Transcript Modern System Engineering

NUAA-Control System Engineering
Chapter 5
Frequency-Domain Analysis
Control System Engineering-2008
Content in Chapter 5
 5-1 Frequency Response (or Frequency Characteristics)
 5-2 Nyquist plot and Nyquist stability criterion
 5-3 Bode plot and Bode stability criterion
Control System Engineering-2008
5-1 Frequency Response
Control System Engineering-2008
A Perspective on the Frequency-Response
Design Method
The design of feedback control systems in industry is
probably accomplished using frequency-response methods
more than any other.
Advantages of frequency-response design:
-It provides good designs in the face of uncertainty in the
plant model
-Experimental information can be used for design purposes.
Raw measurements of the output amplitude and phase of a plant
undergoing a sinusoidal input excitation are sufficient to design a
suitable feedback control.
-No intermediate processing of the data (such as finding
poles and zeros) is required to arrive at the system model.
Control System Engineering-2008
Frequency response
The frequency response of a system is defined
as the steady-state response of the system to a
sinusoidal input signal.
r (t )  R sin 0t 
G(s)
y (t )  Y sin(0t   )
H(s)
For a LTI system, when the input to it is a sinusoid signal,
the resulting output , as well as signals throughout the
system, is sinusoidal in the steady-state;
The output differs from the input waveform only in
amplitude and phase.
Control System Engineering-2008
The closed-loop transfer function of the LTI
system:
Y ( s)
G( s)
M ( s) 

R( s ) 1  G ( s ) H ( s )
For frequency-domain analysis, we replace s
by jω:
Y ( j )
G( j )
M ( j ) 

R( j ) 1  G( j ) H ( j )
The frequency-domain transfer function M(jω)
may be expressed in terms of its magnitude and
phase:
M ( j )  M ( j ) M ( j )
magnitude
phase
Control System Engineering-2008
M (M
j( )j )   A
The magnitude of M(jω) is
G ( j )
M ( j ) 
1  G ( j ) H ( j )
G ( j )

1  G ( j ) H ( j )
Gain characteristic

0
  c
  c
A
0
c


0
The phase of M(jω) is
M ( j )  M ( j )
 G( j )   1  G( j ) H ( j )
Phase characteristic
MGain-phase
( j )
characteristics
of an ideal low-pass filter
Control System Engineering-2008
Example. Frequency response of a Capacitor
Consider the capacitor described by the equation
dv
iC
dt
where v is the input and i is the output. Determine the
sinusoidal steady-state response of the capacitor.
Solution. The transfer function of the capacitor is
I ( s)
 M ( s )  Cs
V ( s)
So
M ( j)  Cj
Computing the magnitude and phase, we find that
M ( j )  Cj  C
M ( j )  M  90
Control System Engineering-2008
Gain characteristic:
M ( j )  Cj  C
Phase characteristic:
M ( j )  M  90
Output:
I ( j)  M ( j)V ( j)
For a unit-amplitude sinusoidal input v, the output i will be a
sinusoid with magnitude Cω, and the phase of the output
will lead the input by 90°.
Note that for this example the magnitude is proportional to
the input frequency while the phase is independent of
frequency.
Control System Engineering-2008
Frequency-Domain Specifications
M ( j )
Resonant peak M r
Mr
Cutoff
rate
0.707
0
0
r
BW
Resonant frequency r
Bandwidth BW


M ( j )
Typical gain-phase characteristic of a control system
d M ( j )
0
d

r
Control System Engineering-2008
Frequency response of a prototype
second-order system
Closed-loop transfer function:
n2
Y ( s)
M ( s) 
 2
R( s ) s  2n s  n2
Its frequency-domain transfer function:
n2
Y ( j )
M ( j ) 

R( j ) ( j ) 2  2n ( j )  n2
Define u   n
M ( ju ) 
1
1  j 2 u  u 2
Control System Engineering-2008
The magnitude of M(ju) is
1
M ( ju ) 
[(1  u 2 )2  (2 u)2 ]1/2
The phase of M(ju) is
2 u
M ( j )  M ( j )   tan
1  u2
1
The resonant frequency of M(ju) is
d M ( ju )
0
du
Resonant peak
1
Mr 
2 1   2
ur  1  2 2
With ur  r n , we have r  n 1  2 2
Since frequency is a real quantity, it requires 1  2 2  0
So
  0.707
Control System Engineering-2008
According to the definition of Bandwidth
1
1
M ( ju) 

