Chapter11 SM[1]

8/12/2019 Chapter11 SM[1]

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Chapter 11

11.1 Analysis yields{ } ],[][][,]1[]1[][ 3122211 nwnwnwnwn x nw α β

{ } ],[][][,]1[]1[][534422213

nwnwnwnwnwnw αα β

{ } ].[][][,]1[]1[][ 01534235 n x nwn ynwnwnw ααα β In matrix form the above equations can be written as

1 1 1 1

2 2

3 3 2 2 2 2

4 4

5 5 2 3 3 3

[ ] 0 0 0 0 0 0 [ ] 0 0 0 0 0

[ ] 1 0 1 0 0 0 [ ] 0 0 0 0 0 0

[ ] 0 0 0 0 0 0 [ ] 0 0 0 0

[ ] 0 0 1 0 1 0 [ ] 0 0 0 0 0 0

[ ] 0 0 0 0 0 0 [ ] 0 0 0 0

[ ] 1 0 0 0 0 0 [ ] 0 0 0 0 0

w n w n

w n w n

w n w n

w n w n

w n w n

y n y n

α β

α β α β

α β α β

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥

= +⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

11

2

3

4

5

0

[ 1][ 1]

0[ 1]

0[ 1]

0[ 1]

0[ 1]

[ ]0 [ 1]

x nw n

w n

w n

w n

w n

x n y n

β

α

−− ⎡ ⎤⎡ ⎤⎡ ⎤⎢ ⎢ ⎥⎢ ⎥−⎢ ⎢ ⎥⎢ ⎥⎢ ⎢ ⎥⎢ ⎥− + ⎢ ⎢ ⎥⎢ ⎥− ⎢ ⎢ ⎥⎢ ⎥⎢ ⎢ ⎥⎢ ⎥−⎢ ⎢ ⎥⎢ ⎥− ⎢ ⎣ ⎦⎣ ⎦ ⎣ ⎦

Here the F matrix is given by: .

000001000000010100000000000101 000000

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=F

Since the F matrix contains nonzero entries above the main diagonal, the above set ofequations are not computable.

11.2 A computable set of equations are given by:{ },]1[]1[][ 2111 nwn x nw α β { },]1[]1[][ 422213 nwnwnw αα β

{ },]1[]1[][ 534235 nwnwnw αα β ],[][][ 312 nwnwnw ][][][ 01 n x nwn y α ,][][][ 534 nwnwnw .

In matrix form the above equations can be written as

1 1 1 1

3 3 2 2 2 2

5 5 2 3 3 3

2 2

4 4

[ ] [ ]0 0 0 0 0 0 0 0 0 0 0

[ ] [ ]0 0 0 0 0 0 0 0 0 0

[ ] [ ]0 0 0 0 0 0 0 0 0 0

[ ] [ ]1 0 1 0 0 0 0 0 0 0 0 0

[ ] [ ]1 0 0 0 0 0 0 0 0 0 0 0

[ ] [ ]0 0 1 0 1 0 0

w n w n

w n w n

w n w n

w n w n

y n y n

w n w n

α β

α β α β

α β α β

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥

= +⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

1 1

3

5

2

0

4

[ 1] [ 1]

[ 1] 0

[ 1] 0

[ 1] 0

[ 1] [ ]

[ 1]0 0 0 0 0 0

w n x n

w n

w n

w n

y n x n

w n

β

α

− −⎡ ⎤⎡ ⎤ ⎡⎢ ⎥

⎤⎢ ⎥ ⎢−

⎢ ⎥⎥

⎢ ⎥ ⎢⎢ ⎥

⎥⎢ ⎥ ⎢− +⎢ ⎥

⎥⎢ ⎥ ⎢−⎢ ⎥

⎥⎢ ⎥ ⎢

⎢ ⎥⎥

⎢ ⎥ ⎢−⎢ ⎥

⎥⎢ ⎥ ⎢−⎢ ⎥

⎥⎣ ⎦ ⎣⎣ ⎦ ⎦


010100000001000101000000000000000000

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=F

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Since the F matrix has no nonzero entries above the main diagonal, the new set of equationsare computable.

