Lecture 6

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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

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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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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


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)


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


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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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

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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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9

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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s-Domain Circuit Element Models Including Non-Zero Initial Conditions: Inductor Current

Inductor

s-domain

ort-domain

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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


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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Eliminating I1 between the two equations:

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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:


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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


2 nodes

Node V1

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Lecture 6

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