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z transform

1. Copyright © 2018 Andreas Antoniou
Victoria, BC, Canada
Email: [email protected]

August 5, 2018

“Dr Dr <E>«E>r E HAG

2. m= The most fundamental method for the inversion of a z
transform is the general inversion method which is based on
the Laurent theorem.

3. Z-Transform Inversion Techniques

= The most fundamental method for the inversion of a z
transform is the general inversion method which is based on
the Laurent theorem.

= In this method, the inverse of a z transform X(z) is given by

x(nT) = 5 e dz

where [ is a closed contour in the counterclockwise sense
enclosing all the singularities of function X(z)z"-1.

Frame # 2 Slide # 3 A. Antoniou Digital Filters — Sec. 4.8

4. ...

x(nT) = re dz

= At first sight, the above contour integration may appear to be
a formidable task.

«Dr Dr <E>«E>r 2 HAG

5. Z-Transform Inversion Techniques Cont'd

x(nT) = 5 px dz

= At first sight, the above contour integration may appear to be
a formidable task.

m However, for most DSP applications, the z transform turns
out to be a rational function and for such functions the
contour integral can be easily evaluated by using the residue
theorem.

Frame #3 Slide # 5 A. Antoniou Digital Filters - Sec. 4.8

6. Z-Transform Inversion Techniques Cont'd

m According to the residue theorem,

pP
x(nT) = En fx dz = y res 2-sp; [X(z)z” ]

27) Jr i=1

where res ¿+p, [X(z)z”!] and P are the residue of pole p; and the

number of poles of X(z)z”!, respectively.

Frame # 4 Slide # 6 A. Antoniou Digital Filters - Sec. 4.8

7. Z-Transform Inversion Techniques Cont'd

m According to the residue theorem,

pP
1 n— n=
on d xx ldz= y res 2 4p; [X(z)z al

r i=1

x(nT) =

where res ¿+p, [X(z)z”!] and P are the residue of pole p; and the
number of poles of X(z)z”!, respectively.

= For a pole of order m;,

n-i 1 de m; n-1
re zap, [XI] = as [ER X 2]

Frame #4 Slide # 7 A. Antoniou Digital Filters - Sec. 4.8

8. Z-Transform Inversion Techniques Cont'd

m According to the residue theorem,

pP
En fx dz = y res 2-sp; [X(z)z” ]

x(nT) = -
(nT) onl À 2

where res zp, [X(z)z
number of poles of X(z)z”

| and P are the residue of pole p; and the

1, respectively.
= For a pole of order m;,

ny 1 gmt

res ¿=p, [X(z)z” '] = mous nt

[(z - p)"X(2)z""]

= For a simple pole,

res ¿=p; [X(2)z"71] [(z - pi)X(2)2"*]

= lim
Zp;

Frame # 4 Slide # 8 A. Antoniou Digital Filters — Sec. 4.8

9. Example — General Inversion Method

Using the general inversion method, find the inverse z transform of

1

Md = E

Solution We note that the factor z”~! introduces a pole in
X(z)z"~1 at the origin for the case n= 0, which must be taken
into account in the evaluation of x(0).

Note: For n > 0, the pole at the origin disappears.

Frame #5 Slide #9 A. Antoniou Digital Filters - Sec. 4.8

10. Example Cont'd

Thus for n = 0, we have

zei

2(z — 1)(2 + 3)

1

X(2)2""|,_¿= "DEF

n=0
1 1
Hence x(0) = —————_——— + ==
Lu 2(z-1)(z+3) +0 2z(z +3) a
1 Lal
+ zen; 1+1+3=0

Actually, this follows from the initial-value theorem (Theorem 3.8)
without any calculations.

Frame #6 Slide # 10 A. Antoniou Digital Filters - Sec. 4.8

11. Example Cont'd

For n>0

x(nT)

z=1

and from the initial-value theorem, x(nT) = 0 for n < 0.

Therefore, for any value of n, we have

x(nT) = u(nT — T)[3-3(-3)""] =

Frame #7 Slide # 11 A. Antoniou Digital Filters - Sec. 4.8

12. Example — General Inversion Method

Using the general inversion method, find the inverse z transform of

(2z — 1)z
x NE

D = 3 - 1)(z +3)
Solution We can write

(2z — 1)z : z"t (2z — 1)z”

DO PR NEE
RT DE AY

We note that X(z)z”~1 has simple poles at z= 1 and —3.

