Chapter 7
Stability Analysis Techniques
7.2 Stability
7.3 Bilinear Transformation
7.4 Routh-Hurwitz Criterion
7.5 Jury Stability Test
7.6 Root Locus
7.7 Nyquist Criterion
7.8 Bode Diagram
7.9 Interpretation of Frequency Response
7.10 Closed-Loop Frequency Response
C.L. Phillips, H. T. Nagle, and A. Chakrabortty,
, 4th Edition, Pearson, 2015
7.2 z-Plane Stability
Consider the LTI system:
Fig. 7-1
(7-1)
(7-2)
Combining complex pairs of poles:
A(r )k cos(qk +f)
(6-7)
3
Transient Response
(Fig. 6-11)
A(r )k cos(qk +f)
4
Stability
System stability may be determined from the location of
the roots of the characteristic equation.
The system characteristic equation is given in various ways:
Classic Characteristic Equation: 1+ GH(z) = 0 (7-3)
Denominator of pseudo-control function, independent of R(z)
|zI - A| = 0
Keep in mind that stability is only a boundary condition in control.
Tests for stability:
Do not give relative stability
Do not give system performance.
For relative stability, performance and design we will use:
Root locus
Nyquist plots
Bode plots
5
z-Plane Stability
(characteristic equation from the transfer function)
R(s)
E(s) E*
D(z)
+
-
T
H(s)
ZOH
Gp(s)
C(s)
Fig. 7-1 with
added D(z) &
ZOH
6
z-Plane Stability
(characteristic equation from pseudo-transfer function)
R(s)
E1*
E1(s)
+
T
+
G1(z)
–
–
C(s)
G2(s)
Fig. 7-2(a) From Example 5.3
é G1G2 é
é
é(z)
2 + G2 é
é R é
é(1+ G2 )R é
é
C(z) = é
é(z) +
é
é(z)
é G1G2 é é 2 + G2 é
é2 + G2 é
1+ é
é(z)
é2 + G2 é
The denominator term, independent of the input R(z),
is the characteristic equation:
7
é G1G2 é
1+ é
é(z) = 0
é2 + G2 é
z-Plane Stability
The characteristic equation may also be found by:
Opening the loop at a sampler:
• Set the input R(s)=0 (stability is independent of the input).
• Open the loop at any sampler (label the sampler output as Ei*(s)
and its input as Eo(s)).
• Solve for the sampler input Eo(s) using Ei*(s) as the source.
• Solve for the open-loop transfer fcn (Gop(z) = Eo(z)/Ei(z)).
• Characteristic equation is 1 + F(z) = 1 – Gop(z) = 0 (7-5)
(Example 7.1: next slide)
State space:
x(k+1) = A x(k) + B r(k)
y(k) = C x(k) + D r(k)
Characteristic equation is |zI - A| = 0
8
(7-6)
Example 7.1
opening a sampler
Open Loop:
Eo = -G2 E1*
E1 = G1 Ei* - G2 HE1*
E1* =
G1* Ei*
1+ G2 H
Fig. 7-3
E =
*
o
-G1*G2*
1+ G2 H
Close the loop ( Eo* = Ei*
Þ Gop (z) =1):
é G (z)G (z) é
2
é1+ 1
éEo (z) = 0
é 1+ G2 H(z) é
1+ F(z) = 1- Gop (z) = 0
(7-6)
*
Ei*
Eo (z) = Gop (z)Ei (z)
Gop (z) = -
1+ G1 (z)G2 (z) + G2 H(z) = 0
*
G1 (z)G2 (z)
1+ G2 H(z)
Gop(z) is the open-loop transfer function.
F(z) is the open-loop function.
9
z-Plane Stability
A discrete-time system is stable if its closed-loop roots are
inside the unit circle.
The text presents two ways to evaluate root locations of the
z-plane characteristic equation:
The Routh-Hurwitz criterion (Section 7.4): a s-plane method
(z-plane mapped to the w-plane which mimics the s-plane):
Jury’s Stability Test (Section 7.5): a z-plane method
10
7.3 Bilinear Transformation
z-Plane to s-Plane Mapping
The obvious question: What mapping method should we use? Consider
the mapping we used for s-plane to z-plane mapping in Chapter 6.
sT
z
e
z = e sT
Thus,
Þ sT
,
z- a e - a
Please note that a polynomial method is preferred!
