# Automatics

**Chapter. 20 Hurwitz Stability Criterion**

**Chapter. 20.1 Introduction**The

**Nyquist**, known earlier, was an example of a

**frequency**-type stability criterion. I don’t think I need to explain why. The

**second**approach is the analysis of the transmittance

**G(s)**. The problem is related to the study of the equation

**M(s)=0**, where the polynomial

**M(s)**is the

**denominator**of the transmittance

**G(s)**

Fig. 20-1

Fig. 20-1

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

Conclusion

The automatics engineer should be well versed in

**M(s)=0**type equations. When

**M(s)**is, for example, a degree

**5**

**polynomial**, then the equation

**M(s)=0**can be, for example.

**An example of a degree**

Fig. 20-2

Fig. 20-2

**5**equation.

**Chapter. 20.2 Quadratic and higher equations and complex numbers**

We are well acquainted with equations of the **2nd** degree, i.e. **quadratic**, not to mention equations of the **1st** degree. There are also formulas for the roots of equations of degree **3**, maybe even **4**. Above, there are no such** formulas**. Or they are, just waiting for their **Columbus**. By the way, maybe someone knows the exact **degree** of the equation for which **there** is no longer a formula? I will be grateful for your response. In any case, equations of **higher** degrees can only be **solved** numerically – that is, by **approximation**. Let me remind you of the well-known formula for a **quadratic** equation.

Fig. 20-3

**Fig. 20-4**Here

**Δ**is

**non-negative**and therefore the equation has

**2**real roots

**x1=3**and

**x2=2**. Knowing the

**roots**of the

**quadratic equation**, we can also write the

**product**form of the

**binomial**. It is comfortable because you can see the

**roots**in it. Let’s solve one more equation.

Fig. 20-5

Fig. 20-5

**Negative**delta! In high school we would say “the equation has no solutions”. However, if we assume that there is such a strange number

**j**that its square is

**-1**, (or is equal to the

**root**of –

**1**, i.e.

**√-1**), then there are

**2**roots:

**x1=-2+1*j**and

**x2=- 2-1*j**. In this way, we entered the

**country**of

**complex numbers**.

Where does that name come from? Because they form a

**set**of

**two numbers**. The

**first**is the

**real part**, the

**second**is the

**imaginary part**. You have been using

**complex numbers**since the first grade of

**primary**school. What you didn’t know was, that you only used the

**real**part of this

**number**. The imaginary part was always

**zero**. Therefore, when buying

**3 kg**or

**(3.0)**kg of potatoes, you will always get the same thing. The

**second**time you treated

**(3.0)**

**kg**or otherwise

**(3+j*0)kg**as a

**complex number**.

**A side note**.

For mathematicians, an

**imaginary**number is

**i**. Electricians fell in love with

**complex numbers**at first sight, especially when

**alternating current**, i.e.

**sinusoids**shifted in phase, appeared. From the previous

**chapter**, you know that these

**sine waves**are beautifully presented as

**vectors**. Then, for example, the sum of

**2**sinusoids is easily represented as a

**vectors**sum. Not only that,

**differentiating**or

**integrating**sinusoids is simply

**shifting**them by an

**angle**of

**+90°**or

**-90°**. For vectors, it’s a banality. And a vector is almost a complex number!

Returning to mathematicians and electricians, the latter use the symbol

**j**instead of

**i**. The justification is simple.

The

**i**symbol would clash too much with the

**current**symbol.

**Geometric interpretation of**

Fig. 20-6

Fig. 20-6

**complex numbers**as equations

**roots**.

**Fig. 20-6a**

Equations

**roots**as

**real**numbers

**x1=3**and

**x2=1**(They are

**complex**numbers too!)

**Fig. 20-6b**

Equations

**roots**as

**complex**numbers

**x1=-2+1j**and

**x2=-2-1j**

**Polynomial as a product**

**How to factorize a degree**

Fig. 20-7

Fig. 20-7

**4**polynomial?

It is enough to know its

**4**roots->

**x1, x2, x3**and

**x4**. These can be

**real**or

**complex**numbers. The methods for higher degrees

**5…n**are analogous. There are no general formulas for roots when

**M(s)**is of high degree. Numerical methods of limited accuracy are then used.

**Chapter 20.3 Roots of the denominator M(s) of trasmittance G(s) real only – What stability?**

It will turn out that the **denominator** of the transmittance **G(s)** determines the **stability**.**Chapter 20.3.1 Stable-All roots of the denominator negative**

