StabilityRouth-Hurwitz Criterion

Unit 5: Stability

Engineering 5821:Control Systems I

Faculty of Engineering & Applied ScienceMemorial University of Newfoundland

February 26, 2010

ENGI 5821 Unit 5: Stability

StabilityRouth-Hurwitz Criterion

1 Stability

1 Routh-Hurwitz Criterion

Special Case: Zero in First Column

Special Case: Row of Zeros

Stability Design Example

ENGI 5821 Unit 5: Stability

Stability

System stability can be defined w.r.t. the stability of the naturalresponse, or of the total response (natural + forced).

Stability of the natural response:

If the natural response approaches 0 as t →∞ the system isstable. If the response approaches ∞ the system is unstable.

If the natural response remains constant or oscillates thesystem is marginally stable.

Stability of the total response. Sometimes referred to asbounded-input, bounded-output (BIBO) stability:

If every bounded input yields a bounded output, the system isstable.

If any bounded input yields an unbounded output, the systemis unstable.

Considering the natural response, the system is stable as long as allclosed-loop system poles lie only in the left half-plane (LHP):

Pure sinusoids

DC

Growing sinusoidsDecaying sinusoids

Decaying sinusoids

Growing exponentialsDecaying exponentials

Growing sinusoids

ωj

σ

STABLEMARGINALLY

STABLE UNSTABLE

Poles in the right half-plane (RHP) yield exponentialgrowth—hence instability.

Pairs of poles on the imaginary axis lead to undamped sinusoids:

A cos(ωt + φ)

The presence of such poles make the natural response (at best)marginally stable. If a pair of poles is repeated we obtainresponses of the form:

Atn cos(ωt + φ)

where n = 1 for the second repeated pair, 2 for the third . . . Thus,if poles on the imaginary axis are repeated the system is unstable.

System stability is determined entirely by the location of the rootsof the closed-loop transfer function’s denominator. For theexample above this polynomial was,

s3 + 3s2 + 2s + k

where k was either 3 or 7. With k = 3 the system was stable, butbecame unstable for k = 7. The roots of the denominatorpolynomial govern the system’s stability. These roots are locatedwherever the following characteristic equation holds,

s3 + 3s2 + 2s + k = 0

If the denominator polynomial has either of the followingcharacteristics, then it will have RHP roots and the system will beunstable:

If the polynomial coefficients differ in sign (i.e. not allpositive or all negative). We have stability iff thepolynomial can be factored as follows where the real part ofeach root ri is positive:

(s + r1)(s + r2) · · · (s + rn).

This will always generate a polynomial with all positivecoefficients (or all negative if we multiply everything by -1).So differently signed coefficients indicate instability.

If the polynomial has missing coefficients.

However, there are unstable polynomials with all cofficients ≥ 0.

One option is to determine the locations of the roots with Matlab.However, we will need to establish analytical criteria for stability...

Routh-Hurwitz CriterionThe Routh-Hurwitz criterion allows us to obtain a count of thenumber of poles in the LHP, RHP, and on the jω-axis. It does nottell us where the poles are, only how many there are of eachcategory.

We must first generate a Routh table. Assume our system is asfollows (we care only about the denominator):

Each row of the Routh table is labelled from the highestdenominator power down to s0. Then start with the highest powerand list every other coefficient in the first row of the table. In thesecond row list the skipped coefficients in order.

s4 a4 a2 a0

s3 a3 a1 0s2

s1

s0

We continue by filling in each entry as a negative determinantdivided by the first column of the row above. The left column ofthe determinant comes from the first entries of the two rowsabove. The right column comes from the elements of columnsabove and to the right.

s4 a4 a2 a0

s3 a3 a1 0

s2

−

˛̨̨̨˛̨ a4 a2

a3 a1

˛̨̨̨˛̨

a3= b1

−

˛̨̨̨˛̨ a4 a0

a3 0

˛̨̨̨˛̨

a3= b2

−

˛̨̨̨˛̨ a4 0

a3 0

˛̨̨̨˛̨

a3= 0

s1

−

˛̨̨̨˛̨ a3 a1

b1 b2

˛̨̨̨˛̨

b1= c1

−

˛̨̨̨˛̨ a3 0

b1 0

˛̨̨̨˛̨

b1= 0

−

˛̨̨̨˛̨ a3 0

b1 0

˛̨̨̨˛̨

b1= 0

s0

−

˛̨̨̨˛̨ b1 b2

c1 0

˛̨̨̨˛̨

c1= d1

−

˛̨̨̨˛̨ b1 0

c1 0

˛̨̨̨˛̨

c1= 0

−

˛̨̨̨˛̨ b1 0

c1 0

˛̨̨̨˛̨

c1= 0

e.g. G (s) = 1000s3+10s2+31s+1030

s3 1 31s2 10 1030

1 103

s1

−

˛̨̨̨˛̨ 1 31

1 103

˛̨̨̨˛̨

1 = −72

−

˛̨̨̨˛̨ 1 0

1 0

˛̨̨̨˛̨

1 = 0

s0

−

˛̨̨̨˛̨ 1 103−72 0

˛̨̨̨˛̨

−72 = 103

−

˛̨̨̨˛̨ 1 0−72 0

˛̨̨̨˛̨

−72 = 0

Note: For convenience, we can multiply any row by a positiveconstant. This was done in the s2 row above.