Furthermore, the zero in X(z) at the origin cancels the pole at the
origin introduced by z"-1 for the case n= 0.

Frame # 8 Slide # 12 A. Antoniou Digital Filters — Sec. 4.8

13. Example Cont'd

(2z — 1)z”
2(z-1)(z+4)

Hence for any n > 0, the general inversion formula gives

X(z)z"! =

x(nT) = res ¿1 [X(2)2"*] + res 1 [X(2)2"7*]
_ (2z-1)z” _ (2z -1)z”
2243) |,, Az-Dl-1

Frame #9 Slide # 13 A. Antoniou Digital Filters - Sec. 4.8

14. Example Cont'd

Since the numerator degree in X(z) does not exceed the
denominator degree, it follows that x(n7) is a right-sided signal,
i.e., x(nT) = 0 for n < 0, according to the Corollary of Theorem
3.8.

Therefore, for any value of n, we have
x(nT) = u(nT) [3 +3 (-3)"] =

where u(nT) is the unit-step function.

Frame #10 Slide # 14 A. Antoniou Digital Filters - Sec. 4.8

15. Z-Transform Inversion Techniques Cont'd

Since
— the z transform is a particular type of Laurent series, and
— the Laurent series in a given annulus of convergence is unique

it follows that any technique that can be used to generate a power
series for X(z) that converges in the outermost annulus of
convergence can be used to obtain the inverse z transform.

Frame # 11 Slide # 15 A. Antoniou Digital Filters - Sec. 4.8

16. Z-Transform Inversion Techniques Cont'd
Consequently, several inversion techniques are available, as follows:

— using the binomial theorem,

— using the convolution theorem,

— performing long division,

— using the initial-value theorem, or

— expanding X(z) into partial fractions.

Some of these techniques are illustrated by examples in the next
few slides.

Frame #12 Slide # 16 A. Antoniou Digital Filters - Sec. 4.8

17. Example — Binomial Theorem

Using the binomial method, find the inverse z transform of
kz”
(2 — w)*

where m and k are integers, and K and w are constants, possibly
complex.

X(z) =

Solution The inverse z transform can be obtained by obtaining a

binomial series for X(z) that converges in the outside annulus of
X(z).

Frame #13 Slide # 17 A. Antoniou Digital Filters — Sec. 4.8

18. Example Cont'd

Such a binomial series can be obtained by expressing X(z) as

X(z) = Kz™ *f1 + (=w22)]7*

De) ce (Je
(here |

where ( ‘) (HACK... (k= 0 +1)

Hence

u ... n + w)"z~? m—k
X(z) = y Ku(nT)$ k)( k 1) ( k 1)( ) ij

nl
n=—00

Frame #14 Slide # 18 A. Antoniou Digital Filters — Sec. 4.8

19. Example Cont'd

xt) = Y Kun) CRICK-N (Ck + ew)

nl
n=—00

Now if we let n= n' + m—k and then replace n' by n, we have

oo

X(z)= Y [Kulln+m-k)T]

n=—00

(-k)(-k = 1). (-n-m+1)(-w)rtmk)
mee SP

Frame #15 Slide # 19 A. Antoniou Digital Filters - Sec. 4.8

20. Example Cont'd

X(z)= Y) {Kulln+m- MT]
(—k)(—k — 1)... (-n-m+1)(-w)m _,
* (n+ m-—k)! \z

Hence the coefficient of z~” is obtained as

Kz"
x(nT) = Z-1 | ———_.
en = 29 |

= Ku[(n + m- k)T] (n+m—k)!

By assigning different values to constants k, K, and m a variety of
z-transform pairs can be deduced as shown in the next slide.