The bilinear transformation is another powerful method.
1+ (T / 2)w
2 z- 1
z=
or w =
1- (T / 2)w
T z+ 1
(7-7) & (7-8)
This transformation maps the z-plane to the w-plane.
How is the w-plane related to the s-plane?
11
z-Plane to w-Plane Mapping
If we let
z= e
sT
, we find from the 1st term of the expansion:
1
1
s = ln z =
T
T
é z-1 1 (z-1)3
ù
2ê
+
+ .......ú
ë z+1 3 z+1
û
2 z-1
2 e sT -1
2 e jw T -1
w=
=
=
sT
T z+1 z=e sT T e +1 s= jw T e jw T +1
2e
=
Te
- e - jw T /2
2
wT
=
j
tan
= jw w
jw T /2
- jw T /2
+e
T
2
Bilinear
Transform
jw T /2
(7-9)
To interpret this, we observe that the imaginary or frequency axis in the wplane is related to the imaginary or frequency axis in the s-plane by:
2 wT
ww = tan
T
2
(7-10)
when
wT
<<1,
2
ww » w
(7-11)
12
Bilinear Transformation
w-plane to z-plane
2 z-1
1)
w=
T z+1
z-plane to w-plane
2)
2
w+
T
z= 2
wT
=
éT é
1+ é é w
é2 é
éT é
1- é é w
é2 é
s-plane frequency,  , to w-plane frequency,  w
2
wT
wT
wT wT
3) w w = tan
, when
<< 1, tan
»
, ww » w
T
2
2
2
2
13
z-Plane to w-Plane Mapping
jw
Fig. 7-4
s-Plane
D
C
ws / 4
B
∞
frequency
ws / 2
A
0
ww =

For
G -w s / 4
E
-w s / 2
F
2
wT
tan
T
2
wT
<< 1,
2
w-Plane
j w
ww » w
C
C
∞
B j2/T
C
z = e sT
D
E -2/T
F
Im z
z-Plane
2 z- 1
w=
T z+ 1
B
1
C
F
D
E 0
A
Re z
z = 1 gives w = 0
z = -1 gives w = ∞
z = 0 gives w = –2/T
G
14
A
0
G -j2/T
∞
F
F
w
Why map to the w-plane?
1) This is a first-order approximation of the mapping from the
z-plane to the s-plane.
2) We can apply the Routh-Hurwitz criterion for stability analysis.
3) We can do compensator design on the vertical j w axis using
continuous-time Bode plot design methods (Chapter 8).
(lag, lead, lag/lead, PI, PD, PID, etc.)
4) Accurate for small  w [tan ( T/2) ≅  T/2]
For example, for:  T/2 < /10
or
 < 2/10T
or
 <  s/10
(error < 4%)
At ω = ωs/10, a ZOH introduces 18º phase lag.
15
(7-12)
Example 7.2
Fig. 7-5
(from Example 6.7): The characteristic equation is a function
of the sampling interval T. As T increases beyond 0.5, the
system goes unstable.
(
G(z) = K
) (
e -T + T - 1) z+ 1- (1+ T)e -T
(z- 1)(z- e
1+ wT / 2
Let z =
,
1- wT / 2
-T
)
)
which leads to
æ1+ wT / 2 æ
-T
e + T - 1) æ
+
1(1+
T)e
æ1- wT / 2 æ
æ
G(w) = K
ææ1+ wT / 2 æ æææ1+ wT / 2 æ -T æ
- 1æææ
-e æ
ææ
æ
æ
ææ1- wT / 2 æ æææ1- wT / 2 æ
æ
(
-T
)
(
16
)
Bilinear Transformation
1+ G(w) = 0,
Next form
getting
ù
éT 2
-T ùù
-T
-T
é
é
w ê ë2(1+ e ) + K ë2(1- e - T(1+ e )ûûú +
û
ë 4
wéëT (1- e -T ) - KT éë1- e -T (1+ T )ùûùû + KT (1- e -T ) = 0
2
or
aw + bw+ c = 0
2
-b± b2 - 4ac -b
±j
=
w1,2 =
2a
2a
4ac - b2
2a
We can determine stability for a specific gain, say K = 4.25, by
increasing T and examining the “-b/2a” term to test for stability.
