Counterexamples. ⢠Albert Collins, Jeff Beck (Yardbirds),. Pete Townshend (The Who), Jimi Hendrix,. Tom Morello (Rage Against the Machine),. Kurt Cobain ...
EE 3CL4, §4 1 / 38 Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example
EE3CL4: Introduction to Linear Control Systems Section 4: Stability and Routh-Hurwitz Condition
Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Tim Davidson McMaster University
Winter 2014
EE 3CL4, §4 2 / 38
Outline
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
1 Stability
Condition in terms of poles Condition in terms of denominator coefficients
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
2 Routh Hurwitz condition
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
3 Using Routh Hurwitz for design
Turning control of a tracked vehicle
EE 3CL4, §4 4 / 38
Stability
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
A systems is said to be stable if all bounded inputs r (t) give rise 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
EE 3CL4, §4 5 / 38 Tim Davidson
Conditions for stability
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example
Z
∞
g(τ )r (t − τ ) dτ
y (t) = −∞
Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Let r (t) be such that |r (t)| ≤ ¯r Z ∞ |y (t)| = g(τ )r (t − τ ) dτ −∞ Z ∞ g(τ )r (t − τ ) dτ ≤ −∞ Z ∞ g(τ ) dτ ≤ ¯r −∞
Using this: system G(s) is stable iff
R∞ g(τ ) dτ is finite −∞
EE 3CL4, §4 6 / 38 Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column
Condition in terms of poles? R∞ We want −∞ g(τ ) dτ to be finite Can we determine this from G(s)? We can write a general rational transfer function in the form Q K i (s + zi ) Q G(s) = N Q 2 2 2 s k (s + σk ) m (s + 2αm s + (αm + ωm ))
Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Poles: 0, −σk , −αm ± jωm Assuming N = 0 and no repeated roots, the impulse response is X X g(t) = Ak e−σk t + Bm e−αm t sin(ωm t + θm ) k
R
m
Stability requires |g(t)| dt to be bounded; that requires σk > 0, αm > 0 In fact, system is stable iff poles have negative real parts
EE 3CL4, §4 7 / 38 Tim Davidson
Marginal stability
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column
• Consider integrator: G(s) = 1/s; simple pole at origin R∞ • 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
Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
• Consider G(s) = 1/(s 2 + 1), simple poles at s = ±j1 • Unit step response: u(t) − cos(t). Bounded • What if r (t) is a sinusoid of frequency 1/(2π) Hz? Not bounded
If G(s) has a pole with positive real part, or a repeated pole on jω-axis output is always unbounded
EE 3CL4, §4 8 / 38 Tim Davidson
Routh-Hurwitz condition
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
We have seen how to determine stability from the poles.
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Much easier than havingR to find impulse response and then determining if |g(τ )| dτ < ∞ Can we determine stability without having to determine the poles? Yes! Routh-Hurwitz condition
EE 3CL4, §4 9 / 38
Routh-Hurwitz condition
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
Let G(s) =
p(s) q(s) ,
where
q(s) = an sn + an−1 sn−1 + . . . a1 s + a0 = an (s − r1 )(s − r2 ) . . . (s − rn )
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
where ri are the roots of q(s) = 0. By multiplying out, q(s) = 0 can be written as q(s) = an sn − an (r1 + r2 + · · · + rn )sn−1 + an (r1 r2 + r2 r3 + . . . )sn−2 − an (r1 r2 r3 + r1 r2 r4 + . . . )sn−3 + · · · + (−1)n an (r1 r2 r3 . . . rn ) = 0 If all ri are real and in left half plane, what is sign of coeffs of sk ? the same!
EE 3CL4, §4 10 / 38 Tim Davidson
Routh-Hurwitz condition
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics
That observation leads to a necessary condition.
