Outline
•
•
•
•
•
Stability of Digital
Control Systems
M. Sami Fadali
Professor of Electrical Engineering
UNR
Asymptotic stability.
Input-output stability.
Stability conditions.
Routh-Hurwitz criterion.
Jury test.
1
2
Bounded-Input-Bounded-Output
(BIBO) Stability
Asymptotic Stability
Response due to any initial conditions
decays to zero asymptotically in the steady
state
Lim
The response due to any bounded input
Marginal Stability: response due to any
initial conditions remains bounded but does
not decay to zero.
remains bounded.
→
3
4
Stable Z-Domain Pole
Locations
Time Sequence
Z
Sampled exponential and its z-transform
with real or complex
Z
 Bounded sequence for poles in the closed unit
disc (i.e. on or inside the unit circle).
 Sequence decays exponentially for poles in the
open unit disc (i.e. inside the unit circle).
 Unbounded sequence for repeated poles on the
unit circle.
 For real time sequences poles and partial fraction
coefficients are either real or complex conjugate
pairs.
_|Å
→
5
Proof
Theorem: Asymptotic Stability
• Response of the system due to ICs
In the absence of pole-zero cancellation, a
LTI digital system is asymptotically stable if
its transfer function poles are in the open unit
disc and marginally stable if the poles are in
the closed unit disc with no repeated poles
on the unit circle.
LTI Model:
1
⋯
1
1
⋯
6
1
0,1,2, …
7
0 ,
1 ,…,
1. Assume no pole-zero cancellation.
denominator=denominator of z-transfer function.
= poles of transfer function.
3. Poles of
due to ICs: bounded for system poles in the
4.
closed unit disc unit (no repeated poles on the unit
circle) & decays exponentially for system poles in the
open unit disc (inside the unit circle).
8
Example
Solution (
)
Determine the asymptotic stability of the
following systems:
• Pole outside the unit circle  unstable
•
•
Use Theorem 1 (a) and (b) without pole-zero
cancellation.
Ignore zeros, (do not affect response due to ICs)
• All poles inside the unit circle 
asymptotically stable.
9
10
BIBO Stability
Solution
• All poles inside the unit circle 
asymptotically stable.
Is system BIBO stable if its impulse response
is
bounded? NO
Counterexample: bounded & strictly positive impulse
response
Bounded input:
Unbounded output:
• One pole on the unit circle  marginally
stable: not asymptotically stable.
11
12
Theorem 4.2:BIBO Stability
Proof of Necessity (Only if)
A discrete-time linear system is BIBO
stable if and only if its impulse response
sequence is absolutely summable i.e.
• Assume the system is BIBO stable but the
impulse response is not absolutely summable.
• Input
• Contradiction: unbounded output with a bounded
input.
13
14
Proof of Sufficiency (If)
Theorem 4.3: BIBO Stability
• Assume an absolutely summable impulse
response and show that the system is BIBO
stable.
in the convolution
• Use the input bound
summation
A discrete-time linear system is BIBO
stable if and only if the poles of its transfer
function lie inside the unit circle.
15
16
Proof of Sufficiency
Proof of Necessity
Z
• Impulse response is bounded if the poles of the
transfer function are in the closed unit disc and
decays exponentially if the poles are in the open
unit disc.
• Assume exponentially decaying impulse
response (i.e. poles inside the unit circle).
is the coefficient of largest magnitude
is the largest pole magnitude.
• The impulse response is bounded by
• Systems with a bounded impulse response that
does not decay exponentially are not BIBO
stable.
17
∞
Example 4.2
Solution
Investigate the BIBO stability of systems
with the impulse response
BIBO stable since the impulse response
satisfies
where
is a finite constant.
18
• Let = upper bound for any impulse
response of finite duration.
• Any FIR system is BIBO stable.
19
20
Example 4.3
Solution
Investigate BIBO stability for Example 4.1
After pole-zero cancellation
a) BIBO stable, all poles inside unit circle.
b) BIBO stable, all poles inside unit circle.
c) BIBO stable, all poles inside unit circle.
• Use Theorem 4.3 with pole-zero cancellation.
21
Solution
22
Z-plane Stable Pole Locations
Im[z]
Unit Circle
• Not BIBO stable, a pole on unit circle.
STABLE
23
Re[z]
24
MATLAB: ddamp
MATLAB Stability Determination
Obtain roots of polynomial:
» roots(den) % denominator coeffts. den
» zpk(g) %g = transfer function
Gives the pole locations,
and
Stable for roots inside the unit circle.
• Closed-loop transfer function
>> H = feedback(gforward, gfeedback, 1)
• Default: negative feedback
Eigenvalue
Magnitude Equiv. Damping
0.2306 + 0.7428I
0.7778
0.1941
0.2306 - 0.7428I
0.7778
0.1941
-0.6612
0.6612
0.1306
Equiv. Freq. (rad/sec)
12.9441
12.9441
31.6871
25
26
Internal Stability
Theorem 4-4
If all the transfer functions that relate the system
inputs ( and ) to the possible system outputs
( and ) are BIBO stable, then the system is
internally stable.
1
1
1
R(z)
R(z)
1
U(z)
C(z)
D(z
E(z)
+
U(z)
C(z)
Y(z)
GZAS(z)