T
K
- b / 2a
0.1
4.25 - 0.395
0.2
4.25 - 0.294
0.5
4.25 - 0.013
0.5145 4.25
0
This result shows that
instability results when T is
greater than 0.5145.
We can also set the sampling
period (say T = 0.1) and examine
stability for values of K.
17
7.4 The Routh-Hurwitz Criterion
• Find the characteristic equation (discrete).
• Use the bilinear transformation to map the function
from the z-plane to the w-plane.
• Construct the Routh array.
• Examine the entries to determine stability.
18
Routh-Hurwitz
The Routh-Hurwitz criterion (Table 7-1) identifies the number of rhp
poles from the number of sign changes in the leftmost column of the
Routh array.
F(w) = bnwn + bn-1wn-1 + bn-2 wn-2 +
Form the array:
n
w :
n-1
w
n-2
w
a1 a2
wn :
a3
: b1
b2
b3
: c1
c2
c3
+ b1w+ b0
1 a1 ai 1
ci  b1 b1 bi 1
19
bn
bn-2
bn-4
wn-1 : bn-1
bn-3
bn-5
wn-2 :
c1
c2
c3
:
d1
w1 :
j1
w0 :
k1
d2
d3
% MATLAB®
Fw=[1 9 8 12 21 30]
roots(Fw)
Routh Array
ans =
-8.1687
0.6302
0.6302
-1.0458
-1.0458
For the an example open-loop function:
F(w) = w5 + 9w4 + 8w3 + 12w2 + 21w+ 30
Forming the array:
wn :
a1 a2
+
+
-
1.2968i
1.2968i
0.8204i
0.8204i
a3
wn :
1
8
21
wn-1 : b1 b2 b3
1 a1 ai+1
ci = wn-1 :
9
12
30
wn-2 : c1 c2 c3
b1 b1 bi+1
wn-2 : 6.667 17.667 0
wn-3 : -11.85
30
0
1
60
(1*12
9
*8)
=
= 6.667
wn-4 :
9
9
1
159 53
wn-5 :
- (1*30 - 9 * 21) =
= =17.667
9
-
9
3
1
79
(9 *17.667 - 6.667*12) = = -11.85
6.667
6.667
The appearance of the minus sign indicates rhp root(s), and thus instability.
20
% MATLAB®
Fw=[1 10 38 67 52 12]
roots(Fw)
Routh Array
A second example:
F(w) = w5 + 10w4 + 38w3 + 67w2 + 52w+ 12
wn :
1
38 52
wn-1 : 10
67 12
wn-2 : 31.3 50.8 0
wn-3 : 50.8 12
0
wn-4 : 43.4
0
0
wn-5 : 12
0
0
1 a1 ai 1
ci  b1 b1 bi 1
-
ans =
-3.0000
-2.6180
-2.0000
-2.0000
-0.3820
1
313
(1* 67 -10 * 38) =
= 31.3
10
10
The process continues to a single number in the bottom two
rows. No minus signs appear so the system is stable.
The number of rhp poles is equal to the number of sign
changes in the left most column. The system is stable.
21
Example 7.2
(Again)
Fig. 7-5
Find the bilinear transform:
>> T=0.1; numz=[0.00484 0.00468]; denz=[1 -1.905 0.905];
Gz=tf(numz,denz,T);
Gw=d2c(Gz,'tustin')
Gw =
-4.199e-05 s^2 - 0.04913 s + 0.9995
----------------------------------s^2 + 0.9974 s + 4.429e-15
22
Example 7.2 (continued)
The characteristic equation for any value of K
The Routh array
No sign changes in the first column. K must be in the range
0 < K < 20.3 for stability.
23
7.5 Jury’s Stability Test
• Can be applied directly to z-domain functions.
