Advanced Electrical Machines and Drives

Running head: ADVANCED ELECTRICAL MACHINES AND DRIVES
ADVANCED ELECTRICAL MACHINES AND DRIVES
Name of the Student
Name of the University
Author Note

DVANCED ELECTRICAL MACHINES AND DRIVES1
Question 1:
Given number of poles P = 4, rated speed or rotor speed N = 1706 rpm.
Now, when supplied by 230 volt, 50 Hz source the synchronous speed is
Ns = 120f/P = (120*50)/4 = 1500 rpm.
Slip frequency fs = (N – Ns)*P/2 = (1706 – 1500)*4/2 = 412 rad/sec.
Slip = (N-Ns)/Ns = (1706-1500)/1500 = 0.1373 or 13.73%
Question 2:
The speed loop transfer function is given below,
Fω(s) =
1
1+ s ( 2 D
ωn ) + s2
ωn
2
∗( 1
s )
Here, D = 0.5 and ωn = 314 rad/sec.
i. Now, if the gain kp = 60 is used as proportional controller then the speed loop transfer
function is given by,
60
1+ s ( 2 D
ωn ) + s2
ωn
2
∗( 1
s ) =
60
s +s2
( 2 D
ωn )+ s3
ωn
2
MATLAB code:
wn = 314; D = 0.5;
sys = tf(60,[1/wn^2,2*D/wn,1,0]);
bode(sys)

DVANCED ELECTRICAL MACHINES AND DRIVES2
Plot:
-100
-80
-60
-40
-20
0
Magnitude (dB)
101 102 103 104
-270
-225
-180
-135
-90
Phase (deg)
Bode Diagram
Frequency (rad/s)
ii. The bandwidth of the closed loop system is the frequency at which the closed loop
magnitude is -3 dB. Now, the approximate point where the closed loop magnitude is -3 dB is
pointed in the magnitude response of the bode plot as given below.
MATLAB code:
wn = 314; D = 0.5;
sys = tf(60,[1/wn^2,2*D/wn,1,0]);
bode(sys)
zoom(1)
grid on