M. J. Roberts - 8/17/04

11-1

Chapter 11 - The z Transform

Selected Solutions1. Using the definition of the z transform and/or the transform pairs,

α

α ααn n

z

z zzu ,[ ] ← →

−=

−>−

Z 1

1 1

and

sin u

sin

cos

sinΩ

ΩΩ

Ω0

02

02 1n n

z

z z( ) [ ] ← →

( )− ( ) +

=Z 001

01 21 2

1( )

− ( ) +>

−

− −

z

z zz

cos,

Ω ,

find the z transforms of these DT signals.

(a) x un n[ ] = [ ]

Using α

α ααn n

z

z zzu ,[ ] ← →

−=

−>−

Z 1

1 1, let α =1. Then

u ,nz

z zz[ ] ← →

−=

−>−

Z

1

1

11

1

(b) x un e nn[ ] = [ ]−10

(c) x sin un e n nn[ ] = ( ) [ ]

x u un ee e

jn

e e

jnn

jn jn j n j n

[ ] =− [ ] =

− [ ]− +( ) −( )

2 2

1 1

(d) x n n[ ] = [ ]δ

2. Sketch the region of convergence (if it exists) in the z plane, of the bilateral z transformof these DT signals.

(a) x u un n n[ ] = [ ] + −[ ]

x u u un n n z n zn

n

n

n

[ ] ← → [ ] + −[ ]( ) = [ ]−

=−∞

∞−

=−∞

∞

∑ ∑Z ++ −[ ] −

=−∞

∞

∑ u n z n

n

x n z z z zn

n

n

n

n

n

n

n

[ ] ← → + = +−

=

∞−

=−∞

−

=

∞

=

∞

∑ ∑ ∑ ∑Z

0

0

0 0

M. J. Roberts - 8/17/04

11-2

zz

zn

n

−

=

∞

−∑ =−

>0

1

11

1, and zz

zn

n =

∞

∑ =−

<0

11

1,

The two regions of convergence do not overlap. Therefore the bilateral transformdoes not exist.

(b) x u un n n[ ] = [ ] − −[ ]10

u un n

z z z z z z z z z[ ] − −[ ] ← →+ + + + + + + + +

1019 8 7 6 5 4 3 2

Z

zz9

3. Using the time-shifting property, find the z transforms of these signals.

(a) x un n[ ] = −[ ]5

(b) x un n[ ] = +[ ]2

Using g G g ,n n z z m z nn m

m

n

+[ ] ← → ( ) − [ ]

−

=

−

∑00

1

00

0Z >> 0

u un z

z

zm z z

z

zm

m

+[ ] ← →−

− [ ]

=−

−−

=∑2

1 12

0

12Z 11

111

2

+( )

=−

− −

−z zz

zz

u ,n zz z z z

z

z

zz+[ ] ← →

− −( ) − −( )−

=−

>21 1

1 11

2Z

Which is the same as the z transform of the unit step, as it should be.

(c) x un nn

[ ] =

+[ ]2

32

4. Draw system diagrams for these transfer functions.

(a) H zz

z( ) =

+

2

12

Y X Yz z zz

z( ) = ( ) − ( )−1

2

This system is non-causal and cannot be built as a real-time system. But itcan be diagrammed for use as an “off-line” system in which all the past andfuture excitation values are available.

M. J. Roberts - 8/17/04

11-3

Y(z)X(z) z

z-112

This is not a unique solution. Many other correct diagrams could be drawn.

(b) H zz

z z( ) =

+ +2 1

5. Using the change-of-scale property, find the z transform of

x sin cos unn n

n[ ] =

[ ]2

322

8π π

.

x cos une e

j

nn

jn

jn

[ ] =−

[ ]

−2

32

2

32

22

8

π π

π

The algebra is in this exercise is straightforward , but long and tedious.

sin cos u

.2

32

2

8

0 1379π πn nn z

z

[ ] ← →Z

22

4 3 2

0 3827 0 1379

2 7741 3 8478 2 774

− +− + −

. .

. . .

z

z z z 11 1z +

6. Using the z-domain-differentiation property find the z transform of

x un n nn

[ ] =

[ ]5

8 .

7. Using the convolution property, find the z transforms of these signals.

(a) x . u un n nn[ ] = ( ) [ ] ∗ [ ]0 9

(b) x . u . un n nn n[ ] = ( ) [ ] ∗ ( ) [ ]0 9 0 6

8. Using the differencing property and the z transform of the unit sequence, find the ztransform of the DT unit impulse and verify your result by checking the z-transformtable.

9. Find the z transform ofx u un n n[ ] = [ ] − −[ ]10

and, using that result and the differencing property, find the z transform of

M. J. Roberts - 8/17/04

11-4

x n n n[ ] = [ ] − −[ ]δ δ 10 .