11.3 Analysis yields { } ],[][][,]1[]1[]1[][ 1263422111 n x nwnwnwnwnwnw ααα β

{ } ],[][][,]1[]1[][ 324634223 nwnwnwnwnwnw αα β { },]1[][ 6335 nwnw α β

],[][][ 546 nwnwnw ].[][][][][ 0634221 n x nwnwnwn y αααα In matrix form the above set of equations are given by

1 1 1 1 1 2

2 2

3 3

4 4

5 5

6 6

1 2 3

0 0 0 0 0 0 0[ ] [ ] 0 0 0

1 0 0 0 0 0 0[ ] [ ]

0 0 0 0 0 0 0[ ] [ ]

0 1 1 0 0 0 0[ ] [ ]0 0 0 0 0 0 0[ ] [ ]

0 0 0 1 1 0 0[ ] [ ]

0 0 0 0[ ] [ ]

w n w n

w n w n

w n w n

w n w n

w n w n

w n w n

a a a y n y n

β α β α β ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= +⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦

1 3 1

2

2 2 2 3 3

4

3 3 5

6

0

00 [ 1]

[ ]0 0 0 0 0 0 0 [ 1]

00 0 0 0 0 [ 1]

00 0 0 0 0 0 0 [ 1]00 0 0 0 0 0 [ 1]

00 0 0 0 0 0 0 [ 1]

[ ]0 0 0 0 0 0 0 [ 1]

w n

x nw n

w n

w n

w n

w n

x n y n

α

β α β α

β α

α

− ⎡ ⎤⎡ ⎤⎡⎢ ⎥⎢ ⎥⎢ −⎢ ⎥⎢ ⎥⎢⎢ ⎥⎢ ⎥⎢ −⎢ ⎥⎢ ⎥⎢ +− ⎢ ⎥⎢ ⎥⎢⎢ ⎥⎢ ⎥⎢ −⎢ ⎥⎢ ⎥⎢ − ⎢ ⎥⎢ ⎥⎢⎢ ⎥⎢ ⎥⎢ −⎣ ⎦⎣ ⎣ ⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦


0000001100000000000000110000000000000010000000

321 ⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

ααα

F

Since the diagonal of the F matrix has all zeros, and no nonzero entries above the maindiagonal, the new set of equations are computable.

11.4

x [n] y[n]w [n]1 w [n]2 w [n]3 w [n]4

w [n]5

α 0

1β_ 1 z

1α 11β

_ 1 z

1

1 1

α 12β_ 1 z

α 22β_ 1 z

α 23β_ 1 z

α 33β_ 1 z

1 Reduced signal flow graph obtained by removing the branches going out of the input nodeand the delay branches is:

y[n]w [n]1w [n]2 w [n]3 w [n]4

w [n]5

1 1

1 1

1

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8/12/2019 Chapter11 SM[1]


8/12/2019 Chapter11 SM[1]


.

⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢

⎣

⎡

][

][][][][][

00000

0000000000000000000100000010

0

00

][00

][

][][][][][

32321

3

21

32321

n y

nsnsnwnwnw

k

k k

n x

n y

nsnsnwnwnw

4 4 4 4 34 4 4 4 21F

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

]1[]1[]1[]1[]1[]1[

010000001000000001000000000000000

32321

32

1

n ynsnsnwnwnw

k k

k

4 4 4 4 4 34 4 4 4 4 21G

As the diagonal elements of F -matrix are all zeros, there are no delay-free loops. However,

the above set of equations are not computable as there are non-zero elements above thediagonal of F .

(b) The reduced signal flow-graph representation of Figure P11.3 is shown below:

From the above flow-graph we observe that the set composed of nodes with only outgoing branches is The set of nodes with incoming branches from and

outgoing branches is The set of nodes with incoming branches fromand and outgoing branches is

]}.[{ 31 nwN 1N

]}.[{ 22 nwN 1N

2N ]}.[{ 13 nwN Finally, the set of nodes with only

incoming branches from , and , and no outgoing branches

Therefore, one possible ordered set of equations that is

computable is given by

1N 2N 3N

]}.[],[],[{ 324 n ynsnsN

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

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

000000000000000000010000001000000

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

12123

32

1

12123

n ynsnsnwnwnw

k k

k

n ynsnsnwnwnw

4 4 4 4 34 4 4 4 21F

.

⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢

⎣

⎡

⎥

⎥⎥⎥

⎥

⎦

⎤

⎢

⎢⎢⎢

⎢

⎣

⎡

]1[]1[]1[]1[

]1[]1[

01000000100000010000000

0000000000

121

23

32

3

n ynsnsnw

nwnw

k

k k

4 4 4 4 4 34 4 4 4 4 21G

Note: All elements on the diagonal and above diagonal of F are zeros.

