Part IB Paper 6: Information Engineering
LINEAR SYSTEMS AND CONTROL
Glenn Vinnicombe
HANDOUT 6
“Feedback stability and the Nyquist diagram”
If L(s) is stable , then:
(either marginally or asymptotically)
"
"
! −1
! −1
−1
L(jω)
L(s)
is
1 + L(s)
asymptotically
stable
=⇒
"
!
L(jω)
L(s)
is
1 + L(s)
marginally
stable
=⇒
L(jω)
L(s)
is
1 + L(s)
unstable
=⇒
That is, the closed-loop system is stable if the Nyquist diagram of the
return ratio doesn’t enclose the point “−1”.
Summary
The Nyquist diagram of a feedback system is a plot of the
frequency
part
!
" response of the return ratio, with
! the imaginary
"
! L(jω) plotted against the real part " L(jω) on an Argand
diagram (that is, like the Bode diagram, it is a plot of an open-loop
frequency response).
The Nyquist stability criterion states that, if
the open-loop system is asymptotically stable (i.e. the return
ratio L(s) has all its poles in the LHP) and
the Nyquist diagram of L(jω) does not enclose the point “−1”,
then the closed-loop system will be asymptotically stable (i.e. the
#
$
closed-loop transfer function L(s)/ 1 + L(s) will have all its poles
in the LHP)
The real power of the Nyquist stability criterion is that it allows you
to determine of the stability of the closed-loop system from the
behaviour of the open-loop Nyquist diagram. This is important
from a design point of view, as it relatively easy to see how
changing K(s) affects L(s) = H(s)G(s)K(s), but difficult to see
how changing K(s) affects L(s)/(1 + L(s)) directly, for example.
In addition, the Nyquist diagram also allows more detailed
information about the behaviour of the closed-loop system to be
inferred. For example
Gain and phase margins measure how close the Nyquist locus
gets to −1 (and hence how close the closed loop system is to
instability).
Contents
6 Feedback stability and the Nyquist diagram
1
6.1 The Nyquist Diagram . . . . . . . . . . . . . . . . . . . . . .
4
6.1.1 Sketching Nyquist diagrams . . . . . . . . . . . . . .
7
6.2 Feedback stability . . . . . . . . . . . . . . . . . . . . . . . .
8
6.2.1 Significance of the point “−1”
6.2.2 Example:
. . . . . . . . . . . . .
9
. . . . . . . . . . . . . . . . . . . . . . . . .
9
6.3 Nyquist Stability Theorem . . . . . . . . . . . . . . . . . . . . 12
6.4 Gain and Phase Margins . . . . . . . . . . . . . . . . . . . . . 13
6.4.1 Gain and phase margins from the Bode plot . . . . . 14
6.5 Performance of feedback systems . . . . . . . . . . . . . . . 15
6.5.1 open and closed-loop frequency response . . . . . . 16
6.6 The Nyquist stability theorem (for stable L(s))
. . . . . . . 19
6.6.1 Notes on the Nyquist Stability Theorem: . . . . . . . 20
6.1 The
TheNyquist
NyquistDiagram
Diagram
6.1
The
Nyquist
diagramofofaasystem
systemG(s)
G(s)isisaaplot
plotofofthe
thefrequency
frequency
The
TheNyquist
Nyquistdiagram
response
G(jω)
onan
anArgand
Arganddiagram.
diagram.
response
responseG(jω)
G(jω)on
That
is:
plotofof!(G(jω))
!(G(jω))vsvs"(G(jω)).
"(G(jω)).
That
Thatis:
is:itit
itisis
isaaplot
!!
""
G(jω)
!!G(jω)
∠G(jω
∠G(jω11))
!!
""
G(jω)
""G(jω)
The
Nyquist
locus
The
The
Nyquist
locus
The
Nyquist
locus
|G(jω11)|)|
|G(jω
ω==ω
ω11
ω
Examples:
Examples:
Examples:
Integrator:
Integrator:
Integrator:
111
G(s)
G(s)
=
G(s)=
=
sss
−j
−j
111
G(jω)
=
G(jω)
=
=
G(jω)=
=
=
jω
jω
jω
ω
ω
|G(jω)|
|G(jω)|
=
|G(jω)|=
=1/ω
1/ω
◦
∠G(jω)
=
∠G(jω)
∠G(jω)=
=−90
−90◦
20dB
2
20dB
20dB
0dB
0dB
0dB
0dB
0.1
0.1
0.1
0.1
00◦◦◦◦0.1
00
0.1
0.1
0.1
◦◦
−90
−90
−90
−90◦◦
11
11
ω
ω
ω
10
10
10
11
11
ω
ω
ω
10
10
10
""
!!
% G(jω)
G(jω)
%
ω
=∞
∞
ω=
−j
−j
!!!
"""
$
$
G(jω)
$ G(jω)
G(jω)
ω
ω=
= 11
!!!!
