steady and transient.pdf

8/17/2019 STEADY AND TRANSIENT.pdf

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The One-Side z-Transform

Z. Aliyazicioglu

Electrical and Computer Engineering DepartmentCal Poly Pomona

ECE 308 -12

ECE 308-12 2


The one-sided z-transform of a signal x(n) is defined as

The one-sided z-transform has the following characteristics:

1. It does not contain information about the signal x(n) for negative

values of time (i.e., for n


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ECE 308-12 3


Example:

1 2 3 5

1 1( ) {1,2,5,7,0,1} ( ) 1 2 5 7

x

x n X z z z z z +

+ − − − −= ↔ = + + + +

↑

1 3

2 2( ) {1, 2, 5, 7, 0,1} ( ) 5 7 z

x n X z z z

+

+ − −= ↔ = + +

↑

2 3 4 5 7

3 3( ) {0,0,1,2,5,7,0,1} ( ) 1 2 5 7 x

x n X z z z z z z

+

+ − − − − −= ↔ = + + + +

↑

,

4 4

( ) ( ) ( ) 1 x

x n n X z δ

+

+= ↔ =5 5( ) ( ) ( )

xk

n n k X z z δ

+

+ −

= − ↔ =

6 6( ) ( ) ( ) 0 x

x n n k X z δ

+

+= + ↔ = 4 4 11

( ) ( ) ( )1

xn

x n a u n X z az

+

+

−= ↔ =

−

ECE 308-12 4


if

Shifting Property

Case 1:Time Delay

In case x(n) is causal, then

Example:

The z-transform of

,. …

( ) ( ) x

x n X z

+

+↔

Then1

( ) ( ) ( ) 0k x

k n

n

x n k z X z x n z k

+

− +

=

− ↔ + − >

∑

( ) ( ) 0 x

k x n k z X z k

+

− +− ↔ >

1

1( ) ( ) ( )

1

xn

x n a u n X z az

+

+

−= ↔ =

−

1( ) ( 2) x n x n= −

2 2 2

1

21

1

1( 2) ( 2) ( 1) ( 2)

1

( 1) ( 2)1

xn

x n a u n z x z x z az

z x z x

az

+

− −

−

−−

−

− = − ↔ + − + − −

= + − + −−


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ECE 308-12 5


Case 2: Time advance

Example:

if ( ) ( ) x

x n X z

+

+

↔

Then1

0

( ) ( ) ( ) 0k x

k n

n

x n k z X z x n z k

+ −+ −

=

+ ↔ − >

∑

1

1( ) ( ) ( )

1

xn

x n a u n X z az

+

+

−= ↔ =

−

1( ) ( 2) x n x n= +The z-transform of

2 2 1

1

2 2

1

22

1

1( 2) ( 2) (0) (1)

1

(0) (1)1

1

+

+ −

−

−

−

+ = + ↔ + + −

= + +−

= + +−

xn x n a u n z x x z

az

z x z x z az

z z az

az

ECE 308-12 6


Final Value Theorem:

This is exits if the ROC of includes the unit circle.

1lim ( ) lim( 1) ( )

x

n z x n z X z

+

+

→∞ →↔ −

if ( ) ( ) x

x n X z

+

+↔

Then

( 1) ( ) z X z +−,

( ) ( )n u n= ( ) ( )nh n u nα =Example: ( ) ( ) * ( ) y n h n x n=

2

1 1

1 1( ) ( ) ( )

1 1 ( )( 1)

z Y z H z X z

z z z z α α − −= = =

− − − − ROC z α >

2 2

( 1) ( ) ( 1)( )( 1) ( )

z z z Y z z

z z z α α − = − =

− − − z α >ROC

2

1

1lim ( ) lim

1

x

n z

z x n

z α α

+

→∞ →↔ =

− −


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ECE 308-12 7


Solution of Difference Equation:

Example:

The one-sided z-transform is a very efficient tool for the solution of

difference equation.

The initial condition

( ) ( 1) ( 2) y n y n y n= − + −,

( 1) 0 y − = ( 2) 1 y − =

1 2 2( ) ( ) ( 1) ( ) ( 1) ( 2)Y z z Y Z y z z Y z y z y z + − + − + = + − + + − + −

1 2 1( ) ( ) ( 1) ( ) ( 1) ( 2)Y z z Y Z y z Y z z y y− + − + −= + − + + − + −

1 2( )(1 ) 1Y z z z − −− − =2

1 2 2

1( )

1 1

z Y z

z z z − −= =

− − − −

ECE 308-12 8


,,,

Example:(cont)

( ) ( ) ( ) ( )1 2

1 1 1 1

1( )

1 1.618 1 0.618 1 1.618 1 1.618− − − −= = +

− + − −

A AY z

z z z z

( )( )1 1

1

1 1 11

1.618 1.618

11 1.618 ( )

1 .618

10.7236

11 .618

1.618

− −

−

−= == − =

+

= =

+

z z A z Y z

z

( )( )1 1

1

2 1 11

0.618 0.618

11 0.618 ( )