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11.7 .5.131

)(21

22

110

= z z

z p z p p z H From Eq. (11.18) we get,

Hence,.

0.5

0.4

2.3

5.1

3

1

2.36.50.7

02.36.5

002.3

2

1

0

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

=

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

p

p

p

.

5.131

542.3)(

21

21

=

z z

z z z H

11.8 From Eqn (11.15),

0

1

12

2

2 0 0

12 2 0

4 2 2

0 8 4 2

0 12 8 4

p

p

d p

d

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− ⎡ ⎤⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥= −

⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥− ⎣ ⎦⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

and from Eqn (11.20),

. Finally, from Eqn (11.21),

Hence,

.

5.3

25.0

12

8

48

24 1

2

1⎥

⎦

⎤⎢

⎣

⎡ =

⎦

⎤⎢

⎣

⎡

⎦

⎤⎢

⎣

⎡

⎦

⎤⎢

⎣

⎡

d

d

.

5.11

5.2

2

5.3

25.0

1

224

022

002

2

1

0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

p

p

p

.5.325.01

5.115.22)(

21

21

= z z

z z z H

11.9 From Eqn. (11.15)

0

1

2

13

2

3

2 0 0 0

4 2 0 0

18 4 2 0

8 8 4 2

0 12 8 8 40 16 12 8 8

0 8 16 12 8

p

p

p

d p

d d

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− ⎡ ⎤⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥−

⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥= − −⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦

⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

. Next, from Eqn. (11.20),

Finally, from Eqn. (11.21),

Hence,

.

4.7

8.4

4.0

8

16

12

81216

8812

488 1

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

d

d

d

.

4.02.3

2.3

0.2

4.78.4

4.0

1

24880248

0024

0002

3

2

1

0

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

=⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

=⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

p p

p

p

.4.78.44.01

4.02.32.32)(

321

321

= z z z

z z z z H

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11.10 . Hence,

0

1

2

3

2 0 0 0 1 2

4 2 0 0 0.6 5.2

4 4 2 0 0.2 6.8

6 6 4 2 1.8 5.6

p

p

p

p

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

( ) 1 22 5.2 6.8 5.6 P z z z z − −= − + − 3− .

11.11

0

2 23 31

232

233

5 13 34

2 0 0 0 0 1 2

2 2 0 0 0

4 2 2 0 0 0 2

6 4 2 2 0 2 4

8 6 4 2 2 3

p

p

p

p

p

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =−⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦⎣ ⎦

. { } { }2 2 2 13 3 3 32, , 2 ,4 , 3i p = − −

11.12 The –th sample of an –point DFT is given by Thus, the

computation of requires

k N .][][ 1

0

=

N n

nk N W n x k X

][k X N complex multiplications and 1 N complex additions. Now, each complex multiplication, in turn, requires 4 real multiplications and 2 realadditions. Likewise, each complex addition requires 2 real additions. As a result, thecomplex multiplications needed to compute require a total of real multiplicationsand a total of real additions. Therefore, each sample of the –point DFT involves

real multiplications and

N ][k X N 4

22 N N N 4 24 N real additions. Hence, the computation of all DFT

samples thus requires real multiplications and24 N N N )24( real additions.

11.13 Let the two complex numbers be b ja α and .d jc β Thus, ))(( d jcb ja αβ which requires 4 real multiplications and 2 real additions. Consider

the product and which require 3 real multiplications and 2 realadditions. The imaginary part of can be formed from

),()( bcad jbd ac

,),)(( acd cba ,bd αβ bd acd cba ))(( ,bcad which now requires 2 real additions. Likewise, the real part of can be formed from

requiring an additional real additions. Hence, the complex multiplication can becomputed using 3 real multiplications and 5 real additions.

