Laplace Transform Analysis of
Signals and Systems
Transfer Functions
• Transfer functions of CT systems can be
found from analysis of
– Differential Equations
– Block Diagrams
– Circuit Diagrams
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2
Transfer Functions
A circuit can be described by a system of differential
equations
d
d

R1 i1(t ) + L  ( i1(t )) − ( i2 (t )) = vg (t )
dt
 dt

t
1
d
d


L  ( i2 (t )) − ( i1(t )) + ∫ i2 ( λ )dλ + vc ( 0 − ) + R2 i2 (t ) = 0
dt
 dt
 C 0−
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3
Transfer Functions
Using the Laplace transform, a circuit can be described by
a system of algebraic equations
R1 I1( s ) + sL I1( s ) − sL I2 ( s ) = Vg ( s )
1
sL I2 ( s ) − sL I1( s ) +
I2 ( s ) + R2 I2 ( s ) = 0
sC
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Transfer Functions
A circuit can even be described by a block diagram.
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Transfer Functions
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 s 2 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 s 2 X2 ( s ) = m2 s 2 X2 ( s )
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6
Transfer Functions
The mechanical system can also be described by a block
diagram.
Time Domain
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Frequency Domain
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7
System Stability
• 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
Cascade
Parallel
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System Interconnections
E(s)
Error signal
Feedback
H1( s ) Forward path
transfer function or the
“plant”
H 2 ( s ) Feedback path
transfer function or the
“sensor”.
T( s ) = H1( s ) H 2 ( s )
Loop transfer function
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E( s ) = X( s ) − H 2 ( s ) Y( s )
Y( s ) = H1( s ) E( s )
Y( s )
H1 ( s )
H( s ) =
=
X( s ) 1 + H1( s ) H 2 ( s )
T( s ) = H1( s ) H 2 ( s )
H1 ( s )
H( s ) =
1 + T( s )
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Analysis of Feedback Systems
Beneficial Effects
K
H( s ) =
1 + K H2 ( s)
If K is large enough that K H 2 ( s ) >> 1 then H( s ) ≈
1
H2 ( s)
. This
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.
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Analysis of Feedback Systems
A very important example of feedback systems is an
electronic amplifier based on an operational amplifier
Let the operational amplifier gain be
Vo ( s )
A0
H1 ( s ) =
=−
s
Ve ( s )
1−
p
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Analysis of Feedback Systems
The amplifier can be modeled as a feedback system with this
block diagram.
The overall gain can be written as
− A0 Z f ( s )
V0 ( s )
=
s
s


Vi ( s ) 
 1 − + A0  Z i ( s ) +  1 −  Z f ( s )