 0.707
2 2
2 1/2
[(1  u )  (2 u) ]
2
u 2  (1  2 2 )  4 4  4 2  2
With u   n , we have
BW  n [(1  2 2 )  4 4  4 2  2]1/2
Control System Engineering-2008
For a prototype second-order system (  0.707)
Resonant peak
Mr 
1
2 1  
2
Resonant frequency
r  n 1  2 2
M r depends on  only.
For   0, the system is unstable;
For 0<  0.707,   M r ;
For   0.707, M r  1
r depends on both  and n .
For 0<  0.707, n fixed,   r ;
For   0.707, r  0.
2
4
2
1/2
Bandwidth BW  n [(1  2 )  4  4  2]
BW is directly proportional to n , n  BW 
  BW n 
For 0    0.707, n fixed,   BW ;
Control
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Correlation between pole locations,
unit-step
response and
the magnitude of the frequency response
j
r (t )

s 2  2n s  n2
y (t )
2
n

n
0
0   1
y (t )
  cos1 
max overshoot  e
 / 1 2

M ( j )
1.0
0dB
0.3dB
0.9
t
0.1
0
tr 
1  0.4167  2.917
n
2
0
BW

BW  n [(1  2 2 )  4 4  4 2  2]1/2
Control System Engineering-2008
Example. The specifications on a second-order unityfeedback control system with the closed-loop transfer
function
n2
Y ( s)
M ( s) 
 2
R( s ) s  2n s  n2
are that the maximum overshoot must not exceed 10
percent, and the rise time be less than 0.1 sec. Find the
corresponding limiting values of Mr and BW analytically.
Solution. Maximum overshoot:
%  e
Rise time: tr 

1 2
 100%  10%
1  0.4167  2.917 2
n
  0.6
 0.1 (0    1)
2.917 2  0.4167  1  0.1n  0
 1,2 
0.4167  0.4167 2  4  2.917  (1  0.1n )
2  2.917
n  18
System
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0.6 and n  18
Based on time-domain analysis,Control
we obtain
Frequency-domain specifications:
1
Resonant peak M r 
(  0.707)
2
2 1  
For 0<  0.707,   M r ;
  0.6
For   0.707, M r  1
1  M r  1.04
2
4
2
1/2
Bandwidth BW  n [(1  2 )  4  4  2]
BW is directly proportional to n , n  BW 
  BW n 
For 0    0.707, n fixed,   BW ;
0.6    0.707
1  BW n 1.15
n  BW  1.15n
n  18
BW  18
Control System Engineering-2008
Effects of adding a zero to the OL TF2

R(s)
n2
s( s  2n )
n
G
(
s
)

Open-loop
TF：
Y ( s)
s( s  2n )
Closed-loop TF：
Y ( s)
G( s )
n2
M ( s) 

 2
R( s) 1  G( s) s  2n s  n2
Adding a zero at s  1 Tz
R(s)
1  Tz s
n2
s( s  2n )
Y ( s)
Open-loop TF：
(1  Tz s)n2
G( s) 
s( s  2n )
2

n (1  Tz s )
Closed-loop TF： ( s)  2
s  (2n  Tzn2 ) s  n2
The additional
zero changes
both numerator
and denominator.
Control System Engineering-2008
As analyzing the prototype second-order system, using
similar but more complicate calculation, we obtain
Bandwidth
BW  ( b  1 / 2 b2  4n4 )1/2
where b  4 2n2  4n3Tz  2n2  n4Tz2
For fixed ωn and ζ, we analyze the effect of
Tz.
Control System Engineering-2008
The general effect of adding a zero the open-loop
transfer function is to increase the bandwidth of the
closed-loop system.
Bode Diagram
Magnitude (dB)
0
-20
Tz=0
Tz=0.2
-40
-60
-80
0
Phase (deg)
n  1
  0.2
20
Tz=1
Tz=5
-45
-90
-135
-180
-1
10
10
0
10
Frequency (rad/sec)
1
10
2
Control System Engineering-2008
Effects of adding a pole to the OL TF 2

R(s)
n2
s( s  2n )
n
G
(
s
)

