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- Forced/Natural/Transient/Steady-StateSystem Responses (U&Y 3-11; EE98)
- s-Domain Circuit Analysis (U&Y 4-1, 4-2)
(Last revised 2/9/2015)
EE110, S15: Circuits and Systems, Lecture 6
Prof. Ping Hsu
1
LTI System Response Partitions2
A pole of X(s) represents a signal component of x(t). The value of the pole is the frequency of this signal component. If this frequency is a real number, the component is an exponential signal. If this frequency is complex in conjugate pair, this pair of poles represents a sinusoidal signal with exponentially decaying or growing (or constant, if the real part is 0) magnitude. The imaginary part of this complex frequency is the frequency of the sinusoidal part of the signal and the real part of the complex frequency is the rate of decaying and growing of the magnitude. Consider the following example:
4 3 2
5 4 3 2 2 2 2
32 167 2390 3870 20 10 1( )3 119 317 1900 1700 2 17 100 1
s s s sX ss s s s s s s s s
1,2 3,4 51 4 , 10 and 1.p j p j p
There are three components in x(t): a decaying sinusoidal (p1,2) , a pure sinusoidal (p3,4), and an exponential decaying function (p5).
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LTI System Response Partitions3
4 3 2
5 4 3 2 2 2
32 167 2390 3870 20 10 1( )3 119 317 1900 1700 2 17 100 1
s s s sX ss s s s s s s s s
1,2 3,4 51 4 , 10 and 1.p j p j p
5131 2
sin(( ) ( ) ( )( )( )
105( )
si (4 ) )n tx t X s u tx tx t x
t tt
ee L
x2(t)
x3(t)x1(t)
0 1 2 320
2
4
t0 1 2 3
20
2
4
t
x(t)
LTI System Response Partitions4
A LTI (linear time invariant) system (a RLC circuit, for example) responds to each component in the input signal x(t) and produces a corresponding component output and introduces some additional component from the natural behavior of the system. For example, if the signal X(s) on the previous slide, i.e.,
is the input to the transfer function , , the output
will contain the following components (or poles). The extra component at -2 is introduced by the system itself.
4 3 2
5 4 3 2 2 2
32 167 2390 3870 20 10 1( )3 119 317 1900 1700 2 17 100 1
s s s sX ss s s s s s s s s
1( )
2H s
s
6 1,2 3,4 5
(Natural response) (Forced re
2 and 1 4 , 10 , 1 D
spoue to the system iteselft Components due t
no
se) the input
p p j p j p
1( ) ( ) ( ) ( )2
Y s H s X s X ss
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LTI System Response Partitions5
2013 National Technology and Science Press. All rights reserved.
Force Response: Output containing only the pole frequencies of the input.
Natural Response: Output containing only the pole frequencies of the system itself (called modes of the system).
Zero State Response: Output due to only the input (with zero initial condition).
Zero Input Response: Output due to only initial condition (with zero input).
Transient Response: All output terms that eventually decay to zero.
Steady State Response: All output terms that remain after the transient response decayed to nearly zero.
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Example : R=1 and C=0.5 F, ( ) 17cos(8 ), and (0) 10.i Cv t t v VoVi
R
C
2
2
(1/ ) 17 1 ( ) (10)(1/ ) 64 (1/ )
Total due to the input due to initial Response
condition32 1 1 (10)64 2 2
Zero StateResponse
cRC sV s
s RC s s RC
ss s s
Zero inputResponse
Zero State Response: Output due to the input (with zero initial condition).
Zero Input Response: Output due to only initial condition (with no input).
0 1 2 3 412
84
0
4
8
12
t
Zero State
Zero InputTotal
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7
2
2
32 1 1 ( ) 1064 2 2
Zero State Zero inputResponse Response
32 1 1 1064 2 2
Forced NaturalResponse Response
csV ss s s
ss s s
Force Response: The part of the output containing only the pole frequencies of the input. For this example, the input signal X(s) has poles at j8.
Natural Response: Output containing only the pole frequencies of the circuit. For this example, pole=-2 which is part of the system.
