routh hurwitz handout

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Tim Davidson

StabilityCondition in terms ofpoles

Condition in terms ofdenominatorcoefficients

Routh HurwitzconditionBasics

Disk drive example

Dealing with zeros

Zeros in first column

Zero rows

Using RouthHurwitz fordesignTurning control of atracked vehicle

EE3CL4:Introduction to Linear Control SystemsSection 4: Stability and Routh-Hurwitz Condition

Tim Davidson

McMaster University

Winter 2015

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Tim Davidson




Disk drive example

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Outline

1 StabilityCondition in terms of polesCondition in terms of denominator coefficients

2 Routh Hurwitz conditionBasicsDisk drive exampleDealing with zeros

Zeros in first columnZero rows

3 Using Routh Hurwitz for designTurning control of a tracked vehicle

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Stability

A systems is said to be stable if all bounded inputs r(t) giverise to bounded outputs y(t)

Counterexamples Albert Collins, Jeff Beck (Yardbirds),

Pete Townshend (The Who), Jimi Hendrix,Tom Morello (Rage Against the Machine),Kurt Cobain (Nirvana)

Tacoma Narrows

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Conditions for stability

y(t) =

g()r(t )d

Let r(t) be such that |r(t)| r

|y(t)| =

g()r(t )d

g()r(t )d r

g()dUsing this: system G(s) is stable iff

g()d is finite

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Condition in terms of poles?We want

g()d to be finite

Can we determine this from G(s)?

We can write a general rational transfer function in the form

G(s) =K

i(s + zi)sN

k (s + k )

m(s2 + 2ms + (2m + 2m))

Poles: 0, k , m jmAssuming N = 0 and no repeated roots, the impulse response is

g(t) =k

Akek t +m

Bmem t sin(mt + m)

Stability requires |g(t)|dt to be bounded;

that requires k > 0, m > 0In fact, system is stable iff poles have negative real parts

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Marginal stability

Consider integrator: G(s) = 1/s; simple pole at origin y(t) =

r(t)dt

if r(t) = cos(t), which is bounded,then y(t) = sin(t). Bounded

If r(t) = u(t), which is bounded,then y(t) = t . Not bounded

Consider G(s) = 1/(s2 + 1), simple poles at s = j1 Unit step response: u(t) cos(t). Bounded What if r(t) is a sinusoid of frequency 1/(2pi) Hz?

Not bounded

If G(s) has a pole with positive real part,or a repeated pole on j-axisoutput is always unbounded

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Routh-Hurwitz condition

We have seen how to determine stability from the poles.

Much easier than having to find impulse responseand then determining if

|g()|d

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Let G(s) = p(s)q(s) , where

q(s) = ansn + an1sn1 + . . . a1s + a0= an(s r1)(s r2) . . . (s rn)

where ri are the roots of q(s) = 0.

By multiplying out, q(s) = 0 can be written as

q(s) = ansn an(r1 + r2 + + rn)sn1+ an(r1r2 + r2r3 + . . . )sn2

an(r1r2r3 + r1r2r4 + . . . )sn3+ + (1)nan(r1r2r3 . . . rn) = 0

If all ri are real and in left half plane, what is sign of coeffs of sk?the same!

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That observation leads to a necessary condition.

Hence, not that useful for design

A more sophisticated analysis leads to the Routh-Hurwitzcondition, which is necessary and sufficient

Hence, can be quite useful for design

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R-H cond: A first lookConsider G(s) = p(s)q(s) . Poles are solutions to q(s) = 0; i.e.,

ansn + an1sn1 + an2sn2 + + a1s + a0 = 0

Construct a table of the formsn an an2 an4 . . .sn1 an1 an3 an5 . . .sn2 bn1 bn3 bn5 . . .sn3 cn1 cn3 cn5 . . .

......

...... . . .

s0 hn1

where

bn1 =an1an2 anan3

an1=1an1

an an2an1 an3

bn3 =1an1

an an4an1 an5 cn1 = 1bn1

an1 an3bn1 bn3

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R-H cond: A first look

Now consider the table that we have just constructed

sn an an2 an4 . . .sn1 an1 an3 an5 . . .sn2 bn1 bn3 bn5 . . .sn3 cn1 cn3 cn5 . . .

......