Open-loop
TF：
Y ( s)
s( s  2n )
Closed-loop TF：
Y ( s)
G( s )
n2
M ( s) 

 2
R( s) 1  G( s) s  2n s  n2
Adding a pole at s  1 Tp
R(s)
1
1  Tp s
n2
s( s  2n )
Closed-loop TF：  (s) 
Y ( s)
Open-loop TF：
n2
G( s ) 
s(s  2n )(1  Tp s)
n2
Tp s3  (1  2nTp )s 2  2n s  n2
The effect of adding a pole the open-loop
transfer
function is
Control System
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to make the closed-loop system less stable, while decreasing
the bandwidth.
Bode Diagram
Tp=0
Tp=0.5
Tp=1
Tp=5
0
-50
-100
-150
0
-45
Phase (deg)
n  1
  0.707
Magnitude (dB)
50
-90
-135
-180
-225
-270
-2
10
10
-1
10
0
Frequency (rad/sec)
10
1
10
2
Control System Engineering-2008
5-2 Nyquist Plot and Nyquist
Criterion
Control System Engineering-2008
Nyquist Criterion
What is Nyquist criterion used for?
R( s)
Y ( s)

G(s)
H(s)
Nyquist criterion is a semigraphical method that
determines the stability of a closed-loop system;
Nyquist criterion allows us to determine the stability of a
closed-loop system from the frequency-response of the
loop function G(jw)H(j(w)
Control System Engineering-2008
Review about stability
Closed-loop TF:
G( s)
M ( s) 
1  G( s) H ( s)
Characteristic equation (CE):
( s)  1  G( s) H ( s)  0
Stability conditions:
Open-loop stability: poles of the loop TF G(s)H(s) are all in
the left-half s-plane.
Closed-loop stability: poles of the closed-loop TF or roots
of the CE are all in the left-half s-plane.
Control System Engineering-2008
Definition of Encircled and Enclosed
Encircled: A point or region in a complex function plane is
said to be encircled by a closed path if it is found inside
the path.
Enclosed: A point or region in a complex function plane is
said to be encircled by a closed path if it is encircled in the
countclockwise(CCW) direction.
B
A
Point A is encircled in the
closed path;
Point A is also enclosed in the
closed path;

Control System Engineering-2008
Number of Encirclements and Enclosures
A
B
Point A is encircled once;
Point B is encircled twice.
D
C
Point C is enclosed once;
Point D is enclosed twice.
Control System Engineering-2008
Mapping from the complex s-plane to the
Δ(s) -plane
Exercise 1: Consider a function Δ(s) =s-1, please map a
circle with a radius 1 centered at 1 from s-plane to the
Δ(s)-plane .
j
1
s3
0
1
s-plane
j Im
Mapping
 ( s2 )
s2
1
s1
2
s4
Δ( s)-plane
 ( s3 )

0
 ( s1 )
Re[( s)]
 ( s4 )
s1  2; s2  1  j
( s1 )  1; ( s2 )  j
s3  0; s4  1  j
( s3 )  1; ( s4 )   j
Control System Engineering-2008
Principle the Argument
Let ( s ) be a single-valued function that has a finite
number of poles in the s-plane.
Suppose that an arbitrary closed path  s is chosen in the splane so that the path does not go through any one of the
poles or zeros of ( s ) ;
The corresponding   locus mapped in the ( s )-plane will
encircle the origin as many times as the difference
between the number of zeros and poles (P) of ( s ) that are
encircled by the s-plane locus  s .
In equation form:
N Z P
N - number of encirclements of the origin by the ( s ) -plane locus
Z - number of zeros of ( s ) encircled by the s-plane locus
P - number of poles of ( s ) encircled by the s-plane locus
Control System Engineering-2008
Nyquist Path  s
A curve composed of the imaginary axis and an arc of
infinite radius such that the curve completely encloses the
right half of the s-plane .
j
s-plane
R

0
Nyquist path is in the
CCW direction
Since in mathematics, CCW
is traditionally defined to be
the positive sense.
s
Note Nyquist path does not pass through any poles or
zeros of Δ(s); if Δ(s) has any pole or zero in the righthalf plane, it will be encircled by  s .
Control System Engineering-2008
Nyquist Criterion and Nyquist Diagram
j
Δ( s)-plane
s-plane
( s)  1  G( s) H ( s)
R
1