**Fig. 20-8**We study the transmittance

**G(s)**, whose

**denominator**is a

**polynomial**of degree

**2**

**M(s)=s²+4s+3**. The

**top**is the same

**G(s)**, but the denominator is a

**product**. You can check. And why in the form of a product? Because it shows the roots

**s1=-3**and

**s2=-1**of the quadratic equation

**M(s)=0**. Both are negative. Well, so what? The

**Dirac**impulse brought the system out of equilibrium, but after about

**3**seconds it returned to

**equilibrium**. In addition, without any adjustments. i.e. the course of

**y(t)**was positive all the time.

**Chapter 20.3.2 Unstable – the positive root of the denominator appears**

Here one positive root **s2=+0.075**

**F****ig. 20-9**As expected, the system is

**unstable**. The output of

**y(t)**tends to

**+ ∞.**This was caused by the

**positive root**of the denominator

**s2=+0.075**.

We examined

**2**transmittances

**G(s)**(

**Fig. 20-8**and

**20-9**) whose denominator

**M(s)**had

**real roots**. Or in other words, the

**Δ**of the

**denominator**was not

**negative**. Even just

**one positive**root is i

**nstability**! This can be generalized to

**M(s)**of any

**nth**degree.

**Conclusion**:

**G(s) is stable when all roots of the denominator M(s)=0 are negative**.

And when the

**Δ**is

**negative**, that is, when the

**roots**are

**complex numbers**?

**Chapter 20.4 Roots of the denominator M(s) of G(s) transmittance are complex numbers- What stability?****Chapter 20.4.1 Stable-All roots of the denominator have a negative real part**

**Fig. 20-10**

**Negative Δ**–> the

**complex roots**of

**M(s)=0**will appear. You can calculate similarly as in

**Fig.**

**20-5**. The

**real parts**of these roots are negative

**-1**. After a few seconds from the

**dirac**impulse, the system returned to

**equilibrium**. However, unlike

**Fig. 20-8**, this return occurred with

**oscillations**. Now we can modify the stability

**theorem**we learned earlier.

**The transmittance G(s) is stable when all the real parts of the roots of M(s)=0 are negative**

The expression “real parts” refers to elements that are

**complex numbers**and have a

**non-zero**imaginary part. Therefore, the above theorem is a generalization of the previous theorem.

**M(s)**polynomials are of course of any

**degree**.

You can also add how the

**stability**in

**Fig. 20-8**differs from the

**stability**in

**Fig. 23-10**:

**–Fig. 20-8**– Return to

**equilibrium**without oscillation. Then the roots are

**real**numbers.

**–Fig. 23-10**– Return with

**oscillation**. Then the roots are

**complex numbers**.

**Chapter 20.4.2 Stable- Roots of the denominator have a positive real part**

And when the **real parts** of **complex roots** are **positive**? Can you guess?

**Fig. 20-11**The real parts of these

**roots**with

**M(s)=0**are

**+0.05**. They are

**positive**.

As we assumed, the system is

**unstable**. The amplitude of

**oscillations**grows to infinity. The reason was a pair of

**complex roots**with

**positive real**parts.

**Chapter 20.5 Hurwitz Criterion – You don’t need to know the M(s) roots to assess stability!**

The transmittance **G(s)** of an open system (without negative feedback) is often in the **product** form, i.e. we know its **roots**. It is also **stable**. Including the transmittance **G(s)** in the **negative** feedback** loop** will change its value to **Gz(s)**. As you can see in the figure below, the denominator** M(s)** is no longer in the **product** form. For** M(s) second** degree this is not a problem, but for **higher** degrees it is not so nice. Formulas, if they exist at all, are no longer as simple as for a **quadratic equation**.