D(z
E(z)
+
The system of Figure 4-3 is internally stable if
and only if all the closed-loop poles are in the
open unit disc.
Y(z)
GZAS(z)

27
28
Transfer Functions
Proof (Necessity: only if)
• Substitute
• Coprime polynomials (no common factors)
• Internally stable (all transfer functions are
asymptotically stable) + coprime
R(z)
D(z
E(z)
+
U(z)
C(z)
characteristic polynomial
no zeros on or outside the unit circle.
Y(z)
GZAS(z)

has
29
30
Theorem 4-5
Proof (Sufficiency: if)
The system is internally stable if and only if:
• Substitute
1. The characteristic polynomial
has no zeros on or outside the unit circle.
has no pole-zero
2. The loop gain
cancellation on or outside the unit circle .
• Characteristic polynomial
• No zeros on or outside the unit circle
R(z)
all the transfer functions are asymptotically
stable system is internally stable.
31
D(z
E(z)
+
U(z)
C(z)
Y(z)
GZAS(z)

32
Proof: Necessity
Proof: Sufficiency
: no zeros on or outside the
unit circle for internal stability (Theorem 4.4)
cancels unstable
(ii)
is both an open-loop and a closed-loop pole
i. No zeros on or outside the unit circle
ii. No unstable pole-zero cancellation for the
closed-loop system if there is no unstable
pole-zero cancellation for
Sufficient for internal stability (Theorem 4.4)
33
34
Solution
Example 4-4
• Discretized process transfer function
Transfer function of isothermal chemical
reactor
Z
o Determine
with
o Verify that the resulting feedback system is
not internally stable with the feedback
controller
• Transfer function from reference input to output
• All its poles are inside the unit circle (some
cancel)
35
36
Solution: Control Variable
Solution: Theorem 4-5
• Violates condition (ii) of Theorem 4-5:
Unstable pole at 1.334 cancels in the loop
gain.
• Pole at 1.334 outside the unit circle.
• The control variable is unbounded even
when the reference input is bounded.
37
Routh-Hurwitz Criterion
38
Geometric Interpretation
Im(s)
1. Transform the inside of the unit circle to
the LHP (bilinear transformation).
2. Use the Routh-Hurwitz criterion for the
investigation of discrete-time system
stability.
z3
Im(z)
z3  1
z2  1
X s3
z3  1
z2
s2 X
Re(s)
Re(z)
1
s1 X
z2  1
z1  1
1
z1
z1  1
s   z  1   z  1
39
 90  inside unit circle

s  90  outside unit circle
 90  on unit circle

40
Advantages/Disadvantages
Example 4.4
• Easy stability test for low-order polynomials.
• Difficult for high order z-polynomials.
Find stability conditions for
a) The first order polynomial
• For high order polynomials, use symbolic
manipulation.
b) The second order polynomial
41
Solution:
1st
42
Solution: 2nd order
order
• Solve for the root.
,
• Stability conditions
 Stability determination by solving for roots is
difficult.
 Monic polynomial
constant term = product of poles
 For pole magnitudes < 1
Necessary stability condition
Sufficient for complex conjugate poles
43
44
Bilinear Transformation
Comments
• Routh-Hurwitz criterion: poles of 2nd order wpolynomial remain in the LHP iff its coefficients are
all positive.
• Recall:
Sufficient for complex conjugate roots and
only necessary for real roots.
• Real roots: substituting the three conditions
in the z-domain characteristic polynomial
gives roots between 1 and +1.
• Adding the first and third conditions gives the
condition obtained earlier
45
Stable Parameter Range for 2nd
Order z-polynomial
Conditions with
:
Jury Test
The roots of the polynomial
are inside the unit circle if and only if
a0
(1)
a0= 1
1
stable
1
a0= a11
46
1
(2)
a0= a11
(3)
a1