0
s
Nyquist Path
Critical point:
(-1+j0)
G( s)H(s)-plane
1 0
Nyquist Diagram:
Plot the loop
function to
determine the
closed-loop
stability
Control System Engineering-2008
Nyquist Criterion and G(s)H(s) Plot
j
G( s)H(s)-plane
s-plane
G( s ) H ( s )
R

0
1 0
s
Nyquist Path
G(s)H(s) Plot
The Nyquist Path is shown in the left figure. This path is
mapped through the loop tranfer function G(s)H(S) to
the G(s)H(s) plot in the right figure. The Nyquist
Creterion follows:
N Z P
Control System Engineering-2008
Nyquist Criterion and Nyquist Plot
j
G( s)H(s)-plane
s-plane
G( s ) H ( s )
R

0
1 0
s
Nyquist Path
Nyquist Plot
N - number of encirclements of (-1,j0) by the G(s)H(s) plot
Z - number of zeros of ( s ) that are inside the right-half plane
P - number of poles of ( s ) that are inside the right-half plane
The condition of closed-loop stability according to the
Nyquist Creterion is:
N  P
Control System Engineering-2008
K  i 1 ( s  zi )
m
G( s) H (s) 

n
j 1
( s  pi )

( s )  1  G ( s ) H ( s ) 
n
j 1
( s  pi )  K  i 1 ( s  zi )
m

n
(
s

p
)
i
j 1
( s ) has the same poles as G(s) H (s) , so P can be obtained
by counting the number of poles of G(s) H (s) in the right-
half plane.
Control System
Engineering-2008
Nyquist Diagram
4
An example
3
Imaginary Axis
Consider the system with
the loop function
5
G( s) H ( s) 
( s  1)3
Matlab program for
Nyquist plot
(G(s)H(s) plot)
>>num=5;
>>den=[1 3 3 1];
>>nyquist(num,den);
Question 1: is the
closed-loop system
stable?
2
1
0
-1
-2
-3
-4
-2
-1
0
1
2
3
Real Axis
Question 2: what if
5K
G( s) H ( s) 
?
3
( s  1)
N=0, P=0,
N=-P, stable
4
5
1. With root locus technique:
5K
1
*
G( s) H ( s) 
K
3
( s  1)
( s  1)3
3
For K* varies from 0
to ∞, we draw the RL 2
>>num=1;
>>den=[1 3 3 1];
>>rlocus(num,den);
Root Locus
K*  
1
Imaginary Axis
When K*=8
(K=1.6), the RL
cross the jw-axis,
the closed-loop
system is
marginally stable.
Control System Engineering-2008
*