""""
$
G(jω)
$
$$G(jω)
G(jω)
G(jω)
Time
Delay:
Time
Time
Delay:
TimeDelay:
Delay:
−sT
−sT
G(s)
=
G(s)
G(s)
==eee−sT
G(s)=
e−sT
−jωT
−jωT
−jωT
G(jω)
=
G(jω)
G(jω)=
e
G(jω)
==eee−jωT
111
|G(jω)|
=
|G(jω)|
|G(jω)|=
|G(jω)|
==1
arg
G(jω)
=
−ωT
arg
G(jω)
arg
G(jω)=
arg
G(jω)
== −ωT
−ωT
−ωT
(radians)
(radians)
(radians)
(radians)
20dB
20dB
20dB
2
20dB
0dB
0dB
0dB
0dB
0.1
0.1
0.1
0.1
TTTT
◦◦
000◦◦00.1
0.1
0.1
0.1
TTTT
◦
◦
◦
−90
−90◦
−90
−90
1111
TTTT
1111
TTTT
ω
ω
ω
ω
10
10
10
10
TTTT
ω
ω
ω
ω
10
10
10
10
TTTT
!!!!
""""
#
#
G(jω)
#
G(jω)
#G(jω)
G(jω)
ω
ω
==0
00
ω=
π
ππ
ω
ω=
ω
==
2T
2T
2T
First-order
lag:
First-order
lag:
First-order
First-order
lag:
First-order
lag:
First-order lag:
lag:
111
1
G(s)
=
1
G(s)
=
G(s)
=
G(s)
=
G(s)
=
G(s) =111+
sT
+
sT
sT
+sT
sT
11++
111
1
1
G(jω)
==
G(jω)
G(jω)
G(jω)
= 11+
G(jω)
G(jω)==
jωT
+
jωT
jωT
+jωT
jωT
111++
111
1
1
1
!
|G(jω)|
=
|G(jω)|
=
!
!
|G(jω)|
=
|G(jω)|
=
!
!
|G(jω)|
|G(jω)|== 11+
222
222
ω
T
2
+
ω
T
2
2
1
+
ω
T
2
1
+
ω
T
11++ωω TT 22
∠G(jω)
=
−
arctan(ωT
∠G(jω)
=
−
arctan(ωT
∠G(jω)
=
−
arctan(ωT
∠G(jω)
=
−
arctan(ωT
∠G(jω)
∠G(jω)=
=−
−arctan(ωT
arctan(ωT)))))
20dB
20dB
20dB
20dB
2
20dB
20dB
0dB
0dB
0dB
0dB
0.1
0.1
0.1
0dB
0.1
0dB
0.1
0.1
T
T
TTT
T
◦◦◦◦◦0.1
◦
0
0
0
0
0.1
0 00.1
0.1
0.1
T
T
◦TTT
T
◦
◦
−90
◦
◦
−90
−90
−90
−90
−90
1
1111
1
T
T
T
TT
1
1111
TTTTT
ω
ω
ω
ω
ω
10
10
10
10
10
10
T
T
TTT
ω
=
0
ω
=
0
ω
ω=0
1
1
1
ω
=
ω=
=
ω
ω
=T
T
T
T
ω
=→
∞
ω
=→
∞
ω =→
=→ ∞
∞
ω
"
#
""
##
"
#
(
G(jω)
(
G(jω)
'
G(jω)
(
(
( G(jω)
G(jω)
ω
=
∞
ω
ω
ω=
=∞
∞
ω
ω
ω
ω
ω
10
10
10
10
10
T
T
TTT
−0.5j
−0.5j
−0.5j
|G(jω)|
|G(jω)|
|G(jω)|
|G(jω)|
1
1
1
1
1
1
√
√
√
2
2
2
0
0
0
∠G(jω)
∠G(jω)
∠G(jω)
∠G(jω)
∠G(jω)
0
0
0
◦
◦
−45
◦
◦
−45
−45
◦
◦◦
◦
−90
−90
−90
"
###
"
#
'
&
G(jω)
G(jω)
'
G(jω)
G(jω)
0.5
0.5
ω
=
0
ω=
=0
0
ω
ω
ω=
= 1/T
1/T
Time
with
Lag
and
Integrator:
Time Delay
Delay with
with Lag
Lag and
and Integrator:
Integrator:
Time
Delay
−sT11
−sT
ee−sT
1
G(s)
G(s)
=
G(s)
=
G(s)
s(1+
+sT
sT22))
s(1
−jωT11
−jωT
1
ee−jωT
G(jω)=
=
G(jω)
G(jω)
jω(1+
+jωT
jωT222)))
jω(1
jωT
1
11
−jωT
−jωT
−jωT
1
×
|G(jω)|=
=|e
|e
1|||×
1
×
|G(jω)|
|G(jω)|
!!
"#"#
11
1
11
×
×
×
$$$ |jω|
|jω|
|1+
+jωT
jωT22
|jω| |1
2
|1
+
jωT
2|||
−jωT1 − ∠(jω) −∠(1 + jωT )
−jωT
−jωT
∠G(jω)
=
∠e
11$−
∠(jω)
−∠(1
∠G(jω)
=
∠e
jωT2222)))
∠G(jω)
−∠(1+
+jωT
jωT
∠G(jω) = !∠e
"#
!
"#
!! "#
"#
$
"# $$ −!∠(jω)
!
$
"# $
−ωT11
−ωT
90◦◦
90
1
Clearly, as
as ω
ω→
→0
then |G(jω)|
|G(jω)|→
→∞.