p
p
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Analysis of Feedback Systems
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 f ( s)
V0 ( s )
≅−
Vi ( s )
Zi ( s)
the gain formula based on an ideal operational amplifier.
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Analysis of Feedback Systems
If A0 = 10 7 and p = −100 then H( − j100 ) = −9.999989 + j 0.000011
If A0 = 10 6 and p = −100 then H( − j100 ) = −9.99989 + j 0.00011
The change in overall system gain is about 0.001% for a change
in open-loop gain of a factor of 10.
The half-power bandwidth of the operational amplifier itself is
15.9 Hz (100/2π). The half-power bandwidth of the overall
amplifier is approximately 14.5 MHz, an increase in bandwidth
of a factor of approximately 910,000.
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15
Analysis of Feedback Systems
Feedback can stabilize an unstable system. Let a forward-path
transfer function be
1
H1 ( s ) =
, p>0
s− p
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
H( s ) =
1
s− p+K
and, if K > p, the overall system is now stable.
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16
Analysis of Feedback Systems
Feedback can make an unstable 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.
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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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).
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Public Address System
18
Analysis of Feedback Systems
Stable Oscillation Using Feedback
Prototype
Feedback
System
Feedback
System
Without
Excitation
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Analysis of Feedback Systems
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
ωaxis, then the transfer function 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 nonlinear).
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20
Analysis of Feedback Systems
Stable Oscillation Using Feedback
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
Laser action begins when a photon is spontaneously emitted from
the pumped medium in a direction normal to the mirrors.
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Analysis of Feedback Systems
Stable Oscillation Using Feedback
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
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.
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Analysis of Feedback Systems
The Routh-Hurwitz Stability Test
The Routh-Hurwitz Stability Test is a method for
determining the stability of a system if its transfer function
is expressed as a ratio of polynomials in s. Let the
numerator be N(s) and let the denominator be
D( s ) = aD s D + aD −1s D −1 + L + a1s + a0
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Analysis of Feedback Systems
The Routh-Hurwitz Stability Test
The first step is to construct the “Routh array”.
D
aD
D − 1 aD −1
D − 2 bD − 2
D − 3 cD − 3
M
M
2
d2
1
e1
0
f0
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aD − 2
aD − 3
bD − 4
cD − 5
M
d0
0
0
D even
aD − 4
aD − 5
bD − 6
cD − 7
M
0
0
0
L a0
L 0
L 0
L 0
M M
0 0
0 0
0 0
D
aD
D − 1 aD −1
D − 2 bD − 2
D − 3 cD − 3
M
M
2
d2
1
e1
0
f0
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aD − 2
aD − 3
bD − 4
cD − 5
M
d0
0
0
D odd
aD − 4
aD − 5
bD − 6
cD − 7
M
0
0
0
L a1
L a0
L 0
L 0
M M
0 0
0 0
0 0
26
Analysis of Feedback Systems
The Routh-Hurwitz Stability Test
The first two rows contain the coefficients of the denominator
polynomial. The entries in the following row are found by the
formulas,
aD
bD − 2
aD − 2
aD −1 aD − 3
=−
aD −1
bD − 4
aD aD − 4
aD −1 aD − 5
=−
aD −1
...
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).
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Analysis of Feedback Systems
Root Locus
Common Type of Feedback System
System Transfer Function
K H1 ( s )
H( s ) =
1 + K H1 ( s ) H 2 ( s )
Loop Transfer Function
T( s ) = K H1( s ) H 2 ( s )
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Analysis of Feedback Systems
Root Locus
Poles of H(s)
T is of the form
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Zeros of 1 + T(s)
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
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Analysis of Feedback Systems
Root Locus
K can range from zero to infinity. For K approaching zero,
using
Q( s ) + K P( s ) = 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.
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Analysis of Feedback Systems
Root Locus
K
Let H1( s ) =
and let H 2 ( s ) = 1 .
( s + 1)( s + 2 )
K
Then T( s ) =
( s + 1)( s + 2 )
Root
Locus
No matter how large K gets
this system is stable because
the poles always lie in the
left half-plane (although for
large K the system may be
very underdamped).
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Analysis of Feedback Systems
Root Locus
Let H1( s ) =
K
( s + 1)( s + 2 )( s + 3)
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.
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Analysis of Feedback Systems
Root Locus
Four Rules for Drawing a Root Locus
1.
Each root-locus branch begins on a pole of T and
terminates on a zero of T.
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 on
the real axis is odd, is a part of the root locus.
3.
.
.
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The root locus is symmetrical about the real axis.
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33
Analysis of Feedback Systems
Root Locus
.
.
4.
If the number of finite poles of T exceeds the number of finite
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 the angles,
kπ
, k = 1, 3, 5,...
m
with respect to the positive real axis. These asymptotes intersect on
the real axis at the location,
1
σ = ∑ finite poles − ∑ finite zeros
m
(
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)
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Analysis of Feedback Systems
Root Locus Examples
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Analysis of Feedback Systems
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( jω ) = −1
a number with a magnitude of one and a phase of -π
radians.
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Analysis of Feedback Systems