K
0
K*  8
K*  0
-1
-2
K*  
-3
-4
-3
-2
-1
0
1
Real Axis
For K*>8 (K>1.6), the closed-loop system has two roots in
the RHP and is unstable.
2
2. With Nyquist plot and
Nyquist criterion:
5K
G( s) H ( s) 
( s  1)3
K=1
Nyquist plot
does not
encircle (-1,j0),
so N=0
>>num=5*K;
>>den=[1 3 3 1];
>>nyquist(num,den);
Nyquist Diagram
4
3
2
Imaginary Axis
No pole of
G(s)H(s) in RHP,
so P=0;
Control System
Engineering-2008
>>K=1;
1
0
-1
-2
-3
-4
-2
-1
0
Thus N=-P
The closed-loop system is stable
1
2
Real Axis
3
4
5
Control System
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>>K=1.6;
2. With Nyquist plot and
Nyquist criterion:
5K
G( s) H ( s) 
( s  1)3
K=1.6
The Nyquist plot
just go through
(-1,j0)
Nyquist Diagram
6
4
2
Imaginary Axis
No pole of
G(s)H(s) in RHP,
so P=0;
>>num=5*K;
>>den=[1 3 3 1];
>>nyquist(num,den);
0
-2
-4
-6
-2
-1
0
1
2
3
Real Axis
The closed-loop system is marginally stable
4
5
6
7
8
Control System
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>>K=4;
2. With Nyquist plot and
Nyquist criterion:
5K
G( s) H ( s) 
( s  1)3
Nyquist Diagram
K=4
Nyquist plot
encircles (-1,j0)
twice, so N=2
15
10
5
Imaginary Axis
No pole of
G(s)H(s) in RHP,
so P=0;
>>num=5*K;
>>den=[1 3 3 1];
>>nyquist(num,den);
0
-5
-10
-15
-5
Thus Z=N+P=2
Real Axis
The closed-loop system has two poles in RHP and is unstable
0
5
10
15
20
Control System Engineering-2008
Nyquist Criterion for Systems with
Minimum-Phase Transfer Functions
What is called a minimum-phase transfer function?
A minimum-phase transfer function does not have poles
or zeros in the right-half s-plane or on the jw-axis,
except at s=0.
Consider the transfer functions
10( s  1)
G1 ( s) 
( s  10)
10( s  1)
( s  10)
Both transfer functions have the same magnitude for all
frequencies
G ( j)  G ( j)
1
G2 ( s ) 
2
But the phases of the two transfer functions are drastically
different.
Control System Engineering-2008
Bode Diagram
Magnitude (dB)
20
15
G1 ( j )  G2 ( j )
10
5
Phase (deg)
0
180
G2 ( j )
135
90
45
0
-2
10
G1 ( j )
10
-1
10
0
10
1
10
2
10
3
Frequency (rad/sec)
A minimum-phase system (all zeros in the LHP) with a
given magnitude curve will produce the smallest change
in the associated phase, as shown in G1.
Control System Engineering-2008
Consider the loop transfer function:
L( s)  G( s) H ( s)
If L(s) is minimum-phase, that is, L(s) does not
have any poles or zeros in the right-half plane or
on the jw-axis, except at s=0
Then P=0, where P is the number of poles of
Δ(s)=1+G(s)H(s), which has the same poles as L(s).
Thus, the Nyquist criterion (N=-P) for a system
with L(s) being minimum-phase is simplified to
N 0
Control System Engineering-2008
Nyquist criterion for systems with minimum-phase loop
transfer function
N 0
For a closed-loop system with loop transfer function L(s)
that is of minimum-phase type, the system is closedloop stable , if the Nyquist plot (L(s) plot) that
corresponds to the Nyquist path does not enclose (-1,j0)
point. If the (-1,j0) is enclosed by the Nyquist plot, the
system is unstable.
The Nyquist stability can be checked by plotting the
segment of L(jw) from w= ∞ to 0.
Control System Engineering-2008
Example Consider a single-loop feedback system with the
loop transfer function
L( s)  G( s) H ( s) 
K
s( s  2)( s  10)
Analyze the stability of the closed-loop system.
Solution.
Since L(s) is minimum-phase, we can analyze the closed-loop
stability by investigating whether the Nyquist plot enclose
the critical point (-1,j0) for L(jw)/K first.
j Im
L( j )
1

Im[ L( j) K ]  0
K
j( j  2)( j  10)

L( j  )
Real
0
 0  270
w=∞:
K
L( j 0)
   90
w=0+:
0
K
Control System Engineering-2008
1
Im[ L( j) K ]  Im[
]0
j( j  2)( j  10)
   20 rad / s
The frequency is positive, so   20 rad / s
L( j 20) K 
1
 0.004167
j 20( j 20  2)( j 20  10)
1. K  240  L( j 20)  1
the Nyquist plot does not enclose (-1,jw);
2. K  240  L( j 20)  1
stable
the Nyquist plot goes through (-1,jw); marginally stable
3. K  240  L( j 20)  1
the Nyquist plot encloses (-1,jw).
unstable
Control System>>z=[]
Engineering-2008
By root locus technique
>>p=[0, -2, -10];
>>k=1
>>sys=zpk(z,p,k);
>>rlocus(sys);
1
L( s)  G( s) H ( s )  K
s( s  2)( s  10)
Root Locus
20
15
Imaginary Axis
10
5
K  240
0
-5
-10
-15
-20
-30
-25
-20
-15
-10
Real Axis
-5
0
5
10
Control System Engineering-2008
Relative Stability
Gain Margin and Phase Margin
For a stable system, relative stability describes how stable
the system is.
In time-domain, the relative stability is measured by
maximum overshoot and damping ratio.
In frequency-domain, the relative
stability is measured by resonance
peak and how close the Nyquist plot
of L(jw) is to the (-1,j0) point.
The relative stability of the
blue curve is higher than the
green curve.
j Im