1 1.618

10.2764

11 1.618( )

0.618

− −

−

−=− =−= + =

−

= =

− −

z z A z Y z

z

( ) ( )( ) 0.7236 1.618 0.2764 0.618 ( ) = − n n

y n u n

Finally


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ECE 308-12 9


Find the step response of the system

The initial condition

Example:

Taking one-sided z-transform

( ) ( 1) ( ) y n y n x nα = − + 1 1α − < <

( 1) 1 y − =

1( ) ( ) ( 1) ( )Y z z Y Z y X z α + − + + = + − +

1

1

1( ) ( ) ( 1)

1Y z z Y Z yα α

+ − +

−= + − +

−

1 1 1

1( )

1 (1 )(1 )Y z

z z

α

α α

+

− − −= +

− − −

11

2

1( ) ( ) ( )

1

1(1 ) ( )

1

nn

n

y n u n u n

u n

α α

α

α α

++

+

−= +

−

= −−

The inverse transform

ECE 308-12 10


Response of System with Rational System Function

Analysis of LTI Systems in the z-Domain

Let us assume that the input signal x(n) and the corresponding

system function h(n) have rational z-transform X(z) and H(z) of

the form

( )( )

( )

N z X z

D z =

( )( )

( )

B z H z

z =and

If the system initially relaxed (initial conditions for differenceequation are zero ), the z-transform of system output has

( ) ( )( ) ( ) ( )

( ) ( )

B z N z Y z H z X z

A z D z = =

Partial fraction expansion of Y(z) will be in the following form if no

pole-zero cancellation

1 11 1

( )1 1

N Lk k

k k x x

A QY z

p z q z − −= == +

− −∑ ∑ where system poles,

the input signal poles, and1, 2 ,... N p p p

1, 2 ,... Lq q q k m p q≠


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ECE 308-12 11

natural response


Natural response is different than zero-input response. If

X(z) is zero, then Y(z) is zero

The inverse transform of Y(z) yields

where Ak and Qk are functions of both sets of poles pk and

qk.

We can separate the y(n) into two parts

1 1

( ) ( ) ( ) ( ) ( ) N Ln n

k k k k

k k

y n A p u n Q q u n= =

= +∑ ∑

1

( ) ( ) ( ) N

n

nr k k

k

y n A p u n=

= ∑

1

( ) ( ) ( ) L

n

fr k k

k

y n Q q u n=

= ∑ forced response of the system

ECE 308-12 12


Response of Pole-zero System with Nonzero Initial Condition.

The input signal x(n) is assumed to be causal. The effect of all

previous input signals to the system are reflected in the initial

conditions y(-1), y(-2), …..y(-N). We will look at the one-sided z-

transform

Since x(n) is causal, we can set

1 1 1

( ) ( ) ( ) ( ) N k M

k n k

k k

k k k

Y z a z Y z y n z b z X z + − + − +

= = =

= − + − +

∑ ∑ ∑

( ) ( ) X z X z + =

1 1 1

1 1

( )

( ) ( )

1 1

M N k k k n

k k

k k k

N N k k

k k

k k

b z a z y n z

Y z X z

a z a z

− −

+ = = =

− −

= =

−

= −

+ +

∑ ∑ ∑

∑ ∑

0 ( )( ) ( ) ( )( )

N z Y z H z X z

z

+ = −0

1 1

( ) ( ) N k

k n

k

k k

N z a z y n z −

= =

= −∑ ∑


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ECE 308-12 13


The output of the system can subdivide into two parts.

the zero-input response

The zero-state response( ) ( ) ( ) zsY z H z X z =

0 ( )( )( )

zi

N z Y z

A z

+ =

Therefore, the total of the inverse z-transform of responses

gives us y(n)

( ) ( ) ( ) zs zi y n y n y n= +

ECE 308-12 14


Determine the unit step response of the system of the

following equation.

Condition: a.

b.

Example:

( ) 0.9 ( 1) 0.81 ( 2) ( ) y n y n y n x n= − − − +

( 1) ( 2) 0 y y− = − =

( 1) ( 2) 1 y y− = − =

Solution:

1 2 1( ) 0.9 ( ) ( 1) 0.81 ( ) ( 1) ( 2) ( )Y z z Y z y z Y z z y y X z − − − = + − − + − + − +

1 2 1( )(1 0 .9 0.81 ) 0.9 ( 1) 0.81 ( 1) 0.81 ( 2) ( )Y z z z y z y y X z − − −− + = − − − + − +

1 2

1( ) ( )

(1 0.9 0.81 )Y z X z

z z − −=

− +

The system function is

1 2

1( )

(1 0.9 0.81 ) H z

z z − −=

− +


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ECE 308-12 15


The system poles are

and

The z-transform of input sequence is

31 0.9

j

p e

π

= 32 0.9

j

p e

π −

=

Example: (cont)