Forced Response
Natural Response
8
Transient Response: All terms that eventually decay to zero after a transient period.
Steady State Response: All terms that remain after the transient response decayed to nearly zero.
0 1 2 3 412
84
0
4
8
12
t
Steady State ResponseTransient period
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Remarks:
Zero-input response (output due to only initial condition) is always part of the Natural Response.
Zero-state responses contains poles of the system so it is NOT the entire Forced Response.
Forced Response is the Steady State response if all natural responses converge to zero (which is often the case).
Unbounded signal x(t)10
If X(s) contains one or more poles on the right-half of the s-plane (complex plane), x(t) is unbounded.
Recall that the real part of a pole determine its rate of exponential diverge (if positive) or converge (if negative). A pole on the right-half of the s-plane has a positive real part. The following two signals are unbounded.
3 1,
1 22 2
1,2
1
2 3
9 30 9 30( ) ( )( 2 17)( 2) ( 2 17)( 2)
1 4 , , 2
(
1
)
2 4p p
s sX s X ss s s s s s
p j p
x t
j
2 2212 12 cos(4 ) 16 cos(4 ) ( ) 4 4 cos(4 ) 22 sin(4 )t t t t t te e t e t x t e e t e t
0 2.5 50
5 104
1 105
x1 t( )
t0 2.5 5
1 103
0
1 103
x2 t( )
t
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Unstable system11
A system represented by a transfer function containing one or more poles on the right-half of the s-plane (complex plane) is an unstable system.
Since the natural response of the system contains the poles of the transfer function itself, such a systems output is unbounded even if the input is bounded. In fact, the output can be unbounded due to just the initial condition (i.e., with zero input).
Example: The following two systems are unstable.
1 22 2
9 30 9 30( ) , ( )( 2 4)( 2) ( 2 4)( 2)
s sH s H ss s s s s s
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Circuits with non-zero initial condition
For LTI circuits with zero initial conditions, H(s) (transfer function) can be found from an s-domain circuit model in which each inductance L is replaced by its impedance sL and each capacitance C is replaced by its impedance 1/(sC). Circuit analysis techniques developed in EE98 are then applied to compute H(s) =Y(s)/X(s).Examples:
Y(s)
I(s)
1sC
( ) 1/( ) 1/( )( ) ,( ) 1/( ) 1/( )
Y s sC RCH sX s R sC s RC
X(s)
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Example: For the circuit, R=1 and C=0.5 F, ( ) 17cos(8 ), and (0) 10.
Find ( ) for 0.
i C
o
v t t v
v t t
VoVi
R
C
1
: ( )
( ) since ( )
Take the Laplace transform:
[ ( ) (0 )] ( ) ( ) since ( ) (0 )
c i
c cc i
cc c c i c c
KVL iR v v tdv dvC R v v t i t Cdt dt
dvRC sV s v V s V s L sV s vdt
Transfer function
(1/ ) 1( ) ( ) (0)(1/ ) (1/ )c i c
RCV s V s vs RC s RC
Note that: You cannot use voltage division formula in this case.
s-Domain Circuit Element Models Including Non-Zero Initial Conditions: Capacitor Voltage
Capacitor
or
s-domaint-domain
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2013 National Technology and Science Press. All rights reserved.
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s-Domain Circuit Element Models Including Non-Zero Initial Conditions: Inductor Current
