the z -transform

The Z - Transform

1

• Let 𝑧 is a complex variable and 𝑧 = 𝑟𝑒𝑗Ω. The z-Transform of an arbitrary signal 𝑥[𝑛] is defined as

(1) ---- ][)(

n

nznxzX

and the inverse z-Transform is

c

n dzzzxj

nx (2) ---- )(2

1][ 1

• The z-Transform exists when the sum in (1) converges. A necessary condition for convergence is absolute summability. Since 𝑥 𝑛 𝑧−𝑛 = |𝑥 𝑛 𝑟−𝑛| we must have,

n

nrnx |][|

• The range of r for which this condition is satisfied is termed the region of convergence (ROC) of the z-transform.

(Bilateral or 2-Sided)

[Unilateral if 𝑛: 0 ⟶ ∞ ]

)(][ : zXnxNotationz

Example 1

2 Check complete solution in class

Determine the z-transform of 𝑥 𝑛 = 𝑎𝑛𝑢 𝑛 , a is real.

For the convergence, we require that

0

1)(][)(n

n

n

nn azznuazX

0

1

n

naz

From ENG TECH 2MA3, We know that the series 1 if ,1

1

0

rr

rn

n

0

1

n

naz Converges if 𝑎𝑧−1 < 1 ⟹ 𝑧 > |𝑎|

az

z

azazzX

n

n

1

0

1

1

1)()( ROC: 𝑧 > |𝑎|

a 1

0 < 𝑎 < 1

1 a

𝑎 > 1

a -1

−1 < 𝑎 < 0

-1 a

𝑎 < −1

Right-Sided Signal

Example 2


Determine the z-transform of 𝑥 𝑛 = −𝑎𝑛𝑢 −𝑛 − 1 , a is real.

0

1

1

1

11

)(1)(

)(]1[)(

n

n

n

n

n

n

n

nn

zaaz

azznuazX

⟹ 𝑋(𝑧) converges if 𝑎−1𝑧 < 1 ⟹ 𝑧 < |𝑎|

az

z

zazazX

n

n

1

0

1

1

11)(1)( ROC: 𝑧 < |𝑎|

a 1

0 < 𝑎 < 1

1 a

𝑎 > 1

a -1

−1 < 𝑎 < 0

-1 a

𝑎 < −1

Left-Sided Signal

Example 3


Determine the z-transform of 𝑥 𝑛 = −𝑢 −𝑛 − 1 +1

2

𝑛𝑢[𝑛].

n

nn

n

n znuznuzX ][)(]1[)(21

converges if 𝑧 < 1

))(1(

)2(

1

1

1

11)(

21

23

1

21

zz

zz

zzzX

0n

nz

0

1

21 )(

n

nz converges if 1

2𝑧−1 < 1 ⟹ 𝑧 >

1

2

⟹ 𝑅𝑂𝐶: 1

2< 𝑧 < 1

; 𝑅𝑂𝐶: 1

2< 𝑧 < 1

1 0.5

0

1

21

1021

1

)()(n

n

n

n

n

nn

n

n zzzz

0

1

21

0

)(1 n

n

n

n zz

Example 4


Determine the z-transform of 𝑥 𝑛 =1

2

𝑛𝑢 𝑛 + −

1

3

𝑛𝑢[𝑛].

n

nn

n

nn znuznuzX ][)(][)()(31

21

converges if 1

2𝑧−1 < 1 ⟹ 𝑧 >

1

2

))((

)2(

1

1

1

1)(

31

21

61

1

311

21

zz

zz

zzzX

0

1

21 )(

n

nz

0

1

31 )(

n

nz converges if −1

3𝑧−1 < 1 ⟹ 𝑧 >

1

3

⟹ 𝑅𝑂𝐶: 𝑧 >1

2

; 𝑅𝑂𝐶: |𝑧| >1

2

0

31

021 )()(

n

nn

n

nn zz

0

1

31

0

1

21 )()(

n

n

n

n zz

0.5

Example 5


Determine the z-transform of 𝑥 𝑛 = −1

2

𝑛𝑢 −𝑛 + 2

1

4

𝑛𝑢[−𝑛].

)41)(21(

3)(

zzzX

; 𝑅𝑂𝐶: 𝑧 <

1

4

0.25

Example 6


Consider a sequence 𝑥 𝑛 = 5, 3, −2, 0, 4, −3 . Find X(z).

Red number represents the value at n = 0

3

2

][][)(n

n

n

n znxznxzX

3212 ]3[]2[]1[]0[]1[]2[ zxzxzxxzxzx

322 34235 zzzz

𝑋(𝑧) converges if 𝑧 ≠ 0, 𝑧 ≠ ∞

⟹ 𝑅𝑂𝐶: 0 < 𝑧 < ∞

Example 7


(a) Determine the unilateral z-transform of the signal 𝑥 𝑡 = 3sin (𝜋𝑡) sampled after every 0.05 seconds starting at 𝑡 = 0.