Compare this result with the z transform found directly by applying the time-shiftingproperty to a DT impulse.

u un n

z

zz

z

z[ ] − −[ ] ← →

−−

−−10

1 110Z

δ δn n

z

zz

z

zz

z

zz

z[ ] − −[ ] ← →−

−−

−−

−− − −101 1 1

1 10 11Z

zz −

1

Simplify.

10. Using the accumulation property, find the z transforms of these signals.

(a) x rampn n[ ] = [ ]

Use x un mm

n

[ ] = −[ ]=

∑ 10

and u n

z−[ ] ← →

−1

1

1Z

(b) x u un m mm

n

[ ] = [ ] − −[ ]( )=

∑ 50

11. Using the final-value theorem, find the final value of functions that are the inverse ztransforms of these functions (if the theorem applies).

(a) X zz

z( ) =

−1

The inverse transform is the unit sequence and its limit exists as t approaches infinity.Applying the final value theorem,

lim g lim Gn z

n z z→∞ →

[ ] = −( ) ( )1

1 ,

we get

lim x limn z

n zz

z→∞ →[ ] = −( )

−=

11

11

and this checks with our understanding of the unit sequence function.

(b) X z zz

z z( ) =

−

− +

274

74

34

2

12. Find the inverse z transforms of these functions in series form by synthetic division.

M. J. Roberts - 8/17/04

11-5

(a) X zz

z( ) =

−12

)z z

z

z

z

z z zk

−

−

−

+ + + +( )

+12

12

1212

14

14

112

14

1

22

L

L L

(b) X zz

z z( ) =

−− +

12 12

Notice that X x uzz

z zz

z

zn n( ) =

−−( )

=−

=−

⇒ [ ] = −[ ]−1

1

11 1

121

13. Find the inverse z transforms of these functions in closed form using partial fractionexpansions, a z transform table and the properties of the z transform.

(a) X zz z

z z z z( ) =

−

= − +−

=−

−

1

12

2 212

21

12

1

x un n nn

[ ] =

− −[ ]

−[ ]−

21

21 1

1

δ

Alternate Solution:

X x uz zz

zn n

n

( ) =−

⇒ [ ] =

−[ ]−−

22

12

1

22

These solutions look different but are they? Check a few values to see.

(b) X zz

z z( ) =

−

−

2

12

34

M. J. Roberts - 8/17/04

11-6

(c) X. .

zz

z z( ) =

+ +

2

2 1 8 0 82

Recognize this as being similar to the forms,

α

αα α

n n nz

z zsin u

sin

cosΩ

ΩΩ0

02

022

( ) [ ] ← →( )

− ( ) +Z ,, z > α

and

α

αα

n n nz z

z zcos u

cos

cosΩ

ΩΩ0

0

2 2( ) [ ] ← →

− ( ) −

Z

002( ) +

>α

α, z

where α = 0 9055. and Ω0 3 031= . .

x . cos . . sin . un n n nn[ ] = ( ) ( ) − ( )[ ] [ ]0 9055 3 031 9 03 3 031

14. Using the z transform, find the total solutions to these difference equations with initialconditions, for discrete time, n ≥ 0.

(a) 2 1216

y y sin un nn

n+[ ] − [ ] =

[ ]π

, y 0 1[ ] =

Z-transform the equation.

2 08

28

12

z z zz

z zY y Y

sin

cos( ) − [ ]( ) − ( ) =

−

+

π

π

Then solve for Y and find the inverse transform of Y.

Y.

.. .

. .z

z

z

z z

z

z( ) =

−−

−− +

+−

0 29340 5

0 2934 0 58681 8478 1 0 52

Use sin u

sin

cos,Ω

ΩΩ00

202 1

1n nz

z zz( ) [ ] ← →

( )− ( ) +

>Z

and

cos u

cos

cosΩ

ΩΩ0

0

202

n nz z

z z( ) [ ] ← →

− ( ) − ( ) +

Z

111, z >

and identify Ω0 0 3926= . . Y can be manipulated into the form,

M. J. Roberts - 8/17/04

11-7

Y.

. ..

..

...

zz

z

zz

z z

z z

z

z z( ) =

−+

−−

−− +

−− +

−0 29340 5 0 5

0 29340 9239

1 8478 12 812

0 38271 8478 1

12

2 2

Then

y . u u . cos . sin un n n n n nn n

[ ] =

−[ ] +

[ ] − −( )

− −( )

−[ ]−

0 293412

112

0 29348

1 2 8128

1 11 π π

(b) 5 2 3 1 0 8y y y . un n n nn+[ ] − +[ ] + [ ] = ( ) [ ] , y , y0 1 1 10[ ] = − [ ] =

Y.

..

.. .

..