• It is not necessary to perform the z-plane to w-plane
transformation.
• The array is larger and the computations are more complex than
in the Routh-Hurwitz test.
• First, determine the characteristic equation:
Q(z) = 1+ D(z)GH(z)
= an z + an-1z
n
n-1
+
+ a1 z+ a0 = 0 for an > 0
(7-13)
• For the system to be stable, all of the roots of this polynomial
must lie within the unit circle.
• From the characteristic equation, match terms to the polynomial
(and form the array on the next slide):
24
Jury’s Array
Q(z) = anz + an-1z +
n
+ a1z+ a0 = 0 for an > 0
n-1
Row
z0
z1
z2

z n -k

z n -1
1
a0
a1
a2

an - k

an -1 an
2
an
an -1 an -2

ak

a1
a0
3
4
b0
bn -1
b1
bn -2
b2
bn -3


bn -k
bk -1


bn -1
b0
0
0
5
c0
c1
c2
c3

cn -2
0
0
6
cn -2
cn -3
cn -4
cn -5

c0
0
0






0
0
0
2n - 5
2n - 4
l0
l3
l1
l2
l2
l1
l3
l0
0
0
0
0
0
0
0
0
2n - 3
m0
m1
m2
0
0
0
0
0
Table 7-2
25
zn
Q(z) = anzn + an-1zn-1 +
Jury’s Array
+ a1z+ a0 = 0
for an > 0
The elements of the array are calculated:
bk =
m0 =
a0
an-k
an
ak
l0
l3
l3
l0
ck =
m1 =
The Jury criterion
states that the
necessary and
sufficient conditions
for stability, i.e. that
all roots lie inside the
unit circle, are:
b0
bn-k-1
bn-1
bk
l0
l2
l3
l1
m2 =
Q(1) > 0
( -1) Q(-1) > 0
n
dk =
l0
l1
l3
l2
c0
cn-k-2
cn-2
ck
(7-14)
order of the
polynomial
and the following (n- 1) constraints
a0 < an b0 > bn-1
first
row
26
c0 > cn-2
last
row
(7-15)
m0 > m2
% MATLAB®
Qz=[17 29 27 7]
abs(roots(Qz))
Jury Example
Q(z) = 17z3 + 29z2 + 27z+ 7 = 0
a0 < a3
Þ
7 < 17
Q(1) = 80 > 0
Þ OK
ans =
1.0415
1.0415
0.3796
Þ OK
(-1)3 Q(-1) = (-1)(-8) = 8 > 0 Þ OK
The array becomes
1
7
2 17
3 -240
b0 =
27
29
-304
29 17
27
7
-256 0
b1 =
m0 > m2
?
Þ Fails
-240 > -256
We thus conclude that the system is
unstable.
27
a0
an-0
an
a0
7 29
17 27
=
7 17
17 7
b2 =
7 27
17 29
Note, the roots are:
- 0.380, - 0.663  j 0.803.
The magnitude of the
complex roots is 1.045.
Example 7.4
Step input
R(s) = 1/s
R(s)
E(s)
+
-
E*
T=1sec
1--Ts
s
The open-loop transfer function is given by:
. K .
s(s+1)
C(s)
Fig. 7-5
The unity feedback closed-loop characteristic equation (with open-loop
gain K) is given by:
(0.368z+ 0.264)K
1+ KG(z) = 1+ 2
=0
z -1.368z+ 0.368
or
z2 + (0.368K -1.368)z+ (0.368 + 0.264K ) = 0
30
Example 7.4
(cont.)
The Jury array is:
z0
z1
0.368 + 0.264 K
0.368 K – 1.368
z2
1
Applying the constraints:
a) Q(1) > 0,
[z2 +(0.368K − 1.368)z + (0.368 + 0.264)K)] |z = 1
1 +(0.368K − 1.368) + (0.368 + 0.264 K) = 0.632K > 0 => K > 0
b)(-1)2 Q(-1) > 0
1− 0.368K + 1.368 + 0.368 + 0.264K > 0 => K < .632/.104 = 26.3
c)|a0| < an
0.368 + 0.264K < 1
=> K < 0.632/0.264 = 2.39
31
% MATLAB®
Qz=[1 -0.488 1];
r=roots(Qz)
abs(r)
angle(r)
Example 7.4 (cont.)
