315hw9sol

7/23/2019 315hw9sol

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5.3-3.

E E

E t S

H W

?

S . , l , l r . : n s

(a)

The

system

equation

n delay

form

is

zul"l

-

3a[n

rl

+

aln

-

2l

-

4x

n]

3c

n

1]

AIso

y ln l

<+

Y l , l

y l n - l l

o l rA

a{ -2 lo \ r l " \ * t

,

[']

(+

X

lzl:

;+fr

c

n

l] **

]ou

The

z-transforrn

of

the equatiotr

s

3 I

, ,

4 z

3

4 z - 3

2y

z]

:Y

[zl

+

ZY

kl

+

|

:

;a25

-,

_

02b

=

;

_

U2s

vl

( r_ +

l )

v lz l

- r

+

4' ;

1=

\ -

z ' 2 2 ) - ' - '

-

z - 0 ' 2 5

and

3z

2.75

z

-

0.25

v L l

'

l ' 1

:

z

z(32

2.75)

alnl

=

( 2 2 2 - 3 2 + 1 ) ( z - 0 . 2 5 )

z(32

2.75)

2 (z

0 . 5 ) ( z

1 ) ( z 0 . 25 )

512

r l3

413

; :Tn-

z1-

, -c .25

l ' r 3 a

I

I i

+

(o

s)"

j(o.zs)"1

n]

L 3

2 '

3 '

J

[1

rrt-"

lrnt-"1

1,,1

LJ ' 2 ' - , /

3 ' ' '

J

(b) From part (a), we

have

/ 1 1 \

4 z - 3

\ r - ;++)YlA: +

+;=a%

I - C . t e r m

,

u *

2 2 2 - 3 2 * 1 , , , ,

4 z - 3

t , [z ] :

- l

+ ;=u2b

and

and

V l r l

'

L' l

z

:

vlnl

:

-z

z(42

3)

2 - i

0 5 1 2 4 I 4 |

,

-a5- ,

-

l -

; :

os

=

i ,

-

|

-

i '

-

u ro

z z ^ z 4 z 4 z

y lz l=

0 .5 ; :os

,_- t+2 ;_os*

5r_ t -

i r4 .2s

[ ] ,0

' ,"

V : i

r

:

lz.r(o.s)"

L

-

f

to

ttl

"l"t

-.1rorl"]" 't .U

and

Ut "

9 r r

7/23/2019 315hw9sol


5.3-10.

Taking

rhe z-transform

of D[nr. ]

(1.01)b[nr.

l]

+

p[rn]

and solving o r B(z)

yiclds

B(z):

f (z);afo-.

Thus,

P(z)

is required

o solve or

b[rn] .

One

way o represent

Sally'sdeposit

schedule

s

p[m]

:

100(u[rn]

ILo

6[m

-

(l2k

+

11)]). Defined

hi s

way, Sally

depositsone

hundreddollars

on

the

first

day

of

every month rn.except

or

Decembers,

m

:

l2k

*

11

or ,b

=

{0,

,2,

. . . }).

Taking

he

z-transform

ields

P(z)

=

roo

(i-i-

-

If=-*

DLo

dlrn

-

(l2k+

ll)]z-*)

=

100

l:i=r

-

ILo

I;=--

6lm

(r2k

+

11)lz--)

:

/ '

-m

^ -(12/t+rr)) . Substi tut ingP(r) into the expres-00(i-::=r

-

Lr=o z

'--

'

').

sion

for

B(")

yields B(r)

=

100

r}r

-

DEo

z-o2k+rL))

i:T#F

=

/ ,

roo

6mc$o:;=l

+

ILo

t#+)

:

roo

i=i-13i;-

cy+

+

ILo

##3)

The hrst

two terms

are easily

nverted using a

table o[ z-transform

pairs,

while the

last

sum

is inverted

using

ables

and

the

shifting

property.

/ @ \

b[m] 100 101(1.01)-u[m]

l00u[rn]

f

{ r .or)-- (12&+11)u[m

(r2k+

11)]

]

\ t = o /

5 . 5 -1 .

(a )

Hlz l=

H

[rjn]

=

ern+ l

:

1

(b )

lH

[. jnl

'

=

H

H'

-

and

Therefore

la

.rn] l :

nn'

=

and

, l

l l l l o l " l l : - :

l " t "

r r

r r t . t o _ 0 . g c o s o

LH

lejnl:

-tan-l

#$7

t l

H

4:

; : ,A:

;

s ,n - ,

T

I

1

: :

1

_

0.4e-ro

0.4rrt,

y' I .16

0.8

os

]

0.4

in

O

I

-

0.4cos

O

t a l

322

I .8z

zz

z

*0 .16

e'z:n

1.3sjo

(3cos O

1.8cos

[lr ( : in

2O

1'8sin

O)

and

H[ern1

ff i .16:

rareinllz

t#=ffi]

t;##*]

