Controls Assignment: Open Loop and Closed Loop Transfer Functions

Controls 1
Controls assignment
By (name)
Tutor
Course
College
Date
Question 1

Controls 2
a) Calculate K and K2
The open loop transfer function is given by,
G ( s )= K
s (s+1)= K
s2 + s
The closed loop transfer function is given by,
C( s)
R( s)= K
s (s +1)= K
s2 + s+ K
Comparing with the standard second order transfer function we have,
ωn
2
s2+ 2ξ ωn s+ωn
2
K=ωn
2 =122=144
Simplifying the block diagram,

Controls 3
-
+ E(s)
C(s)
For the inner loop,
1
s2+ s+ K2 s = 1
s2 +(1+ K 2) s
R(s) + E(s)
C(s)
-
The closed loop transfer function then becomes,
K
s2+ ( 1+ K2 ) s+ K
2 ξ ωn= ( 1+ K2 )
K2=2 ξ ωn −1
K2=2( 12)(0.65)−1
K2=14.6
C( s)
R( s)= 144
s2+ 15.6 s+144b) Simulink model
K2 s
1
s (s +1)
1
s2+(1+ K2 )s
K

Controls 4
c) Response
Without tachometer
With tachometer

Controls 5
The system without a tachometer is unstable. It oscillates with the amplitudes of
the oscillations getting smaller with time. On adding the tachometer, the system
stabilizes with a fast rise time and a small overshoot.Question 2
a) Open loop transfer function
(G¿¿ 1 D) G2 D ¿=( 1
s +1 )( 1
3 s +2 )=G(D)

Controls 6
G ( s ) = G( D)
1+ G ( D ) H ( D ) −G(D)
G ( s )=
1
3 s2+ 5 s+ 2
1+ 2 ( 1
3 s2+5 s+2 )− 1
3 s2 +5 s+2
=
1
3 s2+5 s+2
1+ ( 2
3 s2 +5 s+2 )− 1
3 s2 +5 s +2
G ( s ) =
1
3 s2+ 5 s+ 2
1+ 1
3 s2 +5 s +2
=
1
3 s2 +5 s +2
3 s2+5 s+2+1
3 s2 +5 s +2
=
1
3 s2 +5 s +2
3 s2 +5 s +3
3 s2 +5 s +2
G ( s ) = 1
3 s2+5 s+2 × 3 s2+5 s+2
3 s2+5 s+3 = 1
3 s2 +5 s +3
b) Closed loop transfer function
C( D)
R( D) = G(s)
1+ H (s) G( s)= 1
3 s2 +5 s+ 4c) Step response of the closed loop system
d) Polar plot
G ( s )= 1
3 s2+5 s+3
G ( jω )= 1
3( jω)2 +5( jω)+3