Then for stability:
0 < K < 2.39
and marginal stability is at K = 2.39, for which:
[z2 +(0.368K−1.368)z + (0.368 + 0.264) K)]|K=2.39
r =
0.2440 + 0.9698i
0.2440 - 0.9698i
ans =
1.0000
1.0000
ans =
1.3243
-1.3243
= z2 – 0.488 z + 1
with roots:
z = 0.244 ± j0.970 = 1Ð (± 75.9o) = Ð
1
= 1 Ð (±  T),
(± 1.32 rad)
for T = 1 sec =>  = 1.32 rad/sec
32
Jury’s Method
“Quick” tests for low-order systems:
First-order:
a0
<1
a1
Second-order: a2 + a1 + a0 > 0
a2 - a1 + a0 > 0
a0 - a2 < 0
Third-order:
a3 + a2 + a1 + a0 > 0
a3 - a2 + a1 - a0 > 0
a0 < a3
2
2
a
a
( 0 3 ) < ( a0 a2 - a1a3 )
33
7.6 Root Locus
Determines the location of roots of the characteristic
equation of a closed loop control system as the overall system
gain (often denoted by K) varies. The characteristic equation
may be written in any of the following forms:
1+ KGH = 0
(7-16)
Roots of this equation must meet two conditions:
KGH(z) = 1,
(
)
Ð KGH(z) = (1+ 2k)(180°)
A root locus is a plot of the roots of this equation in the z-Plane
as K varies. While construction rules for the z-Plane are identical
with those for the s-Plane, the interpretation is different.
34
Root Locus
DESIGN METHOD:
1. Derive the open loop function KGH(z) .
2. Factor numerator and denominator, linear & quadratic factors.
3. Plot roots in z-Plane.
4. Determine the root loci geometrically,
analytically, by computer-any convenient method.
5. Calibrate in terms of K.
6. Find time response, if needed.
7. Refine, redesign, relocate, etc., to meet specifications.
8. Synthesize or realize any necessary compensation
networks or digital filters.
35
Root Locus Construction
1+ KGH (z) = 0
GH (z) =
N(z)
D(z)
1. Loci originate on the poles of KGH(z) and terminate on its zeros.
2. For K > 0, the real axis loci lie to the left of an odd number of poles
and zeros.
3. The loci are symmetrical with respect to the real axis.
4. The number of asymptotes is equal to the number of poles of KGH(z),
np, minus the number of its zeros, nz. The angles of the asymptotes
are found by
(1+ 2k)(180 o )
qA =
,
np - nz
k = 0, 1, 2, ... (np - nz -1)
5. The origin of the asymptotes on the real axis is given by:
poles of GH(z) - å zeros of GH(z)
å
s=
np - nz
36
Root Locus Construction
6. The breakaway point for the locus between two poles
(or the break-in point for the locus between two
zeros) is found by
break-in
dGH(z)
= 0. Solve for the roots.
dz
num(z)
d(num(z))
d(den(z))
GH (z) =
,
den(z)
- num(z)
=0
den(z)
dz
dz
breakaway
7. The unit circle crossings of the root locus may be determined
by the Jury test (the value of K and location of roots).
37
G(z) =
Root Locus Example
(z+ 0.9700)
(z- 1)(z- 0.9100)
(z+ 0.9700)
N(z)
1+ K
= 1+ KG(z) = 1+ K
=0
(z-1)(z- 0.9100)
D(z)
To find the breakaway points:
den(z) = (z-1)(z- 0.9100)
num(z) = z+ 0.9700
d(num(z))
d(den(z))
- num(z)
= z2 +1.9400z- 2.7627 = 0.
dz
dz
Solving yields z = 0.9471, - 2.9171
Use the Jury test to find that the locus crosses
den(z)
the unit circle for K = ?.
z2 + (K -1.9100)z+ (0.9700 K + 0.9100) = 0
Q(1) > 0 Þ K > 0; (-1) n Q(-1) > 0 Þ K < 127.33
a2 > a0 Þ 1 > 0.97 K + 0.91,
or K < 0.0928.