. .z

zz

z z

z z

z

z z( ) =

−− −

−− +

+− +

−0 4444

0 81 9 5556

0 30 6 0 2

0 93250 3317

0 6 0 21

2

2 2

y . . u . .cos .

. sin .un n n

n

nn

n n[ ] = ( ) [ ] − [ ] − ( )−( )( )

+ −( )( )

−[ ]

−

0 4444 0 8 9 5556 0 44720 8355 1

0 9325 0 8355 11

1δ

This is not the only way to find the inverse transform and other solution forms canbe correct, but they should be equivalent to this one.

15. From each block diagram, write the difference equation and find and sketch theresponse, y n[ ] , of the system for discrete time, n ≥ 0, assuming no initial energystorage in the system and impulse excitation, x n n[ ] = [ ]δ .

(a)

D

x[n] y[n]

y yn n n[ ] + −[ ] = [ ]1 δ

Y y uzz

z

zn n

n( ) =+

=+

⇒ [ ] = −( ) [ ]−1

1 111

n-5 20

y[n]

-1

1

(b)

D

x[n] y[n]

0.8

M. J. Roberts - 8/17/04

11-8

(c)

D

Dx[n]

y[n]

0.9

-0.5

Designate the lower input to the last summer as w n[ ] . Then the excitation of the delayfeeding that point is w n +[ ]1 and

x . w wn n n[ ] − −[ ] = +[ ]0 9 1 1or

w . w xn n n[ ] + −[ ] = −[ ]0 9 2 1

y . x . w wn n n n[ ] = − [ ] − −[ ]( ) + [ ]0 5 0 9 1

Then z-transform, eliminate W, solve for Y and inverse transform to find y.

y . . . sin u . sin un nn

nn

nn[ ] = − [ ] + ( )

[ ] +

−( )

−[ ]

0 5 0 9486 1 05412

0 51

21δ π π

n-5 20

y[n]

-1

1

16. Sketch regions in the z plane corresponding to these regions in the s plane.

(a) 01

0< < < <σ ω πT Ts s

, , z e esT j Ts s= = +( )σ ω

First find some corresponding points. Then the region should be obvious.

M. J. Roberts - 8/17/04

11-9

s j e z

e

T Te

jT

jT

e

sT

s s

s s

j

sσ ω

π π π

0 0 0 1

1 10 2 718

0 1

0

1 .

−

11 12 718

20

2

1

2

TjT T

jT

e e

jT

jT

e j

s s s s

j

s s

j

+ −π π

π π

π

π

.

11

2

1

22 7181 2

Tj

T Tj

Te e j

s s s s

j+

π π π

.

[z]

Re(z)

Im(z)

1

2.718

(b) − < < − < <1

0 0T Ts s

σ π ω,

(c) −∞ < < ∞ < <σ ω π, 0

2Ts

17. Find the bilateral z transforms and ROC’s of these signals.

(a) x un n[ ] = −[ ]

u u X ,n n z zc[ ] −[ ] ← → = ( )Z 1 any

u u X ,n n

z

z zzac[ ] [ ] ← →

−=

>Z

1

11

u X ,−[ ] ← →

−=

−= ( ) <n z

zz

z zacZ

1

11

1

11

X X x X ,z z zz z

zc ac( ) = ( ) − [ ] + ( ) = − +−

=−

<0 1 11

11

11

M. J. Roberts - 8/17/04

11-10

Alternate Solution:

X x u ,z n z n z z zz

zn

n

n

n

n

n

n

n

( ) = [ ] = −[ ] = = =−

<−

=−∞

∞−

=−∞

∞−

=−∞ =

∞

∑ ∑ ∑ ∑0

0

11

1

(b) x un nn[ ] = −[ ]α

(c) x . u . un n nn n[ ] = ( ) −[ ] + ( ) [ ]0 5 0 3

(d) x . cos unn

nn[ ] = −( )

−[ ]1 5

28π

18. Using the definition of the z transform verify the z transforms of the followingfunctions:

(a)

z-Transform Table Entry: u n

z

z z[ ] ← →

−=

− −Z

1

1

1 1

X uz n z zz z

n

n

n

n

( ) = [ ] = = + + +−

=

∞−

=

∞

∑ ∑0 0

211 1

L

)1

1 1

1

1

1

11

1

2+ + +

−−

−

z z

z z

z

z

z

L

Check.