1
0
0
Real
Control System Engineering-2008
Gain Margin (GM)
(for minimum-phase loop transfer functions)
Phase crossover
Phase crossover frequency ωp
L( j p )  180
For a closed-loop system with
L(jw) as its loop transfer
function, it gain margin is
defined as
gain margin (GM) = 20log10
j Im
p

L( j p )
0
1
L( j p )
 20log10 L( j p ) dB
L(jw)-plane
0
Real
gain margin (GM) = 20log10
1 Control System Engineering-2008
L( j p )
 20log10 L( j p ) dB
When L( j )  1 (stable), log10 L( j )  0  GM  0
L( j)  (closer to the origin)  GM  (more stable)
L( j)  (closer to -1)  GM  (less stable)
When L( j )  1 (marginally stable), log10 L( j )  0  GM=0
When L( j)  1 (unstable), log10 L( j )  0  GM<0
Gain margin represents the amount of gain in decibels (dB)
that can be added to the loop before the closed-loop
system becomes unstable.
Control System Engineering-2008
Phase Margin (PM)
(for minimum-phase loop transfer functions)
Gain margin alone is inadequate to indicate relative
stability when system parameters other the loop gain are
j Im
subject to variation.
With the same gain margin,
system represented by plot A is
more stable than plot B.
1
Gain crossover frequency ωg
0
PM
L( j g )  1
L( jg )  1
Phase margin:
phase margin (PM) =L( jg )  180
L( jg )
B
A
Real
Control
System Engineering-2008
Example Consider the transfer
function
10
G (s) 
s ( s  1)
Draw its Nyquist plot when w varies from 0 to ∞.
Solution. Substituting s=jw into G(s) yields: G( j ) 
10
j ( j  1)
The magnitude and phase of G(jw) at w=0 and w=∞ are
computed as follows.
lim G ( j )  lim
 0
 0
10
10
 lim  
j ( j  1)  0 
lim G ( j )  lim 
 0
 0
lim G ( j )  lim
 
 
10
0
j ( j  1)
lim G ( j )  lim 
 
 
10
10
 lim 
 90


0
j ( j  1)
j
10
10
 lim  2  180
j ( j  1)   
Thus the properties of
the Nyquist plot of G(jw)
at w=0 and w=∞ are
ascertained.
Next we determine the
intersection…
Control System Engineering-2008
If the Nyquist plot of G(jw) intersects with the real axis, we have
Im[G( j)]  0
G ( j ) 
10
10
10
 4

j
j ( j  1)    2
4  2
2
10
0
4
2
 
 
This means that the G(jw) plot intersects only with the real axis of the
G(jw)-plane at the origin.
Similarly, intersection of G(jw) with the imaginary axis:
Re[G( j)]  0
 
which corresponds to the origin of the G(jw)-plane.
The conclusion is that the Nyquist
plot of G(jw) does not intersect any
one of the axes at any finite
nonzero frequency.
At w=∞, Re[G( j)]  0
At w=0, Re[G( j)]  10
Control
System
Engineering-2008
Example Consider a system with
a loop
transfer
function as
2500
L( s ) 
s( s  5)( s  50)
Determine its gain margin and phase margin.
Solution. Phase crossover frequency ωp:
j Im
Im[ L( j )]  0   p  15.88 rad/sec
Gain margin: L( j p )  0.182
GM =  20log10 L( j p )  14.80 dB
31.72
Gain crossover frequency ωg:
L( jg )  1  g  6.22 rad/sec
Phase margin:
PM =L( jg )  180  31.72
1
 p  15.88
0