∞. But
But this
this
is
not
enough
Clearly,
Clearly,
then
thisis
isnot
notenough
enough
Clearly,
as
ω
→
00 then
|G(jω)|
→
∞.
But
this
is
not
enough
information to
to sketch
sketch the
the Nyquist
Nyquist diagram.
diagram.
Precisely
how
does
information
information
sketch
the
Nyquist
diagram.Precisely
Preciselyhow
howdoes
does
information
to
diagram.
Precisely
how
does
|G(jω)|→
→∞?
∞? To
To answer
answer this,
this, we
we use
use
Taylor
series
expansion
|G(jω)|
→
∞?
|G(jω)|
To
answer
this,
we
useaaaTaylor
Taylorseries
seriesexpansion
expansion
|G(jω)|
Taylor
series
expansion
aroundω
ω=
= 0.
0.
around
ω
=
around
0.
around
−jωT
and
11
→ 11−
−jωT
jωT11 and
and
1→
ee−jωT
1/(jωT22++1)
1)→
→ 11−
−jωT
jωT222,,,
1/(jωT
00
(1−−jωT
jωT111)(1
)(1−
−jωT
jωT222)))
)(1
−
jωT
(1
111
jωT111T
=
−(T
(T111 +
+T
T222)))+
+jωT
G(jω)→
→
=
−
jωT
TT222...
(T
+
T
+
==
⇒⇒ G(jω)
=
−
jω
jω
jω
jω
jω
jω
!!
"""
G(jω)
G(jω)
%% G(jω)
−(T11++TT22))
−(T
""""
!!!!
$
G(jω)
$
G(jω)
$ G(jω)
G(jω)
$
Second-order
lag:
Second-orderlag:
lag:
Second-order
Second-order
lag:
1
1
1
1
G(s)
=
G(s)=
=
G(s)
G(s)
=
(1
+
sT
)(1
+
sT
(1+
+sT
sT1111)(1
)(1+
+sT
sT2222))))
(1
(1
+
sT
)(1
+
sT
11
1
1
G(jω)
=
G(jω)=
=
G(jω)
G(jω)
=
(1
+
jωT
)(1
+
jωT
(1+
+jωT
jωT11
)(1+
+jωT
jωT2222))))
(1
(1
+
jωT
)(1
+
jωT
11)(1
11
1
1
!
!
!
!
|G(jω)|
=
|G(jω)|=
=!
!
|G(jω)|
!
!
|G(jω)|
=
222 11
2222
2222T
+ω
ω22
1
+
ω
T
22T
+
ω
T
+
ω
T
11
+
ω
11+
+
ω
TT11
1
+
ω
T
11
2222
""
##
G(jω)
'
&' G(jω)
∠G(jω)=
=−
−arctan(ωT
arctan(ωT111))))−
−arctan(ωT
arctan(ωT222))
∠G(jω)
=
−
arctan(ωT
−
arctan(ωT
∠G(jω)
∠G(jω)
=
−
arctan(ωT
1 − arctan(ωT
2
ω=
=∞
∞
ω
|G(jω)| ∠G(jω)
∠G(jω)
|G(jω)|
∠G(jω)
|G(jω)|
ω=
=00
ω
=
ω
=
00
ω
ω=→
=→∞
∞
ω
=→
∞
ω
=→
∞
ω
111
000
000
◦
◦
◦
−180
−180
−180
"""
###
& G(jω)
G(jω)
&
%
&
G(jω)
ω
=00
0
ω
ω==
6.1.1
Sketching Nyquist diagrams
Unlike the Bode diagram, there are no detailed rules for sketching
Nyquist diagrams. It suffices to determine the asymptotic behaviour as
ω → 0 and ω → ∞ (using the techniques we have seen in the
examples) and then calculate a few points in between. Note that if
G(0) is a finite and non-zero, then the Nyquist locus will always start
off by leaving the real axis at right angles to it.
1 If G(0) is infinite,
due to the presence of integrators, then we must explicity find the
first two terms of the Taylor series expansion of G(jω) about ω = 0,
as in the example with a time delay, a lag and an integrator.
1
This is since G(j") = G(0) + j"G# (0) − "2 G## (0) − · · · ≈ G(0) + j"G# (0)
6.2 Feedback stability
r̄ (s)+
−
ē(s)
Σ
K(s)
G(s)
ȳ(s)
r̄ (s)+
≡
Closed-loop poles
−
ē(s)
Σ
L(s)
ȳ(s)
G(s)K(s)
≡ poles of
1 + G(s)K(s)
≡ roots of 1 + G(s)K(s) = 0
It is difficult to see how K(s) should be chosen to ensure that all the
closed-loop poles are all in the LHP. But . . .
Nyquist’sStability
Stability
Theorem
allowsTheorem
usto
todeduce
deduce
closed-loop
Nyquist’s
Stability
Theorem
allows
us
to
deduce
closed-loop
Nyquist’s
Theorem
allows
us
closed-loop
Nyquist’s
Stability
allows
us to deduce closed-loo
properties:
properties:
properties:
properties:
G(s)K(s)
G(s)K(s)
G(s)K(s)
,
the location
location of
of
the
poles
of
,
the
the
poles
of
,
the location of11the
poles of
+ G(s)K(s)
G(s)K(s)
+
1 + G(s)K(s)
fromopen-loop
open-loopproperties
properties
from
open-loop
properties
from
from
open-loop properties
frequency response
response
of the
the
return ratio
ratio
frequency
of
return
frequency
response
of
the return ratio
L(jω) =
= G(jω)K(jω).
