02 discrete time signals systems release

8/13/2019 02 Discrete Time Signals Systems Release

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Curtin University, Australia 2-1YH Leung (2005, 2006, 2007, 2012)

DISCRETE-TIME SIGNALS AND SYSTEMS

1 Discrete-Time Signals

Recall we are interested in discrete-time continuous-amplitudesignals

Fig. 1 Classification of signals

Strictly, a discrete-time signal is simply a sequenceof numbers, represented

mathematically as follows

{ }[ ]x x n= (1)

where n and [ ]x n is the nth number of the sequence

In practice, x is often derived by sampling an analog signal ( )cx t periodically,

i.e.

= [ ] ( ),cx n x nT n (2)

where T is sampling period

Continuous amplitude Discrete amplitude

Continuous-time

Discrete-time

Analog signal

Digital signal


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Remarks

(i) As stated, strictly, [ ]x n means the nth number of the sequencex

However, by corruption of use, it can also refer to the entire sequence

itself

(ii) [ ]x n is defined only for integer values of n

(iii) [ ]x n is assumed to be always finite in value, i.e.

[ ] ,x n n< " (3)


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2 Some Basic Sequences

Unit sample sequence(aka discrete-time impulse, or simply, impulse)

0, 0

[ ] 1, 0

n

n nd

= = (4)

Unit step sequence

0, 0[ ]

1, 0

nu n

n


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Alternating sequence (exponential sequence with a= -1)

{ }

0

[ ] ( 1) , 1, 1, 1, 1, 1,n

n

x n=

= - = + - + - + (7)

Periodic sequence

The sequence [ ]x n is said to beperiodicwithperiod N if

+ = " [ ] [ ],x n N x n n (8)

The fundamental periodof [ ]x n is smallest positive Nthat satisfies Eq. (8)1

Sinusoidal sequence

[ ] cos( ), , ,o ox n A n Aq f q f= + (9)

Ais amplitude, oq is frequency, f isphase

The sinusoidal sequence [ ]x n is periodic with period N if and only if

2 , for someoN k kq p= (10)

Complex exponential sequence

[ ] on j nn jx n A A e e

qfa a= =

{ }cos( ) sin(] )[ n on

ojx n n nA A a q f fa q== + + + (11)

where , , , , ,ojj

oA A e e Aqf a a a f q= =

1 Thus alternating sequence has fundamental period = 2N

n

1

x[n]

0 1 2 3 4-1-2-3


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3 Basic Sequence Operations

Suppose { }[ ]x x n= and { }[ ]y y n= are two sequences

Scaling: { }[ ] ,a x a x n a = (12)

Addition: { }[ ] [ ]x y x n y n+ = + (13)

Multiplication2: { }[ ] [ ]x y x n y n = (14)

Convolution: [ ] [ ] [ ] [ ]k k

x y x k y n k x n k y k

=- =-

* = - = - (15)

Example 1

(a)0, 0

[ ], 0n

nx n

A na


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4 Discrete-Time Fourier Transform (DTFT)

Discrete-time Fourier transformof [ ]x n is defined by

{ } q q

q

-

=-= =

DTFT [ ] ( ) [ ] ,

j jn

nx n X e x n e (17)

q( )jX e is periodic in q with period p2 since " m

q p q p q+ = =( 2 ) 2( ) ( ) ( )j m j j m jX e X e e X e (18)

Therefore, we consider only [ ],q p p -

We call q digital frequency. Its unit is radians/sample

Inverse DTFTis given by

{ }q q qp

q

p

- += =

1

2

1DTFT ( ) [ ] ( )

2

j j jnX e x n X e e d (19)

where integral is evaluated over any p2 interval of q

For following examples, next result on geometric seriesis useful

111

0

, 1

, 1

NaNn a

n

aa

N a

---

=

= =

(20)


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Example 2

(a) d=[ ] [ ]x n n

q qd

-

=-

= =( ) [ ] 1j jnn

X e n e

(b) [ ] [ ], 1nx n a u n a= <

( )0

1( ) [ ]

1

nj n jn j

n n

X e a u n e aeae

q q q

q

- -

-=- =

= = =-

0.75a=

0.75a= -

-1

1 a

+1

1 a

- -1 2tan ( 1 )a a

+1

1 a

-11 a

- -1 2tan ( 1 )a a


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0.75a=

(c) [ ] , 1n

x n a a= <

( ) ( )1

0 0 1

2

2

( )

1

1 1

1

1 2 cos

n mj n jn n jn n jn j j

n n n n m

j

j j

X e a e a e a e ae ae

ae

ae ae

a

a a

q q q q q q

q

q q

q

- - - - - -

=- = =- = =

-

= = + = +

= +- -

-=

- +

0.75a=

(d) [ ] [ 1], 1nx n a u n a= - - - <

( ) ( )

( )

1

1

0

1

1

( ) [ 1]

1

n nj n jn j j

n n n

n

j

n

X e a u n e ae a e

a e

q q q q

q

- - -

=- =- =

=

-

-

= - - - = - = -

= -

-

Last summation converges only if 1 1ja e q- < or 1 1a- < which contradicts the given condition

1a < . Therefore, given sequence does not have a DTFT

?

