System type, steady state tracking,
& Bode plot
R(s)
C(s)
G s   C ( s )G p ( s ) 
Gp(s)
Y(s)
K T a s  1 T b s  1 
s
N
T1 s  1T 2 s  1
At very low frequency:
gain plot slope = –20N dB/dec.
phase plot value = –90N deg
Type = N
Type 0: gain plot flat at very low frequency
phase plot approached 0 deg
Kv = 0
Ka = 0
Low freq phase = 0o
Type 1: gain plot -20dB/dec at very low frequency
phase plot approached 90 deg
Low frequency
tangent line
Kp = ∞
=Kv
Ka = 0
Low freq phase = -90o
Back to general theory
N = 2,
type = 2
Bode gain plot has –40 dB/dec
slope at low freq.
Bode phase plot becomes flat
at –180° at low freq.
Kp = DC gain → ∞
Kv = ∞ also
Ka = value of straight line at ω = 1
= ws0dB^2
20 log G  j w   20 log K  N 20 log w
A sym ptotic straight line:
y  20 log K  N 20 log w
A t w = 1: y  20 log K  20 log K a
W hen y= 1, straight line cross hor. axis.
T he crossing frequency is:
0 dB  20 log K  N 20 log w s 0 dB
 20 log K a  2  20 log w s 0 dB
 20 log K a  20 log w s20 dB

K a  w s20 dB
Type 1: gain plot -40dB/dec at very low frequency
phase plot approached 180 deg
Low frequency
tangent line
Kp = ∞
Kv = ∞
Low freq phase = -180o
Example
Ka
ws0dB=Sqrt(Ka)
How should the phase plot look like?
Example continued
At low freq :
straight
G  jw  
G  jw  
line :
it is 0
dB at w  4
||
1

K
4
2
 1,
K  16
 K a  16
Kv  
Kp 
K
 jw 
K
w
2
2
Example continued
Suppose the closed-loop system is stable:
If the input signal is a step,
ess would be =
If the input signal is a ramp,
ess would be =
If the input signal is a unit acceleration,
ess would be =
System type, steady state tracking,
& Bode plot
At very low frequency:
gain plot slope = –20N dB/dec.
phase plot value = –90N deg
If LF gain is flat, N=0, Kp = DC gain,
Kv=Ka=0
If LF gain is -20dB/dec, N=1, Kp=inf,
Kv=wLFg_tan_c , Ka=0
If LF gain is -40dB/dec, N=2, Kp=Kv=inf,
Ka=(wLFg_tan_c)2
System type, steady state tracking,
& Nyquist plot
C(s)
G  jw  
Gp(s)
K  jT a w  1  jT b w  1 
 j w  N  jT1w
As ω → 0
 1  jT 2 w  1 
G  jw  
K
 jw N
Type 0 system, N=0
Kp=lims0 G(s)
=G(0)=K
Kp
G(jw)
w0+
Type 1 system, N=1
Kv=lims0 sG(s)
cannot be determined
easily from Nyquist
plot
winfinity
w0+
G(jw)  -j∞
Type 2 system, N=2
Ka=lims0 s2G(s)
cannot be determined
easily from Nyquist
plot
winfinity
w0+
G(jw)  -∞
System type on Nyquist plot
Kp
System relative order
Examples
System type =
System type =
Relative order =
Relative order =
Margins on Bode plots
G(s)
In most cases, stability of this closed-loop
can be determined from the Bode plot of G:
– Phase margin > 0
– Gain margin > 0
w gc : gain cross - over freq.
at w gc , G  j w   1 or 0 dB
PM : phase margin
 180    G  j w gc 
w
pc
: phase cross - over freq.
 G  j w pc   180 
GM : gain margin
  20 log G  j w pc  dB
 1 G  j w pc  in value
If G  j w  never cross 0 dB line (always below 0
dB line), then PM = ∞.
If  G  j w  never cross –180° line (always
above –180°), then GM = ∞.
If  G  j w  cross –180° several times, then
there are several GM’s.
If G  j w  cross 0 dB several times, then there
are several PM’s.
Example:
G s  
100  s  1
 s  2  s  5 
 10
Bode plot on next page.
s 1
 12 s  1 15 s  1
1 . G  j w  cross 0 dB line near w  100
 w gc  100
PM  _______
2 .  G  j w  cross  180  at ω pc  ______
 GM  _______
Example:
G s  

s s  4 s  25

Bode plot on next page.
25
s

2
1
1
25
s 
2
4
25
s 1
1 . G  j w  cross 0 dB line near ______
 w gc  ______
 G  j w  at ω gc is about ______
 PM  _______


1. Where does  G  j w  cross the –180° line
Answer: __________
 w pc  ________
at ωpc, how much is G  j w   _______
 GM  ________
2. Closed-loop stability: __________
Example
:
G s  
40
s s  2 
 20
1
s  12 s  1 
1. G  j w  crosses 0 dB at __________
 w gc  ________
at this freq,  G  j w   _______
 PM  ________
2. Does  G  j w  cross –180° line?
________
 GM  ________
3. Closed-loop stability: __________
Margins on Nyquist plot
Suppose:
• Draw Nyquist plot
G(jω) & unit circle
• They intersect at point A
• Nyquist plot cross neg.
real axis at –k
Then : PM  angle indicated
GM  1 in value
k
Nyquist Diagram
150
1.5
100
1
50
0
-50
0.5
0
-0.5
-100
-1
-150
-1.5
-200
-100
-50
0
Real Axis
Nyquist Diagram
2
Imaginary Axis
Imaginary Axis
200
50
-2
-2
-1.5
-1
Real Axis
-0.5
0
Nyquist Diagram
10
Nyquist Diagram
2
1.5
1
Imaginary Axis
Imaginary Axis
5
0
-5
0.5
0
-0.5
-1
-1.5
-10
-4
-2
0
Real Axis
2
4
-2
-2
-1.5
-1
Real Axis
-0.5
0