G(jω)K(jω).
L(jω)
L(jω) = G(jω)K(jω).
Thebasic
basicidea
ideais
is
asfollows:
follows:
Negative
feedback
isused
usedfeedback
toreduce
reduce
the
The
basic
idea
is
as
follows:
Negative
feedback
used
to
reduce
the
The
as
Negative
feedback
isis
to
The
basic
idea
is as follows:
Negative
isthe
used to re
sizeof
ofthe
theerror
errore(t)
e(t)in
inthe
theabove
abovefigures.
figures.IfIf
Ify(t)
y(t)isis
istoo
toolarge
large(i.e
(i.e
size
of
the
error
e(t)
in
the
above
figures.
y(t)
too
large
(i.e
size
size
of
error
e(t)
in the above
figures.
If y(t)
is too larg
greaterthan
thanrrr(t))
(t))then
thene(t)
e(t) is
isnegative,
negative,which
whichwill
willtend
tendto
toreduce
reducey(t)
y(t)
greater
than
(t))
then
e(t)
negative,
which
will
tend
to
reduce
greater
greater
thanis
r (t))
then e(t)
is negative,
which
willy(t)
tend to re
(providedthe
thesigns
signsof
ofK(s)
K(s)and
andG(s)
G(s)have
havebeen
beenchosen
chosenappropriately).
appropriately).
(provided
the
signs
of
K(s)
and
G(s)
chosen
appropriately).
(provided
(provided
the
signs
ofhave
K(s)been
and G(s)
have
been chosen appr
However,for
forany
anyreal
realsystem
systemthe
thephase
phaselag
lagfrom
fromthe
theinput
inputto
tothe
the
However,
for
any
real
system
the
phase
lag
from
input
the
However,
However,
for any
real system
the the
phase
lagto
from
the input to
output(−∠L(jω))
(−∠L(jω))will
willtend
tendto
toincrease
increasewith
withfrequency,
frequency,eventually
eventually
output
(−∠L(jω))
will
tend
to
increase
with
frequency,
output
output
(−∠L(jω))
will tend
to
increase eventually
with frequency, even
◦◦.. When
◦
reaching
180
Whenthis
thishappens,
happens,
thenegative
negativefeedback
feedbackisis
isturned
turned
◦ . When
reaching
180
this
happens,
the
negative
feedback
turned
reaching 180 . When
the
reaching
180
this
happens,
the negative
feedback is
intopositive
positivefeedback.
feedback.IfIfIfthe
thegain
gain|L(jω)|
|L(jω)|has
hasnot
notdecreased
decreasedto
toless
less
into
positive
feedback.
the
gain
|L(jω)|
has
not
decreased
less
into
into positive
feedback.
If the
gain
|L(jω)|
hastonot
decreased
than111by
bythis
thisfrequency
frequencythen
theninstability
instabilityof
ofthe
theclosed-loop
closed-loopsystem
systemwill
will
than
by
this
frequency
the
closed-loop
will
than
than 1 bythen
this instability
frequency of
then
instability
of system
the closed-loop
s
result.
result.
result.
result.
r
If L(s) is stable , then:
(either marginally or asymptotically)
"
"
! −1
L(jω)
L(s)
is
1 + L(s)
asymptotically
stable
=⇒
L(s)
"
! −1
−1
Σ
e
!
L(jω)
L(s)
is
1 + L(s)
marginally
stable
=⇒
L(jω)
L(s)
is
1 + L(s)
unstable
=⇒
That is, the closed-loop system is stable if the Nyquist diagram of the
return ratio doesn’t enclose the point “−1”.
y
6.2.1
r
Significance of the point “−1”
Σ
e
y
L(s)
If the Nyquist locus passes through the point “−1”,
i.e. L(j
1)
=
1 for some
1
!
"
then the closed-loop frequency response L(jω)/ 1 + L(jω) becomes
infinite at that frequency, ie
!
"
L(jω1 )/ 1 + L(jω1 ) → ∞
This is not a good thing!
In this case, if e(t) = cos( 1 t) then in steady-state we have
#
$
y(t) = |L(jω1 )| cos ω1 t + ∠L(jω1 )
= cos(⇥1 t +
)=
cos(
1 t)
However e(t) = r (t) − y(t), which means that
r (t) = e(t) + y(t) = cos(ω1 t) − cos(ω1 t)
=0
That is, there is a sustained oscillation of the feedback system even
when there is no external input!
6.2.2
Example:
6.2.2
Example:
6.2.2
6.2.2 Example:
Example:
Let
Let
Let
Let
Let
1
11
G(s)
=
K(s)
=
k,
G(s)
G(s)=
= s 3333+ s 222+ 2s + 1 ,,, K(s)
K(s)=
=k,
k,
k,
ss +
+ss +
+2s
2s+
+11
k
kk
=
⇒
L(s)
=
...