+-

1

1

a

a

-+

1

1

a

a


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Properties of DTFT

Let q( )jX e and q( )jY e be the DTFTs of [ ]x n and [ ]y n respectively:

qDTFT[ ] ( )jx n X e

qDTFT[ ] ( )jy n Y e

Linearity q q+ +DTFT[ ] [ ] ( ) ( )j jax n by n aX e bY e

Time shiftingqq -- DTFT[ ] ( ) oj njox n n X e e

Frequency shiftingq q q-DTFT ( )[ ] ( )o oj n jx n e X e

Conjugation q* * -DTFT[ ] ( )jx n X e

Time reversal q-- DTFT[ ] ( )jx n X e

Time expansionDTFT

( )

multiple of

multiple of

[ ],[ ] ( )

0,

jkk

n k

n k

x n kx n X e q

=

=

Convolution q q* DTFT[ ] [ ] ( ) ( )j jx n y n X e Y e

Multiplication x q xp

xp

- DTFT ( )21

[ ] [ ] ( ) ( )2

j jx n y n X e Y e d

Time differencing q q-- - -DTFT[ ] [ 1] (1 ) ( )j jx n x n e X e

Accumulation qq

p d q p

-=- =-

+ --

DTFT 01[ ] ( ) ( ) ( 2 )1

nj j

jk k

x k X e X e ke

Frequency differentiation q

qDTFT[ ] ( )j

dn x n j X e

d

[ ]x n real q q- *=( ) ( )j jX e X e

[ ]x n real and even q( )jX e real and even

[ ]x n real and odd q( )jX e purely imag and odd

Parsevals relation22

2

1[ ] ( )

2

j

n

x n X e dqp

qp

=-

=


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Table of DTFT pairs

Sequence [ ]x n Transform q( )jX e

1. d[ ]n 1

2. [ ]u n q

pd q p

-=-

+ --

1 ( 2 )1 j k

ke

3. d -[ ]n m q-j me

4. [ ]na u n q--

1

1 jae

5. ( 1) [ ] , 1nn a u n a+ < q-- 21

(1 )jae

6.( 1)!

[ ] , 1!( 1)!

nn r a u n an r

+ -

( )( )

11 2

sin ( )

sin 2

Nq

q

+

8.sin

sincc c cn n

n

q q q

p p p

=

1,

0 ,

c

c

q q

q q p

<

9. 1 p d q p

=-

-2 ( 2 )k

k

10.qoj ne p d q q p

=-

- -2 ( 2 )ok

k

11. qcos on { }( 2 ) ( 2 )o ok

k kp d q q p d q q p

=-

- - + + -

12. qsin on { }( 2 ) ( 2 )o ok

k kj

pd q q p d q q p

=-

- - - + -

13. Periodic square wave

1

1

1,0 , 2

n NN n N

<

+ =[ ] [ ]x n N x n

( )22 kk Nk

a pp d q

=-

-

( )2 11 222

sin

sin

kN

k kN

Na

N

p

p

+ =

14. Impulse train

d

=-

- ( )k

n kN ( )2

2 kN

kNpp d q

=-

-


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5 z-Transform

z-transform of [ ]x n is defined by

{ }

-

=-= =

[ ] ( ) [ ] ,

n

nx n X z x n z z (21)

Recall from Eq. (17)

q q q

-

=-

= ( ) [ ] ,j jnn

X e x n e

Therefore, z-transform generalises DTFT:3

( )jX e q is simply ( )X z evaluated around the unit circle jz e q=

Region in z-plane where ( )X z is finite is its region of convergence (ROC)

Clearly, if q( )jX e exists, then ROC of ( )X z must include unit circle

ROCs consist of rings in z-plane centered about the origin

Inverse z-transformis given by

{ }p

- -= = 1 11

( ) [ ] ( )2

n

CX z x n X z z dz

j (22)

where Cis a closed contour in ROC of ( )X z

3 In the same way Laplace transform generalises the Fourier transform


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Example 3

In the following, we re-work Example 2 for z-transforms

(a) d=[ ] [ ]x n n

d

-

=-

= =( ) [ ] 1nn

X z n z and converges everywhere except = 0z

(b) =[ ] [ ]nx n a u n

- -

=- =

-=

-

= =

=

=-

=-

0

1

0

1

( ) [ ]

( )

1

1

n n n n

n n

n

n

X z a u n z a z

az

az

z

z a

provided1 1az- < or z a>

(c) [ ] nx n a=

1

0

1

0 1

1

2

1

( )

( ) ( )

1 11

11

(1 )

( )( )

n n n n n n

n n n

n n

n n

a

X z a z a z a z

az az

azaz

a z

a z a z

-- - - -

=- = =-

-

= =

-

= = +

= +

= + -

- -

-=

- - -

provided1 1az- < and 1az< ,

i.e., 1a

a z< <

a 1 Re

Im

z-plane

unit circle

1 Re

Im

z-plane

a

1/a


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(d) = - - -[ ] [ 1]nx n a u n