Routh Table Interpretation: The number of RHP roots equalsthe number of sign changes in the first column. (The proof of thisstatement is beyond the scope of this course)

For the example above there are two sign changes. Therefore thesystem has two RHP poles and is unstable.

StabilityRouth-Hurwitz Criterion

Special Case: Zero in First ColumnSpecial Case: Row of ZerosStability Design Example

Note that the width of the table is governed by the formula dn+12 e

where n is the order of the polynomial and n + 1 is the number ofcoefficients.

e.g. 7s7 + 6s6 + 5s5 + 4s4 + 3s3 + 2s2 + s + k requires 4 columns.

e.g. 8s8 + 7s7 + 6s6 + 5s5 + 4s4 + 3s3 + 2s2 + s + k requires 5columns.

You can forget this width rule as long as you remember to stopfilling in entries when a column of zeros is obtained.

ENGI 5821 Unit 5: Stability

Special Case: Zero in First Column

If the first column of any row becomes 0 we have to handle this asa special case. Replace the 0 with a small positive constant ε.Continue as normal...

e.g. s5 + 2s4 + 3s3 + 6s2 + 5s + 3

s5 1 3 5s4 2 6 3s3 ε 7

2 0s2 6ε−7

ε 3 0

s1 42ε−49−6ε2

12ε−14 0 0

s0 3 0 0

Signs: +, +, +, -, +, +

Thus, there are two sign changes and therefore two RHP poles.The system is unstable.

StabilityRouth-Hurwitz Criterion

Special Case: Zero in First ColumnSpecial Case: Row of ZerosStability Design Example

Special Case: Row of Zeros

A row of zeros (ROZ) will occur if there is an even polynomial(EP) that is a factor of the polynomial we are working on. Thecoefficients of this EP are found in the row above the zero row.The power of the EP is given by the row above the zero row. Wecontinue the Routh table by replacing the zero row with thecoefficients of the derivative of the EP.

e.g. s5 + 7s4 + 6s3 + 42s2 + 8s + 56

ENGI 5821 Unit 5: Stability

e.g. s5 + 7s4 + 6s3 + 42s2 + 8s + 56

s5 1 6 8s4 7 42 56 Multiply by 1/7

1 6 8s3 0 0 0 ROZ! Form EP = s4 + 6s2 + 8 from above

row’s cofficients. Differentiate to obtain4s3 + 12s.

4 12 0 Multipy by 1/41 3 0

s2 3 8 0s1 1/3 0 0s0 8 0 0

Interpretation: We must split our interpretation and consider firstthe part above the ROZ separately from the part below.Everything below the ROZ is a test of EP. In this case we see nosign changes above or below. Therefore there are no RHP poles. Isthe system stable or only marginally stable?

In the previous example the EP s4 + 6s2 + 8 was a factor of theoriginal polynomial. EP’s have the property that their roots arealways symmetric about the origin. These roots can be purely real,purely imaginary, or complex, but each root will have a matchlocated 180o opposite.

This means that a system with an even polynomial factor willeither be unstable (A or C) or marginally stable.

In our example s4 + 6s2 + 8 has four purely imaginary roots (twopairs of roots as in B). Therefore the system is marginally stable.

e.g. s8 + s7 + 12s6 + 22s5 + 39s4 + 59s3 + 48s2 + 38s + 20

s8 1 12 39 48 20s7 1 22 59 38 0s6 -10 -20 10 20 0 Multiply by 1/10

-1 -2 1 2 0s5 20 60 40 0 0 Multipy by 1/20

1 3 2 0 0s4 1 3 2 0 0s3 0 0 0 0 0 ROZ! EP = s4 + 3s2 + 2

4 6 0 0 0 Multiply by 1/42 3 0 0 0 Multiply by 1/2

s2 3/2 2 0 0 0 Multipy by 23 4 0 0 0

s1 1/3 0 0 0 0s0 4 0 0 0 0

Interpretation: Two sign changes above the line, implies two RHP poles. The system is unstable, but we may still

want to know where the poles are. Below the line there are no sign changes. Since this part comes from the EP

which has symmetrical roots, we know that all four of its roots must be on the jω-axis. The final two poles belong

to the original polynomial and must be in the LHP.

Stability Design Example

e.g. Find the range of gain K for the closed-loop transfer functionT (s) that will lead to stable, unstable, and marginally stablebehaviour. Assume K > 0.

T (s) =K

s3 + 18s2 + 77s + K

s3 1 77s2 18 K

s1 1386−K18 0

s0 K 0

If K < 1386 there are no sign changes and the system is stable.

If K > 1386 we have two sign changes and the system is unstable.

If K = 1386 we have a ROZ. The EP is 18s2 + 1386. Differentiateto obtain 36s. Replacing the ROZ with [36, 0] the next rowbecomes [1386, 0]. There are no sign changes so the EP must haveboth roots on the jω axis. Thus, the system is marginally stable.