dsp n06 chapter6

8/10/2019 Dsp n06 Chapter6

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66 Chapter 6.

Figure 6.3: Pole-zero plot and region of convergence for Example 6.3.

Property 1 The ROC is a ring or disk in the Z-plane centered at the origin, i.e. 0 rR |z|< rL< .

Property 2 The DTFT exists if and only if the ROC includes the unit circle.

Property 3 The ROC cannot contain any poles.

Property 4 Ifx[n] is a finite-length sequence, then the ROC is the entire Z-plane withthe possibleexception ofz = 0 or z = .

Property 5 If x[n] is a right-sided sequence, i.e. x[n] = 0 for n < N1 < , for someN1 Z, then the ROC extends outwards from the outermostfinite pole in X(z).

Property 6 If x[n] is a left-sided sequence, i.e. x[n] = 0 for n > N2 > , for someN2 Z, then the ROC extends inwards from the innermostfinite pole in X(z).

Property 7 A two-sided sequence is an infinite-duration sequence that is neither right-sided nor left-sided. If x[n] is a two-sided sequence, then the ROC will consist ofa ring in the Z-plane bounded on the interior and exterior by a pole and does not

contain any poles.Property 8 The ROC must be a connected region.

6.2 The inverse Z-transform

By considering

X

rej

= F

x[n]rn

,


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6.3. Partial fraction expansion 67

one can obtain

x[n]rn =F1

X

rej

,

or

x[n]rn = 1

2

X

rej

ejn d,

or

x[n] = 12

Xrej rejn d. (6.8)That is, we can recoverx[n] from integrating its Z-transform along a contour z = rej

in its ROC, by fixing r and varying from to . Withz = rej , for fixed r we seethat dz= jrej d= jz d or d= dz

jz.

Since the integral in (6.8) is over a 2 interval of, it corresponds to a traversal aroundthe circle |z|= r. Consequently

x[n] = 1

2j

X(z)zn1 dz, (6.9)

where the symbol denotes integration around a counter-clockwise closed circular contourcentered at the origin and with radius ofr. The value ofr can be chosen such that |z|= ris in the ROC.

6.3 Partial fraction expansion

X(z) =

M1k=0 bkz

kN1k=0 akz

k=

b0M1

k=1

1 ckz

1

a0N1

k=1 (1 dkz1)

. (6.10)

If M < N and all poles are simple, i.e. unique dks, we can do a partial fraction

expansion to get

X(z) =N1k=0

Ak1 dkz1

, (6.11)

where

Ak =

1 dkz1

X(z)

z=dk

. (6.12)


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68 Chapter 6.

IfM > N, we can obtain the right form by first doing a long division of the numerator bythe denumerator. This gives gives a general form

X(z) =MNr=0

Brzr +

N1k=0

Ak1 dkz1

, (6.13)

where the Brs are obtained from the long division and the Aks are as before.IfX(Z) has a pole of order s atdi and all other poles are simple we get

X(z) =

MNr=0

Brzr +

N1k=0,k=i

Ak1 dkz1 +

sm=1

Cm(1 diz1)m , (6.14)

where

Cm= 1

(s m)!(di)sm

dsm

dwsm

(1 diw)s X

w1

w=d1

i

(6.15)

and the Aks andBrs are obtained as before.Using (6.14) and the ROC, we can invert the Z-transform using the inspection method,

i.e.

anu[n] Z

1

1 az1

, |z|> |a|. (6.16)

anu[n 1] Z

1

1 az1 , |z|< |a|. (6.17)

6.4 Z-transform properties

1. Linearity:

a1x1[n] + a2x2[n] Z a1X1(z) + a2X2(z), ROC containsRx1 Rx2

2. Time-shifting:

x[n n0] Zzn0X(z), ROC =Rx

(except for the possible addition or deletion ofz = 0 or z = )

3. Multiplication by exponential sequence:

zn0

x[n] Z X(z/z0) , ROC =|z0|Rx


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6.4. Z-transform properties 69

4. Differentiation ofX(z):

nx[n] Z z

dX(z)

dz , ROC =Rx

Example: Inverse of a non-rationalZ-transform

X(z) = log

1 + az1

, |z|> |a|.

dX(z)

dz =

1

1 + az1.

az2

Hence

zdX(z)

dz =

az1

1 + az1Z1

a(a)n1u[n 1] |z|> |a|.

Hencenx[n] =a(a)n1u[n 1], or x[n] = an

(a)n1u[n 1]. Hence

(1)

n1an

nu[n 1]

Z

log 1 + az15. Conjugation of a complex sequence:

x[n] Z X(z) ROC =Rx

6. Time-reversal:

x[n] Z X

1

z

, ROC =

1

Rx

7. Convolution of sequences:

x1[n] x2[n] Z X1(z)X2(z), ROC containsRx1 Rx2

8. Initial value theorem: Ifx[n] is zero for n


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70 Chapter 6.

6.5 Analysis and characterization of LTI systems using Z-transform

The Z-transform plays a particularly important role in the analysis and representation ofdiscrete-time LTI systems. From the convolution property, for an LTI system with impulse

response h[n] Z H(z) and an inputx[n]

Z X(z), we have,

y[n] = (h x)[n] =h[n] x[n] Z H(z)X(z), (6.18)

where X(z), Y(z) and H(z) are the Z-transforms and the ROC ofY(z) is

Ry =Rx Rh. (6.19)

H(z) is referred to as the system function or transfer function of the system.

6.5.1 Causality

A causal LTI system has an impulse response that is h[n] = 0 for n


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6.5. Analysis and characterization of LTI systems using Z-transform 71

6.5.3 LTI systems and linear constant-coefficient difference equations

If

y[n] =N1k=1

aky[n k] +M1k=0

bkx[n k],

then

Y(z) =N1

k=1

akzkY(z) +

M1

k=0

bkzkX(z).

Hence

Y(z)

1

N1k=1

akzk

= X(z)

M1k=0

bkzk

or

Y(z)

X(z)=

M1k=0 bkz

k

1 N1

k=1 akzk

=H(z).

Therefore the system function for such systems are easy to write. This form ofH(z) =Y(z)X(z) is also called the transfer function of a linear constant-coefficient difference equationsystem.


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dsp n06 chapter6

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