11

1

1 1

( ) ( )

1 11

1 1

n n n

n n

X z a z a z

a z az

z

z a

- - -

=- =

- -

= - = -

= - - =

- -

=-

provided1 1a z- < or z a<

Comparing Examples 3(b) and 3(d), we see that

( )X z is not unique different sequences can have the same ( )X z

( )X z s are distinguishable only by their ROCs

a 1 Re

Im

z-plane


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Sequence ROC

Fig. 2 Regions of convergence of z-transforms

n

x[n]

n

x[n]

n

x[n]

n

x[n]

All z

except z= 0

All z

except z=

All z

except z= 0 and

R< |z|

n

x[n]

n

x[n]

|z|


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Properties of z-transform

Let ( )X z with ROC XR and ( )Y z with ROC YR be z-transforms of [ ]x n and [ ]y n

respectively:

[ ] ( ), Xx n X z R

[ ] ( ), Yy n Y z R

Linearity + + [ ] [ ] ( ) ( ), contains X Yax n by n aX z bY z R R

Time shifting-- [ ] ( ) onox n n X z z , XR except for possible addition or deletion

of origin or

Frequency shifting ( )[ ] ,o

n zo o Xz

x n z X z R

Conjugation * * *[ ] ( ), Xx n X z R

Time reversal -- 1[ ] ( ), 1 Xx n X z R

Time expansion 1( )[ ],

[ ] ( ),0,

kkk X

n rk

n rk

x rx n X z R

=

=

Convolution * [ ] [ ] ( ) ( ), contains X Yx n y n X z Y z R R

Multiplication ( ) 11[ ] [ ] ( ) , contains , in ROC2 z X YCx n y n X Y d R R Cj x xx xp

Time differencing { }1[ ] [ 1] (1 ) ( ), contains 0Xx n x n z X z R z-- - - >

Accumulation { }1

1[ ] ( ), contains 1

1

n

Xk

x k X z R zz-=-

>-

Differentiation in z -[ ] ( ), Xd

n x n z X z R dz

[ ]x n real * *=( ) ( )X z X z

Parsevals relation2 11[ ] ( ) (1 ) , in

2 X

Cn

x n X X d C Rj

x x x x p

* * -

=-

=


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Table of z-transform pairs

Sequence Transform ROC

1. d[ ]n 1 All z

2. [ ]u n -- 11

1 z 1z >

3. - - -[ 1]u n -- 11

1 z 1z <

4. d -[ ]n m -mz All zexcept 0 (if > 0m )

or (if < 0m )

5. [ ]na u n -- 11

1 az z a>

6. - - -[ 1]na u n -- 11

1 az z a<

7. [ ]nna u n

-

--

1

1 2(1 )

az

az z a>

8. - - -[ 1]nn a u n -

--

1

1 2(1 )

az

az z a<

9. [ ]cos [ ]on u nq q

q

-

- -

-

- +

1

1 2

1 [cos ]

1 [2cos ]

o

o

z

z z

1z >

10. [ ]sin [ ]on u nq q

q

-

- -- +

1

1 2

[sin ]

1 [2cos ]

o

o

z

z z 1z >

11. cos [ ]n

or n u nq

q

q

-

- -

-

- +

1

1 2 2

1 [ cos ]

1 [2 cos ]

o

o

r z

r z r z z r>

12. sin [ ]n

or n u nq

q

q

-

- -- +

1

1 2 2

[ sin ]

1 [2 cos ]

o

o

r z

r z r z z r>

13. , 0 10, otherwise

n

a n N -

-

--- 111

N N

a zaz

0z >


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6 Discrete-Time Systems

Systems can be: continuous-time or discrete-time

linear or non-linear

time-invariant or time-varying

stable or unstable

causal or non-causal

with memory or memoryless

invertible or non-invertible

We focus on discrete-timelinear time-invariant stable causalsystems

Let be a discrete-time system with input [ ]x n and output [ ]y n

{ }[ ] [ ]y n x n= (23)

(i) is linearif, for any ,a b { } { } { }1 2 1 2[ ] [ ] [ ] [ ]ax n bx n a x n b x n+ = + (24)

(ii) is time-invariantif, for all on

{ }[ ] [ ]o ox n n y n n- = - (25)

(iii) is stable in the bounded-input-bounded-output (BIBO) sense if andonly if for all input sequences satisfying

[ ] ,xx n B n < " (26)

we can find a yB such that

[ ] ,yy n B n < " (27)

(iv) is causalif, for all on , [ ]oy n depends only on { }[ ], ox n n n

x[n] y[n]


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Suppose is linear time-invariant(LTI). Define

{ }[ ] [ ]h n nd= (28)

[ ]h n is the unit sample response, or impulse response, of

(i) It follows from Eqs. (16), (24) and (25) that

{ } { }d d

=- =-

=-

= = - = -

= -

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ]

k k

k

y n x n x k n k x k n k

x k h n k

i.e. = * = *[ ] [ ] [ ] [ ] [ ]y n x n h n h n x n (29)