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
Gain and Phase Margin
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Analysis of Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
A very common type of feedback system is the unity-gain
feedback connection.
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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Analysis of Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
The Laplace transform of the error signal is
X( s )
E( 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
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
Steady-State Tracking Errors in Unity-Gain Feedback Systems
If the forward transfer function is in the common form,
bN s N + bN −1s N −1 + Lb2 s 2 + b1s + b0
H1 ( s ) =
aD s D + aD −1s D −1 + L a2 s 2 + a1s + a0
then
1
a0
lim e(t ) = lim
=
N
N −1
2
t →∞
s→0
bN s + bN −1s + Lb2 s + b1s + b0 a0 + b0
1+
aD s D + aD −1s D −1 + L 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
bN s N + bN −1s N −1 + Lb2 s 2 + b1s + b0
H1 ( s ) =
s( aD s D −1 + aD −1s D − 2 + L a2 s + a1 )
which has a pole at s = 0.
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Analysis of Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
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.
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Analysis of Feedback Systems
Steady-State Tracking Errors in Unity-Gain Feedback Systems
The steady-state error with ramp excitation is
Infinite for a stable type 0 system
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
It is possible, by a series of operations, to reduce a
complicated block diagram down to a single block.
Moving a Pick-Off Point
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Block Diagram Reduction
Moving a Summer
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Block Diagram Reduction
Combining Two Summers
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Block Diagram Reduction
Move Pick-Off Point
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Block Diagram Reduction
Move Summer
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Block Diagram Reduction
Combine Summers
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Block Diagram Reduction
Combine Parallel Blocks
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Block Diagram Reduction
Combine Cascaded Blocks
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Block Diagram Reduction
Reduce Feedback Loop
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Block Diagram Reduction
Combine Cascaded Blocks
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53
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 )
Loop transfer Function of ith Feedback Loop - Ti ( s )
NL
∆( s ) = 1 + ∑ Ti ( s ) +
i =1
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∑
ith loop and
jth loop not
sharing a signal
Ti ( s ) Tj ( s ) +
∑
Ti ( s ) Tj ( s ) Tk ( s ) + L
ith, jth,kth
loops not
sharing a signal
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54
Mason’s Theorem
The overall system transfer function is
Np
H( s ) =
∑ P ( s)∆ ( s)
i
i =1
i
∆( s )
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.
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Mason’s Theorem
10
P1( s ) =
s
1
P2 ( s ) =
s+3
Np = 2
1 3
3
∆( s ) = 1 +
= 1+
ss+8
s( s + 8 )
Np
H( s ) =
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∑ Pi ( s)∆ i ( s)
i =1
∆( s )
NL = 1
∆1 ( s ) = ∆ 2 ( s ) = 1
10
1
+
( s + 8 )(11s + 30 )
s
s
+
3
=
=
2
3
s
+
s
+ 8 s + 3)
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 )
of the unit step response is
N( s ) N1( s ) K
Y( s ) = H −1( s ) =
=
+
s D( s ) D( s ) s
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
response is
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H( 0 )
.
s
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57
System Responses to Standard Signals
N( s )
Let H( 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
N( s )
N( s ) N x ( s ) N1( s ) N x1( s )
Y( s ) =
X( s ) =
=
+
D( s )
D( s ) D x ( s ) D( s )
D x ( s)
123 12
3
same poles
as system
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same poles
as excitation
58
System Responses to Standard Signals
Unit Step Response
Let H( s ) =
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A
1−
s
p
. Then the unit step response is y(t ) = A(1 − e pt ) u(t )
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59
System Responses to Standard Signals
Unit Step Response
Let
Aω 02
Η( s ) = 2
s + 2ζω 0 s + ω 02
(pp. 710-712)
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System Responses to Standard Signals
Unit Step Response
Aω 02
Let Η( s ) = 2
s + 2ζω 0 s + ω 02
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61
System Responses to Standard Signals
N( s )
Let H( s ) =
be proper in s. If the excitation is a suddenlyD( s )
applied, unit-amplitude cosine, the response is
N( s ) s
Y( s ) =
D( s ) s 2 + ω 02
which can be reduced and inverse Laplace transformed into
(pp. 713-714)
 N1( s )
y(t ) = L 
 + H( jω 0 ) cos(ω 0t + ∠ H( jω 0 )) u(t )
 D( s ) 
−1
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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62
Pole-Zero Diagrams and
Frequency Response
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,
s − z1 )( s − z2 )L( s − z N )
(
H( s ) = A
( s − p1 )( s − p2 )L( s − pD )
Therefore H(jω) is
jω − z1 )( jω − z2 )L( jω − z N )
(
H( jω ) = A
( jω − p1 )( jω − p2 )L( jω − pD )
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63
Pole-Zero Diagrams and
Frequency Response
3s
Let H( s ) =
s+3
jω
H( jω ) = 3
jω + 3
The numerator, jω, and the
denominator, jω + 3, can be
conceived as vectors in the
s plane.
jω
H( jω ) = 3
jω + 3
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∠ H( jω ) = ∠
{3 + ∠jω − ∠( jω + 3)
=0
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64
Pole-Zero Diagrams and
Frequency Response
jω
lim− H( jω ) = lim− 3
=0
ω →0
ω →0
jω + 3
jω
lim+ H( jω ) = lim+ 3
=0
ω →0
ω →0
jω + 3
jω
lim H( jω ) = lim 3
=3
ω →−∞
ω →−∞
jω + 3
jω
lim H( jω ) = lim 3
=3
ω →+∞
ω →+∞
jω + 3
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65
Pole-Zero Diagrams and
Frequency Response
π
π
lim− ∠ H( jω ) = − − 0 = −
ω →0
2
2
π
π
lim+ ∠ H( jω ) = − 0 =
ω →0
2
2
π  π
lim ∠ H( jω ) = − − −
=0
ω →−∞