0.1820
g  6.22
Control System Engineering-2008
Advantages of Nyquist plot:
-By Nyquist plot of the loop transfer function, the
closed-loop stability can be easily determined with
reference to the critical point (-1,j0).
-It can analyze systems with either minimum phase or
nonminimum phase loop transfer function.
Disadvantages of Nyquist plot:
-By Nyquist plot only, it is not convenient to carry out
controller design.
Control System Engineering-2008
5-3 Bode Plot
Control System Engineering-2008
Bode Plot
The Bode plot of the function G(jw) is composed of two
plots:
-- the amplitude of G(jw) in decibels (dB) versus log10w
or w
-- the phase of G(jw) in degrees as a function of
log10w or w.
Without loss of generality, the following transfer function
is used to illustrate the construction of the Bode Plot
G( s) 
K (1  T1s)(1  T2 s)
s j (1   1s)(1  2 s / n  s 2 / n2 )
where K, T1, T2, τ1, ζ, ωn are real constants. It is assumed
that the second-order polynomial in the denominator has
complex conjugate zeros.
Control System Engineering-2008
Substituting s=jw into G(s) yields
G( j ) 
K (1  jT1 )(1  jT2)
( j)(1  j 1 )(1  j 2 / n   2 / n2 )
The magnitude of G(jw) in dB is obtained by multiplying the
logarithm (base 10) of |G(jw)| by 20; we have
G ( j ) dB  20log10 G ( j )
 20log10 K  20log10 1  jT1  20log10 1  jT2
 20log10 j  20log10 1  j 1  20log10 1  j 2 / n   2 / n2
The phase of G(jw) is
G( j )  K  (1  jT1 )  (1  jT2 )  j
 (1  j 1 )  (1  j 2 / n   2 / n2 )
Control System Engineering-2008
In general, the function G(jw) may be of higher order and
have many more factored terms. However, the above two
equations indicate that additional terms in G(jw) would
simply produce more similar terms in the magnitude and
phase expressions, so the basic method of construction of
the Bode plot would be the same.
In general, G(jw) can contain just four simple types of
factors:
1. Constant factor: K
2. Poles or zeros at the origin of order p: (jw)±p
3. Poles or zeros at s =-1/T of order q: (1+jwT )±q
4. Complex poles and zeros of order r:
(1 + j2ζω/ωn-ω2/ω2n)
Control System Engineering-2008
1. Real constant K
K dB  20log 20 K
 constant
0
K  
180
K 0
K 0
Control System Engineering-2008
p
2. Poles or zeros at the origin, ( j )
Magnitude of ( j )  p in dB:
( j )  p
dB
 20log 20 ( j )  p  20 p log 20  dB
For a given p, it is a straight line with the slope:
d
 20 p log10    20 p dB/decade
d log10 
Thus a unit change in log10  corresponds to a change of
±20 dB in the magnitude.
At  =1, ( j )  p
dB
 0.
So these lines pass through the 0dB axis at ω =1.
Control System Engineering-2008
Magnitude of ( j )  p :
20 p log 20  dB
Phase of ( j )  p :
 p  90
Control System Engineering-2008
3. (a) Simple zero 1+jwT
Consider the function
G( j)  1  jT
where T is a positive real constant.
The magnitude of G(jw) in dB is
G ( j ) dB  20log10 G ( j )  20log10 1   2T 2
At very low frequencies, T  1
G( j ) dB  20log10 1  0 dB
(  2T 2 is neglected when compared with 1.)
At very high frequencies, T  1
G ( j ) dB  20log10  2T 2  20log10 T dB
The two lines
intersect at:
 1/ T
(corner
frequency)
This represents a straight line with a slope of 20dB
jT dB Engineering-2008
The steps of making of sketchControl
of 1 System
Step 1: Locate the corner frequency w=1/T on the
frequency axis;
Step 2: Draw the 20dB/decade line and the horizontal line
at 0 dB with the two lines intersecting at w=1/T.
Step 3: Sketch a smooth curve by locating the 3-dB point at
the corner frequency and the 1-dB points at 1 octave above
and below the corner frequency.
Control System Engineering-2008
The phase of G(jw)=1+jwT is
G ( j )  tan 1 T
At very low frequencies, G ( j )  0
At very high frequencies, G ( j )  90
3. (b) Simple pole, 1/(1+jwT)
Control System Engineering-2008
1
G( j ) 
1  jT
The magnitude of G(jw) in dB is
Consider the function
G ( j ) dB  20log10 G( j )  20log10 1   2T 2
At very low frequencies, T  1
G( j) dB  20log10 1  0 dB
At very high frequencies, T  1
G ( j ) dB  20log10  2T 2  20log10 T dB
The two lines
intersect at:
 1/ T
(corner
frequency)
This represents a straight line with a slope of -20dB
The phase of G(jw): G ( j )   tan 1 T
For w varies from 0 to ∞, G( j) varies from 0°to -90°.
Control System Engineering-2008
Control System Engineering-2008
4. Complex poles and zeros
Consider the second-order transfer function
n2
1
G( s)  2