Eq. (29) shows an LTIsystem is completely characterised by [ ]h n

(ii) z-transform of Eq. (29) is given by

=( ) ( ) ( )Y z H z X z (30)

where { }=( ) [ ]H z h n 4 (31)

Hence an LTI system is also completely characterised by ( )H z . Indeed,the system is often identified as ( )H z

( )H z is called the transfer function of the system. It is also called the

system function

Fig. 3 Time- and transform-domain representations of a system

(iii) Frequency responseof system is given by the DTFT

( ) ( ) jj

z eH e H z q

q== (32)

4 Alternatively, since ( ) 1X z = if [ ] [ ]x n nd= , so by Eq. (30),

[ ] [ ]( ) ( )

x n nY z H z

d= = . But [ ] [ ]( )x n nY z d= = { }[ ] [ ][ ]x n ny n d= and by definition, [ ] [ ][ ] [ ]x n ny n h nd= = . Therefore, it follows that { }[ ] ( )h n H z =

h[n]x[n] y[n] H(z)X(z) Y(z)


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(iv) It can be shown an LTI system is stableif and only if5

[ ]n

h n

=-

< (33)

But1

[ ] [ ] nz

n n

h n h n z -

==- =-

=

Therefore, a stable ( )H z must include the unit circle in its ROC

(v) As can be seen from Eq. (29), an LTI system (stable or unstable) is

causalif and only if =[ ] 0h n for < 0n

(vi) If is LTIand causal, and [ ]x n is also causal, then Eq. (29) simplifies to

0 0

[ ] [ ] [ ] [ ] [ ]n n

k k

y n x k h n k h k x n k = =

= - = - (34)

5See A. V. Oppenheim and R. W. Schafer, Discrete-Time Signal Processing, 3rd ed., Prentice-Hall,

2010, pp. 59-60


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7 Interconnected Systems

It can be shown (exercise) from Eqs. (29) and (30) that:

Cascaded Systems Parallel Systems

Fig. 4 System interconnections

Example 4

Defining [ ]w n as shown and taking z-transforms of [ ]x n , [ ]y n and [ ]w n , we get, at summer output

= +( ) ( ) ( ) ( ) ( ) ( )W z A z X z B z C z W z

\ =-

( )( ) ( )

1 ( ) ( )

A zW z X z

B z C z

But =( ) ( ) ( )Y z B z W z

\ =-

( )( ) ( ) ( )

1 ( ) ( )

A zY z B z X z

B z C z

Hence = =-

( ) ( ) ( )( )

( ) 1 ( ) ( )

Y z A z B z H z

X z B z C z

h1[n] h2[n]

h1[n] *h2[n]

h2[n] h1[n]

H1(z)H2(z)

h1[n]

h2[n]

h1[n] + h2[n]

H1(z) + H2(z)

B(z)

C(z)

A(z) y[n]x[n]w[n]


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8 Linear Constant-Coefficient Difference Equations

Linear constant-coefficient difference equations (LCCDEs) describe an

important class of discrete-time LTI systems6

An Nth order LCCDE is defined by

0 0

[ ] [ ]N M

k mk m

a y n k b x n m= =

- = - (35)

where ka and mb are constants, and 0a normally equals 1

Taking z-transform of Eq. (35), we get

0 0

( ) ( )N M

k mk m

k m

a Y z z b X z z - -

= =

=

or

-

=

-

=

=

0

0

( ) ( )

Mm

mm

Nk

kk

b z

a z

Y z X z (36)

Transfer functionof LCCDE system (see Eq. (30)) is given by

- -

= =

- -

= =

= = =

+

00 0

00 1

( )( )

( )1

M Mm mm

mm m

N Nk kk

kk k

bb z z

aY zH z

X z aa z z

a

(37)

6 LCCDE systems form a special class of LTI systems. Can you think of an LTI system that cannot be

described by an LCCDE?


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Re-write ( )H z as follows

00 0

0 0 0

( )

MM MN pk N M k

M pk kpk k

N N N

k M N k M pk k N pk k p

z b zb z z b z

H z

a z z a z z a z

- --

== =

- - -= = =

= = =

(38)

Roots ofnumerator

denominator

polynomial are called

zeros

polesof ( )H z

Hence ( )H z has Mzeros and Npoles at non-zero locations in z-plane

andzeros

poles

N M

M N

- - at = 0z if

N M

M N

> >

Beware!