=
⇒
=
⇒ L(s)
L(s)=
= s3
=
⇒
2
3
2
+
+
2s
+
11
+sss 2+
+2s
2s+
+1
ss 3+
The
closed-loop
poles
are
the
roots
The
closed-loop
The
closed-loop
poles
Theclosed-loop
closed-looppoles
polesare
arethe
theroots
roots of
of
The
of
kk
3
2
k
3
3
22 +
1
+
s
+
s
2s
+
1
+
k
=
0
=
0
⇐
⇒
+
+ 333
s
+
s
+
2s
+
1
+
k
=
0
=
0
⇐
⇒
111+
s
+
s
+
2s
+
1
+
k
=
0
=
0
⇐
⇒
!
"#
$$$
2
2
!
"#
!
"#
2
s
+
s
+
2s
+
1
3
2
+
+sss ++2s
2s++11
sss +
CLCE
CLCE
CLCE
and
the
frequency
response
of
the
loop
and
the
frequency
andthe
thefrequency
frequencyresponse
responseof
ofthe
the loop
loop is:
is:
and
is:
k
k
k
L(jω)
==
L(jω)
L(jω)=
L(jω)
33 + 2ω) + (−ω22 + 1)
j(−ω
j(−ω3
+2ω)
2ω)+
+(−ω
(−ω +
+ 1)
1)
j(−ω
+
√
√
√
√
At
ω
=
2,
L(jw)
isispurely
purely
real.
That
At
ω
=
2,
L(jw)
Atω
ω=
= 2,
2,L(jw)
L(jw)is
purelyreal.
real. That
That is
is
At
is
√
√
k
√
kk
√
√
2
j)
=
L(
√
√
=
L(
2
j)
=
−k
=−k
L(
√
√
=
L( 2 j) =
j(−2
22+
+
j(−2 2
+2
2)−
−2
+1
j(−2
22 2)
2)
−
22+
+
11
1L(
2 j)
0
Imag Axis
k=1
0
−1 Closed-loop step response
0
10
20
30
40
50
Nyquist diagram, k=1
2
2
−1
L(j 2)
Imag Axis
1
0
−2
−2
L(j
)
L−1
1
1
L(s)
·
1 + L(s) s
0
−1
0
Real Axis
1
2
−1
=
0
10
20
30
2
40
Closed-loop poles are at the roots of s 3 + s 2 + 2s + 2 = 0, i.e.
−1
−2
−2
−1
0
Real Axis
1
2
Closed-loop poles are at the roots of s 3 + s 2 + 2s + 2 = 0, i.e.
50
s = −0.0000 + 1.4142j,
−0.0000 − 1.4142j,
−1.0000
⇥
=⇒
(because L(j 2) = 1,
and so 1 + L(s) = 0 at s = j 2)
X
X
X
Closed-loop poles
closed-loop system is marginally stable
kk =
= 0.8
0.8
22
Closed-loop
Closed-loop step
step response
response
Nyquist
Nyquist diagram,
diagram, k=0.8
k=0.8
22
11
ImagAxis
Axis
Imag
11
00
−1
−1
00
00
20
20
30
30
40
40
50
50
Closed-loop is
asymptotically stable
−1
−1
−2
−2
−2
−2
10
10
−1
−1
00
11
Real
Real Axis
Axis
22
3
2
Closed-loop
Closed-loop poles
poles are
are at
at the
the roots
roots of
of ss 3 +
+ss 2+
+2s
2s +
+1.8
1.8 =
=0,
0, i.e.
i.e.
ss =
= −0.0349
−0.0349+
+1.3906j,
1.3906j, −0.0349
−0.0349−
−1.3906j,
1.3906j, −0.9302
−0.9302
k = 1.2
k = 1.2
2
2
Nyquist diagram, k=1.2
Nyquist diagram, k=1.2
2
2
1
1
ImagAxis
Axis
Imag
1
1
0
0
−1
−10
0
0
0
10
10
20
20
30
30
40
40
50
50
Closed-loop is
unstable
−1
−1
−2
−2
−2
−2
Closed-loop step response
Closed-loop step response
−1
−1
00
11
Real
Real Axis
Axis
22
3 + s 2 + 2s + 2.2 = 0, i.e.
Closed-loop
poles
are
at
the
roots
of
s
Closed-loop poles are at the roots of s 3 + s 2 + 2s + 2.2 = 0, i.e.
ss =
0.0319++1.4377j,
1.4377j, 0.0319
0.0319
− 1.4377j,
−1.0639
= 0.0319
− 1.4377j,
−1.0639
kk =
= 0.8
0.8
22
r
e
Σ
Closed-loop
Closed-loop step
step response
response
Nyquist
Nyquist diagram,
diagram, k=0.8
k=0.8
22
11
ImagAxis
Axis
Imag
11
00
−1
−1
00
00
20
20
30
30
40
40
50
50
Closed-loop is
asymptotically stable
−1
−1
−2
−2
−2
−2
10
10
−1
−1
00
11
Real
Real Axis
Axis
22
3
2
Closed-loop
Closed-loop poles
poles are
are at
at the
the roots
roots of
of ss 3 +
+ss 2+
+2s
2s +
+1.8
1.8 =
=0,
0, i.e.
i.e.
ss =
= −0.0349
−0.0349+
+1.3906j,
1.3906j, −0.0349
−0.0349−
−1.3906j,
1.3906j, −0.9302
−0.9302
L(s)
y
k = 1.2
k = 1.2
2
2
r
Nyquist diagram, k=1.2
Nyquist diagram, k=1.2
2
2
ImagAxis
Axis
Imag
e
Closed-loop step response
Closed-loop step response
1
1
1
1
0
0
−1
−10
0
0
0
10
10
20
20
30
30
40
40
50
50
Closed-loop is
unstable
−1
−1
−2
−2
−2
−2
Σ
−1
−1
00
11
Real
Real Axis
Axis
22
3 + s 2 + 2s + 2.2 = 0, i.e.
