Transfer Functions
• Transfer functions of CT systems can be
found from analysis of
– Differential Equations
– Block Diagrams
– Circuit Diagrams
Laplace Transform Analysis of
Signals and Systems
Chapter 10
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Transfer Functions
Using the Laplace transform, a circuit can be described by
a system of algebraic equations
d
"d
%
R1 i1( t ) + L $ ( i1( t )) ! ( i2 (t )) ' = v g ( t )
dt
# dt
&
R1 I1( s) + sL I1( s) ! sL I2 ( s) = Vg ( s )
t
d
"d
% 1
L $ ( i2 (t )) ! ( i1( t ))' + ) i2 ( ( )d( + vc ( 0 ! ) + R2 i2 ( t ) = 0
dt
# dt
& C 0!
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2
Transfer Functions
A circuit can be described by a system of differential
equations
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sL I2 ( s) ! sL I1( s) +
3
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Transfer Functions
1
I ( s) + R2 I2 ( s) = 0
sC 2
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4
Transfer Functions
A circuit can even be described by a block diagram.
A mechanical system can
be described by a system of
differential equations
f( t ) ! K d x1" ( t ) ! K s1[ x1 (t ) ! x 2 ( t )] = m1 x1""( t )
K s1 [x1 (t ) ! x 2 (t )] ! K s2 x 2 (t ) = m2 x "2"( t )
or a system of algebraic equations.
F( s) ! K d s X1 ( s ) ! K s1[ X1 ( s) ! X2 ( s )] = m1s 2 X1( s)
K s1 [ X1 ( s ) ! X2 ( s)] ! K s2 X2 ( s ) = m2 s 2 X2 ( s )
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1
System Stability
Transfer Functions
The mechanical system can also be described by a block
diagram.
Frequency Domain
Time Domain
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• System stability is very important
• A continuous-time LTI system is stable if its
impulse response is absolutely integrable
• This translates into the frequency domain as
the requirement that all the poles of the
system transfer function must lie in the open
left half-plane of the s plane (pp. 675-676)
• “Open left half-plane” means not including
the ω axis
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System Interconnections
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8
System Interconnections
Cascade
E(s)
Feedback
Error signal
H1( s ) Forward path
transfer function or the
“plant”
H 2 ( s ) Feedback path
transfer function or the
“sensor”.
Parallel
T( s) = H1( s) H 2 ( s)
Loop transfer function
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Analysis of Feedback Systems
Y( s)
H1( s)
=
X( s) 1+ H1( s) H2 ( s)
T( s) = H1( s) H 2 ( s)
H ( s)
H( s ) = 1
1+ T( s)
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10
K
1+ K H2 ( s)
1
. This
H 2 ( s)
Let the operational amplifier gain be
means that the overall system is the approximate inverse of the
system in the feedback path. This kind of system can be useful
for reversing the effects of another system.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
H( s ) =
A very important example of feedback systems is an
electronic amplifier based on an operational amplifier
If K is large enough that K H2 ( s) >> 1 then H( s ) !
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Y( s) = H1( s) E( s )
Analysis of Feedback Systems
Beneficial Effects
H( s ) =
E( s) = X( s) ! H 2 ( s) Y( s )
H1( s ) =
11
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Vo ( s)
A
=! 0
s
Ve ( s)
1!
p
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2
Analysis of Feedback Systems
Analysis of Feedback Systems
The amplifier can be modeled as a feedback system with this
block diagram.
If the operational amplifier low-frequency gain, A0 , is very
large (which it usually is) then the overall amplifier gain
reduces at low-frequencies to
Z ( s)
V0 ( s)
!" f
Vi ( s )
Zi ( s )
The overall gain can be written as
!A0 Z f ( s)
V0 ( s)
=
Vi ( s ) "
s
%
"
s%
$ 1 ! + A0 ' Zi ( s ) + $ 1 ! ' Z f ( s )
#
&
#
p
p&
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the gain formula based on an ideal operational amplifier.
13
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Analysis of Feedback Systems
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14
Analysis of Feedback Systems
Feedback can stabilize an unstable system. Let a forwardpath
transfer function be
1
H 1( s ) =
, p>0
s! p
The change in overall system gain is about 0.001% for a change
in open-loop gain of a factor of 10.
This system is unstable because it has a pole in the right halfplane. If we then connect feedback with a transfer function,
K, a constant, the overall system gain becomes
1
H( s ) =
s! p+ K
The half-power bandwidth of the operational amplifier itself is
15.9 Hz (100/2π). The half-power bandwidth of the overall
amplifer is approximately 14.5 MHz, an increase in bandwidth
of a factor of approximately 910,000.
and, if K > p, the overall system is now stable.