Fig. 20-12**-a** In an **open** system, the denominator **M(s)=(s+1)(s+2)** is in the product form. The roots** s1=-1, s2=-2** are visible. It is easy to judge that an open system is **stable****-b** closed system transmittance **Gz(s)** – intermediate result**-c** transmittance of a closed system **Gz(s)** – the final result. Here **M(s)** is in the form of a **polynomial**. Roots are not visible.

You have to study the **denominator** of a **closed** system **c**. So it’s not like in **Nyquist**, in which you study an **open** system. Again, let me remind you of the stability theorem.**The transmittance G(s) is stable when all the real parts of the roots of M(s)=0 are negative.**Otherwise

The

**roots**lie in the

**left**half-plane of

**complex numbers**, as, for example, in

**Fig. 20-6b**. So you don’t need to know the

**exact**values of the

**roots**. It is enough to know whether their

**real**parts are

**negative**. This is what the

**Hurwitz theorem**deals with.

**This is what it looks like Hurwitz’s**

Fig. 20-13

Fig. 20-13

**theorem**for degree

**5**

**polynomials**. The mechanism for a polynomial of

**degree 6.7,…n**works in a similar way.

I advise you to use this approach for all formulas “which, for

**any**n”, are simply

**not very transparent**. I personally check them for a specific

**n**, as above for

**n=5**. Then everything “sees better”.

If, Dear Reader, you are a high school student, you have the right to have a problem with

**W1, W2, W3**and

**W4**

**determinants**.

Approach this hedgehog as follows:

– Each array (or matrix) of

**degree n**is assigned a specific number called the

**determinant**of degree n i.e.

**Wn**

– The determinants

**W1**and

**W2**are calculated as in the above-mentioned drawing.

As for the next determinants, i.e.

**W3, W4, … Wn**is a monkey job. They are given in every

**matrix calculus textbook**.

**Chapter 20.6. Checking the stability of the “stable” transmittance using the Hurwitz Criterion**

**On**

Fig. 20-14

Fig. 20-14

**a**we have the transmittance

**G(s)**covered by

**negative**feedback

On

**b**we have the transmittance

**Gz(s)**taking into account the

**negative**feedback – immediately after applying the formula

On

**c**we have the transmittance after the final transformations.

How to check the stability of

**Gz(s)**using the

**Hurwitz criterion**?

We remember that

**Mr. Hurwitz**is only interested in the Gz(s)

**denominator.**

**It’s good that you don’t have to calculate the**

Fig. 20-15

Fig. 20-15

**W3**determinant according to the

**2nd**

**Hurwitz**condition, a little less work.

**Both**Hurwitz conditions are met -> the

**real**parts of the

**roots**of the

**Gz(s)**denominator in

**Fig. 20-15**are

**negative**-> the

**transmittance**is

**stable**. How about we check? It is exactly

**Gz(s)**from

**Fig. 20-14c**. We will check the stability by tapping lightly with a

**dirac**hammer. Will it return to a steady state, i.e. to

**y(t)=0**, as promised by Hurwitz?

**Fig. 20-16**It swinged and the system returned to a stable state where

**y(t)=0**.

One more thing.

**Hurwitz**says nothing about whether the

**roots**of

**M(s)=0**are,

**real**or

**complex**. More precisely, do they have imaginary parts. From the

**oscillations**it follows that some have. It also doesn’t say how stable the system is. How far from unstable.

**Chapter 20.7 Checking the stability of the “unstable” transmittance using the Hurwitz Criterion****Fig. 20-17**The

**Second Hurwitz**condition is not met, i.e. the system is

**unstable**. Let’s check it. Let’s try to unbalance the system with the Dirac hammer?

**Fig. 20-18**

**Hurwitz**was right.