Although ( )H z is often expressed as a ratio of two polynomials in 1z- , its poles

and zeros are found from ( )H z expressed as a ratio of two polynomials in z

Example 5

Consider-= - 1

1( )

1H z

az

It is often said ( )H z has no zeros and only one pole at

-- =11 0az

or=z a

Above statement is not entirelycorrectsince re-writing ( )H z as follows

-= =

-- 11

( )1

z zH z

z z aaz

we see, apart from the pole at =z a , ( )H z actually has a zero at = 0z


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Let [ ]x n be a known input sequence. Output sequence [ ]y n can be found by:

(a) solving LCCDE formally through finding the homogeneousand particular

solutions

(b) using z-transforms

{ } { }1 1[ ] ( ) ( ) ( )y n Y z H z X z - -= = (39)

(c) determining the linear convolution

= * = *[ ] [ ] [ ] [ ] [ ]y n x n h n h n x n (40)

(d) assuming [ 1], , [ ]o oy n y n N - - are known, computing the forwardrecursion

0 01 0

[ ] [ ] [ ], , 1,

N M

k m o ok m

a by n y n k x n m n n na a= =

= - - + - = + (41)

or assuming [ ], , [ 1]o oy n y n N - + are known, computing the backwardrecursion

1

0 0

[ ] [ ] [ ], , 1,N M

k mo o

N Nk m

a by n N y n k x n m n n n

a a

-

= =

- = - - + - = - (42)

Clearly, to exercise Eq. (39), ROC of ( )H z must be known

Likewise, to exercise Eq. (40), ROC of ( )H z must also be known since actual

form of [ ]h n depends on ROC of ( )H z

In next section, it will be shown how, given an LCCDE, we can determine all

possible [ ]h n

Above comment implies knowing LCCDE is not enough more information is

required

Therefore, for example, suppose LCCDE system is causaland stable

(i) [ ]h n causal ROC of ( )H z has form R z-<

(ii) [ ]h n stable ROC includes unit circle

\ [ ]h n causal and stable ( )H z must have all its poles inside unit circle


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Example 6

Consider the LCCDE

[ ] [ 1] [ ]y n a y n x n= - +

As can be readily verified

1

1( )

1H z

az-=

-

Clearly, ( )H z has a pole at z a= . Therefore, there are two possible ROCs: { }z a> and { }z a< ,

and from z-transform table

11

1[ ] [ ] ( ) ,n

azh n a u n H z z a--

= = >

11

1[ ] [ 1] ( ) ,n

azh n a u n H z z a--

= - - - =


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9 Partial Fraction Expansion (PFE)

Recall Eq. (37). Factorising ( )H z

1

0 0 1

0 1

10

(1 )( )

(1 )

M Mk

k kk k

N Nk

kkkk

b z c zbH z

ad za z

- -

= =

--

==

-= =

-

(43)

we see: non-zero zeros given by 1, , Mc c

non-zero poles given by 1, , Nd d

If = = =1 2 sk k k

d d d for some 1 2 sk k k , then we say 1kd is an sthorder pole, or a repeated polewith multiplicitys

Suppose ( )H z has only one repeated pole (with multiplicity s), and poles are

so indexed that -1, , N sd d are simple poles while - +1N sd is the repeated

pole

PFE of ( )H z is given, with - + = =1N s Nd d , by

10

0 1

10

10

(1 )( )

( )( )

(1 )

M Mk

k kk k

N Nk

kkkk

b z b c zN z

H zD z

a d za z

- -

= =

--

==

-

= = =

-

- --

- -

= = = - +

= + +

- -

= 1 10 1 1 11

( )( )

1( ) ( )

M N N s sr n m

r m

r n mn N s

A CB z

N zH z

D z d z d z

(44)

where residues rB , nA and mC are given by

1

10

1

( )(1 ) ( )

(1 )n

n

n k Nz d

kkk n z d

N zA d z H z

a d z

-=

-

= =

= - =

- (45)


26/37


11

11

1

1(1 ) ( )

( )!( )N s

s ms

m N ss m s mN s w d

dC d w H w

s m d dw -- +

--

- +- -- + =

= - - -

11

1

10

1

1 ( )

( )!( ) (1 )

N s

s m

s m s m N sN sk

k

m

w d

C d N w

s m d dw a d w-

- +

- -

- - -- +

= =

= - - -

(46)

and rB are found by long division of ( )N z by ( )D z with division process

stopping when degree of the remainder is less than N

Can readily extend above procedure to ( )H z s with more than one repeated

pole

[ ]h n is found with following z-transforms

1( 1)!

1( 1)

1

1 2

1

!

1( 1)!

1( 1)!

1( 1)!2

3 1 11

2

[ ]

[ ]1

1 [ 1]

( 1) [ ]1

(1 ) ( 1) [ 2]

( 1)( 2) [ ]1

(1 ) ( 1)( 2) [ 3]

r

n

n

n

n

n

k

k

k

n

k

k

z n r

a u n

az a u n

n a u n

az n a u n

n n a u n

az n n a u n

d

-

-

-

-

-

-

-

-

-

-

- - - -

= =

= =

+ - - + - -

+ +

= =

= =- - + + - -

1

( 1)!1 1

( 1

( 1)!

)!