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Analysis of Feedback Systems
As the amplifier gain is
increased, any sound
entering the microphone
makes a stronger sound
from the speaker until, at
some gain level, the
returned sound from the
speaker is a large as the
originating sound into the
microphone. At that point
the system goes unstable
(pp. 685-689).
A familiar example of this kind of instability caused by
feedback is a public address system. If the amplifier gain
is set too high the system will go unstable and oscillate,
usually with a very annoying high-pitched tone.
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16
Analysis of Feedback Systems
Feedback can make an ustable system stable but it can also
make a stable system unstable. Even though all the poles
of the forward and feedback systems may be in the open left
half-plane, the poles of the overall feedback system can be
in the right half-plane.
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Public Address System
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3
Analysis of Feedback Systems
Analysis of Feedback Systems
Stable Oscillation Using Feedback
Stable Oscillation Using Feedback
Can the response be non-zero when the
excitation is zero? Yes, if the overall
system gain is infinite. If the system
transfer function has a pole pair on the
ω axis, then it is infinite at the frequency
of that pole pair and there can be a response without an excitation.
In practical terms the trick is to be sure the poles stay on the ω axis.
If the poles move into the left half-plane the response attenuates
with time. If the poles move into the right half-plane the response
grows with time (until the system goes non-linear).
Prototype
Feedback
System
Feedback
System
Without
Excitation
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Analysis of Feedback Systems
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Analysis of Feedback Systems
Stable Oscillation Using Feedback
Stable Oscillation Using Feedback
Laser action begins when a photon is spontaneously emitted from
the pumped medium in a direction normal to the mirrors.
A real example of a system that oscillates stably is a laser.
In a laser the forward path is an optical amplifier.
The feedback action is provided by putting mirrors at each
end of the optical amplifier.
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Analysis of Feedback Systems
Stable Oscillation Using Feedback
A laser can be modeled by a block diagram in which the K’s
represent the gain of the pumped medium or the reflection or
transmission coefficient at a mirror, L is the distance between
mirrors and c is the speed of light.
If the “round-trip” gain of the
combination of pumped laser
medium and mirrors is unity,
sustained oscillation of light will
occur. For that to occur the
wavelength of the light must fit into
the distance between mirrors an
integer number of times.
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Analysis of Feedback Systems
Stable Oscillation Using Feedback
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4
Analysis of Feedback Systems
Analysis of Feedback Systems
The Routh-Hurwitz Stability Test
The Routh-Hurwitz Stability Test
The first step is to construct the “Routh array”.
The Routh-Hurwitz Stability Test is a method for determing
the stability of a system if its transfer function is expressed
as a ration of polynomials in s. Let the numerator be N(s)
and let the denominator be
D
aD aD!2 aD!4
D ! 1 a D!1 aD!3 aD!5
D ! 2 bD!2 bD!4 bD!6
D ! 3 cD!3 cD!5 cD!7
"
"
"
"
2
d2
d0
0
1
e1
0
0
0
f0
0
0
D even
D( s) = a D s D + aD!1 s D!1 +! + a1s + a0
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25
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Analysis of Feedback Systems
Root Locus
bD!4
a D aD!4
aD!1 a D!5
=!
aD!1
...
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27
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K H1 ( s)
1+ K H1 ( s) H2 ( s )
System Transfer Function
H( s ) =
Loop Transfer Function
T( s) = K H1 ( s) H2 ( s )
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Analysis of Feedback Systems
Analysis of Feedback Systems
Root Locus
Root Locus
T is of the form
P( s)
T( s) = K
Q( s)
Poles of H(s)
P( s)
Zeros of 1 + K
Q( s )
Poles of H(s)
Q( s) + K P( s) = 0
or
Q( s)
+ P( s) = 0
K
28
K can range from zero to infinity. For K approaching zero,
using
Q( s) + K P( s) = 0
Zeros of 1 + T(s)
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26
Common Type of Feedback System
The entries on succeeding rows are computed by the same
process based on previous row entries. If there are any zeros
or sign changes in the aD column, the system is unstable. The
number of sign changes in the column is the number of poles
in the right half-plane (pp. 693-694).