Replace 𝑡 = 𝑛𝑇, where 𝑇 = 0.05 ⟹ 𝑥 𝑛 = 3sin (0.05𝜋𝑛)

1|| ,1)cos(2

)sin()sin(

2

z

bzz

bzbn

z

198.1

47.0

1)cos(2

)sin(3)(

2

20

220

zz

z

zz

zzX

(b) Determine the the z-transform of 𝑥 𝑛 = 2𝑛𝑢[𝑛 − 2] using real shifting property and verify your result by definition.

]2[24]2[2][ 2)22( nununx nn

After applying real shifting property #2

2|| ,)2(

4

24)(][ 2

z

zzz

zzzXnx

z

Now use definition for verification …….

Example 8


A function 𝑥[𝑛] has the unilateral z-transform 𝑋 𝑧 =8𝑧2

4𝑧2−1 ; |𝑧| > 0.5.

(a) Find the z-transform of 𝑦[𝑛] = 𝑥[𝑛 − 2]

14

8)()(][

2

2

zzXzzYny

z Using real shifting property #2

(b) Find the z-transform of 𝑦[𝑛] = (2)𝑛𝑥[𝑛]

1

2)()(][

2

2

2

z

zXzYny zz

Using complex shifting property #4

(c) Find the z-transform of 𝑦[𝑛] = 𝑛𝑥[𝑛]

22

2

22 )14(

16

)14(

16)()(][

z

z

z

zzzX

dz

dzzYny

z Using property #5

(d) Find the z-transform of 𝑦 𝑛 = 𝑥 𝑛 ∗ 𝑥[𝑛 − 4]

22

2

2

244

)14(

64

14

8)()()(][

zz

zzzXzzXzYny

z

Using property #5 & #7

Example 9


Determine the inverse z-transform of 𝑋 𝑧 =𝑧

(𝑧−1)(𝑧−2)2 ; |𝑧| > 2.

2)2)(1(

1)(

zzz

zX

222 )2(

1

2

1

1

1

)2(21)2)(1(

1

zzzz

C

z

B

z

A

zz

Using Partial Fractions

2)2(

1

2

1

1

1)(

zzzz

zX

2)2(21)(

z

z

z

z

z

zzX

Since ROC 𝑧 > 2 is a right sided signal

][)221(][ 1 nunnx nn

2)(][ , ][

az

aznuna

az

znua

znzn

⟹ x n = {0, 0, 1, 5, … . . }

Example 10


Determine the inverse z-transform of 𝑋 𝑧 =𝑧

2𝑧2−3𝑧+1 ; 𝑧 <

1

2.

))(1(2

1)(

21

zzz

zX

21

21

21

1

1

1

1))(1(2

1

zzz

B

z

A

zzUsing Partial Fractions

21

1

1

1)(

zzz

zX

211

)(

z

z

z

zzX

Since ROC 𝑧 <1

2 is a left sided signal

]1[)(]1[][21 nununx n

⟹ x n = { … . . , 15, 7, 3, 1, 0}

]1[1)(][21 nunx n

]1[az

znua

zn

Example 11


Determine the inverse z-transform of 𝑋 𝑧 =𝑧3−𝑧2+𝑧

(𝑧−1)(𝑧−2)(𝑧−1

2)

; 1 < 𝑧 < 2.

))(2)(1(

1)(

21

2

zzz

zz

z

zX

21

21

21

2 1

2

2

1

2

21))(2)(1(

1

zzzz

C

z

B

z

A

zzz

zz


21

1

2

2

1

2)(

zzzz

zX

212

2

1

2)(

z

z

z

z

z

zzX

21

212

1 ][)(

, |z|z

znu

znSince ROC is greater than the pole 𝑧 =

1

2⟹

11

2][2

, |z|

z

znu

zSince ROC is greater than the pole 𝑧 = 1 ⟹

22

2]1[)2(2

, |z|

z

znu

znSince ROC is less than the pole 𝑧 = 2 ⟹

][)(]1[)2(2][2][21 nunununx nn

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

21 nununx nn

2 1

Example 12


Determine the inverse z-transform of 𝑋 𝑧 =𝑧3−10𝑧2−4𝑧+4

2𝑧2−2𝑧−4 ; 𝑧 < 1.