0
(4)
1
(5)

(n+1)
47
48
Jury Table
Table Entries
z0
a0
an
z1
a1
an  1
z2
a2
an  2



zn  k
an  k
ak



zn  1
an  1
a1
3
4
b0
bn  1
b1
bn  2
b2
bn  3


bn  k
bk


bn  1
b0
5
6
c0
cn  2
c1
cn  3
c2
cn  4




cn  2
c0
.
.
.
.
.
.
.
.
.
.
.
.






2 n  5 s0
2 n  4 s3
s1
s2
s2
s1
s3
s0
2 n  3 r0
r1
r2
Row
1
2
49
Remarks
1. First row of Jury table: list the coefficients of the
polynomial
in order of increasing power of .
2. Number of rows of table 2
3 is always odd and the
coefficients of each even row are the same as the odd
row directly above it with the order of the coefficients
reversed.
3. There are
1 conditions for
1coefficients of
.
4. Conditions 3 through
1 are calculated using the
coefficient of the first column of the Jury table, together
with the last coefficient of the last row.
5. The middle coefficient of the last row is never used and
need not be calculated.
6. Conditions (1) and (2) are calculated from
directly. If
one of the first two conditions is violated,
has roots
on or outside the unit circle (no need to construct the Jury
table or test the remaining conditions).
51
zn
an
a0
50
Remarks (cont.)
7- Condition (3) with
1, requires the constant term of
the polynomial to be less than unity in magnitude. The
constant term is simply the product of the roots and
must be smaller than unity for all the roots to be inside
the unit circle.
8 For first and second order systems, the Jury stability
conditions reduce to the conditions derived earlier.
9- For higher order systems, applying the Jury test by
hand is laborious and it is preferable to test the stability
of a polynomial
using a CAD package.
10- If the coefficients of the polynomial are functions of
system parameters, the Jury test can be used to obtain
the stable ranges of the system parameters.
52
Solution
Example
(1)
(2)
Test the stability of the polynomial
Row
1
0.35
2
1
0.0775
2.6
2.05
0.56
0.56
2.05
2.6
0.0775
0.8325
0.8775
3
4
0.8775
0.8325
2.5729
1.854
0.1575
0.1575
1.854
2.5729
5
6
0.0770
0.5151
0.7143
0.2693
0.2693
0.7143
0.5151
0.0770
7
0.2593
0.0837
0.3472
1
1
1
2.6
1
0.56 2.05 0.0775 0.35 1.4175 0
1
1 2.6 0.56 2.05 0.0775 0.35
0.3825 0
(3) 0.35
1
(4) 0.8775
0.8325
(5) | 0.0770 | < | 0.5151 |
(6) | 0.2593 | < | 0.3472 |
• Conditions (2), (5) & (6) violated .
• Condition (2) is sufficient to conclude instability.
No. of conditions violated  No. of roots outside the unit circle.
• Polynomial has roots on or outside the unit circle.
0.7
0.5
0.5
0.8
2.5
• Root at 2.5 outside the unit circle.
1
0.35
53
54
Solution
Example 4.6
• Transfer function of analog subsystem, ADC and DAC
Find the stable range of the gain for
the unity feedback digital control system
with analog plant
Z
Z
• Partial fraction expansion
with DAC and ADC if the sampling
period is 0.02 s.
• Transfer function (
55
)

56
Closed-loop System
Example
• Unity feedback, closed-loop characteristic
equation
Find the stable range of the gain
for the digital
position control system with the analog plant
transfer function
• Stability conditions