Closed-loop
poles
are
at
the
roots
of
s
Closed-loop poles are at the roots of s 3 + s 2 + 2s + 2.2 = 0, i.e.
ss =
0.0319++1.4377j,
1.4377j, 0.0319
0.0319
− 1.4377j,
−1.0639
= 0.0319
− 1.4377j,
−1.0639
L(s)
y
6.3 Nyquist Stability Theorem (informal
version)
We can now give an informal statement of Nyquist’s stability theorem:
“If a feedback system has an asymptotically stable return ratio L(s),
then the feedback system is also asymptotically stable if the Nyquist
diagram of L(jω) leaves the point −1 + j0 on its left”.
This is unambiguous in most cases, and usually still works if L(s) has
poles at the origin or is unstable.
For completeness, a full statement of this theorem will be given later.
Definition: We say that the feedback system (or closed-loop system)
L(s)
is asymptotically stable if the closed-loop transfer function
1 + L(s)
L(s)
is asymptotically stable, that is if all the poles of
(i.e. the
1 + L(s)
roots of 1 + L(s) = 0) lie in the LHP.
6.4 Gain and Phase Margins
L(jω) encirling or going through the −1 point is clearly bad, leading
to the closed-loop not being asymptotically stable. However, L(jω)
coming close to −1 without encircling it is also undesirable, for two
reasons:
It implies that a closed-loop pole will be close to the imaginary axis
and that the closed-loop system will be oscillatory.
If G(s) is the transfer function of an inaccurate model, then the
“true” Nyquist diagram might actually encircle −1.
Gain and phase margins are widely used measures of how close the
return ration L(jω) gets to −1.
The gain margin measures how much the gain of the return ratio can
be increased before the closed-loop system becomes unstable.
The phase margin measures how much phase lag can be added to the
return ratio before the closed-loop system becomes unstable.
"
!
−1
L(j
Gain Margin =
1
)
Phase Margin =
In this example we have θ = 35◦ and −α = −0.75. Hence
Phase Margin = 35◦ and Gain Margin = 1/0.75 = 4/3.
6.4.1
"
Gain and phase margins from the Bode plot
|L(jω)| (dB)
−1 −α
0dB
L(jω)
Gain Margin
−10dB
Gain Margin =
∠L(jω)
1
α
Phase Margin = θ
In this example we have θ = 35◦ and −α = −0.75. He
Phase Margin = 35◦ and Gain Margin = 1/0.7
0◦
−180◦
θ
log10 ω
Phase Margin
Gain Margin = 20log10 4/3 = 2.5dB. Phase Margin = 35◦ (as before)
Hint: Given a Nyquist diagram of L(s) = kG(s) for k = 1, it is easy
to find gain and phase margins for k ≠ 1 (just look at the “−1/k”
point instead of “-1”).
“−1/k” −0.75
−1
#
"
θ (=phase margin when k = 0.8)
L(jω)
If k = 0.8, as here, then Gain Margin= −1.25
−0.75 = 5/3 (= 4.4dB), and
Phase Margin=80◦ .
6.5 Performance of feedback systems
Good
feedback
properties
⇐⇒
“Small”
sensitivity
!
!
!
!
1
!
!
!
!
! 1 + L(jω) !
For 1) rejection of disturbances.
d̄(s)
L(s)
ȳ(s)
−
1
Transfer function
Transfer function
=
with
f/b
withoutfunction
f/b
Transfer function
Transfer
1 +1L(s)
=
×
with f/b
without f/b
1 + L(s)
Plus, 2) reducing the effects of uncertainty.
– if L(s) depends on an uncertain parameter λ (eg
1
L(s) = 2
) then
s + 2λs + 1
d L
dλ 1 + L
L
! 1+
"# L $
relative change
in closed-loop
=
dL
dL %
(1 + L) × dλ
− (L) × dλ
(1 + L)2
L
1+L
1
=
!1 +
"# L$
S
d
L
dλ
L $
! "#
relative change
in open-loop
Good design aims for sensitivity reduction over an appropriate
range of frequencies
&
&
&
&
1
&
&
& # 1 for ω < ω1 where
Typically, by requiring that &
& 1 + L(jω) &
ω1 here denotes the desired control bandwidth.
Fundamental limits on performance
As described in Paper 5 (Linear Circuits) operational amplifiers are
typically compensated so that their frequency response is similar to
that of a pure integrator. Ideally they would have a transfer function
G(s) = A/s
or G(jω) = A/jω.
With a feedback gain of B, this would mean that the feedback system
has a phase margin of 90◦ , for any A and B (see page 4).
