’s easy to calculate the roots of the quadratic equation. There is problem with the M(s)=0 when degree n is higher.
23-16
Fig. 23-16
-a M(s)=(s+1)*(s+2) is in the product form. The roots  s1=-1, s2=-2 are visible and negative –>G(s) is stable.
-b Closed loop Gz(s) – intermediate result
-c  Closed loop Gz(s) – finite result
I remind the stability statement again.

G(s) is stable, when all M(s)=0 equation real parts of the roots are negative.

The roots values are difficult to calculate when degree of  M(s)=0 equation is higher than 2 You must only know the sign of the real part! More strictly-all roots real parts must be negative–>system is stable.
There is a job for Adolf Hurwitz –german matematician 19/20 century.
The algorithm is for all n degree of course. The particular n=5 will be used here.
23-17a
Fig. 23-17
You know the Hurwitz for n=5. The action is analogical for 6,7…n. By the way. I recommend this method for all statements with
general n. The real matematician will be disgusted a little, but all is much visible here.
Dear reader. Are you secondary school student? You have a right not to know discriminants W1, W2, W3 and W4 .
-Every matrix n degree has his own number- discriminant Wn.
W1 and W2 are easy to calculate as on Fig. 23-17.
– The W3, W4,…Wn calculation is a “monkey job” given in all books about matrix.

Chapter 23.6 The Hurwitz statement check when G(s) is stable

23-18a
Fig. 23-18
aG(s) with the negative feedback
b– Gz(s) as a open loop – intermeadiate result
c Gz(s) as a open loop – finite result
The a, b and c objects are absolutely the same. Hurwitz may be used to c form only.
23-19a
Fig. 23-19
Two
conditions of Hurwitz statement are fulfilled–>G(s) is stable! Let’s test it.
Call Desktop/PID/09_kryterium_Hurwitza/05_Hurwitz_stabilny.zcos.
23-20
Fig. 23-20
Click “start”
23-21
Fig. 23-21
System is stable and Mr Hurwitz is right.
We know that system is stable. But how deeply stable? Maybe on near the border? Are the roots real or complex numbers?
Mr Hurwitz is silent here.
The oscillations proclaim that there are complex numbers in the equation M(s)=0 roots.

Chapter 23.7 The Hurwitz statement check when G(s) is instable

23-22a
Fig. 23-22
23-23a
Fig. 23-23
The Hurwitz condition 2 isn’t fulfilled. The system is instable. Let’s test it.
Call Desktop/PID/09_kryterium_Hurwitza/06_Hurwitz_niestabilny.zcos
23-24
Fig. 23-24
Click “start”
23-25
Fig. 23-25
System is instable. Mr Hurwitz is ok.

Leave a Reply

Your email address will not be published. Required fields are marked *