We conclude that the system is stable iff:
38
0 < K < 0.0928
G(z) =
Root Locus Example (cont.)
(z+ 0.9700)
(z- 1)(z- 0.9100)
K=0.0928
0
-2.9171
-0.97
x
x1
0.91
0.9471
39
num=[1 0.970];den=[1 -1.91 0.91];
Gz=tf(num,den,0.1);rlocus(Gz)
axis([-4 1.5 -2 2])
G(z) =
40
(z+ 0.9700)
(z- 1)(z- 0.9100)
Root Locus Example (cont.)
Find the breakaway point.
rlocus(Gz)
axis([.75 1.1 -1.1 1.1])
Breakaway point
(Damping = 1.0)
is 0.954
41
G(z) =
Root Locus Example (cont.)
(z+ 0.9700)
(z- 1)(z- 0.9100)
Suppose we add some compensation, with a digital filter D(z):
D(z) =
z- 0.9100
,
z- 0.4000
D(z)G(z) =
(z+ 0.9700)
(z-1)(z- 0.4000)
K=0.6200
x
-0.97
0 0.40
0
x
1
0.6700
While the root locus has a very similar shape, analysis of the data shows
that the system is stable for 0 < K < 0.62. This greatly reduces the
steady-state errors in the system.
42
Compensated Example
T=0.1;
num=[1 0.970];
den=[1 -1.91 0.91];
Gz=tf(num,den,T);
Dz=tf([1 -0.91],[1 0.4],T);
DzGz=zpk(Dz*Gz)
rlocus(Dz*Gz)
axis([-1.5 1.5 -1.5 1.5])
DzGz =
(z+0.97) (z-0.91)
---------------------(z-1) (z-0.91) (z+0.4)
43
Compensated Example
44
7.7 The Nyquist Criterion
Let GH (z) be the open-loop function,
Then the characteristic equation for the closed-loop system is
1+ GH (z) = 0
(7-21)
Consider GH (z) = -1
We can examine the behavior of GH (z) in relation to the -1 point
as we traverse the frequency response contour in the z-Plane.
Let
Zi = number of zeroes of characteristic
equation inside the unit circle
Zo = number of zeroes of characteristic
equation outside the unit circle
Pi = number of poles of the open-loop
function that are inside the unit circle.
Fig. 7-15
Po = number of poles of the open-loop
function that are outside the unit circle.
57
The Nyquist Criterion
We can examine the behavior of GH (z) in relation to the -1 point
as we traverse the frequency response contour in the z-Plane.
The number of encirclements of the –1 point is
(7-24)
Fig. 7-15
In general, the numerator and denominator of (7-21)
are the same, say n
(7-25) & (7-26)
Zi = number of zeroes of
characteristic equation inside the
unit circle
Zo = number of zeroes of
characteristic equation outside the
unit circle
Pi = number of poles of the openloop function that are inside the
unit circle.
Po = number of poles of the openloop function that are outside the
unit circle.
So
Dropping the subscripts
(7-27)
N = number of encirclement of the –1 point.
Z = number of zeroes of characteristic equation outside the unit circle
P = number of poles of the open-loop function (or characteristic
equation) that are outside the unit circle.
58
Nyquist Diagrams (Stability)
gain margin =
1
a
phase margin = fm
Fig. 7-18
59
Nyquist Diagrams (Example 7.10)
0.368z+ 0.264
Open-loop function: G(z) = 2
z -1.368z+ 0.368
T =1
Frequency contour is plotted in G(z) plane. Closed-loop system is
unstable if -1 point in G(z) plane is encircled.
Fig. 7-17
60
Nyquist Diagrams (dnyquist)
dnyquist: calculates
the frequency response
of discrete-time LTI
systems.
