Feedback Control Systems (FCS)
Lecture-32-33
Closed Loop Frequency Response
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
Introduction
• One of the important problems in analyzing a control
system is to find all closed-loop poles or at least those
closes to the jω axis (or the dominant pair of closed-loop
poles).
• If the open-loop frequency-response characteristics of a
system are known, it may be possible to estimate the
closed-loop poles closest to the jω axis.
Closed Loop Frequency Response
• For a stable, unity-feedback closed-loop system, the closed-loop
frequency response can be obtained easily from that of the open
loop frequency response.
• Consider the unity-feedback system shown in following figure. The
closed-loop transfer function is
C(s)
R( s )

G(s)
1  G(s)
Closed Loop Frequency Response
• Following figure shows the polar plot of G(s).
• The vector OA represents
G(jω1), where ω1 is the
frequency at point A.
• The length of the vector
OA is G ( j  1 )
• And the angle is  G ( j  1 )
Closed Loop Frequency Response
• The vector PA, the vector
from -1+j0 point to
Nyquist locus represents
1+G(jω1).
• Therefore, the ratio of
OA, to PA represents the
closed loop frequency
response.
OP
PA

G ( j1 )
1  G ( j 1 )

C ( j 1 )
R ( j 1 )
Closed Loop Frequency Response
• The magnitude of the closed
loop transfer function at ω=ω1
is the ratio of magnitudes of
vector OA to vector PA.
• The phase of the closed loop
transfer function at ω=ω1 is the
angle formed by OA to PA (i.e
Φ-θ).
• By measuring the magnitude
and phase angle at different
frequency points, the closedloop frequency-response curve
can be obtained.
Closed Loop Frequency Response
• Let us define the magnitude of the closed-loop frequency
response as M and the phase angle as α, or
C ( j )
R ( j )
 Me
Z e
i
j
Closed Loop Frequency Response
• Let us define the magnitude of the closed-loop frequency
response as M and the phase angle as α, or
C ( j )
R ( j )
 Me
j
• From above equation we can find the constant-magnitude
loci and constant-phase-angle loci.
• Such loci are convenient in determining the closed-loop
frequency response from the polar plot or Nyquist plot.
Constant Magnitude Loci (M circles)
• To obtain the constant-magnitude loci, let us first note
that G(jω) is a complex quantity and can be written as
follows:
G ( j  )  X  jY
• Then the closed loop magnitude M is given as
M 
• And M2 is
M
2

X  jY
C(s)
1  X  jY
X
2
Y
2
R( s )
2
(1  X )  Y
2

G(s)
1  G(s)
Constant Magnitude Loci (M circles)
M
• Hence
M
M
2
2
2

1  2 X
2
X
2
• If M=1 then,
2
2
 X
2
2
Y
2
(1  X )  Y
 2M X  M X
(1  M ) X
2
2

2
Y
2
M Y
2
2
X
2
2
2
Y
 X
 2 M X  (1  M )Y
2
2
2
Y
M
2
2
0
0
2 X  1  0
X  
1
2
• This is the equation of straight line parallel to y-axis
and passing through (-0.5,0) point.
Constant Magnitude Loci (M circles)
2
2
2
2M
(1  M ) X
2
2
 2 M X  (1  M )Y
2
M
2
0
• If M≠1 then,
X
• Add
X
2
M
M
2
M
X
2
M
2
1
2M


2


2
2
2
1
X Y
2

M
2
M
2
1
 0
to both sides
2
1
X Y

M
2M
M
2
M
2
2
1
X 
2
2
M
M
2
1

M
M
4
1

2
Y
2
2
2
1


2

M
M
2
M
M
2
2
1

2
2
1

2
Constant Magnitude Loci (M circles)
X
2

2M
M
2
2
1

X


2

X 
M
M
2
M
M
2
4
1

2
2

2

 Y
 1 
2
Y
2

M
M
2
• This is the equation of a circle with
2


M
centre   
,0 
 M 2 1 


M
radius 
M
2
1
M
M
2
2
1

2
2
1

2
Constant Magnitude Loci (M circles)
• The constant M loci on the G(s) plane are thus a
family of circles.
• The centre and radius of the circle for a given value of
M can be easily calculated.
• For example, for M=1.3, the centre is at (–2.45, 0) and
the radius is 1.88.
Constant Phase Loci (N circles)
C ( j )
R ( j )

X  jY
1  X  jY
• The phase angle of closed loop transfer function is
e
j
 
• The phase angle α is
X  jY
1  X  jY
   
  tan
1
(
Y
X
)  tan
1
(
Y
1 X
)
Constant Phase Loci (N circles)
  tan
1
(
Y
)  tan
1
X
• If we define
(
Y
1 X
)
tan   N
• then

N  tan  tan

1
(
• We obtain
Y
)  tan
X
Y
N 
1
X
1

Y
1 X
Y
Y
X 1 X

(
)
1 X 
Y
Constant Phase Loci (N circles)
Y

X
N 
1
X
X
2
1 X
Y
Y
X 1 X
Y
N 
N(X
Y
2
2
 X Y
2
2
 X Y )Y
 X Y
2

1
N
Y  0
Constant Phase Loci (N circles)
2
X
Adding
1
4

X

1
Y  0
N
1
4N
 X Y
2
2
2
to both sides
 X 
1
4
Y
2

1
Y 
N
1
4N
2
1
1 


 X    Y 

2
2N 



2
2
1

4
This is an equation of circle with
 1 1 
centre    ,

 2 2N 
radius 
1
4

1
4N
2
1
4


1
4N
1
4N
2
2
Closed Loop Frequency Response
Closed Loop Frequency Response
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END OF LECTURES-32-33