Disk drive example Dealing with zeros Zeros in first column
Hence, not that useful for design
Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
A more sophisticated analysis leads to the Routh-Hurwitz condition, which is necessary and sufficient Hence, can be quite useful for design
EE 3CL4, §4 11 / 38 Tim Davidson
Consider G(s) = Stability Condition in terms of poles
R-H cond: A first look Poles are solutions to q(s) = 0; i.e.,
an sn + an−1 sn−1 + an−2 sn−2 + · · · + a1 s + a0 = 0
Condition in terms of denominator coefficients
Routh Hurwitz condition
p(s) q(s) .
Construct a table of the form
Basics
sn
Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
sn−1 sn−2 sn−3 .. .
an an−1 bn−1 cn−1 .. .
s0
hn−1
an−2 an−3 bn−3 cn−3 .. .
an−4 an−5 bn−5 cn−5 .. .
... ... ... ... ...
where
bn−3
an−1 an−2 − an an−3 −1 an an−2 = bn−1 = an−1 an−1 an−1 an−3 −1 an −1 an−1 an−4 = cn−1 = an−1 an−1 an−5 bn−1 bn−1
an−3 bn−3
EE 3CL4, §4 12 / 38
R-H cond: A first look
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Now consider the table that we have just constructed sn sn−1 sn−2 sn−3 .. .
an an−1 bn−1 cn−1 .. .
s0
hn−1
an−2 an−3 bn−3 cn−3 .. .
an−4 an−5 bn−5 cn−5 .. .
... ... ... ... ...
Loosely speaking: • Number of roots in the right half plane is equal to the number of sign changes in the first column of the table • Stability iff no sign changes in the first column Now let’s move towards a more sophisticated statement
EE 3CL4, §4 14 / 38
Stability (Revision)
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column
Let G(s) =
p(s) q(s) ,
where
Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
q(s) = an sn + an−1 sn−1 + . . . a1 s + 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 depends only on ak so that we don’t have to solve q(s) = 0?
EE 3CL4, §4 15 / 38
Routh-Hurwitz condition
Tim Davidson Stability
1
Condition in terms of poles
an sn + an−1 sn−1 + an−2 sn−2 + . . . a1 s + a0 = 0
Condition in terms of denominator coefficients
Routh Hurwitz condition Basics
Consider q(s) with an > 0
2
Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Construct a table of the form Row n Row n − 1 Row n − 2 Row n − 3 .. .
an an−1 bn−1 cn−1 .. .
Row 0
hn−1
an−2 an−3 bn−3 cn−3 .. .
an−4 an−5 bn−5 cn−5 .. .
... ... ... ... ...
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
EE 3CL4, §4 16 / 38 Tim Davidson
Constructing RH table
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics
an sn + an−1 sn−1 + an−2 sn−2 + . . . a1 s + a0 = 0
Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Step 2.1: Arrange coefficients of q(s) in first two rows Row n Row n − 1
an an−1
an−2 an−3
an−4 an−5
... ...
EE 3CL4, §4 17 / 38 Tim Davidson
Interlude
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Determinant of a 2 × 2 matrix: a b c d = ad − cb
EE 3CL4, §4 18 / 38
Constructing RH table
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
Step 2.2: Construct 3rd row using determinants of 2 × 2 matrices constructed from rows above
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design
Row n Row n − 1 Row n − 2
an an−1 bn−1
an−2 an−3
an−4 an−5
Turning control of a tracked vehicle
bn−1
−1 an = an−1 an−1
an−2 an−3
... ...
EE 3CL4, §4 19 / 38
Constructing RH table
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
Step 2.2, cont: Construct 3rd row using determinants of 2 × 2 matrices constructed from rows above
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design
Row n Row n − 1 Row n − 2
an an−1 bn−1
an−2 an−3 bn−3
an−4 an−5 ...
Turning control of a tracked vehicle
bn−3
−1 an = an−1 an−1
an−4 an−5
... ...