%Example 7.10
num=[0 .368 .264];
den=[1 -1.368 .368];
T=1;
dnyquist(num,den,T)
axis([-1.2 0.2 -0.7 0.7])
61
7.8 The Bode Diagram - Discrete System Example
A review of Bode diagrams for continuous systems is available at the end
of this slide set. These continuous-time procedures can be used on
sampled-time open-loop transfer functions that have been mapped to the
w-plane (which mimics the frequency behavior of the s-plane if the
sampling period is chosen appropriately).
Example 7.12
Consider again the system of Example 7.3
Numerator break frequencies: 2.0, 12.14
Denominator break frequencies: 0, 0.924
62
Gm = 7.58 dB
Pm = 30.4 deg
Example 7.12
Straightline
Bode
diagram
Fig. 7-21
63
Example 7.12 (via MATLAB & G(w))
>>Gw = -0.0381*tf([1 -2],[1 0])*tf([1 12.14],[1 0.924]);
margin(Gw), grid % margin generates a labeled Bode diag.
Gm = 7.58 dB
Pm = 30.4 deg
64
Example 7.12 (via MATLAB & G(s) to G(z))
>>Gs=tf([1],[1 1 0]); T=1; Gz=c2d(Gs,T); margin(Gz), grid
Gm = 7.58 dB
Pm = 30.4 deg
65
Gain-Phase Plots
Another way of displaying the relationship between the gain and phase of a
frequency response is to plot them on different axes of the same diagram. For
Example 7.12, such a gain-phase plot is shown below. This plot is called a
Nichols chart and will be examined in more detail in Section 7.10.
Origin is 0 dB and –180º
Fig. 7-22
66
Example: Bode diagram, root locus, Nyquist plot
Compare the gain margins.
Gp (s) =
10
10
=
s(s2 + 4s+ 8) s3 + 4s2 + 8s
>> T=0.1; num=[10]; den=[1 4 8 0];
Gp=tf(num,den); Gz=c2d(Gp,T); Gz=zpk(Gz)
[numz,denz]=tfdata(Gz,'v');
figure(1),margin(Gz) % Bode with margins
figure(2),rlocus(Gz),grid
figure(3),nyquist(numz,denz,T),grid
Gz =
0.0015067 (z+3.373) (z+0.2427)
-----------------------------(z-1) (z^2 - 1.605z + 0.6703)
67
)
68
Gz =
0.0015067 (z+3.373)
(z+0.2427)
----------------------------(z-1) (z^2 - 1.605z + 0.6703)
69
Gz =
0.0015067 (z+3.373)
(z+0.2427)
----------------------------(z-1) (z^2 - 1.605z + 0.6703)
70
Gz =
0.0015067 (z+3.373)
(z+0.2427)
----------------------------(z-1) (z^2 - 1.605z + 0.6703)
K = 1.0
71
Gz =
0.0015067 (z+3.373)
(z+0.2427)
----------------------------(z-1) (z^2 - 1.605z + 0.6703)
K = 2.7
>> Gm=2.7; Gmdb=20*log10(Gm)
Gmdb =
8.6273
72
Gz =
0.0015067 (z+3.373)
(z+0.2427)
----------------------------(z-1) (z^2 - 1.605z + 0.6703)
>> A0=-0.365;
Gmdb=20*log10(-1/A0)
Gmdb =
8.7541
73
7.9 Interpretation of the Frequency Response
Consider a discrete-time system described by C(z) whose input is a
sampled sine wave E(z)
(7-35)
Cg(z) is the response originating from the poles of G(z), which
tends to zero over time for stable systems. Let Css(z) be the
steady-state response.
(7-37)
(7-36)
(7-38)
74
Interpretation of the Frequency Response
The steady-state time response css(t) becomes
(7-39)
If the input is a sinusoid of frequency ω, the steady-state response
is also sinusoidal at the same frequency. The amplitude of the
response is equal to the amplitude of the input multiplied by the
magnitude of
and the phase of the response is equal to the
phase of the input plus the angle of
.
75
7.10 Closed-Loop Frequency Response
Graphically relating a system’s closed-loop frequency response to
its open-loop frequency response.