Frame # 16 Slide # 20 A. Antoniou Digital Filters - Sec. 4.8

21. Example Cont'd

Frame # 17 Slide # 21

x(nT) X(z)
=.
y
u(nT) ag
kz UD)
u(nT — kT)K =—
u(nT)Kw" EA
EN
unt — kT) Kwa | Kel
zw
T —anT 2
u(nT e Ze et
TZ
r(nT) Ey 17
u Te-aTz
r(nT)e="T Em

A. Antoniou

Digital Filters — Sec. 4.8

22. E From the real-convolution theorem

Z $ x(kT)o(nT — kT) = X1(z)X2(z)

k=-00

<O> Dr <E>«E>r E HAG

23. Use of Real Convolution

= From the real-convolution theorem

Z Y x(kT)m(nT — kT) = X1(2)X2(2)

k=—00

u If we take the inverse z transform of both sides, we get

ES

Y x(«T)(nT = kT) = ZX (2)X2(2)]
k=-oo
or oo
ZX (z)X2(z)] = y xa(kT)o(nT — kT)
k=—00

Thus, if a z transform can be expressed as a product of two z
transforms whose inverses are available, then performing the
convolution summation will yield the desired inverse.

Frame #18 Slide # 23 A. Antoniou Digital Filters - Sec. 4.8

24. Example — Real Convolution

Find the inverse z transform of
Z
X I
(2) = Gap

Solution We note that

X3(z) = X1(2)X2(2)

where

Z
X1(z) = Z-1 and Xo(z) =

—1

Frame #19 Slide # 24 A. Antoniou Digital Filters - Sec. 4.8

25. Example Cont'd

z 1
Xi(z) = = and X2(z) = 1

From the table of standard z transforms, we can write
xi(nT)=u(nT) and x2(nT) = u(nT — T)

Hence for n > 0, the real convolution yields

x3(nT) = 5 ureter ir) = Y u(kT)u(nT — T — kT)
k=—00 k=—00
k=-1 k=0 k=1

SSS |S
= +++ +u(-T)u(nT)+ u(O)u(nT — T)+u(T)u(nT — 2T)+---

k=n- k=n
AR pe en
+u(nT — T)u(0) + u(nT)u(—T) +--+
=0+1+1+..+1+0=n

Frame # 20 Slide # 25 A. Antoniou Digital Filters — Sec. 4.8

26. Example Cont'd

For n < 0, we have
x3(nT) = Y u(kT)u(nT — T ~kT)
k=—00
k=-1 k=0 k=1

Ps E SES
= ++ +u(-T)u(nT)+ u(O)u(nT — T)+u(T)u(nT — 2T) +---

k=n-1 k=n

SS eS
+u(nT — T)u(0)+u(nT)u(-T)+---
and since all the terms are zero, we get
x(nT) =0

(This result also follows from the initial value theorem.)

Frame # 21 Slide # 26 A. Antoniou Digital Filters - Sec. 4.8

27. Summarizing, for n > 0,
x(nT)=n

and forn <0,
x3(nT) =0

Therefore, for any value of n, we have

x3(nT)=u(nT)n m

“Dr Dr <E>«E>r E HAG

28. Example — Real Convolution

Using the real-convolution theorem, find the inverse z transforms of

zZ

=p

Solution For this example, we can write

1
and Xa(z) = —

Az) = z-1

z
(2-17?
and from the previous example, we have

xi(nT) = u(nT)n and x2(nT)= u(nT — T)

Frame # 23 Slide # 28 A. Antoniou Digital Filters - Sec. 4.8

29. Example Cont'd

From the initial value theorem, for n < 0, we have
x3(n) = 0

For n > 0, the convolution summation gives
co
Y ku(kT)u(nT — T —kT)

k=-oo

x3(nT)

k=0 k=1

re, PA o,
+0: [u(nT — T)]+1-[u(nT -2T)]+--
k=n-1 k=n

——— NS
+(n-— 1)u(0) + nu(—T)
+0+1+4+2+---+n—-14+0

n-1
Sok
k=1

Frame # 24 Slide # 29 A. Antoniou Digital Filters - Sec. 4.8

ll

Il

30. m A closed-form solution can be obtained by using an old trick
of algebra.

“Dr Dr <E>«E>r E HAG

31. Example Cont'd

= A closed-form solution can be obtained by using an old trick
of algebra.

m The story goes that Gauss’ mathematics teacher had
something to attend to and wanted to keep his class busy. So
he asked the class to find the sum:

1+2+3+...99

Frame # 25 Slide # 31 A. Antoniou Digital Filters — Sec. 4.8

32. Example Cont'd

= A closed-form solution can be obtained by using an old trick
of algebra.

m The story goes that Gauss’ mathematics teacher had
something to attend to and wanted to keep his class busy. So
he asked the class to find the sum:

1+2+3+...99

m As the teacher was getting ready to leave, Gauss shouted out
“Sir, the answer is 4950!”