L2.24

10.8 os

Q

and

H[ej ' ]

:#rn '=

lH

Leial:

-

tan-r

Hlz l=

t - O . l c o s O - j s i n O

c o s Q - 0 . 4 + j s i n l )

1

:

T:lo

o3;o'

)

2.0256

2^nAs

Q

+

0.32

os

O

7/23/2019 315hw9sol


lH

k'"ll

=

[

12.24 10.8 osO

t t / 2

I

J

i n Q

an d

t

H

ejaj:

an-,

##tsj*#)

-,*-'

(

(a) The

z-transform

of

the

two equations

ield:

/

0.e\

( i ) . { l+= lY [ r l

=X lz l

" \

z /

t i i l .

r

-

on)

YPI:

v7"1

"

\

z /

Hence

he transfer

functions

of

these

ilters

are

( i ) .H [z ]

; *

( i i ) .H [z ]

h

Consider

he

first

system.

(i)

H

Pial=

:*

:

1+oh:ln

=

l a l e ro l l =g , tH [e je '

'l

VI-ET

+

l-slosQ-'

-"

1-

J

( i i

n

leiol

#h:

1-*;ld

=

ls [.,n]l

t*ffi,

tH

eiel:

tan-l

i##*]

Filter(i)

has a

zero at

the

origin

and

a

pole

at

-0.9.

Because

he

only

pole

is

near

O

=

r(z

:

-1),

this is

a highpass

iiter,

as

verified

from the

frequency

response

hown

n

Figure

S5.5-4a.

Filter(ii)

has

a

zeroat

the

origin

and a

pole

at 0.9.

Because

he only

pole

is near

O:0(z

=

1), this

is a lowpass

ilter,

as

verif ied

rom the

frequency

esPonse

shown

n

Figure S5.5-4b.

(b)

(i)

For

Q

=

0.012r

H lpro.o l r l

=

0.5966

-

t -

'

r / t .gt

*

l .8coso.0lzr

2.0256

2.32

os

O

*

0.32

os2Q

c o s 2 Q - c o s O + 0 . 1 6

sin2O s

t r t r A

1

0.9

os

-

j0.9

in

O

, | -

-o.gs ino

:

-

l e r - r l - -

|

l-

+

o.s

os

Ql

I

-

0.9

os

Q

+

j0.9

sin

Q

7/23/2019 315hw9sol


For Q

=

0.99r

( i i )

F o r Q = 0 . 0 1 r

For Q

:

0.994

Filter(i)

gain

at

Os

s

Pilter(ii)

gain

at n

-

Qs s

5)

->

H

lejosstl:

-:-]--

=

9.S8

y ' I .81

+

l .8cos0.99n

H

lejootnf

=

-:i.-

=

g.tS

V l .8 l

-

l . 8cos0.0 lur

H

ldo

eeil

=

o.b9o6

V1.81

l .8cos0.99zr

rHr=-g

y ' l .81

1.8

os

O6

-

-:r-_

/ 1 .81

-

1 .8cos(z rns)

y ' I .8 I

+

l .8cos(O6)

l l l

t

5

t

5 . 9 -1 .

(u )

FigureS5.5-4

z - 0 . 8

z - 2

-1.22

7

- ' L8z

+ t s

0 .8< l z l 2

0 .8< l z l 2

r

[n]

(0.8)"u

[nl

+

2"u

-(n

+

t)]

-\,- --#

c

r

tl

tr

[.1

t

r r [ n ] +=+

,_*

l z l

>0 .8

r2ln\

€+

;-

lrl<2

x l r l

=

(b)

llence

xr

[r ]

XzlrJ

- o

t

:

z - 3

l z l>2

l r l

<3

* - *

2 < [ z l < 3

ff i

2<Pl<3

Hence

x l r l

=

7/23/2019 315hw9sol


(c )

g

r-2

xtlr)

xzl r l

X

lrl

==

rlnl

--

cos2.Zrn

cos

3.3zrn

2 Z

z - 0 . 8

z - U ' 9

6@A1; i6

o8<lz l

o 'e

Il"ors.srrr.,

cos

.lrnl

]

[co,

.srrn

cos

.1nni

z - U . 6

z - 0 . 9

Figure

9.1-2

l z l

>

0,8

lzl

<

0.9

11rrt . t ' .

L,

s-ir '5at

"ir 'Lrn

1

,-j l

lrn1

2 '

I

lait

no

I

"io.5tt

a

, j1'1tn1

sjo'srr1

2 '

The

fundanrental

requency

Qo

=

0.1

and

Ns

=

#o

=

20

There

are

only

5th'

9th'

11th

and

15th

harmonies

with

coefficients

D s = D s -

D t r = O r r = j

All

the

form

coefficients

re

real

(phases

ero).

The

spectrum

s showrr

n Figure

315hw9sol

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