Inductor
s-domain
ort-domain
15
2013 National Technology and Science Press. All rights reserved.
16
Example:
VoVi (t)=17cos(8t)
R=1
C=0.5F
(0) ( ) (0) / (0)1 1( ) ( ) 1/( )
( ) (0) (0) 1 ( 1)
1/( ) 1 ( ) (0)1/( ) 1/( )
c i c co
i c c
i c
v V s v s vV s I ssC s R sC sC s
V s v vsRC sRC s s
RC V s vs RC s RC
VoVi
R
sC1
svc(0)
I(s)
( ) (0) /( )1/( )
i cV s v sI sR sC
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Or use super-position:
1/( ) /( )( ) ( ) (0)1/( ) 1/( )
1/( ) 1 ( ) (0)1/( ) 1/( )
o i c
i c
sC R sCV s V s CvR sC R sC
RC V s vs RC s RC
Vo(s)Vi(s)
R
sC1 Cvc(0)Vo
Vi (t)=17cos(8t)
R=1
C=0.5F
18
2
If R=1 , and C=0.5F, ( )=17cos(8t), and (0) 10,
1/( ) 1 1( ) ( ) (0) where 2, (0) 101/( ) 1/( )
2 1 17 ( ) 10 where ( ) 17cos(8 )2 2 64
i c
o i c c
i i
v t v
RCV s V s v vs RC s RC RC
sV s V s L ts s s
2
2
2 2
1 22 2
2 17 10 Take partial fraction expansion2 64 2
32 1 10 64 2 2
4(8) 9 64 64 2
4(8) 9 cos(8 ) 4sin(8 ) 9 ( )64 64 2
t
ss s s
ss s s
ss s s
sL t t e u ts s s
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EXAMPLE: The circuit is at steady state for t 0
From Irwin & Nelms, 201119
For t < 0 (Cap=open, inductor-=short)
+-1 V
iL(0 )
vC (0 )
2 2
2 1( )2 3 2
sI ss s
Vo(s) 2s
I2 (s) 1s Vo(s) 2s 7
2s2 3s 2
Circuit for t > 0
KVL: solve the two equations below for I2
From Irwin & Nelms, 2011
1 2
1 2
4 ( 1) 1
2 1( 1 ) 1
s I sIs
sI s Is s
20
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1( )V s
Vo(s) 2s 72s2 3s 2From Irwin & Nelms, 2011
Group terms
11 1
1
4 ( 1) 01 1
1 01 2
o
o o
V VV s Vs
V V V ss
21
Alternative way of finding Vo(s):
(Use nodal equation)
1
1
1 32
112 2
o
o
V Vs s
sV V
Remove fration
1
1
2 1 3
2 2 1o
o
s V sV
V s V
1
1
1
Remove
2 1 3 2
+ 2 2 1 (2s+1)
-------------------------------------------------- - 2 (2 1) 2 6 (2s+1)
6 (2s+1) (2 1) 2 2
o
o
o o
o
V
s V sV
V s V
sV s s V
Vs s s
Now determine the inverse transform
Note that as2 bs c 0 has complex conjugate roots if b2 4ac 0
Vo(s) K1s 3
4 j 7
4
K 1*
s 34 j 7
4
K1 s 34 j7
4
Vo(s)
s 34 j 7
4
K1(s j )
K1*
(s j ) 2 | K1 | et cos(t K1)u(t)
vo(t) 4.28 e34
tcos( 7
4t 76.5)
2.14 76.5
22From Irwin & Nelms, 2011
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Example: For the given vin(t), determine vout(t), t > 0-
Circuit in the t-time domain
Circuit in the s-domain: t > 0-Circuit to determine the I.C.s: t = 0-
vC (0 ) 9 V, iL (0 ) 3 A
Input signal x(t) = vin(t)
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2013 National Technology and Science Press. All rights reserved.
Example: For the given vin(t), determine vout(t), t > 0-
Eliminating I1 between the two equations:
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2013 National Technology and Science Press. All rights reserved.
Matlab: roots([42 162 306 300])
-2.0000 -0.9286 + 1.6460i-0.9286 - 1.6460i
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Laplace Transform pairs:
Time-domain current:
Example: For the given vin(t), determine vout(t), t > 0-
25
2013 National Technology and Science Press. All rights reserved.
Example: For the given vin(t), determine vout(t), t > 0-
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2013 National Technology and Science Press. All rights reserved.
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EXAMPLE: Write the node equations in the s-domain
IA(s) i1(0)s C1v1(0)i2 (0)
s
G1 1L1s 1
L2sC1s
V1(s)
1L2s
C1sV2 (s)
IB(s)C2v2 (0)C1v1(0) i2 (0)s
G2 C2s C1s 1L2s
V2 (s) C1s
1L2s
V1(s)
Node V2
Do not increase number of nodes
From Irwin & Nelms, 2011
2 nodes
Node V1
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