Frame # 25 Slide # 32 A. Antoniou Digital Filters - Sec. 4.8

33. Example Cont'd

Frame # 25

A closed-form solution can be obtained by using an old trick
of algebra.

The story goes that Gauss’ mathematics teacher had
something to attend to and wanted to keep his class busy. So
he asked the class to find the sum:

1+2+3+...99

As the teacher was getting ready to leave, Gauss shouted out
“Sir, the answer is 4950!”

“It's very simple, Sir, twice the sum is 100 x 99”.

Slide # 33 A. Antoniou Digital Filters — Sec. 4.8

34. Example Cont'd

Gauss’ reasoning was as follows:

1 2 3 n-1
n-1 n-2 n-3 ++. 1
n n n n
That is,
n—1
k=&n(n-1)
k=1

Using this result, x3(nT) can be obtained as

x3(nT) = ai zu(nT)n(n—1) =

Frame # 26 Slide # 34 A. Antoniou Digital Filters - Sec. 4.8

35. Use of Long Division

m Given a z transform X(z) = N(z)/D(z), a series that
converges in the outermost annulus of X(z) can be readily
obtained by arranging the numerator and denominator
polynomials in descending powers of z and then performing
polynomial division also known as long division.

Frame # 27 Slide # 35 A. Antoniou Digital Filters - Sec. 4.8

36. Example — Long Division

Using long division, find the inverse z transform of
4 + dz 12 123 + 22° + 2°
-141,- 22 3
3432-27 +2

X(z)

Solution The numerator and denominator polynomials can be
arranged in descending powers of z as

5 4_7,3_1,2 1, 1
Z° +22" — 32 32 4 523

3_7241,_1
zZ o

X(z)

Frame # 28 Slide # 36 A. Antoniou Digital Filters - Sec. 4.8

37. Example Cont'd

2 1 1
e

Frame # 29 Slide # 37

243zt1l+z?t+z 743274
2° +4224 - 17? 1224 l2-1
5 41,534 1,2
2’ iz 37 32
32) - 42-32 +32 i
F32* +32 + 377 4 32
2 -22+5z-1
3 ui. gi
Fz tz +F32+;
z

1,-1.1,
z+1F3z 32

3,-1 1,-3
32 + 32
A. Antoniou

Digital Filters — Sec. 4.8

38. Therefore,

X(z) = 2243241422 42734 3244... m

x(-2T)=1, x(-T)=3, x(0)=1, x(T)=0

x(2T)=1, x(3T)=1, x(4T)= 3, gem

«Dr Dr <E>«E>r 2 HAG

39. Use of Long Division Cont'd

= As illustrated by the previous example, the long-division
approach readily yields any nonzero values of the signal for
n <0 but does not yield a closed-form solution.

Frame # 31 Slide # 39 A. Antoniou Digital Filters - Sec. 4.8

40. Use of Long Division Cont'd

= As illustrated by the previous example, the long-division
approach readily yields any nonzero values of the signal for
n <0 but does not yield a closed-form solution.

= On the other hand, the general-inversion method yields a
closed-form solution but presents certain difficulties in z
transforms of two-sided signals because such z transforms
have a higher-order pole at the origin whose residue is difficult
to obtain.

Frame # 31 Slide # 40 A. Antoniou Digital Filters - Sec. 4.8

41. Use of Long Division Cont'd

= As illustrated by the previous example, the long-division
approach readily yields any nonzero values of the signal for
n <0 but does not yield a closed-form solution.

= On the other hand, the general-inversion method yields a
closed-form solution but presents certain difficulties in z
transforms of two-sided signals because such z transforms
have a higher-order pole at the origin whose residue is difficult
to obtain.

m The inverses of such z transforms can be easily obtained in
closed form by finding the values of the signal for n < 0 using
long division and then applying the general inversion method
to the remainder of the long division.