2
s  2n s  n 1  (2 n ) s  (1 n2 ) s 2
We are interested only in the case when ζ ≤ 1, since
otherwise G(s) would have two unequal real poles, and
the Bode plot can be obtained by considering G(s) as the
product of two transfer functions with simple poles.
By letting s=jw, G(s) becomes
1
G( j ) 
[1  ( 2 n2 )]  j 2 ( n )
Control System Engineering-2008
G( j ) 
1
[1  ( 2 n2 )]  j 2 ( n )
The magnitude of G(jw) in dB is
G ( j ) dB  20log10 G ( j )
 20log10 [1  ( 2 n2 )]2  4 2 ( n ) 2
At very low frequencies,  / n  1
G( j) dB  20log10 1  0 dB
At very high frequencies,  / n  1
G ( j ) dB  20log10
The two lines
intersect at:
  n
(corner
( n )  40log10 ( n ) dB frequency)
4
This equation represents a straight line with a slope of
40 dB decade in the Bode plot coordinates.
Control System Engineering-2008
The actual magnitude curve of G(jw) in this case may
differ strikingly from the asymptotic curve.
The reason for this is that the amplitude
and phase curves of the second-order G(jw)
depend not only on the corner frequency wn,
but also on the damping ratio ζ, which does
not enter the asymptotic curve.
Control System Engineering-2008
The phase of G(jw) is given by
2


   

1  2
G( j )   tan 
1     



 n   n   

Control System Engineering-2008
Example Consider the following
transfer function
G( s) 
10( s  10)
s( s  2)( s  5)
Sketch its Bode Plot.
Solution. Letting s=jw, we have
10( j  10)
G( j ) 
j( j  2)( j  5)
Reformulating it into the form for Bode Plot
G( j ) 
K (1  jT1 )
10(1  j 0.1 )

j(1  j 1 )(1  j 2 ) j(1  j 0.5 )(1  j 0.2 )
where K  10, T1  0.1,  1  0.5,  2  0.2
So G(jw) has corner frequencies at w=10,2 and 5 rad/sec.
Control System Engineering-2008
1. Bode plot of K=10
Bode Diagram
Magnitude (dB)
21
20.5
20
19.5
Phase (deg)
19
1
0.5
0
-0.5
-1
-1
10
10
0
10
1
Frequency (rad/sec)
10
2
10
3
Control System Engineering-2008
2. Bode Plot of the component with pole at origin : jw
magnitude curve: a straight line with slope of 20 dB/decade,
passing through the w=1 rad/sec point on the 0-dB axis.
Bode Diagram
Magnitude (dB)
20
0
-20
-40
Phase (deg)
-60
-89
-89.5
-90
-90.5
-91
-1
10
10
0
10
1
Frequency (rad/sec)
10
2
10
3
Control System Engineering-2008
3. Bode plot of simple zero component 1+j0.1w
Corner frequency: w=1/0.1=10 rad/sec
Bode Diagram
Magnitude (dB)
60
50
40
30
Phase (deg)
20
90
45
0
-1
10
10
0
10
1
Frequency (rad/sec)
10
2
10
3
Control System Engineering-2008
4. Bode plot of simple pole componet 1/(1+j0.5w)
Corner frequency: w=1/0.5=2 rad/sec
Bode Diagram
Magnitude (dB)
0
-20
-40
-60
Phase (deg)
-80
0
-45
-90
-1
10
10
0
10
1
Frequency (rad/sec)
10
2
10
3
Control System Engineering-2008
5. Bode plot of simple pole component 1/(1+j0.2w)
Corner frequency: w=1/0.2=5 rad/sec
Bode Diagram
-10
Magnitude (dB)
-20
-30
-40
-50
Phase (deg)
-60
0
-45
-90
-1
10
10
0
10
1
Frequency (rad/sec)
10
2
10
3
|G(jw)|dB is obtained
by adding the
component curves
together, point by
point.
Bode Plot:
Gain crossover
point: |G(jw)|dB
cross the 0-dB
axis
Phase crossover
point: where the
phase curve cross
the -180°axis.
Control System Engineering-2008
Control System Engineering-2008
Nyquist Plot (Polar Plot) :
The gain-crossover point is where G ( j )  1 ,
The phase crossover point is where G ( j )  180 .