Poles of H(s)
! a1
! a0
! 0
! 0
" "
0 0
0 0
0 0
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The Routh-Hurwitz Stability Test
a D aD!2
aD!1 aD!3
bD!2 = !
aD!1
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D
aD aD!2 aD!4
D ! 1 a D!1 aD!3 aD!5
D ! 2 bD!2 bD!4 bD!6
D ! 3 cD!3 cD!5 cD!7
"
"
"
"
2
d2
d0
0
1
e1
0
0
0
f0
0
0
D odd
Analysis of Feedback Systems
The first two rows contain the coefficients of the denominator
polynomial. The entries in the following row are found by the
formulas,
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! a0
! 0
! 0
! 0
" "
0 0
0 0
0 0
the poles of H are the same as the zeros of Q( s) = 0 which
are the poles of T. For K approaching infinity, using
Q( s)
+ P( s) = 0
K
the poles of H are the same as the zeros of P( s) = 0 which
are the zeros of T. So the poles of H start on the poles of T
and terminate on the zeros of T, some of which may be at
infinity. The curves traced by these pole locations as K is
varied are called the root locus.
29
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5
Analysis of Feedback Systems
Analysis of Feedback Systems
Root Locus
Root Locus
K
Let H1( s ) =
and let H 2 ( s ) = 1 .
( s + 1)( s + 2 )
K
Let H1( s ) =
( s + 1)( s + 2 )( s + 3)
K
Then T( s) =
( s + 1)( s + 2 )
and let H 2 ( s ) = 1 .
Root
Locus
At some finite value of K
the system becomes
unstable because two
poles move into the right
half-plane.
No matter how large K gets
this system is stable because
the poles always lie in the
left half-plane.
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31
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Analysis of Feedback Systems
1.
Each root-locus branch begins on a pole of T and
terminates on a zero of T.
the angles,
2.
Any portion of the real axis for which the sum of the
number of real poles and/or real zeros lying to its right is
odd, is a part of the root locus.
k!
, k = 1, 3,5,... with respect to the positive real axis.
m
These asymptotes intersect on the real axis at the location,
!=
The root locus is symmetrical about the real axis.
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32
Root Locus
.
.
4.
If the number of poles of T exceeds the number of zeros of
T by an integer, m, then m branches of the root locus terminate on
zeros of T which lie at infinity. Each of these branches approaches
a straight-line asymptote and the angles of these asymptotes are at
Four Rules for Drawing a Root Locus
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Analysis of Feedback Systems
Root Locus
3.
.
.
Root
Locus
33
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Analysis of Feedback Systems
1
m
(" finite poles # " finite zeros)
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Analysis of Feedback Systems
Root Locus Examples
Gain and Phase Margin
Real systems are usually designed with a margin of error
to allow for small parameter variations and still be stable.
That “margin” can be viewed as a gain margin or a phase
margin.
System instability occurs if, for any real ω,
T( j! ) = "1
a number with a magnitude of one and a phase of -π
radians.
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6
Analysis of Feedback Systems
Analysis of Feedback Systems
Gain and Phase Margin
Gain and Phase Margin
So to be guaranteed stable, a system must have a T whose
magnitude, as a function of frequency, is less than one when
the phase hits -π or, seen another way, T must have a phase, as
a function of frequency, more positive than - π for all |T|
greater than one.
The difference between the a magnitude of T of 0 dB and the
magnitude of T when the phase hits - π is the gain margin.
The difference between the phase of T when the magnitude
hits 0 dB and a phase of - π is the phase margin.
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Analysis of Feedback Systems
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38
Analysis of Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
The Laplace transform of the error signal is
A very common type of feedback system is the unity-gain
feedback connection.
E( s) =
X( s)
1 + H1 ( s)
The steady-state value of this signal is (using the finalvalue theorem)
X( s )
lim e( t ) = lim s E( s) = lim s
t!"
s!0
s!0 1 + H ( s )
1
The aim of this type of system is to make the response
“track” the excitation. When the error signal is zero, the
excitation and response are equal.
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If the excitation is the unit step, A u(t ) , then the steadystate error is
A
lim e( t ) = lim
t!"
s!0 1+ H ( s )
1
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Analysis of Feedback Systems
40
Analysis of Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
If the forward transfer function is in the common form,
b s N + b N !1 s N !1 + !b2 s2 + b1s + b0
H1( s ) = N D
aD s + a D!1 s D!1 +!a2 s 2 + a1 s + a0
then
1
a0
lim e( t ) = lim
=
t!"
s!0
b s N + bN #1s N #1 +!b2 s2 + b1 s + b0 a0 + b0
1+ N D
aD s + aD#1s D#1 + !a2 s 2 + a1s + a0
If a0 = 0 and b0 ! 0 the steady-state error is zero and the
forward transfer function can be written as
H1( s ) =
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
If the forward transfer function of a unity-gain feedback
system has a pole at zero and the system is stable, the
steady-state error with step excitation is zero. This type
of system is called a “type 1” system (one pole at s = 0 in
the forward transfer function). If there are no poles at
s = 0, it is called a “type 0” system and the steady-state
error with step excitation is non-zero.
bN s N + b N !1 s N !1 + !b2 s2 + b1s + b0
s( a D s D!1 + aD!1s D!2 + !a2 s + a1 )
which has a pole at s = 0.