αβ bd ac

11.14 Recall, , / 2 s jceW N j N π where ) / 2cos( N c π and ). / 2sin( N s π Thus,

Now,.122 =sc ( )}Im{}Re{)( r r r N r j s jcW Ψ+Ψ+=Ψ⋅=Ψ +1

.}Re{}Im{}Im{}Re{ r r r r sc jsc ΨΨΨΨ Thus, }Im{}Re{}Re{ 1 r r r sc ΨΨΨ ⋅ and }.Re{}Im{}Im{ 1 r r r sc ΨΨΨ ⋅

Figure P11.4 with internal node label is shown below. Its analysis yields

c 1s

_____ c 1s

_____

s

Re{ ψ }r

Im{ ψ }r

Re{ ψ }r +1

Im{ ψ }r +1

U

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(1): },Im{1

}Re{ r r sc

U ΨΨ

(2): ,}Im{}Im{ 1 sU r r ΨΨ and

(3): }.Im{1

}Re{ 11 r r sc

U ΨΨ Substituting Eqs. (1) in Eq. (2) we get

(4): }.Re{}Im{}Re{}Im{1

}Im{}Im{ 1 r r r r r r scsc

s ΨΨΨΨΨΨ ⋅⎠

⎞

⎜⎝

⎛

Next,substituting Eqs. (1) and (4) in Eq. (3) we get (5):

(5): }Re{}Im{1

}Re{}Im{1

}Re{ 1 r r r r r sc

scs

cΨΨΨΨΨ =

}.Re{Im}Re{}Re{}Im{}Re{}Im{1 22

r r r r r r csccss

cs

cΨΨΨΨΨΨ ⋅==

It thus follows that the structure of Figure P11.4 implements the multiplication of a complexsignal with the twiddle factor using only 3 real multiplications.

.

In the case of multiplication by complex twiddle factor we have, jscW N

−=−1

( )}Im{}Re{)( r r r N r j s jcW Ψ+Ψ−=Ψ⋅=Ψ −+

11

( ) ( ).}Re{}Im{}Im{}Re{ r r r r sc j sc Ψ⋅−Ψ⋅+Ψ⋅+Ψ⋅= Thus, here}Im{}Re{}Re{ r r r sc Ψ⋅+Ψ⋅=Ψ +1 and }.Re{}Im{}Im{ r r r sc Ψ⋅−Ψ⋅=Ψ +1 It follows

then that the real and imaginary parts are simply obtained by reversing the sign of of thereal and imaginary parts derived in the case of multiplication by As a result, thecorresponding structure is obtained by cascading an inverter to the real multipliers in FigureP11.4 as indicated below:

s. N W

c 1s

_____ c 1s

_____s

Re{ ψ }r

Im{ ψ }r

Re{ ψ }r +1

Im{ ψ }r +1

U _1 _

1_1

11.15 The center frequency bin ,)(: N kF

k f k T c = where # N of bins, and T F is the sampling

frequency. Inverting we have .)( ⎥⎦

⎥⎢⎣

⎢=T F

N f f k Therefore, the absolute difference from one of

the given four tones (150 Hz, 375 Hz, 620 Hz, and 850 Hz) to the center of its bin is given by

.),( ⎥⎦⎥⎢

⎣⎢

T

T F N f

N F f f N dist It follows from this equation that the distance goes to zero if

f F N f

N F

T

T =⎦

⎥⎢⎣

⎢ or

T F N f

is an integer.

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The total distance is reduced to zero ifT

i

F

N f is an integer for .4,,1 Ki The minimum value

of for which this true is 500. N However, the total distance can be small, but nonzero, for significantly smaller values of . N

11.16 .1

1)(

1 zW z H

k N

k Hence, ),()(1

1

1

)()( 2 /

1

2 /

1 zV z zV

zW

z

zW

z X zY N

k N

N

k N

=

where .1

1)(

1 zW zV

k N

Or, in other words, ].[][][2

N nvnvn y

Consider Then:1k .1

1)(

11 zW zV

N

This implies,

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=

LL ,,,,,,1][][)1

2(

2 / 21 N

N N

N N N n

N W W W W nW nv µ

{ },,,1,,,,1 121 LL N N N W W W since ,,1 1

)12

(2 /

N

N

N N

N W W W and

son on. Thus, { }.,,,1,,0,0,0][ 212

LL = N N N

W W nv Hence,

.,0,0,0,,,,,1][][][)1

2(

212

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=

LL

N

N N N N

W W W nvnvn y

Now, consider :2

N k = .

