## Preliminary Automatics Course

**Chapter 23. Hurwitz stability criterion
**

**Chapter 23.1 Introduction**

**Nyquist**is a stability criterion 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.

**Fig. 23-1**

**G(s)** as a fraction.

**Conclusion
**Automatics should have a knowledge about the equations type

**M(s)=0**.

**Fig. 23-2**

An example of the

**M(s)=0**equation

**5**degree.

**Chapter 23.2 Quadratic equations, higher degree equations and complex numbers**

**Fig. 23-3
**

**Fig. 23-4
**The discriminator isn’t

**negative**–> there are

**2**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.

**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

**I**

**maginary**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.

**Fig. 23-6**

Geometric interpretation of the

**real**and

**complex**numbers as a quadratic equation solution.

**Product polynomian form**

**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

**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”

**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

**Fig. 23-10**

One root **s2=+0.075** is positive.

Click “start”

**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

**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”

**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

**Fig. 23-14
**The

**real**2 root parts are

**+0.05**, i.e. are positive.

Click “start”

**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.

**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.

**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**

**Fig. 23-18**

**a**– **G(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.

**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.

**Fig. 23-20**

Click “start”

**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**

**Fig. 23-22**

**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

**Fig. 23-24**

Click “start”

**Fig. 23-25**

System is

**instable**. Mr

**Hurwitz**is ok.