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7
Analysis of Feedback Systems
Block Diagram Reduction
Steady-State Tracking Errors in Unity-Gain Feedback Systems
The steady-state error with ramp excitation is
It is possible, by a series of operations, to reduce a
complicated block diagram down to a single block.
Infinite for a stable type 0 system
Moving a Pick-Off Point
Finite and non-zero for a stable type 1 system
Zero for a stable type 2 system (2 poles at s = 0 in
the forward transfer function)
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Block Diagram Reduction
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Block Diagram Reduction
Moving a Summer
Combining Two Summers
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Block Diagram Reduction
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Block Diagram Reduction
Move Pick-Off Point
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Move Summer
47
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8
Block Diagram Reduction
Block Diagram Reduction
Combine Summers
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Combine Parallel Blocks
49
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Block Diagram Reduction
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Block Diagram Reduction
Combine Cascaded Blocks
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Reduce Feedback Loop
51
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Block Diagram Reduction
52
Mason’s Theorem
Definitions:
Number of Paths from Input to Output - N p
Number of Feedback Loops - N L
Transfer Function of ith Path from Input to Output - Pi ( s)
Combine Cascaded Blocks
Loop transfer Function of ith Feedback Loop - Ti ( s)
NL
!( s ) = 1+ " Ti ( s) +
i=1
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"
ith loopand
jth loopnot
sharing a signal
Ti ( s ) Tj ( s ) +
"
Ti ( s) Tj ( s ) Tk ( s ) +!
ith, jth,kth
loops not
sharing a signal
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9
Mason’s Theorem
Mason’s Theorem
P1( s) =
10
s
P2 ( s) =
1
s+3
The overall system transfer function is
Np
H( s ) =
" P ( s )! ( s )
i
i=1
i
!( s )
Np = 2
1 3
3
!( s ) = 1+
=1+
ss+8
s( s + 8)
where ! i ( s ) is the same as !( s ) except that all feedback
loops which share a signal with the ith path, Pi ( s) , are
excluded.
Np
H( s ) =
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System Responses to Standard Signals
" P ( s )! ( s )
i
i=1
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i
!( s )
NL = 1
!1( s ) = ! 2 ( s) = 1
10
1
+
( s + 8 )(11s + 30 )
= s s+3 =
3
( s + 3)( s 2 + 8s + 3)
1+
s( s + 8 )
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56
System Responses to Standard Signals
Unit Step Response
Let H( s ) =
N( s)
be proper in s. Then the Laplace transform
D( s)
Let H( s ) =
of the unit step response is
Y( s) = H !1( s) =
N( s) N1( s) K
=
+
s D( s) D( s ) s
N( s)
be proper in s. If the Laplace transform of the
D( s)
excitation is some general excitation, X(s), then the Laplace
transform of the response is
K = H( 0 )
N (s )
If the system is stable, the inverse Laplace transform of 1
D( s)
is called the transient response and the steady-state
Y( s) =
N( s)
N( s) N x ( s) N1( s) N x1( s)
X( s ) =
=
+
D( s )
D( s) D x ( s ) D( s )
D x (s )
!"# !"
#
same poles
as system
H( 0 )
.
s
response is
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same poles
as excitation
58
System Responses to Standard Signals
System Responses to Standard Signals
Unit Step Response
Unit Step Response
Let H( s ) =
A
1!
s
p
. Then the unit step response is y( t ) = A(1 ! e
pt
) u(t )
Let
!( s ) =
A" 02
s 2 + 2#" 0 s + " 02
(pp. 710-712)
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10
System Responses to Standard Signals
System Responses to Standard Signals
Let H( s ) =
Unit Step Response
Let !( s ) =
A" 02
s 2 + 2#" 0 s + " 02
N( s)
be proper in s. If the excitation is a suddenlyD( s)
applied, unit-amplitude cosine, the response is
Y( s) =
N( s) s
D( s ) s2 + ! 02
which can be reduced and inverse Laplace transformed into
(pp. 713-714)
" N ( s )%
y( t ) = L!1 $ 1 ' + H( j( 0 ) cos(( 0 t + ) H( j( 0 )) u( t )
# D( s) &
If the system is stable, the steady-state response is a sinusoid of
same frequency as the excitation but, generally, a different
magnitude and phase.