( 1)( 2) ( 1) [ ]1(1 ) ( 1)( 2) ( 1) [ ]

k

n

kk n

k

n n n k a u naz n n n k a u n k

--

-

-

+ + + -= - - + + + - -

=-

Since it is necessary that [ ] 0h n as n for ( )H z to be stable, abovez-transform pairs show causal poles of a stable ( )H z must lie strictly inside the

unit circle and anitcausal poles must lie strictly outside the unit circle7

7 This is consistent with a result derived in pp. 19 where it is shown the ROC of a stable LTI system

must include the circle unit


27/37


Example 7

( )( )( )

1 2 3 4 5

21 1 1

1 2 3 4 5

1 2 3 4

8 7.5 4.03 0.926 0.174 0.018( )

1 (0.2 0.4) 1 (0.2 0.4) 1 0.3

8 7.5 4.03 0.926 0.174 0.0181 0.53 0.156 0.018

z z z z z H z

j z j z z

z z z z z z z z z

- - - - -

- - -

- - - - -

- - - -

- + - + -=

- + - - -

- + - + -= - + - +

( )H z has = 5M non-zero zeros, = 4N non-zero poles and - = 1M N pole at = 0z . Two of the

non-zero poles form a complex conjugate pair at = 1,2 0.2 0.4d j , and the other non-zero pole is at

=3 0.3d but with a multiplicity of = 2s . Thus PFE will have form

2

11

1

1

0

10

2

1

1

1

1 1

( )1 (1

(1 0.

)

1 3 )

M N N s sr n m

r m

n

r n

n

mm

n mr

n N s

r

r

m

A CH z B z

d z d z

B z Cz

Ad z-=

- --

- -= = = - +

-=

-=

= + +- -

=--

+ +

(i) 0B and 1B found as follows.

4 3 2 1 5 4 3 2 1

5 4 3 2 1

4 3 2 1

4 3 2 1

3 2 1

0.018 0.156 0.53 1 0.018 0.174 0.926 4.03 7.5 8

0.018 0.156 0.53

0.018 0.396 3.03 6.5 8

0.018 0.156 0.53 1

0.24 2.5 5.5 7

z

z z z z z z z z z

z z z z z

z z z z

z z z z

z z z

- - - - - - - - -

- - - - -

- - - -

- - - -

- - -

-

- + - + - + - + - +

- + - + -

- + - +

- + - +

- + - +

1 1- +

Hence 1 1B =- and 0 1B =

(ii) 1A and 2A found as follows.

( )( )

2

12

1 2 3 4 5

2

2

1 1

0.2 0.4

(1 ) ( )

8 7.5 4.03 0.926 0.174 0.018

1 (0.2 0.4) 1 0.3

1 2

z d

z j

d z H z A

z z z z z j

j z z

-=

- - - - -

- -= -

= -

- + - + -= =

--

- +

21 1 2AA j* += =

(iii) 1C and 2C found as follows.

( )( )

3

2 22 1

3 32 2 2 213

2 3 4 5

1 0.3

2

1(1 ) ( ) , 0.3

(2 2)!( )

8 7.5 4.03 0.926 0.174 0.018(1)

1 (0.2 0.4) 1 (0.2 0.4)

4

w d

w

dd w H w d

d dw

w w w w w

j w j w

C-

-- -

=

=

= - = - -

- + - + - = - + - -

=


28/37


( )( )

3

2 12 1

3 32 1 2 113

2 3 4 5

1 0.3

1

2

1(1 ) ( ) , 0.3

(2 1)!( )

1 8 7.5 4.03 0.926 0.174 0.018

( 0.3) 1 (0.2 0.4) 1 (0.2 0.4)

1 8 7.5 4.03

( 0.3)

w d

w

dd w H w d

d dw

d w w w w w

dw j w j w

d w w

dw

C-

-- -

=

=

= - = - -

- + - + - = - - + - -

- + -= -

3 4 5

21 0.3

2 3 4 5

2 2

2 3 4

2

0.926 0.174 0.018

1 0.4 0.2

1 8 7.5 4.03 0.926 0.174 0.018( 1)( 0.4 0.4 )

( 0.3) (1 0.4 0.2 )

7.5 8.06 2.778 0.696 0.09

1 0.4 0.2

w

w

w w w

w w

w w w w w w

w w

w w w w

w w

=

+ - - + - + - + -= - - +- - +

- + - + - + - + 1 0.31

=

=

Thus

1 1 1 1

1

2

1 2 1 2

1 (0.2 0.4) 1 (0.2 0.4

1 4( )

1 0.3 (1 0.1 ( 1)

3 ))

j j

j z j z z zzH z -

-- --= + + + + +

-

+ -

-

-+ --

-

Suppose ( )H z is causal. Then, impulse response is given by8

{ }

(0.3)

(1 2)(0.2 0.4) [ ] (1 2)(

[ ]

[ ] [ 1] 2Re (1 2)(0.2 0.4) [ ] (4 5

[

)(0.3) [

] 4(

0.2 0.4)

1)(0

[ ]

.3

[ ] [ ]

) ]

]

1

[n

n

n

n

n

n

h n

n n j

j j u

j u n n u

n

u n n

j j u

n

n

n

n

u

nd d

d d= -

+ +

+ +

= - - + + + + +

+

+ + - -

-

Exercise:Write a MATLABprogram to evaluate first 10 terms of [ ]h n .