...... . . .

s0 hn1

Loosely speaking:

Number of roots in the right half plane is equal to the numberof sign changes in the first column of the table

Stability iff no sign changes in the first columnNow lets move towards a more sophisticated statement

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Stability (Revision)

Let G(s) = p(s)q(s) , where

q(s) = ansn + an1sn1 + . . . a1s + a0

System is stable iff all poles of G(s) have negative real parts

Recall, poles are solutions to q(s) = 0

Can we find a necessary and sufficient condition that dependsonly on ak so that we dont have to solve q(s) = 0?

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Routh-Hurwitz condition1 Consider q(s) with an > 0

ansn + an1sn1 + an2sn2 + . . . a1s + a0 = 0

2 Construct a table of the form

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1 bn3 bn5 . . .Row n 3 cn1 cn3 cn5 . . ....

......

... . . .Row 0 hn1

Procedure provided on the following slides

3 Count the sign changes in the first column

4 That is the number of roots in the right half plane

Stability (poles in LHP) iff ak > 0 and all terms in first col. > 0

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Constructing RH table

ansn + an1sn1 + an2sn2 + . . . a1s + a0 = 0

Step 2.1: Arrange coefficients of q(s) in first two rows

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .

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Interlude

Determinant of a 2 2 matrix: a bc d = ad cb

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Step 2.2: Construct 3rd row using determinants of 2 2matrices constructed from rows above

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1

bn1 =1an1

an an2an1 an3

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Step 2.2, cont: Construct 3rd row using determinants of2 2 matrices constructed from rows above

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1 bn3 . . .

bn3 =1an1

an an4an1 an5

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Step 2.3: Construct 4th row using determinants of 2 2matrices constructed from rows above

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1 bn3 . . .Row n 3 cn1 . . .

cn1 =1bn1

an1 an3bn1 bn3

Step 2.4: Continue in this pattern.

Caveat: Requires all elements of first column to be non-zeroWill come back to that. Lets see some examples, first

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RH table, second-order system

q(s) = a2s2 + a1s + a0

Row 2 a2 a0Row 1 a1 0Row 0 b1

b1 =1a1

a2 a0a1 0 = a0

Therefore, second order system is stable iff all three denominatorcoefficients have the same sign

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RH table, third order system

q(s) = a3s3 + a2s2 + a1s + a0

Row 3 a3 a1Row 2 a2 a0Row 1 b1 0Row 0 c1 0

b1 =1a2

a3 a1a2 a0 c1 = 1b1

a2 a0b1 0 = a0

Therefore, if a3 > 0, necessary and sufficient condition forthird-order system to be stable is that a2 > 0, b1 > 0 and a0 > 0.

b1 > 0 is equiv. to a2a1 > a0a3, and this implies a1 > 0.

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Disk drive read controlAdd velocity feedback (switch closed)

Using block diagram manipulation

G1(s) =5000

s + 1000G2(s) =

1s(s + 20)

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Closed loop

T (s) =Y (s)R(s)

=KaG1(s)G2(s)

1 + KaG1(s)G2(s)(1 + K1s)

Hence, char. eqn:s3 + 1020s2 + (20000 + 5000KaK1)s + 5000Ka = 0

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Stabilizing values of K1 and Ka

s3 + 1020s2 + (20000 + 5000KaK1)s + 5000Ka = 0

Routh table

Row 3 1 20000 + 5000KaK1Row 2 1020 5000KaRow 1 b1Row 0 5000Ka

b1 =1020(20000 + 5000KaK1) 5000Ka

1020

For stability we require b1 > 0 and Ka > 0

For example, Ka = 100 and K1 = 0.05.That pair gives a 2% settling time of 260ms

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Reminder: Constructionprocedure

Row k + 2 p1 p3 p5 . . .Row k + 1 q1 q3 q5 . . .Row k r1 r3 r5 . . .

To compute r3, multiply 1first element of previous row = 1q1 by determinant of 22 matrix formed in the following way:

The first column contains the first elements of the tworows above the element to be calculated

The second column contains the elements of the tworows above that lie one column to the right of theelement to be calculated

Therefore

r3 =1q1

p1 p5q1 q5 = 1q1 (p1q5 q1p5)

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RH table, dealing with zeros

The Routh-Hurwitz table encounters trouble when thereis a zero in the first column

The next row involves (1/0) times a determinant When some other elements in that row are not zero, we

can proceed by replacing the zero by a small positivenumber , and then taking the limit as 0 after thetable has been constructed.