Therefore,

1

1( )

(1 ) X z

z −=

−

1 1 13 3

1( )

(1 0.9 )(1 0.9 )(1 ) j j

Y z

e z e z z π π

− − −

=

− − −

95.2 95.2

11 13 3

0.544 0.544 1.099( )

(1 )(1 0.9 ) (1 0.9 )

j j

j j

e eY z

z e z e z

π π

− −

−− −

= + +−

− −

( ) 1.099 1.088(0.9) cos( 95.2 ) ( )3

n

zs y n n u nπ

= + −

The zero-state response is

( ) ( ) zs y n y n=Since the initial condition are zero

ECE 308-12 16


Example(cont) Using initial condition

The total response

The zero input response

1 2 1( )(1 0.9 0.81 ) 0.9 0.81 0.81 ( )Y z z z z X z − − −− + = − + +

1

1 2 1 2

1 0.81 0.81( ) ( )

(1 0.9 0.81 ) (1 0.9 0.81 )

z Y z X z

z z z z

−

− − − −

−= −

− + − +1

0

1 2

0.81 0.81( )( )

( ) (1 0.9 0.81 ) zi

z N z Y z

A z z z

−

− −

−= =

− +

84.8 84.8

1 13 3

0.4956 0.4956

( )1 0.9 1 0.9

j j

zi j j

e e

Y z e z e z

π π

−

− −= +− −

( ) 0.99(0.9) cos( 84.8 ) ( )3

n

zi y n n u n

π = +

( ) ( ) ( ) zs ziY z Y z Y z = +

( ) 1.099 1.44(0.9) cos( 38 ) ( )3

n

zs y n n u nπ

= + +


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ECE 308-12 17


Transient and Steady-State Response

The response of a system to a given input can be separated

into two components

1

( ) ( ) ( ) N

n

nr k k

k

y n A p u n=

= ∑

where , are the poles of the system and are scale factor.k p 1,2,...,i k = k

If for all k , goes zero for n approaches infinity. In this

case, natural response of the system accept as transient response.

1k p < ( )nr y n

The forced response:1

( ) ( ) ( ) L

n

fr k k

k

y n Q q u n=

= ∑

Natural response

where qk, are the poles in the forcing function and are Qkscale factor.

1,2,...,i k =

ECE 308-12 18


If for all k, goes zero for n approaches infinity

If the causal input signal is sinusoidal, then the forced signal

will be sinusoidal for all . In the case, the forced response

is called the steady-state response

Example:

1k q < ( ) fr y n

0n ≥

Find the transient and steady-state response of

when the input signal is

( ) 0.5 ( 1) ( ) y n y n x n= − +

( ) 10 cos( ) ( )4n x n u nπ =

The system is initially relaxed.

Solution: 1( ) 0.5 ( ) ( )Y z z Y z X z −= +

1

( ) 1( )

( ) 1 0.5

Y z H z

X z z −

= =−

The poles at 0.5 z =


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ECE 308-12 19


Example: (cont)

The z-transform of input signal

11

1 21 2

110(1 )1 cos

24( ) 101 21 2 cos

4

z z X z

z z z z

π

π

−−

− −− −

−−= =

− +− +

1

/ 4 1 / 4 1

110(1 )

2( )(1 )(1 ) j j

z

X z e z e z π π

−

− − −

−

=− −

1

1 / 4 1 / 4 1

110(1 )

2( ) ( ) ( )

(1 0.5 )(1 )(1 ) j j

z

Y z H z X z z e z e z π π

−

− − − −

−

= =− − −

31 2

1 / 4 1 / 4 1( )

(1 0.5 ) (1 ) (1 )

j j

A A AY z

z e z e z

π π − − − −= + +

− − −

1

1

1 21 2 2

1 110(1 ) 10(1 2)

2 2 1.907(1 2 ) (1 22 2 )

z

z

A z z

−

−

=− − −

− −

= = = −− + − +

ECE 308-12 20


Example (cont)

And the forced or steady-state response is

The natural or transient response is

1 /4

1

28.7

2 1 / 4 1

110(1 )

2 6.78(1 0.5 )(1 )

j

j

j z e

z

A e z e z

π π − −

−

−

− − − =

−

= =− −

1 /4

1

28.7

3 1 / 4 1

110(1 )

2 6.78(1 0.5 )(1 )

π π −

−

− − =

−= =

− − j

j

j z e

z

A e z e z

28.7 28.7

1 / 4 1 / 4 1

1.907 6.78 6.78( )

(1 0.5 ) (1 ) (1 )

j j

j j

e eY z

z e z e z

π π

−

− − − −

−= + +

− − −

( ) 1.907(0.5) ( )nnr y n u n= −

28.7 / 4 28.7 / 4( ) [6.78 ( ) 6.78 ( )] ( ) j j n j j n fr y n e e e e u nπ π − −= +

( ) 13.56cos( 28.7 ) ( )4

fr y n n u n

π = −

steady and transient.pdf

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