Above PFE procedure is implemented in MATLAB by the function residuez

(in signal processing toolbox)

8Alternatively, since

1

1 2

1 2 1 2 2 2

1 (0.2 0.4) 1 (0.2 0.4) 1 0.4 0.2

j j z

j w j w z z

-

- -+ - -

+ =- + - - - +

it follows (exercise) from entries 11 and 12 of the z-transform tables, that

( ) ( )

11 1

1 2

2 2

2 0.2 cos 2sin [ ], cos 0.21 0.4 0.2

n

o o o

z

n n u nz z q q q

-- -

- -

- = - = - +

It can be readily verified using MATLABthat above sequence equals { }2Re (1 2)(0.2 0.4) [ ]nj j u n+ +


29/37


If ( )H z is causal, [ ]h n can also be found by performing a long division of ( )N z

by ( )D z for as many terms as we wish

However, with above procedure, round-off errors will accumulate as nincreases

and computed [ ]h n will not be accurate for nlarge

Example 7 (revisited)

( )( )( )

1 2 3 4 5

21 1 1

1 2 3 4 5

1 2 3 4

8 7.5 4.03 0.926 0.174 0.018( )

1 (0.2 0.4) 1 (0.2 0.4) 1 0.3

8 7.5 4.03 0.926 0.174 0.018

1 0.53 0.156 0.018

z z z z z H z

j z j z z

z z z z z

z z z z

- - - - -

- - -

- - - - -

- - - -

- + - + -=

- + - - -

- + - + -=

- + - +

We demonstrate here method of finding the causal [ ]h n by long division. It is instructive to compare

and contrast long division shown below with that shown in pp. 27 to find the residues 0B and 1B

1 2 3 4 1 2 3 4 5

1 2 3 4

1 2 3 4 5

1 2 3 4 5

2

1 0.53 0.156 0.018 8 7.5 4.03 0.926 0.174 0.018

8 8 4.24 1.248 0.144

0.5 0.21 0.322 0.03 0.018

0.5 0.5 0.265 0.078 0.009

0.29 0.057

z z z z z z z z z

z z z z

z z z z z

z z z z z

z z

- - - - - - - - -

- - - -

- - - - -

- - - - -

-

- + - + - + - + -

- + - +

- + + -

- + - +

+

1 2 3

3 4 5

2 3 4 5 6

3 4 5 6

3 4

8 0.5 0.29 0.347

0.108 0.027

0.29 0.29 0.1537 0.0452 0.0052

0.347 0.0457 0.0182 0.0052

0.347 0.347

z z z

z z

z z z z z

z z z z

z z

- - -

- - -

- - - - -

- - - -

- -

+ + +

+ -

- + - +

- + -

-

Quotient gives the first 4 terms of [ ]h n , namely, { }8, 0.5, 0.29, 0.347,


30/37


Problems

1. Consider the continuous-time signal

( )( ) cos 8c t A tp=

(a) What is the (fundamental) period of ( )c t ?

(b) Suppose ( )c t is being sampled with a sampling period of1

16secT= to yield

the discrete-time sequence [ ] ( )cx n x nT= . Determine the fundamental period N of [ ]n , and comment on the value NT .

(c) Repeat Part (b) for 118

secT= and 1401

secT= .

2. Suppose { } 21

[ ] == M

n Mx x n , { } 2

1[ ] == N

n Ny y n , and { } 2

1[ ] == * = K

n Kw x y w n .

Show that 1 1 1K M N= +

and 2 2 2K M N= +

3. Given { }42

[ ] 3, 1, 2, 0, 4,1, 2 =-= - - nx n

and { }

3

1[ ] 2,1, 0, 1, 3 =-= - ny n Find the sequences

(a) 1 2w x y= -

(b) 2w x y=

(c) 3w x y= *

4. Use the properties of DTFT and the table of DTFT pairs to find the DTFT of the

sequence[ ] , 1

nx n a a= <

(This problem was solved in pp. 8 but the solution was based on the geometric series

formula Eq. (20).)


31/37


5. Consider the discrete-time signal

{ }1, 0,1, 2,1, 0,1, 2,1, 0, 1 , 3 7[ ]

0, otherwise

nx n

- - - =

Without explicitly evaluating the DTFT ( )

j

X e

q

, find

(a)0

( )qq=

jX e

(b) ( )qq p=

jX e

(c) arg ( )j

X e q

(d) ( )jX e dp

q

pq

-

(e) Determine the signal whose DTFT is ( )j

X e q-

(f) Determine the signal whose DTFT is { }Re ( )jX e q

6. Determine thez-transform, including the ROC, for each of the following sequences. Use

the properties ofz-transforms and the table ofz-transform pairs.

(a) ( )12[ ] [ ]n

n u n=

(b) ( ) ( )12[ ] [ ] [ 10]n

x n u n u n= - -

(c) [ ] 2 [ 1]nx n u n-= - - -

(d) [ ] , 1nx n a a= <

(e) [ ] , 1nx n a a= >

7. Determine the inverse z-transform for each of the following using the properties of

z-transforms, a table of z-transform pairs, and the PFE.