EE 3CL4, §4 20 / 38 Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Constructing RH table Step 2.3: Construct 4th row using determinants of 2 × 2 matrices constructed from rows above Row n Row n − 1 Row n − 2 Row n − 3
Using Routh Hurwitz for design Turning control of a tracked vehicle
cn−1 =
an an−1 bn−1 cn−1
an−2 an−3 bn−3 ...
−1 an−1 bn−1 bn−1
an−4 an−5 ...
... ...
an−3 bn−3
Step 2.4: Continue in this pattern. Caveat: Requires all elements of first column to be non-zero Will come back to that. Let’s see some examples, first
EE 3CL4, §4 21 / 38 Tim Davidson
RH table, second-order system
Stability Condition in terms of poles Condition in terms of denominator coefficients
q(s) = a2 s2 + a1 s + a0
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Row 2 Row 1 Row 0 b1 =
−1 a2 a1 a1
a2 a1 b1
a0 0
a0 = a0 0
Therefore, second order system is stable iff all three denominator coefficients have the same sign
EE 3CL4, §4 22 / 38
RH table, third order system
Tim Davidson Stability Condition in terms of poles
q(s) = a3 s3 + a2 s2 + a1 s + a0
Condition in terms of denominator coefficients
Routh Hurwitz condition
Row 3 Row 2 Row 1 Row 0
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
b1 =
−1 a3 a2 a2
a1 a0
a3 a2 b1 c1
c1 =
a1 a0 0 0
−1 a2 b1 b1
a0 = a0 0
Therefore, if a3 > 0, necessary and sufficient condition for third-order system to be stable is that a2 > 0, b1 > 0 and a0 > 0. b1 > 0 is equiv. to a2 a1 > a0 a3 , and this implies a1 > 0.
EE 3CL4, §4 23 / 38
Disk drive read control
Tim Davidson Stability
Add velocity feedback (switch closed)
Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design
Using block diagram manipulation
Turning control of a tracked vehicle
G1 (s) =
5000 s + 1000
G2 (s) =
1 s(s + 20)
EE 3CL4, §4 24 / 38
Closed loop
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
T (s) =
Y (s) Ka G1 (s)G2 (s) = R(s) 1 + Ka G1 (s)G2 (s)(1 + K1 s)
Hence, char. eqn: s3 + 1020s2 + (20000 + 5000Ka K1 )s + 5000Ka = 0
EE 3CL4, §4 25 / 38 Tim Davidson
Stabilizing values of K1 and Ka
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
s3 + 1020s2 + (20000 + 5000Ka K1 )s + 5000Ka = 0 Routh table Row 3 Row 2 Row 1 Row 0 b1 =
1 1020 b1 5000Ka
20000 + 5000Ka K1 5000Ka
1020(20000 + 5000Ka K1 ) − 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
EE 3CL4, §4 26 / 38 Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
Reminder: Construction procedure Row k + 2 Row k + 1 Row k
p1 q1 r1
p3 q3 r3
p5 q5 r5
... ... ...