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Pole-Zero Diagrams and
Frequency Response
Pole-Zero Diagrams and
Frequency Response
3s
s+3
j!
H( j! ) = 3
j! + 3
Let H( s ) =
If the transfer function of a system is H(s), the frequency
response is H(jω). The most common type of transfer function
is of the form,
H( s ) = A
( s ! z1 )( s ! z2 )!( s ! zN )
( s ! p1 )( s ! p2 )!( s ! pD )
The numerator, jω, and the
denominator, jω + 3, can be
conceived as vectors in the
[s] plane.
Therefore H(jω) is
H( j! ) = A
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( j! " z1 )( j! " z2 )!( j! " z N )
( j! " p1 )( j! " p2 )!( j! " p D )
H( j! ) = 3
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lim H( j! ) = lim# 3
j!
=0
j! + 3
lim H( j! ) = lim 3
j!
=3
j! + 3
! "#$
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! "0
! "#$
j!
=0
j! + 3
lim H( j! ) = lim 3
j!
=3
j! + 3
! "+#
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
! "0
! "+#
! H( j" ) = !3
! + !j" # !( j" + 3)
=0
64
Pole-Zero Diagrams and
Frequency Response
lim H( j! ) = lim+ 3
! "0 +
j!
j! + 3
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
Pole-Zero Diagrams and
Frequency Response
! "0 #
62
lim $ H( j! ) = #
! "0 #
%
%
#0=#
2
2
lim % H( j! ) = #
! "#$
65
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& ' &)
# #
=0
2 ( 2*
lim # H( j! ) =
$
$
%0=
2
2
lim $ H( j! ) =
% %
& =0
2 2
! "0 +
! "+#
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66
11
Butterworth Filters
Butterworth Filters
The squared magnitude of an nth order, unity-gain, lowpass
Butterworth filter with a corner frequency of 1 radian/s is
H( j! ) =
2
A Butterworth filter transfer function has no finite zeros and
the poles all lie on a semicircle in the left-half plane whose
radius is the corner frequency in radians/s and the angle
between the pole locations is always π/n radians.
1
1 + ! 2n
This is called a normalized
Butterworth filter because
its gain is normalized to
one and its corner frequency
is normalized to 1 radian/s.
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Butterworth Filters
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68
Standard Realizations of Systems
Frequency Transformations
A normalized lowpass Butterworth filter can be transformed
into an unnormalized highpass, bandpass or bandstop Butterworth
filter through the following transformations (pp. 721-725).
Lowpass to Highpass
Lowpass to Bandpass
Lowpass to Bandstop
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s!
There are multiple ways of drawing a system block diagram
corresponding to a given transfer function of the form,
"c
s
N
Y( s)
H( s ) =
=
X( s)
s 2 + " L" H
s!
s(" H # " L )
s!
69
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Standard Realizations of Systems
Canonical Form
H1( s ) =
H 2(s ) =
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!a s
k
k
=
bN s N + b N "1 s N "1 + !+ b1s + b0
, aN = 1
s N + a N "1s N "1 +! + a1s + a0
k
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70
Standard Realizations of Systems
Canonical Form
The transfer function can be conceived as the product of two
transfer functions,
and
k
k =0
N
k =0
s(" H # " L )
s 2 + " L" H
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
!b s
The system can then be realized in this form,
Y1( s)
1
=
X( s) s N + a N !1s N !1 +! + a1s + a0
Y( s)
= b s N + bN !1 s N !1 + !+ b1 s + b0
Y1( s) N
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71
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72
12
Standard Realizations of Systems
Standard Realizations of Systems
Cascade Form
Cascade Form
The transfer function can be factored into the form,
H( s ) = A
A problem that arises in the cascade form is that some poles
or zeros may be complex. In that case, a complex conjugate
pair can be combined into one second-order subsystem of the
form,
s ! z1 s ! z2 s ! zN
1
1
1
!
!
s ! p1 s ! p2 s ! p N s ! p N +1 s ! p N +2 s ! p D
and each factor can be realized in a small canonical-form
subsystem of either of the two forms,
and these subsystems can then be cascade connected.
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73
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Standard Realizations of Systems
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl
74
Standard Realizations of Systems
Parallel Form
Parallel Form
The transfer function can be expanded in partial fractions of
the form,
K1
K2
KD
H( s ) =
+
+ !+
s ! p1 s ! p2
s ! pD
Each of these terms describes a subsystem. When all the
subsystems are connected in parallel the overall system is
realized.
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13