)2)(1(2

429

2

1

)422(

4410)( 2

2

23

zzz

zz

zzz

zzz

z

zX

2

3

1

2/11

21)2)(1(2

429 2

zzzz

C

z

B

z

A

zzz

zz

Using long division

2

3

1

2/11

2

1)(

zzzz

zX

2

3

1

2/11

2)(

z

z

z

zzzX , 𝑧 < 1

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

1][]1[

2

1][ nununnnx nn

𝑥 𝑛 = {… … … ,5

4,3

2, −1}

Example 13


Find the impulse response if,

𝑥 𝑛 = (−1

3)𝑛𝑢 𝑛 , 𝑦 𝑛 = 3(−1)𝑛𝑢[𝑛] + (

1

3)𝑛𝑢[𝑛].

31

31

|| ,)(][

zz

zzXnx

z

1|| ,))(1(

4

1

3)(][

31

2

31

zzz

z

z

z

z

zzYny

z

1|| ,))(1(

)(4

)(

)()(

31

31

z

zz

zz

zX

zYzH

31

31

31

31

2

1

2

1))(1(

)(4)(

zzz

B

z

A

zz

z

z

zH

31

2

1

2)(

z

z

z

zzH

][)(2][)1(2][31 nununh nn

Example 14


Find the difference equation description of an LTI system with transfer

function 𝐻 𝑧 =5𝑧+2

𝑧2+3𝑧+2.

23

25

)(

)()(

2

zz

z

zX

zYzH

21

21

231

25

)(

)(

zz

zz

zX

zY

Divide each term by 𝑧2

)(2)(5)(2)(3)( 2121 zXzzXzzYzzYzzY

Apply inverse z-transform with real shifting property #2

𝑦 𝑛 + 3𝑦 𝑛 − 1 + 2𝑦 𝑛 − 2 = 5𝑥 𝑛 − 1 + 2𝑥[𝑛 − 2]

Example 15


Consider the pole/zero diagram of a DTS as shown in Figure. It corresponds to a Z-transform H[z]. Write down all possible time functions h[n] that could have this pole/zero diagram. Choose H[z] so that it has a DC gain of 1. In each case, specify the region of convergence (ROC).

x x o o

−34

13 3

2

))((

)()(

31

43

23

zz

zCzzH Where C is any constant.

For a DC gain of 1, we need 𝐻 1 = 1 ⟹ 1 =𝐶(−

1

2)

7

4∙2

3

⟹ C = −7

3

31

3998

43

1363

31

43

31

43

23

37

))((

)()(

zzz

B

z

A

zz

z

z

zH

31

3998

43

1363

)(

z

z

z

zzH

Possible time functions are:

𝑖 ℎ 𝑛 = −63

13−

3

4

𝑛𝑢 𝑛 + 98

39

1

3

𝑛𝑢 𝑛 for ROC: 𝑧 >

3

4

𝑖𝑖 ℎ 𝑛 = 63

13−

3

4

𝑛𝑢 −𝑛 − 1 + 98

39

1

3

𝑛𝑢 𝑛 for

1

3< 𝑧 <

3

4

𝑖𝑖𝑖 ℎ 𝑛 = 63

13−

3

4

𝑛𝑢 −𝑛 − 1 − 98

39

1

3

𝑛𝑢 −𝑛 − 1 for 𝑧 <

1

3

Example 16


Solve 𝑦 𝑛 + 2 − 𝑦 𝑛 + 1 − 2𝑦 𝑛 = 0 with 𝑦 0 = 0, 𝑦 1 = 1.

zyzyzYzzyyzYznyz

]1[]0[)(]]1[]0[)([]2[ 2212

zyzzYyzYznyz

]0[)(]]0[)([]1[

Using Property #3 0

0

Substitute in to given difference equation

0)(2)()(2 zYzzYzzYz

2)(

2

zz

zzY

1212)1)(2(

1)( 31

31

zzz

B

z

A

zzz

zY

12)( 3

131

z

z

z

zzY

][)1(2][31 nuny nn

Example 17


Solve 𝑦 𝑛 + 2 − 6𝑦 𝑛 + 1 − 55𝑦 𝑛 = −(−3)𝑛 with 𝑦 0 = 0, 𝑦 1 = 0.

zyzyzYzzyyzYznyz

]1[]0[)(]]1[]0[)([]2[ 2212

zyzzYyzYznyz

]0[)(]]0[)([]1[

Using Property #3 0

0

Substitute in to given difference equation

3)(55)(6)(2

z

zzYzzYzYz

3)()556( 2

z

zzYzzz

3511)3)(5)(11(

1)(

z

C

z

B

z

A

zzzz

zY

][)3()5()11(][281

321

2241 nuny nnn

0

3][)3(

z

znu

zn

3511

)( 281

321

2241

zzzz

zY

3511)( 28

1321

2241

z

z

z

z

z

zzY

the z -transform

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