Frame #31 Slide # 41 A. Antoniou Digital Filters - Sec. 4.8

42. m Consider a z transform whose numerator degree exceeds the
denominator degree of the form

_NG) _ ey az’
Diz) Do bizN-i

X(z)

43. Use of Long Division Cont'd

= Consider a z transform whose numerator degree exceeds the
denominator degree of the form
M La
M2) _ Dino az
D(z) NE b;zN-i

X(z)

= The first nonzero value of x(nT) occurs at n = (N — M)T
according to the initial value theorem.

Frame # 32 Slide # 43 A. Antoniou Digital Filters - Sec. 4.8

44. Use of Long Division Cont'd

u Performing long division until the signal values x[(N — M)T],
x[(N — M+1)T], ..., x(0) are obtained, X(z) can be expressed as

X(2)= D = Q(z) + R(z)

where
Q(z) = x[(N- M)T]zM=M 4 x[(N— M41) T](W-N=D +. .-+x(0)

is the quotient polynomial and

is the remainder whose numerator degree is less than the
denominator degree.

Frame # 33 Slide # 44 A. Antoniou Digital Filters - Sec. 4.8

45. Use of Long Division Cont'd

x= Bey = Q(z)+R(z) where R(z) = HE
m Hence
x(nT) = 2 Q(z) + ze
= x{(N— M)T]z\—-™) + xf(N— M4 1) T] UND 4...
+x(0) + 271 xa

Since Z-1N@) represents a right-sided signal, it can be easily
D(z) 8!

evaluated in closed-form by using the general inversion method.

Frame # 34 Slide # 45 A. Antoniou Digital Filters - Sec. 4.8

46. Example — Long Division and General Inversion
Method

Using long division along with the general inversion method, obtain
a closed-form solution for the inverse z transform of

X(z)

Frame #35 Slide # 46 A. Antoniou Digital Filters - Sec. 4.8

47. Example Cont'd

Solution

Hence

Frame # 36 Slide # 47 A. Antoniou Digital Filters - Sec. 4.8

48. Example Cont'd

Applying the inverse z transform, we have
x(nT) = 21 (22432414 =>
en ( z—z24+4z-}
x(-2T)z? + x(-T)z + x(0) + Z7*R(z)
where x(—2T) = 1, x(—T) = 3, x(0) = 1, and

The inverse z transform of R(z) can now be obtained by using the
general inversion method.

Frame # 37 Slide # 48 A. Antoniou Digital Filters - Sec. 4.8

49. Example Cont'd

z z
Bo 2+ 52-4 (z-1)(z+ j3)(z - 43)

R(z)

Since —j$ = $e~/"/2, the residues of R(z)z"-! can be obtained as

z 1 N
Rı = lim =:
zal(244) 142 *
n 1 ea dan /2
Ra lim 2 (3)"
O (CH
1 — jn7
_ 2 (3)" eee 2 (4)? elor/2+r un")
~ V5 ellr=tan=2) ~~ V5 \2

Rs = Ri = 2 (4)" eilnn/2+r-tan”'2)

Frame # 38 Slide # 49 A. Antoniou Digital Filters - Sec. 4.8

50. Example Cont'd

Thus for n > 1, we have

R(z) =) R, + Ro + R3
_ $+4(3)"3 Bez

Hence
r(nT) = &u(nT) + 7 (3)" cos(nn/2 +7 — tan?2)

Since x(—2T) = 1, x(—T) = 3, and x(0) = 1, the value of x(nT)
for any value of n is given by

x(nT) = ö(nT +2T) + 30(nT + T)+ö(nT)
+u(nT - T)[é + 4g (4)" cos(nr/2+7—tan*2)] m

Frame # 39 Slide # 50 A. Antoniou Digital Filters - Sec. 4.8

51. Use of Partial Fractions

m If the degree of the numerator polynomial in X(z) is equal to
or less than the degree of the denominator polynomial and the
poles are simple, the inverse of X(z) can very quickly be
obtained through the use of partial fractions.

Frame # 40 Slide # 51 A. Antoniou Digital Filters - Sec. 4.8

52. Use of Partial Fractions

m If the degree of the numerator polynomial in X(z) is equal to
or less than the degree of the denominator polynomial and the
poles are simple, the inverse of X(z) can very quickly be
obtained through the use of partial fractions.

= Two techniques are available, as detailed next.

