Question

Use MATLAB to draw Bode Plots for a negative unity feedback system with each of the following forward-path transfer functions: 100(s +2) (b) G(s)50s +3 (s+5) Then, using the Bode Plots, s(s +2) (s+4) (s +6) · estimate the transient performance Ts and %OS (step response) . and compare the estimated values with actual values obtained from simulation.

Answer 1

ANSWER:

(a)

MATLAB Script:

%% Problem (a)
clear
clc

num = [100 200];
den = [1 5 4 0];

G = tf(num,den);

T = feedback(G,1);

figure()
bode(T)
grid on

figure()
step(T)

stepinfo(T)

Bode Plot:

Bode Diagram 20 -40 45 0) -90 0 -135 180 0 10 10 10 Frequency (rad/s)

From the Bode plot, the %OS can be estimated as

os = (10^(10.8/20) - 1)/10^(10.8/20) = 71.159%

Step response and system characteristics:

Step Response 1.8 r 1.6 1.4 1.2 E 0.8 0.6 0.4 0.2 0 0.51 1.5 2 2.5 3 3.5 4 Time (seconds)

• RiseTime: 0.1171
• SettlingTime: 2.6186
• SettlingMin: 0.6133
• SettlingMax: 1.6438
• Overshoot: 64.3750
• Undershoot: 0
• Peak: 1.6438
• PeakTime: 0.3113

  

(b)

MATLAB Script:

%% Problem (b)
clear
clc

num = [50 50*8 50*15];
den = [1 12 44 48 0];

G = tf(num,den);

T = feedback(G,1);

figure()
bode(T)
grid on

figure()
step(T)

stepinfo(T)

Bode Plot:

Bode Diagram 20 r 2 -20 E -40 -60 45 Ф -90 0 -135 -180 10 10 10 10 Frequency (rad/s)

From the Bode plot, the %OS can be estimated as

os = (10^(5.94/20) - 1)/10^(5.94/20) = 49.533%

Step response and system characteristics:

Step Response 1.5 0.5 0 0.511.5 2 2.5 33.5 Time (seconds)

• RiseTime: 0.1823
• SettlingTime: 2.0295
• SettlingMin: 0.8251
• SettlingMax: 1.4426
• Overshoot: 44.2624
• Undershoot: 0
• Peak: 1.4426
• PeakTime: 0.4656