• Hence, the stable range of
and with DAC and ADC if
.
is

57
Solution
58
C.L. Characteristic Equation
Z
Z
• Partial fraction expansion
• Transfer function (
)
59
60
Stability Testing
1)
2)
3)


0


Nyquist Criterion
• Given the number of unstable open-loop poles.
• Count the number of unstable closed-loop
poles.
• Follow a closed contour and substitute each
(vector)
point in
• Plot the resulting closed curve.
• Count the number of counterclockwise turns
of the vector
(clockwise is
)
2
3


2
3





• The stable range is
61
Mapping
62
Inside/Outside Contour
1
…
…
closed-loop poles,
open-loop poles
Net rotation: zero for outside, 1 turn for intside
1
Count the number of counterclockwise turns
(clockwise is
)
of the vector
63
64
Proof Outline
Thm. 4-7 Nyquist Criterion
=no. of counterclockwise encirclements of the
when traversing
point (1,0) for a loop gain
clockwise)
the stability contour (i.e.
has open-loop poles inside the contour.
• Then the system has closed-loop poles outside
the unit circle with given by
or (
)
Corollary: An open-loop stable system is closedloop stable iff the Nyquist plot does not encircle
the point (1,0) (i.e.
)
65
•

= number of c.l. (resp. o.l.) poles inside
the contour.
66
Modified contour
• Small semicircle:
1
• Avoid pole at unity
• Pole at 1 mapped to large
counterclockwise semicircle
• Maps to large semicircle.
Denominator angle: net effect =
clockwise semicircles
1
is the frequency response for
∈ 0, /
0, /2 .
complex conjugate of
• Net angle change:
/zero,
/pole inside the
contour and zero for those outside it.
• One encirclement =
Modified Contour
Contour for Stability
Determination
• On the unit circle,
angles
∈ 0, , or
• Traverse contour clockwise and substitute in
to produce a plot.
(mirror image)
• Points on the large circle map to zero or to a single
point.
• For infinitesimal, angle contributions of values on the
straight line portions close to the real axis cancel.

67

68
Example
Shift by (
Nyquist Diagram
0.8
Nyquist Diagram
0.6
0.6
1+ L
0.4
L
0.4
Imaginary Axis
0.2
Imaginary Axis
0.2
0
0
-0.2
-0.4
-0.2
-0.6
-0.4
-0.8
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
-0.6
-1
-0.5
0
0.5
1
1.5
10
2
No. encirclements of origin for
= No. of encirclements of (
of
Real Axis
69
Phase Margin and Gain Margin
4.95
4.901
1.97
0.9704
, no open-loop poles outside the unit
circle: system is closed-loop stable.
70
Nyquist Plot: PM & GM
Gain Margin: gain perturbation that makes
the system marginally stable.
Phase Margin: negative phase perturbation
that makes the system marginally stable.
R(s) +
G(z)
C(z)
G(z)

71
72
Example 4.10
MATLAB Plots and Margins
Digital control system for the furnace model
>> nyquist(gd) % Nyquist plot.
>> bode(gd) % Bode plot.
>> [gm,pm]=margin(gd) % Find PM & GM
>> margin(gd) % GM & PM on Bode plot
Nyquist plot: click on plot and select
Characteristics
All stability margins
Discrete-time first order actuator
1. Determine the closed-loop stability.
2. How does an amplifier of gain
closed-loop stability?
affect
73
Solution
74
GM & PM
Nyquist Diagram
2
stable
1.5
1
Nyquist Diagram
25
0.5
Imaginary Axis
20
System: untitled1
Real: -1.41
Imag: 0.000741
Frequency (rad/sec): -5.33
>>[gm,pm]=margin(gtd)
gm =
3.4817
pm =
37.5426
15
10
5
Imaginary Axis
• Using MATLAB
>> gtd=c2d(gt,0.01)
>> nyquist(gtd)
>> hold on
>> nyquist(5*gtd)
• Change scale
0
-0.5
0
-5
System: gdt
Real: -0.795
Imag: -0.624
Frequency (rad/sec): 2.59
-1
-10
-15
-1.5
-20
-25
-5
-2
-5
0
5
10
15
20
25
Real Axis
30
35
40
45
75
50
-4.5
-4
-3.5
-3
-2.5
Real Axis
-2
-1.5
-1
-0.5
0
76