Frame # 40 Slide # 52 A. Antoniou Digital Filters - Sec. 4.8

53. Use of Partial Fractions, Technique |

= The function X(z)/z can be expanded into partial fractions as

Z— D:
j=1 Pi

where P is the number of poles in X(z) and

Ro= lim X(z) Rj = res 2p, ES
z>0 Z

Frame # 41 Slide # 53 A. Antoniou Digital Filters - Sec. 4.8

54. Use of Partial Fractions, Technique |

= The function X(z)/z can be expanded into partial fractions as

Z— D:
j=1 Pi

where P is the number of poles in X(z) and

Ro= lim X(z) Rj = res 2p, ES
z>0 Z

m Hence
R:

iZ
Z—Pi

P
X(2) = Ro+ Y
i=1

Frame # 41 Slide # 54 A. Antoniou Digital Filters — Sec. 4.8

55. Use of Partial Fractions, Technique | Cont'd

P
x
= Ro + —
232
i=1
= Therefore,

P
x(nT) = 21 (me + =

and from the table of standard z transforms, we get

.-)- a Riz

Z—Pi

P
x(nT) = Rod(nT) + y u(nT)Rip?
i=1

Frame # 42 Slide # 55 A. Antoniou Digital Filters - Sec. 4.8

56. Example — Partial Fractions Method

Using Technique I, find the inverse z transform of

Zz

x) = 2+2+1

Solution On expanding X(z)/z into partial fractions, we get

Xz) 1. _ 1 Rio, R
z 24243 (z — p1)(z — p2) z-pı z-m
PFELYZ) e 37/4
where Pi V5 and p2 = V3
Thus we obtain
X X
Ri = res ¿=p, | 2) =-j and Ry=res ¿-p, | a] =j
z

Note: Complex conjugate poles give complex conjugate residues.

Frame #43 Slide # 56 A. Antoniou Digital Filters - Sec. 4.8

57. From the table of z transforms, we can now obtain

x(nT) = u(nT)( — jpi + jp?)
n 1,; en —j3mn
= Pam im)

=2 aya u(nT) sin om =

58. w An alternative approach is to expand x (z) into partial fractions

R;
as 693

where Ro = ‚im X(z) Rj = res ¿=p,X(2)

and P is the number of poles in X(z).

59. Use of Partial Fractions, Technique Il

m An alternative approach is to expand a into partial fractions

as X(z) =

where Ro = Jim X(z) Ry = res 2=p,X(z)
and P is the number of poles in X(z).

= Thus

x(nT)

P R;
sr]

i=1

= R;
-1 1 2
ZRH DE I}

Frame #45 Slide # 59 A. Antoniou Digital Filters - Sec. 4.8

60. ...

1 a Ri
x(nT) = Z*Ro +) 23

z-p
j=1 Pi

m Therefore, from Table 3.2, we obtain

P
X(nT) = Roë(nT)+ SU u(nT — T) Rip?
i=l

“Dr Dr <E>«E>r E HAG

61. Example — Partial Fractions Method

Using Technique Il, find the inverse z transform of

Solution X(z) can be expressed as

X(z) =

where

Ro = lim X(z) = lim
Z—00 Z—00 (z 3) (z 1) zZ700 Z

Frame # 47 Slide # 61 A. Antoniou Digital Filters - Sec. 4.8

62. Example Cont'd

1
Ro = lim X(z) = lim LG = lim = =0
Zz 00 z—00 (z _ 3) (z- 1) zZ 00 Z
R res X(z) 2 2
= 1X(z) = =
22 =D
T2
Ra = res __1X(z) = 1
d 2 = res _1X(z) = =-
an 2=3 (z- 3) hal
2 =
Hence AD) Fam

and from Table 3.2
x(nT) = 4u(nT — 7) [(3)"-(2)"] =

A. Antoniou Digital Filters — Sec. 4.8

Frame # 48 Slide # 62

63. = The partial fraction method is based on the assumption that the
denominator degree of the z transform is equal to or greater than
the numerator degree.

64. Use of Partial Fractions Cont'd

= The partial fraction method is based on the assumption that the
denominator degree of the z transform is equal to or greater than
the numerator degree.

@ If this is not the case, then through long division the z transform
can be expressed as

x
a
N
—
Il
Il

Q(z) + R(z)

where
Q(z) = x[((N— M)T]z\—) + x[(N— M+1)T]M-N=D 4...4x(0)
is the quotient polynomial and
_N@
R(z) = Diz)

is the remainder polynomial whose denominator degree is greater
than the numerator degree.