2
2
π π
lim ∠ H( jω ) = − = 0
ω →+∞
2 2
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Butterworth Filters
The squared magnitude of the transfer function of an nth order,
unity-gain, lowpass Butterworth filter with a corner frequency of
1 radian/s is
H( jω )
2
1
=
1 + ω 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
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.
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Butterworth Filters
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).
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Lowpass to Highpass
ωc
s→
s
Lowpass to Bandpass
s 2 + ω Lω H
s→
s(ω H − ω L )
Lowpass to Bandstop
s(ω H − ω L )
s→ 2
s + ω Lω H
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69
Standard Realizations of Systems
There are multiple ways of drawing a system block diagram
corresponding to a given transfer function of the form,
N
Y( s )
H( s ) =
=
X( s )
k
b
s
∑k
k =0
N
k
a
s
∑ k
bN s N + bN −1s N −1 + L + b1s + b0
= N
, aN = 1
N −1
s + a N −1s + L + a1s + a0
k =0
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70
Standard Realizations of Systems
Canonical Form
The transfer function can be conceived as the product of two
transfer functions,
H1 ( s ) =
and
1
Y1( s )
= N
X( s ) s + a N −1s N −1 + L + a1s + a0
Y( s )
H2 ( s) =
= bN s N + bN −1s N −1 + L + b1s + b0
Y1( s )
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Standard Realizations of Systems
Canonical Form
The system can then be realized in this form,
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72
Standard Realizations of Systems
Cascade Form
The transfer function can be factored into the form,
1
1
1
s − z1 s − z2 s − z N
L
L
H( s ) = A
s − p1 s − p2 s − p N s − p N +1 s − p N + 2 s − pD
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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Standard Realizations of Systems
Cascade Form
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,
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74
Standard Realizations of Systems
Parallel Form
The transfer function can be expanded in partial fractions of
the form,
K1
K2
KD
H( s ) =
+
+L+
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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Standard Realizations of Systems
Parallel Form
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76
State-Space Analysis
• In larger systems it is important to keep the
analysis methods systematic to avoid errors
• One popular method for doing this is through
the use of state variables
• State variables are signals in a system which,
together with the excitations, completely
characterize the state of the system
• As the system changes dynamically the state
variables change value and the system moves
on a trajectory through state space
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State-Space Analysis
• State space is an N-dimensional space where
N is the order of the system
• The order of a system is the number of state
variables needed to characterize it
• State variables are not unique, there can be
multiple correct sets
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State-Space Analysis
• There are several advantages to state-space
analysis
–
–
–
–
Reduction of the probability of analysis errors
Complete description of the system signals
Insight into system dynamics
Can be formulated using matrix methods and the
system state can be expressed in two matrix
equations
– Combined with transform methods it is very
powerful
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State-Space Analysis
To illustrate state-space methods, let the system be this circuit
Let the state variables be the capacitor voltage and the inductor
current and let the output signals be vout (t ) and i R (t ).
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State-Space Analysis
Two differential equations in the two state variables, called the
state equations, characterize the circuit,
1
i′L (t ) = vC (t )
L
1
G
1
v′C (t ) = − i L (t ) − vC (t ) + iin (t )
C
C
C
Two more equations called the output equations define the
responses in terms of the state variables,
vout (t ) = vC (t )
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i R (t ) = G vC (t )
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81
State-Space Analysis
The state equations can be written in matrix form as
 0
 i′L (t )  
v′ (t ) =  1
 C  −
 C
1 
0