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
To compute r3 , multiply −1 • first element −1 of previous row = q1 by • determinant of 2 × 2 matrix formed in the following way: • The first column contains the first elements of the two rows above the element to be calculated • The second column contains the elements of the two rows above that lie one column to the right of the element to be calculated • Therefore
−1 p1 p5 −1 r3 = p1 q5 − q1 p5 = q1 q1 q5 q1
EE 3CL4, §4 27 / 38 Tim Davidson
RH table, dealing with zeros
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
• The Routh-Hurwitz table encounters trouble when there
is a zero in the first column
Basics Disk drive example Dealing with zeros
• The next row involves (−1/0) times a determinant
Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
• When some other elements in that row are not zero, we
can proceed by replacing the zero by a small positive number , and then taking the limit as → 0 after the table has been constructed. • When a whole row is zero, we need to be a bit more
sophisticated (later)
EE 3CL4, §4 28 / 38
RH table, zero first element in non-zero row
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition
As an example, consider q(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10
Basics Disk drive example Dealing with zeros Zeros in first column
Routh table
Zero rows
Row 5 Row 4 Row 3 Row 2 Row 1 Row 0
Using Routh Hurwitz for design Turning control of a tracked vehicle
c1 =
1 2 0← c1 d1 10
4 − 12 −12 =
2 4 6 10 0 0 d1 =
11 10 0 0 0 0 6c1 − 10 →6 c1
Two sign changes, hence unstable with two RHP poles
EE 3CL4, §4 29 / 38
Zero row
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics
• It is possible that the Routh Hurwitz procedure can
produce a zero row • While this complicates the procedure, it yields useful
information for design
Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
• Zero rows occur when polynomial has either equal and
opposite roots on the real axis, or 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 this? • Routh Hurwitz procedure provides an “auxiliary polynomial” that contains the roots of interest as factors • The coefficients of this polynomial appear in the row above the zero row • We replace the zero row by the coefficients of the derivative of the auxiliary polynomial
EE 3CL4, §4 30 / 38
Zero row example
Tim Davidson Stability
• q(s) = s 5 + 2s 4 + 24s 3 + 48s 2 − 25s − 50 = 0
Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example
• Construct table
Row 5 Row 4 Row 3
1 2 0
24 48 0
−25 −50
Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
˜ (s) = 2s4 + 48s2 − 50 • Auxiliary polynomial: q This is actually a factor of q(s). ˜ (s) are s2 = 1, −25 Using quadratic formula, roots of q ˜ (s) are s = ±1, ±j5 Hence roots of q ˜ • d qds(s) = 8s 3 + 96s.
Replace zero row by these coefficients Row 5 1 24 −25 Row 4 2 48 −50 Row 3 8 96
EE 3CL4, §4 31 / 38
Zero row example, cont
Tim Davidson Stability
• Now complete the table in the usual way
Condition in terms of poles
Row 5 Row 4 Row 3 Row 2 Row 1 Row 0
Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
1 2 8 24 112.7 −50
24 48 96 −50 0
−25 −50
• One sign change in first column.
Indicates one root in right half plane. ˜ (s) is a factor of q(s). • Recall q ˜ (s) Indeed, by polyn division q(s) = (s + 2)q ˜ (s) are ±1 and ±j5. We have seen that roots of q • Hence q(s) does indeed have one root with a positive
real part.
EE 3CL4, §4 33 / 38 Tim Davidson Stability
Turning control of a tracked vehicle
Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
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
EE 3CL4, §4 34 / 38
Deal with stability first
Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics
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
Disk drive example
Row 4 Row 3 Row 2 Row 1 Row 0
Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
b3 =
126 − K 8
1 8 b3 c3 Ka
17 K+10 Ka
c3 =
Ka
b3 (K + 10) − 8Ka b3
For stability we require b3 > 0, c3 > 0 and Ka > 0
EE 3CL4, §4 35 / 38
Stability region
Tim Davidson Stability
These constraints can be rewritten as
Condition in terms of poles
K < 126
Condition in terms of denominator coefficients
Routh Hurwitz condition
Ka > 0 (K + 10)(126 − K ) − 64Ka > 0
Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
For positive K , ) • last constraint becomes a < (K +10)(126−K 64K • Region of stable parameters
EE 3CL4, §4 36 / 38 Tim Davidson
Steady-state error to ramp
Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example
• For a ramp input r (t) = At, we have ess = A/Kv , where
Kv = lim sGc (s)G(s) = Ka/10 s→0
Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
• 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 will
satisfy design constraints
EE 3CL4, §4 37 / 38 Tim Davidson Stability
Set of parameters with desired performance
Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
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)
EE 3CL4, §4 38 / 38 Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients
Routh Hurwitz condition Basics Disk drive example Dealing with zeros Zeros in first column Zero rows
Using Routh Hurwitz for design Turning control of a tracked vehicle
Ramp response Ramp response for case of K = 70 and a = 0.6