Frame # 49 Slide # 64 A. Antoniou Digital Filters - Sec. 4.8

65. Important Notes

m Given a z transform X(z), a partial fraction expansion can be
obtained through the following steps:

— represent the residues by variables,
— generate a system of simultaneous equations, and then
— solve the system of equations for the residues.

Frame #50 Slide # 65 A. Antoniou Digital Filters - Sec. 4.8

66. Important Notes Cont'd

m For example, if

2-2
2) = 16 3) (A)
we can write
Ri Ra
X(z) = Ro + + TS
Ro(z = 1)(z = 2) + Riz — 2R1 + Roz — Ra
(z-1)(z-2)
Ro(z? — 3z + 2) + Riz — 2R + Roz — Ro
@-1)@=2)
Roz? — 3Roz + 2Ro + Riz — 2R1 + Roz — Ro
(z-1)(z-2)
Roz? + (Ry + Ro — 3Ro)z + 2Ry — 2Rı — Ro (8)
(2 - 1)(z -2)

Frame # 51 Slide # 66 A. Antoniou Digital Filters - Sec. 4.8

67. Important Notes Cont'd

m By equating equal powers of z in Eqs. (A) and (B), we get
z?: Ro=1
zi: R + R — 3R5 = 0
2° : 2Ro — 2R, — Ro = -2

Frame # 52 Slide # 67 A. Antoniou Digital Filters - Sec. 4.8

68. Important Notes Cont'd

m By equating equal powers of z in Eqs. (A) and (B), we get
z?: Ro=1
zi: R + R — 3R5 = 0
2° : 2Ro — 2R, — Ro = -2

= Solving this system of equations would give the correct
solution as

Frame # 52 Slide # 68 A. Antoniou Digital Filters - Sec. 4.8

69. Important Notes Cont'd

= For a z transform with six poles, a set of 6 simultaneous
equations with 6 unknowns would need to be solved.

Frame # 53 Slide # 69 A. Antoniou Digital Filters - Sec. 4.8

70. Important Notes Cont'd

= For a z transform with six poles, a set of 6 simultaneous
equations with 6 unknowns would need to be solved.

= Obviously, this is a very inefficient method and it should
definitely be avoided.

Frame # 53 Slide # 70 A. Antoniou Digital Filters - Sec. 4.8

71. Important Notes Cont'd

= The quick solution for this example is easily obtained by
evaluating the residues individually, as follows:

Z7—2 2-2
Ro 1, Ri
(z - 1)(2 - 22-00 (2-2) 7-1
z2-2
Ra = =2
(z-1)|,25

Frame #54 Slide # 71 A. Antoniou Digital Filters - Sec. 4.8

72. Important Notes Cont'd

= The quick solution for this example is easily obtained by
evaluating the residues individually, as follows:

Z7—2 2-2
Ro 1, Ri
(z - 1)(2 - 22-00 (2-27
2_
ui | =)
(z-1)l,->
m Hence
z2—2 R Ro
x R +
(2) (z —1)(z - 2) zT z-2
=1+ 1 + 2
7 z-1 z-2

Frame # 54 Slide # 72 A. Antoniou Digital Filters - Sec. 4.8

73. E In the partial-fraction method, constant Ro must always be included
although it may sometimes be found to be zero.

74. Important Notes Cont'd

= In the partial-fraction method, constant Ro must always be included
although it may sometimes be found to be zero.

u For example, if Ro were omitted in the partial-fraction expansion

2-2 i Ri R
E-De-2 ™*z-i*

-1 z-2
then the right-hand side would assume the form

X(z)

Rı + R (Ri + Ro)z — (2Rı + Ro)
z-1 z-2 (z -1)(z- 2)

which cannot represent the given function whatever the values of Ri
and Ro!

Frame #55 Slide # 74 A. Antoniou Digital Filters - Sec. 4.8

75. Important Notes Cont'd

m By the way, you can always check your work by combining the
partial fractions back into a function, as you can check a
division by multiplying.

Frame #56 Slide # 75 A. Antoniou Digital Filters - Sec. 4.8

76. This slide concludes the presentation.
Thank you for your attention.