Fig. 7-23
(7-40)
At frequency ω1
(7-41)
76
Constant M Circles
The locus of points in the
plane for which the magnitude of the
closed-loop frequency response, M, is a constant is called a constant
magnitude locus, or a constant M circle.
For
(7-42)
(7-44)
This relationship is the
equation of a circle of radius
with center at
(7-43)
77
For
a straight line
Constant M Circles
Fig. 7-24
78
Constant N Circles
The locus of points of constant phase
are also circles.
This relationship is the
equation of a circle of radius
with center at
(7-45)
79
NOTE: Accuracy is limited
near the –1 point in plots of
constant M and constant N
circles. The Nichols chart is
more accurate
Constant N Circles
Fig. 7-25
80
Nichols Chart
These are gain-phase plots the magnitude of
on the vertical axis
and the angle of
on the horizontal. Contours of constant
magnitude and constant phase are also included. Here is the plot format.
0
–180
81
Nichols Chart
Constant M and N circles are shown as contours.
0
–180
82
Example 7.13
Compute the Nichols chart for
>> Gp = tf([1],[1 1 0]); T = 1; Gz=c2d(Gp,T); nichols(Gz), grid
83
Fig. 7-27
84
Example 7.13
(continued)
Using Bode plots and step responses, compare the performance of this system
with T = 1 and T = 0.1 Verify the gain margin via Nyquist plots.
>> num = [1]; den = [1 1 0]; Gp = tf(num,den);
T1 = 1; Gz1=c2d(Gp,T1); [Gm1,Pm1] = margin(Gz1);
Cz1 = feedback(Gz1,1);
T2 = 0.1; Gz2 = c2d(Gp,T2); [Gm2,Pm2] = margin(Gz2);
Cz2 = feedback(Gz2,1);
% Plot the closed-loop frequency response
figure(1), bode(Cz1,'--k',Cz2,'-k'), axis([0.25 2.5 -225 10]), grid
% Plot the step response
figure(2), step(Cz1,'--k',Cz2,'-k'), axis([0 20 0 1.5]), grid
% List the step response and margin parameters
StepT1=stepinfo(Cz1),Pm1,Gm1=20*log10(Gm1) % Convert to dB
StepT2=stepinfo(Cz2),Pm2,Gm2=20*log10(Gm2) % Convert to dB
% Compute Nyquist plots
figure(3),nyquist(Gz1,Gz2), axis([-0.5 0.05 -0.5 0.5]), grid
figure(4),nyquist(Gz1,Gz2), axis([-0.08 0.01 -0.08 0.08]), grid
85
Example 7.13 (Bode plots)
T = 1: peak mag. is
6.06 dB at 0.87 rad/s
T = 0.1: peak mag. is
1.56 dB at 0.74 rad/s
Fig. 7-28
86
Example 7.13
(Step responses)
T = 1: peak mag.
is 1.4 at 3 s
T = 0.1: peak
mag. is 1.18 at
3.5 s
Fig. 7-29
87
Example 7.13
(step response parameters and gain margins)
T = 0.1
T=1
StepT2 =
StepT1 =
RiseTime:
SettlingTime:
SettlingMin:
SettlingMax:
Overshoot:
Undershoot:
Peak:
PeakTime:
RiseTime:
SettlingTime:
SettlingMin:
SettlingMax:
Overshoot:
Undershoot:
Peak:
PeakTime:
1
16
0.8015
1.3996
39.9576
0
1.3996
3
Pm1 =
30.3747
Pm2 =
49.5807
Gm1 =
7.5779
Gm2 =
26.1666
88
1.6000
8.3000
0.9090
1.1837
18.3712
0
1.1837
3.6000
Example 7.13 (Nyquist plots)
T = 1, a = –0.419 @ 1.32 rad/s
Gm = 20*log10(1/0.419) = 7.56 dB ✔︎
*
Fig. 7-30(a)
89
Example 7.13 (Nyquist plots)
T = 0.1, a = –0.0488
Gm = 20*log10(1/0.0488) = 26.2 dB ✔︎
*
Fig. 7-30(b)
90
Homework
Problems:
7.2-6
7.5-1
7.6-4 a,c (using MATLAB)
7.8-3
7.9-1
91