"
#
$ G(jω)
+
−
Σ
A/s
B
"
#
# G(jω)
In this case, the sensitivity function would be given by
dB
0dB
1
s
1
=
×
.
1 + AB/s
AB
(1 + s/AB)
AB
log10 ω
|S(jω)|
However, any real op-amp (and, indeed, any real system) will inevitably
have an attenuation rate of greater 20 dB/decade (and a phase lag of
greater than 90◦ ) at high frequencies. In this case, the Bode sensitivity
integral applies:
This theorem will not be examined.
Theorem: If both L(s) and 1/(1 + L(s)) are asymptotically stable, and
L(jω) rolls off at a rate greater than 20 dB/decade, then
"
"
!∞
"
"
1
"
"
" dω = 0
20log10 "
"
1 + L(jω) "
0
dB
0dB
|S(jω)|
ω
this is sometimes called the “waterbed” effect.
(from Gunter Stein’s Bode Lecture, CDC 1989)
6.5.1
The relationship between open and closed-loop
frequency responses
Ultimately what we are always interested in are properties of the
closed-loop system, such as its frequency response and pole locations.
The following plots are representative of a typical feedback system,
and correspond to a feedback system with a Return Ratio of
2
L(s) =
s(1 + s)
As is typical, the feedback reduces the effect of disturbances at low
frequencies, up to ω1 , as evident from the plot of Sensitivity S(jω).
ω1 is defined here as the lowest frequency at which |S(jω)| = 1. The
closed-loop system will respond to reference inputs at frequencies up
to around ω2 , as evident from the plot of the Complementary
Sensitivity T (jω). ω2 is defined here as the highest frequency at
which |T (jω)| = 1. Between these frequencies both disturbances and
reference signals are amplified (because of the “waterbed” effect).
The actual value of the frequencies ω1 and ω2 , and the size of these
peaks, can be determined directly from the open-loop frequency
response.
1
A: |S(jω)| = |1+L(jω)| = 1 when |1 + L(jω)| = 1, which is when the
distance from the point −1 to the Nyquist locus equals 1 (this is the
point ω = ω1 overleaf).
|L(jω)|
B: |T (jω)| = |1+L(jω)| = 1 when |L(jω)| = |1 + L(jω)|, which is
when the distance from the point −1 to the Nyquist locus equals
the distance from the origin to the Nyquist locus (this is the point
ω = ω2 overleaf).
1
C: |S(jω)| = |1+L(jω)| is maximized when |1 + L(jω)| is minimized,
that is when the distance from the point −1 to the Nyquist locus is
at a minimum.
|L(jω)|
D: The easiest way to find the maximum value of |T (jω)| = |1+L(jω)|
is probably to try a few points around where |1 + L(jω)| is
minimized.
2
Nyquist diagram of the return ratio L(s) = s(1+s)
ω = ω2
−1
"
"1"
−0.5
!
1
"1 + L(j
"L(j
1 )"
ω = ω1
L(jω)
1 )"
ω
1
1.732
L(jω)
−1 − j
−.5 − .289j
1
1+L(jω)
L(jω)
1+L(jω)
j
1.5 + .866j
1−j
−.5 − .866j
L(jω)
1
Closed-loop frequency responses: S(jω) = 1+L(jω) , T (jω) = 1+L(jω)
dB
0dB
|T (jω)|
|S(jω)|
ω1 ω2
r̄ (s) + ē(s)
L(s)
Σ
ω
−
d̄o (s)
+ + ȳ(s)
Σ
Note: this is not a Bode diagram, because it is for a closed-loop
system, and Bode diagrams are always drawn for open-loop systems
(the plant, controller, return ratio etc).
Small gain and/or phase margins correspond to there being
frequencies at which L(jω) comes close to the −1. We now see that
this also corresponds to making |1 + L(jω)| small and hence there
being resonant peaks in the closed-loop transfer functions.
So,
Small gain and/or phase margins
are
bad for robustness, and
bad for performance.
California PATH project
V5
V4
V3
V2
V1
Block diagram
Example: Vehicle Platooning
Example: Vehicle Platooning
Example: Vehicle Platooning
Example: Vehicle Platooning
Example: Vehicle Plat
x
ḡ(jω)
e
California PATH project
V5
Σ
carintrackse distance
Each
carcar
in tracks
to car
carintracks
to car
car intr
xto car
e distance
e distance
e distance
+ xref xEach car tracks
+ xref Each
+ xref xto
+ xref xEach
+ xref Each
Σ
Σ
Σ
Σ
front.ḡ(jω)
front. ḡ(jω)
front. ḡ(jω)
front.ḡ(jω)
front.