When a whole row is zero, we need to be a bit moresophisticated (later)

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RH table, zero first element innon-zero row

As an example, consider

q(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10

Routh table

Row 5 1 2 11Row 4 2 4 10Row 3 0 6 0Row 2 c1 10 0Row 1 d1 0 0Row 0 10 0 0

c1 =4 12

=12

d1 =6c1 10

c1 6

Two sign changes, hence unstable with two RHP poles

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Zero row It is possible that the Routh Hurwitz procedure can

produce a zero row

While this complicates the procedure, it yields usefulinformation for design

Zero rows occur when polynomial has roots that areradially symmetric.

Since the roots must also occur in conjugate pairs, thismeans that there is at least one pair of roots that issymmetric in the imaginary axis.(All roots are symmetric in the real axis.)

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Zero row

Common examples include: equal and opposite roots on the real axis, a pair of complex conjugate roots on the imaginary axis.

The latter is more common, and more useful in design

So how can we deal with the zero row?

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Dealing with a zero row Routh Hurwitz procedure provides an auxiliary polynomial,a(s), that contains the roots of interest as factors

The auxiliary polynomial is a factor of the original polynomial;i.e., q(s) = a(s)b(s); b(s) can be found by polyn. division

Given the symmetry of the roots, the auxiliary polynomial isof even order (or will have all its roots at the origin)

The coefficients of the auxiliary polynomial appear in the rowabove the zero row

Let k denote the row number of the row above the zero row,and let ck,1, ck,2, ck,3 denote the coefficients in that row. Theauxiliary polynomial is constructed as

a(s) = ck,1sk + ck,2sk2 + ck,3sk4 + . . .

Finally, we replace the zero row by the coefficients of thederivative of the auxiliary polynomial; i.e., the zero row isreplaced by kck,1, (k 2)ck,2, (k 4)ck,3

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Zero row example q(s) = s5 + 2s4 + 24s3 + 48s2 25s 50 = 0 Construct table

Row 5 1 24 25Row 4 2 48 50Row 3 0 0

Auxiliary polynomial: a(s) = 2s4 + 48s2 50This is actually a factor of q(s).Using quadratic formula, roots of a(s) are s2 = 1,25Hence roots of a(s) are s = 1,j5

da(s)ds = 8s3 + 96s.Replace zero row by these coefficients

Row 5 1 24 25Row 4 2 48 50Row 3 8 96

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Zero row example, cont Now complete the table in the usual way

Row 5 1 24 25Row 4 2 48 50Row 3 8 96Row 2 24 50Row 1 112.7 0Row 0 50

One sign change in first column.Indicates one root in right half plane.

Recall a(s) is a factor of q(s).Indeed, by polyn division q(s) = (s + 2)a(s)We have seen that roots of a(s) are 1 and j5.

Hence q(s) does indeed have one root with a positivereal part.

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Turning control of a trackedvehicle

Select K and a so that the closed-loop is stable, and the steady-state error due to a ramp is at most 24% of

the magnitude of the command

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Deal with stability first

Transfer function: T (s) = Gc(s)G(s)1+Gc(s)G(s)

Char. equation: s4 + 8s3 + 17s2 + (K + 10)s + Ka = 0

Routh table

Row 4 1 17 KaRow 3 8 K+10Row 2 b3 KaRow 1 c3Row 0 Ka

b3 =126 K

8c3 =

b3(K + 10) 8Kab3

For stability we require b3 > 0, c3 > 0 and Ka > 0

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Stability regionThese constraints can be rewritten as

K < 126Ka > 0

(K + 10)(126 K ) 64Ka > 0For positive K , last constraint becomes a < (K+10)(126K )64K Region of stable parameters

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Steady-state error to ramp

For a ramp input r(t) = At , we have ess = A/Kv , where

Kv = lims0

sGc(s)G(s) = Ka/10

Therefore, ess = 10A/(Ka)

To obtain ess < 0.24A, we need Ka > 10/0.24 41.67,

Any (K ,a) pair in stable region with Ka > 41.67 willsatisfy design constraints

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Set of parameters with desiredperformance

For positive K , stability region is below the blue solid curve desired steady-state error region is above the red

dashed curve and below the blue solid curve Design example: (70,0.6)

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Ramp responseRamp response for case of K = 70 and a = 0.6

StabilityCondition in terms of polesCondition in terms of denominator coefficients

Routh Hurwitz conditionBasicsDisk drive exampleDealing with zeros

Using Routh Hurwitz for designTurning control of a tracked vehicle

routh hurwitz handout

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