(a)

11

2 1 14 21 23 1

4 8

1( ) ,1

zX z zz z

-

- --= <

+ +


32/37


8. Two LTI systems, one with impulse response [ ]h n and the other with impulse response

[ ]g n , are cascaded together as shown below.

Show, using the definition of convolution, that the cascaded system can be represented

by a single system with impulse response [ ] [ ]h n g n* .

9. Are the following systems linear? Are they time-invariant?

(a) { }[ ] [ ] [ ]= x n g n x n with [ ]g n given

(b) { }[ ] [ ]=

=o

n

k nx n x k

(c) { } [ ][ ] = x nx n e

(d) { }[ ] [ ]= + x n ax n b

(e) { }[ ] [ ]= - x n x n

10. Example 3.12 of Oppenheim and Schafer, pp. 152, gives the transfer function of a

discrete-time LTI system that cannot be described a finite order LCCDE. Give other

examples.

11. When the input to an LTI system is

( )12[ ] [ ] 2 [ 1]n nx n u n u n= + - -

the output is

( ) ( )312 4[ ] 6 [ ] 6 [ ]nn

y n u n u n= -

(a) Find the transfer function ( )H z of the system. Where are its poles and zeros and

what is its ROC?

(b) Find the impulse response of the system for all n.

(c) Write the LCCDE that characterises the system.

(d) Is the system stable? Is it causal?

h[n] g[n]x[n] w[n] y[n]

h[n] *g[n]x[n] y[n]


33/37


34/37


15 Given

1

1[ ]

1 -

-= =na u n

az

Use the properties of z-transform to find

(a) ( 1) [ ]nn a u n+

(b) ( 1)( 2) [ ]nn n a u n+ +

(c) ( 1)( 2) ( 1) [ ]nn n n k a u n+ + + -


35/37


Answers

1. (a) Fundamental period is 14p

T =

(b) 4N= , pNT T= (c) For 1

18T= , 9N= and 2 pNT T= . For 1401T= , sequence is not periodic.

2. Start by considering the definition of convolution

[ ] [ ] [ ]

k

w n x k y n k

=-

= -

Next, sketch generic plots for the finite length sequences [ ]x k and [ ]y n k- and observe

what happens as n increases from - .

3. (a) { }42

2 6, 4, 3, 0, 9, 1, 4 =-- = - - - nx y

(b) { }31

2, 2, 0, 4, 3 =- = - - nx y

(c) { }7 36, 1, 3, 1, 18, 1, 3, 6, 11, 5, 6 =-* = - - - nx y

4. { }

2

2

1

1 2 cosDTFT

an

a aa q

-

+ -=

5. (a) 6

(b) 2

(c) 2q-

(d) 4p

(e) { }

3

71, 0,1, 2,1, 0,1, 2,1, 0, 1 n=-- - (f) { }

71 1 1 12 2 2 2 7

, 0, , 1, 0, 0,1, 2,1, 0, 0,1, , 0,n=-

- -

6. (a)11

2

1 121

,z

z-- >

(b)( )

( )10 101

211

2

1 91 91 12 21

1 , 0z

zz z z

-

-

- - --

= + + + >

(c)112

1 1

21

,z

z--

<


36/37


(d)2

2 1

1 1

(1 ) ( ),

a

aa a z z a z-

-

+ - + < <

(e) Transform does not exist.

7. (a) ( ) ( )1 12 44 [ 1] 3 [ ]n n

u n u n- - - - - -

(b) ( ) ( )1 12 4[ ] 4 [ ] 3 [ ]n n

x n u n u n= - - -

8. We begin by noting that [ ] [ ] [ ]

=-= -kw n h k x n k and [ ] [ ] [ ]

=-

= - ly n g l w n l .The stated result follows by substituting the first equation into the second equation.

9.

Linearity Time-Invariance

(a)

(b)

(c)

(d)

(e)

10.1 1 2 31 1

2! 3!( ) 1 , 0

- - - -= = + + + + >zH z e z z z z

11. (a)1

134

1 2 341

( ) ,z

zH z z

-

--

-= >

(b) ( ) ( )13 3

4 4[ ] [ ] 2 [ 1]

n nh n u n u n

-= - -

(c) ( )34[ ] [ 1] [ ] 2 [ 1]y n y n x n x n= - + - -

(d) System is stable and causal.

12. (a)1

1 232

1( ) , 2z

z zH z z

-

- -- -= >

(b) ( ) ( )( )2 2 15 5 2[ ] 2 [ ] [ ]nnh n u n u n= - -

(c)

( ) ( )( )2 2 1

5 5 2[ ] 2 [ 1] [ ]

nnh n u n u n= - - - - -


37/37

13. (a) Cannot be determined

(b) Cannot be determined

(c) False

(d) True

14. (a) 12

z >

(b) ( )H z is stable

(c)11

21

1

1 2( )

z

zX z

-

-

-

-=

15. (a)1 2

1

(1 )az--

(b)1 3

2

(1 )az--

(c)1

( 1)!

(1 )kk

az--

-

02 discrete time signals systems release

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