−
−
−
−
−
California PATH project
V4
V3 V5
V2 V4
California PATH project
V1 V3
V5
V2
California PATH project
V4
V1
V3 V5
California PATH project
V2 V4
V1 V3 V5
V2 V4
V1 V3
11
ȳȳii(s)
(s)== (exp(−sτ
(exp(−sτii))ūūii(s)
(s)−−ūūi+1
(s))
i+1(s))
sα
sα
1
ȳi (s) =
(exp(−sτi )ūi (s) − ūi+1 (s))
sα
ūūii(s)
(s)==CCii(s)ē
(s)ēii(s)
(s)==CCii(s)(r̄
(s)(r̄ii(s)
(s)−−ȳȳii(s))
(s))
20
ūi (s)
= Ci (s)ēi (s) = Ci (s)(r̄i (s) − ȳi (s))
10
0
gain(dB)
gain(dB)
−10
−20
−30
−40
−50
−60 −3
−3
10
10
−2
−2
10
10
−1
10−1
10
rad/min
rad/min
0
0
10
10
Teei+1 →e
i+3
i+2 →eii
6.6 The Nyquist stability theorem (for
asymptotically stable L(s))
On page 12 we gave an informal statement of the Nyquist stability
criterion. The formal statement of the Nyquist stability theorem
requires counting encirclements of the point −1:
As before, we take L(s) to be the return ratio so that the closed-loop
characteristic equation is 1 + L(s) = 0.
We make the following simplifying assumption:
L(s) is asymptotically stable
This also guarantees that L(jω) is finite for all ω ( i.e. it has no
jω-axis poles), and that L(∞) is finite (since L(s) must be proper – see
Handout 4).
Under this condition the “full” Nyquist diagram of L(jω), for
−∞ < ω < +∞, is a closed curve (since L(j∞) = L(−j∞) = L(∞)).
Note that, since (−jω) = (jω)∗ , it follows that L(−jω) = L(jω)∗ . So
the section of the Nyquist locus for ω < 0 is the reflection in the real
axis of the section for ω > 0.
With this assumption we have:
The Nyquist Stability Theorem (for stable L(s))
Consider a feedback system with an asymptotically stable
return ratio L(s). In this case, the feedback system is
L(s)
asymptotically stable (i.e. 1+L(s) is asymptotically stable) if
and only if the point −1 + j0 is not encircled by the “full”
Nyquist diagram of L(jω), for −∞ < ω < +∞.
6.6.1
Notes on the Nyquist Stability Theorem:
1. Encirclements must be ‘added algebraically’. If there is 1 clockwise
and 1 anticlockwise encirclement then they ‘add up’ to 0
encirclements.
2. L(s) often has one or more poles at 0 (due to integrators in the
plant or the controller). The theorem still works, but one has to
worry about what happens to the graph of L(jω) at ω = 0 – as the
locus is no longer a closed curve. It can be shown that, if L(s) has
n poles at the origin, then the Nyquist locus should be completed
by adding a large n × 180◦ arc, in a clockwise direction.
3. If L(s) is unstable, and has np unstable poles, then the theorem
must be modified as follows: “The feedback system is stable if and
only if the ‘full’ Nyquist diagram encircles the point −1 + j0 np
times in an anticlockwise direction.”
4. (This a repeat of the informal statement of the Nyquist stability
criterion from page 12.)
A potentially ambiguous statement of the theorem, but one which
almost always works, is: “The feedback system is stable if the
Nyquist diagram of L(jω) ‘leaves the point −1 + j0 on its left’ ”.
This still works (usually) if there are poles at the origin or even if
L(s) is unstable.
This informal version of the theorem is adequate for the
examples which follow, for any others that you will encounter
on this course and indeed for any you are likely to encounter
in practice.
2
1.5
1
3
ℑ(L(s))
0.5
4
0
30
22
5
16
6
−0.5
7
10
−1
−1.5
−3
−2
−1
0
ℜ(L(s))
MotorPower = -35*WheelAngle - 0.3*WheelVelocity
+ 0.8*LeanAngle + 0.35*GyroValue
3
1.5
1
4
ℑ(L(s))
0.5
30
0
5
−0.5
22
16
6
7
−1
10
−1.5
−3
−2
−1
0
ℜ(L(s))
MotorPower = (-35*WheelAngle - 0.3*WheelVelocity
+ 0.8*LeanAngle + 0.35*GyroValue ) *1.7
1.5
1
2
3
ℑ(L(s))
0.5
4
0
30
5
22
16
6
7 10
−0.5
−1
−1.5
−3
−2
−1
0
ℜ(L(s))
MotorPower = (-35*WheelAngle - 0.3*WheelVelocity
+ 0.8*LeanAngle + 0.35*GyroValue ) *0.6
Examples: Formal application of Nyquist stability theorem.
"
−1
no encirlements of −1
L(s)
is
=⇒
1 + L(s)
asymptotically
stable
"
!
−1
!
2 clockwise encirclements of −1
L(s)
is
=⇒
1 + L(s)
unstable
"
−1
"
!
1 clockwise + 1 anticlockwise
encirclement of −1
i.e. 0 net encirclements (note 1)
L(s)
=⇒
is
1 + L(s)
asymptotically stable
−1
!
no encirlements of −1 (note 2)
L(s)
=⇒
is
1 + L(s)
asymptotically
stable
Examples: Informal application of the Nyquist stability theorem
(based on note 4).
"
−1
-1 to left of locus
L(s)
=⇒
is
1 + L(s)
asymptotically stable
"
!
−1
-1 to right of locus
L(s)
=⇒
is
1 + L(s)
unstable
!
"
−1
-1 to left of locus
L(s)
is
=⇒
1 + L(s)
asymptotically stable
"
!
−1
!
-1 to left of locus
L(s)
is
=⇒
1 + L(s)
asymptotically stable
Hence: the informal application of the Nyquist stability criterion works
for all these cases.