(b) x!

unn

n[ ] = [ ]2

2

z-Transform Table Entry: n

nz z

z

z

z z

2

3

1

12

1

2 1

1

2 1!u[ ] ← →

+( )−( )

=+−( )

−

−Z

(c) x un n nn[ ] = [ ]α

(d) x sin un F n nn[ ] = ( ) [ ]α π2 0

M. J. Roberts - 8/17/04

11-11

19. Sketch the region of convergence (if it exists) in the z plane, of the bilateral z transformof these DT signals.

(a) x un nn

[ ] =

[ ]1

2

Using α

α ααn n

z

z zzu ,[ ] ← →

−=

−>−

Z 1

1 1

1

2 12

1

112

1

21

[ ] ← →

−=

−>

−

n

nz

z zzu ,Z

[z]

ROCROC

ROCROC

Re(z)

Im(z)

12

(b) x u un n nn n

[ ] =

[ ] +

−[ ]5

4107

20. Using the time-shifting property, find the z transforms of these signals.

(a) x un nn

[ ] =

−[ ]

−23

11

2

3 23

[ ] ← →

−

n

nz

zu Z

2

31

23

123

11

−[ ] ← →−

=−

−−

n

n zz

z zu Z

(b) x un nn

[ ] =

−[ ]2

31

M. J. Roberts - 8/17/04

11-12

x un nn

[ ] =

−[ ]

−23

23

11

(c) x sin unn

n[ ] =−( )

−[ ]2 14

1π

21. Draw system diagrams for these transfer functions.

(a) H zz z

z z( ) =

+

+ +

23

23

34

2

(b) H. . .

zz

z z z( ) =

−( ) +( ) −( )2

0 75 0 1 0 3

22. If the z transform of x n[ ] is X zz

( ) =−

134

, and

Y X Xz j e z e zj j( ) =

−

−π π6 6

what is y n[ ] ?

23. Using the convolution property, find the z transforms of these signals.

(a) x sin u unn

n n[ ] =

[ ] ∗ [ ]2

8π

(b) x sin u u unn

n n n[ ] =

[ ] ∗ [ ] − −[ ]( )2

88

π

From part (a),

sin u usin

2

84

2

3

ππ

nn n

z

z z

[ ] ∗ [ ] ← →

−

Z

22 24

1 1 24

cos cosπ π

−

+ +

z −1

24. Find the inverse z transforms of these functions in closed form using partial fractionexpansions, a z transform table and the properties of the z transform.

(a) X. .

zz

z z( ) =

−+ +

11 8 0 822

M. J. Roberts - 8/17/04

11-13

X. . . .

zz

z z z z( ) =

+ +−

+ +2 21 8 0 821

1 8 0 82

The denominators are of the form, z z20

22− ( ) +α αcos Ω where α = 0 9055. andΩ0 3 031= . . Therefore

Xsin

sin

. . sin

sin

. .z

z

z z

z z

z z( ) = ( )

( )+ +

− ( )( )

+ +

−11 8 0 82 1 8 0 820

02

1

0

02α

αα

αΩ

ΩΩ

Ω

or

X.

. ..

. .z

z

z zz

z

z z( ) =

+ +−

+ +−10

0 11 8 0 82

100 1

1 8 0 8221

2

Then, using

α

αα α

n n nz

z zsin u

sin

cosΩ

ΩΩ0

02

022

( ) [ ] ← →( )

− ( ) +Z ,, z > α

x . sin . u . sin . un n n n nn n[ ] = ( ) ( ) [ ] − ( ) −( )( ) −[ ] −

10 0 9055 3 031 0 9055 3 031 1 11

Alternate Solution:

X. .

. .

. .

.z

z

z z

j

z j( ) =

−+ +

=+

+ −+

1

1 8 0 82

0 5 9 5

0 9 0 1

0 52

−−+ +

j

z j

9 5

0 9 0 1

.

. .

x . . . . . . .n j j jn[ ] = +( ) − +( ) + −( ) −−

0 5 9 5 0 9 0 1 0 5 9 5 01

99 0 1 11−( ) −[ ]−

j nn

. u

x. . .

.

.

nj e

n j n

[ ] =+( )( )

+ −

− −( )0 5 9 5 0 9055

0 5

1 3 031 1

jj en

n j n9 5 0 90551

1 3 031 1. .u

.( )( )

−− − −( ) [[ ]

x. . . .

ne e

n j n j n

[ ] =( ) +− −( ) − −0 5 0 9055

1 3 031 1 3 031 1(( )

− −( ) −

( )+ ( ) −j e e

n j n j n9 5 0 90551 3 031 1 3 031. . . . −−( )( )

−[ ]1

1u n

x . cos . .n nn n[ ] = ( ) −( )( ) − ( )−

0 9055 3 031 1 19 0 90551 −− −( )( ) −[ ]1

3 031 1 1sin . un n

Although the two solutions above look different in analytical form, they areequivalent. That is, the graphs of these functions are identical.

(b) X. .

zz

z z z( ) =

−+ +( )

1

1 8 0 822

The answer to this part is the same as the answer to the previous part exceptdelayed by 1 in discrete time.

M. J. Roberts - 8/17/04

11-14

(c) X zz

z z( ) =

− +

2

2 14