t
i
(
)



L
L
+  1 [ iin (t )]



G vC (t )  
− 
C 
C
and the output equations can be written in matrix form as
vout (t ) 0 1   i L (t )  0 
 i (t )  = 0 G  v (t ) + 0 [ iin (t )]
 C   
 R  
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State-Space Analysis
A block diagram for the
system can be drawn
directly from the state
and output equations.
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State-Space Analysis
The state and output equations can be written compactly as
q′(t ) = Aq(t ) + Bx(t )
y(t ) = Cq(t ) + Dx(t )
where
 i L (t ) 
q(t ) = 

vC (t )
 0

A=
1
−
 C
vout (t )
y(t ) = 

t
i
(
)
 R 
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1 
L 
G
− 
C
0
B= 1
 C 
0 1 
C=

0
G


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x(t ) = [ iin (t )]
0 
D= 
0 
84
State-Space Analysis
The solution of the state and output equations can be found
using the Laplace transform.
sQ( s ) − q( 0 + ) = AQ( s ) + BX( s )
or
[sI − A ]Q( s ) = BX( s ) + q(0 + )
−1
Multiplying both sides by[ sI − A ]
[
]
Q( s ) = [ sI − A ] BX( s ) + q( 0 + )
−1
The matrix, [ sI − A ] , is conventionally designated by the
symbol, Φ( s ) . Then
−1
[
]
Q( s ) = Φ( s ) BX( s ) + q( 0 + ) = Φ( s )BX( s ) + Φ( s )q( 0 + )
14243 14243
zero − state
response
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zero − input
response
85
State-Space Analysis
The time-domain solution is then
q(t ) = φ(t ) ∗ Bx(t ) + φ(t )q( 0 + )
14243 1424
3
zero − state
response
zero − input
response
L
where φ(t ) ←
→ Φ( s ) and φ(t ) is called the state transition
matrix.
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State-Space Analysis
To make the example concrete, let the excitation and initial
conditions be
i(t ) = A u(t )
1
and let R = , C = 1 , L = 1
3
s

−1
Φ( s ) = ( sI − A ) = 
1

C
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+

 0 
0
i
(
)
L
+
q( 0 ) = 
= 
+ 
vC ( 0 ) 1 
s + G 1 
 C L
−1
 1

1 
s
− 
−
 C

L
=
G
1
G
2
s+ 
s + s+
C
C
LC
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87
State-Space Analysis
Solving for the states in the Laplace domain,
1
1


+

1 
1 
 2 G
 2 G
L s + s+
 sLC  s + s +



C
LC
C
LC

Q( s ) = 
s
1


+
 C s2 + G s + 1  s2 + G s + 1




C
LC
C
LC 
Substituting in numerical component values,
1
1


+
 s( s 2 + 3s + 1) s 2 + 3s + 1
Q( s ) = 

s
1


+ 2
2
 s + 3s + 1 s + 3s + 1 
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State-Space Analysis
Inverse Laplace transforming, the state variables are
1 − 0.277 e −2.62 t − 0.723e −0.382 t 
u(t )
q(t ) = 
−0.382 t
−2.62 t 
+ 0.277 e

 0.723e
and the response is
−2.62 t
− 0.723e −0.382 t 
0 1 
0 
0 1  1 − 0.277 e
u(t )
y(t ) = 
q +  x = 

−0.382 t
−2.62 t 


+ 0.277 e
0 G 
0 
0 3  0.723e

or
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0.723e −0.382 t + 0.277 e −2.62 t 
u(t )
y(t ) = 
−0.382 t
−2.62 t 
+ 0.831e

 2.169 e
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89
State-Space Analysis
In a system in which the initial conditions are zero (the
zero-input response is zero), the matrix transfer function
can be found from the state and output equations.
sQ( s ) − q( 0 + ) = AQ( s ) + BX( s )
123
or
=0
Q( s ) = [ sI − A ] BX( s ) = Φ( s )BX( s )
−1
The response is
Y( s ) = CQ( s ) + DX( s ) = CΦ( s )BX( s ) + DX( s ) = [CΦ( s )B + D]X( s )
and the matrix transfer function is
H( s ) = CΦ( s )B + D
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State-Space Analysis
Any set of state variables can be transformed into another
valid set through a linear transformation. Let q1(t ) be the
initial set and let q2 (t ) be the new set, related to q1(t ) by
q2 (t ) = Tq1(t )
Then
q′2 (t ) = Tq1′ (t ) = T( A1q1(t ) + B1x(t )) = TA1q1(t ) + TB1x(t )
and using q1(t ) = T −1q2 (t ) ,
q′2 (t ) = TA1T −1q2 (t ) + TB1x(t ) = A 2 q2 (t ) + B2 x(t )
where
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A 2 = TA1T −1
B2 = TB1
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State-Space Analysis
By a similar argument,
y(t ) = C2 q2 (t ) + D2 x(t )
where
C2 = C1T −1
D2 = D1
Transformation to a new set of state variables does not change
the eigenvalues of the system.
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