1. 6. Nyquist Diagram, Bode Diagram, Gain Margin, Phase Margin,
Forward path
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.
Nyquist diagram is obtained by drawing the value of the KGp(s) for different ω (rad/s) values in the complex plane
for K=5

2. Im Re Input amplitude is 1 Output amplitude is 0.7906
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
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
KGp = i
mag =
0.7906
phas = Im Re
-0.75
-0.25
Input amplitude is 1
Output amplitude is

3. KGp(s) plane Im Re Increasing ω Nyquist curve -1
-0.75
-0.25i
KGp(s) plane -1
ω=2.2 rad/s KGp(s)= i
ω=2.0 rad/s KGp(s)= i
ω=1.8 rad/s KGp(s)= i
ω=1.7 rad/s KGp(s)= i i i
-0.78 i
Nyquist curve
Increasing ω
ω=1.7 rad/s
ω=1.8 rad/s
ω=2.0
ω=2.2
Nyquist diagram is drawn by connecting the tip of the vectors representing the KGp(s) for different ω frequencies.

4. When KGp(s)=-1, the closed loop control system is marginally stable.
num=[K];
den=[ ];
nyquist(num,den)
Denominator of the closed loop system
From Routh-Hurwitz
When KGp(s)=-1, the closed loop control system is marginally stable.

5. Bode diagram Nyquist diagram Vector length [KGp(s)]= a
When the vector length of [KGp(s)] is 1, its value in logarithmic scale is zero.
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.
K  KGp(s)=-a
Kc  KGp(s)=-1
Gain Margin (gm)
Gain margin in logarithmic scale
The relation can be used to find the critical value of the proportinal controller
Amplitude 1, decibel value 0
Vector horizontal, phase -180º
What is the value of vector length in decibel when the vector is in horizontal position.?
How many degrees there are when the vectıor length is 1?

6. or KGp(s) GM PM Kcr=K*gm=5*2=10 bode(num,den)
clc;clear
K=5;
num=K*[1];
den=[ ];
sys=tf(num,den);
bode(sys)
[gm,pm,w2,w1]=margin(sys)
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.
bode(num,den)
[gm,pm,w2,w1]=margin(num,den) or
gm =
2.0024
pm =
w2 =
2.2367
w1 =
1.8830 PM GM ω1 ω2
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.
Kcr=K*gm=5*2=10
The gain margin is obtained in decibel scale from Bode diagram (GM). The relationship between GM and gm is
GM=20log10(gm)
gm=10GM/20
tf is a Matlab command to form the transfer function.
The closed loop system is stable if the gain curve is intersected below 0, otherwise control system is unstable.
Frequency for gain margin
Frequency for phase margin
Phase margin (degree)
Gain margin in linear scale
Damping ratio
KGp(s)

7. Corner frequency=a (rad/s) Resonance
Bode diagrams of first order systems
Bode diagrams of second order systems

8. Design of different type controllers:
Phase-Lead Controller:
φM=41.48 : a=4.92
ωM=9.05: T=0.0498

9. Phase-Lag Controller:
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