6. Nyquist Diagram, Bode Diagram, Gain Margin, Phase Margin,
Forward path
R(s) 

K
G p (s )
1
G p (s)  3
s  4s 2  5s  10
C(s)
H (s ) 
KG p (s )
1  KG p (s )
Nyquist diyagram is obtanined by plotting the
response of the forward path [KGp(s)] to
harmonic inputs with different frequencies (ω), in
the complex plane. The amlitude and phase
information of KGp(s) can be seen from the
Nyquist diagram
It is a critical situation for the stability of the closed loop control system, if the
denominator of the closed loop transfer function 1+KGp(s)=0, or in other words
the transfer function of the forward path KGp(s)=-1.
1  KG p (s )  0  KG p (s )  1
Nyquist diagram is obtained by drawing the value of the KGp(s) for different ω
(rad/s) values in the complex plane
1
G p (s)  3
s  4s 2  5s  10
for K=5
5
KG p (s)  3
s  4s 2  5s  10
5
KG p (s)  3
s  4s 2  5s  10
clc;clear
w=2;
s=i*w;
K=5;
KGp=K/(s^3+4*s^2+5*s+10)
mag=abs(KGp)
phas=angle(KGp)*180/pi
Input amplitude is 1
The value of output of KGp(s) to a harmonic input
having ω= 2 rad/s and amplitude of 1 can be
calculated as shown below
Im
KGp =
-0.7500 - 0.2500i
mag =
0.7906
Re
-0.75
phas =
-161.5651
-0.25
Output amplitude is 0.7906
0.7906
KGp(s) plane
Im
-1
Nyquist diagram is drawn
by connecting the tip of the
vectors representing the
KGp(s) for different ω
frequencies.
Re
-0.78 -0.75 -0.5334
-0.5098
-0.0201i
ω=2.2
-0.25i
ω=2.0
Increasing ω
ω=1.8 rad/s
-0.8427i
ω=1.7 rad/s
Nyquist curve
-1.1722i
ω=2.2
ω=2.0
ω=1.8
ω=1.7
rad/s KGp(s)=-0.5334
rad/s KGp(s)=-0.7500
rad/s KGp(s)=-0.7873
rad/s KGp(s)=-0.5098
- 0.0201i
- 0.2500i
- 0.8427i
- 1.1722i
Nyquist Diagram
4
K=5;
num=[K];
3
K=15
den=[1 4 5 10];
K=10
nyquist(num,den)
2
K
KG p (s)  3
s  4s 2  5s  10
1  KG p (s )
Denominator of the closed loop
system
1
Imaginary Axis
H (s ) 
KG p (s )
K=7
K=3
K=1
0
-1
D(s)  s 3  4s 2  5s  (10  K ) -2
From Routh-Hurwitz
 10  K  10
K cr  10
-3
-4
-3
When KGp(s)=-1, the closed loop
control system is marginally stable.
-2.5
-2
-1.5
-1
-0.5
Real Axis
0
0.5
1
1.5
2
KG p (s) plane
Im
r ( t )  cos t
M(dB)
Bode diagram
s  i
Vector length [KGp(s)]= a
1
log 
0
GM (dB )
Re
a
M


Nyquist diagram
 180
Vector length [KGp(s)]= 1
When the vector length
of
[KGp(s)] is 1, its value in
logarithmic scale is zero.
0
M  20 log 10 KG p (s )
M  20 log 10 1
M0
Φ is the phase of KGp(s). When the phase angle is -180º ,
KGp(s) is on the real axis and its amplitude is shown as a. When Nyquist curve intersects the real axis at -1,
control system is marginaly stable.
How many
degrees there are
when the vectıor
length is 1?
What is the value of
vector length in
decibel when the
vector is in
horizontal position.?
log 
M
The relation can be used to find the
critical value of the proportinal
controller
K
Kc


Kc  K
Gain margin in
logarithmic scale
KGp(s)=-a
KGp(s)=-1
1
a
Gain Margin (gm)
1
GM  20 log 10  
a
Bode diagram is the representation of the magnitude and phase of the forward path of a closed loop system
KGp(s) in decibel scale for different harmonic input frequencies. Magnitude is given in decibel and phase is given
in degree. The frequencies for gain and phase margin are represented by ω2 and ω1, respectively and they are
calculated using Matlab as described below.
clc;clear
or
KGp(s)
K=5;
tf is a Matlab command to
clc;clear
form the transfer function.
K=5;
num=K*[1];
den=[1 4 5 10];
sys=tf(num,den);
bode(sys)
[gm,pm,w2,w1]=margin(sys)
1
K 3
s  4 s 2  5s  10
num=K*[1];
den=[1 4 5 10];
bode(num,den)
[gm,pm,w2,w1]=margin(num,den)
Matlab gives the gain margin in
linear scale as gm. Critical gain
value Kcr can be calculated by
multiplying the actual gain value K
by gm.
Bode Diagram
20
0
2.0024
Gain margin in
linear scale
pm =
Magnitude (dB)
gm =
GM
-20
-40
-60
The closed loop system is stable if the
gain curve is intersected below 0,
otherwise control system is unstable.
-80
Kcr=K*gm=5*2=10
-100
33.2155 Phase margin
w2 =
-45
Frequency for gain
margin
w1 =
1.8830
Frequency for
phase margin
Phase (deg)
2.2367
The gain margin is obtained in
decibel scale from Bode diagram
(GM). The relationship between GM
and gm is
-120
0
(degree)
-90
-135
Damping ratio
GM=20log10(gm)
gm=10GM/20
-180
PM
-225
-270
-1
10
0
10
ω1 ω2
Frequency (rad/sec)
1
10
2
10
PM ( o )

100
Bode diagrams of first order systems
Bode diagrams of second order systems
n  1
  0.05
a2
a1
a4
  0.3
  0.7
  0.05
a1
G (s ) 
a
sa
a4
a2
g(t )  ae  at
Corner frequency=a (rad/s)
  0.3
  0.7
n2
G (s )  2
s  2ns  n2
g(t )  Ae  t cos( t   )

 1 Resonance
n
Design of different type controllers:
   z  p
1
aT2
i
s  plane

1

T
2
m

1  aTs
1  Ts
a1
sin  m 
a1
o
1

aT
V1


R1

C
R2


V2

a
φM=41.48 : a=4.92
a1
G c (s ) 
Phase-Lead Controller:
ωM=9.05: T=0.0498
R1  R 2
R2
R1
 3.92
R2
aT  R 1C
R 1C  0.245
a1
Phase-Lag Controller:
V1



G c (s ) 

R1
R2 V
2
C
a
R2
R1  R 2
1 (1  aTs)
a (1  Ts)
aT  R 2C

Phase Lead-Lag Controller:
Phase Lead: PM increases, damping increases, OV decreases, ess
remains same.
Phase Lag: PM decreases, damping decreases, OV increases,
ess decreases