1

1)(

12 / zW

zV N

N

This implies,

.,,,,,,1][][ 2)1

2(

222)2 / (

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=⋅

LL

N N

N

N N

N N

N

N

N n N

N W W W W nW nv µ

Thus, { }.,,,1,0,,0,0,0][ 2 / 2

LL N N

N N

N W W nv =

Now, ),1()1(,1,1,)1( 2 / 2)1

2(

22 / 22 N N N

N N

N

N

N N

N N

N W W W W etc. Hence,

.,,)1(1,)1(1,,1,1,1,1][1

1

2 /

2 /

2

2⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

= LLL N

N

n N n

N n y Therefore, if

is even,

2 / N

,,2,2,2,,1,1,1,1][12 /

2⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

LL N

n N nn y and if is odd,2 / N

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,,0,0,0,,1,1,1,1][12 /

2⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

LL N

n N nn y

11.17

11.18

11.19 ∑=

= 1

0

][][ N

n

nk N W n x k X

∑

=

=

=

1

0

)1(11

1

01

1

01

111

11

11 ]1[]1[][

r N

n

k r nkr N

r N

n

k nkr N

r N

n

nkr N W r nr x W nr x W nr x K

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∑ ∑−

=−

−

=⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+=1

0

1

01

1

1

11

r

i

ki N

r N

r N

n

nkr N W W inr x

DFT point)/(

][ .

Thus, if the -point DFT has been calculated, we need at the first stage an additionalmultiplications to compute one sample of the –point DFT and, as a result,

additional multiplications are required to compute all

) / ( 1r N )1( 1 r N ][k X

N r )1( 1 N samples of the N –pointDFT. Decomposing it further, it follows that additional N r )1( 2 multiplications are neededat the second stage, and so on. Therefore, the total number of multiply (add) operations

.)1()1()1(121 N r N r N r N r

i i ⎟⎠

⎞⎜⎝

⎛ ∑ = νν

νK

11.20 An examination of the flow-graph of the 8-point DIT FFT algorithm shown in Figure 11.24

reveals that in the first stage all twiddle factors are and in the second stage thetwiddle factors are either or Hence, there are no twiddle factors with

nonunity magnitudes in the first two stages. In all succeeding stages, one of the twiddle

factors is and another one is The number of twiddle factors with

nonzero magnitudes in the -th stage, is Hence, the total number of twiddlefactors with nonzero magnitudes in an N -point radix-2 FFT algorithm is

where

10 = N W

10 = N W ./ jW N N =4

10 = N W ./ jW N N =4

i ,3i .22 1 −−i

,),()( 3222R3

1 ≥ν−ν−⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ =ν ∑

ν

=

−

i

i .2ν= N

11.21 Direct computation of M samples of an N -point DFT requires multiplications,

whereas, the Radix-2 FFT algorithm requires

2 M

N N

22log multiplications. In order for a N -

point radix-2 FFT algorithm to be computationally more efficient than a direct computation

of M samples of an N -point DFT, the following inequality must hold: .log 22 ⎥⎦

⎥⎢⎣

⎢> N M N

a) N = 512 , M = 49 , b) N = 1024 , M = 72 , c) N = 2048 , M = 107

11.22 ).()()()( 32

231

130 z X z z X z z X z X Thus, the –point DFT can be expressed as

Hence, the structural

interpretation of the first stage of the radix-3 DFT is as indicated below:

N

].[][][][ 3 / 22

3 / 13 / 0 N k

N N k

N N k X W k X W k X k X

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3

3

3

z

z

N _3

-pointDFT

N _3

-pointDFT

N _3

-pointDFT

+ X [< k > ]1 N /3

+ x [n] x [n]0

x [n]1

x [n]22k

N W

k N W

X [< k > ]2 N /3

X [< k > ]0 N /3 X [k ]

11.23 k k k

n

nk W x W x W x W n x k X 69

39

09

8

09 ]6[]3[]0[][][ ⋅

=∑

k k k k k k W x W x W x W x W x W x 89

59

29

79

499 ]8[]5[]2[]7[]4[]1[

k k k k k k k W W x W x W x W x W x W x 969

39

09

69

39

09 ]7[]4[]1[]6[]3[]0[ ⋅

k k k k W W x W x W x 29

69

39

09 ]8[]5[]2[ ⋅ ,][][][ 2

93293130k k W k GW k Gk G where

,]6[]3[]0[][ 2

330

330k k k

W x W x W x k G ⋅

,]7[]4[]1[][ 2

330

331k k k

W x W x W x k G ⋅

are 3-point DFTs. A flow-graph representation

of this radix-3 DIT FFT computation scheme is shown below,

,]8[]5[]2[][ 233

0332

k k k W x W x W x k G ⋅

where the twiddle factors for computing the DFT samples are indicated below for a typicalDFT sample:

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In the above diagram, the 3-point DFT computation is carried out as indicated below:

11.24 k k k k k

n

nk W x W x W x W x W x W n x k X 1215

915

615

315

015

14

015 ]12[]9[]6[]3[]0[][][ ⋅

=∑

k k k k k W x W x W x W x W x 1315

1015

715

41515 ]13[]10[]7[]4[]1[

k k k k k W x W x W x W x W x 14151115815515215 ]14[]11[]8[]5[]2[ where,][][][ 2

1552155150k k W k GW k Gk G

,]12[]9[]6[]3[]0[][ 45

35

255

0550

k k k k k W x W x W x W x W x k G ⋅

,]13[]10[]7[]4[]1[][ 45

35

255

0551

k k k k k W x W x W x W x W x k G ⋅ and

.]14[]11[]8[]5[]2[][ 45

35

255

0552

k k k k k W x W x W x W x W x k G ⋅

A flow-graph representation of this mixed-radix DIT FFT computation scheme is shown below:

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11.25

11.26 Now, by definition, ]}.[Re{]}[Im{][ n X jn X nq Its -point DFT is given by N

Thus,.][][1

0∑=

= N

n

nk N W nqk Q

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(1): ∑=

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ⎠

⎞⎜⎝

⎛=1

0

2sin][Re{

2cos][Im{]}[Re{

N

m N mk

n X N

mk n X k Q

π π ,

(2): ∑=

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ⎠

⎞⎜⎝

⎛ 1

0

2cos][Re{

2sin][Im{]}[Im{

N

m N

mk n X

N mk

n X k Q π π

.

From the definition of the inverse DFT we observe .][1

][1

0∑

=

= N

m

mk N W m X

N k x Hence,

(3): ,2

sin]}[Im{2

cos]}[Re{1

]}[Re{1

0∑

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ⎠

⎞⎜⎝

⎛= N

m N mk

m X N

mk m X

N k x

π π

(4): .2

sin]}[Re{2

cos]}[Im{1

]}[Im{1

0∑

=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ⎠

⎞⎜⎝

⎛= N

m N mk

m X N

mk m X

N k x

π π

Comparing Eqs. (2) and (3) we get ,]}[Im{1

]}[Re{nk

k Q N

n x =

and comparing Eqs. (1)

and (4) we get .]}[Re{1

]}[Im{ nk k Q N

n x =

11.27 Therefore,

Thus,

⎩ ≠==

.0if ],[,0if ],0[][][

nn N X n X n X nr N

∑

=

=

=

1

1

1

1

1

0

][]0[][]0[][][ N

n

nk N

N

n

nk N

N

n

nk N W n N X X W nr r W nr k R

].[][][]0[][]0[1

0

1

1

1

1

)( k x N W n X W n X X W n X X N

n

nk N

N

n

nk N

N

n

k n N

N ⋅

∑

=

=

=

.][][ 1nk N

k Rn x =

11.28 Let denote the result of convolving a length- L sequence x [n] with a length- N sequence h[n]. The length of y[n] is then L + N – 1. Here L = 16 and N = 9, hence length of y[n] is 24.

][n y

Method #1: Direct linear convolution - For a length- L sequence x [n] and a length- N sequenceh[n],

# of real mult. = .135)1916(92)1(2

9

11 =∑=n

N

n n N L N n

Method # 2: Linear convolution via circular convolution - Since y[n] is of length 24, to getthe correct result we need to pad both sequences with zeros to increase their lengths to 24

before carrying out the circular convolution.# of real mult. = 24 ×24 = 576.

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Method #3: Linear convolution via radix-2 FFT - The process involves computing the 16- point FFT G of the length-16 complex sequence][k ],[][][ nh jn x ng e where is

obtained by zero-padding h[n] to length 16. Then, the 16-point DFTs, and of

][nhe

][k X ],[k H e][n x and , respectively, are recovered from . Finally, the IDFT of the product

yields y[n].

][nhe ][k G

][][][ k H k X k Y e

Now, the first stage of the 16-point radix-2 FFT requires 0 complex multiplications, thesecond stage requires 0 complex multiplications, the third stage requires 4 complexmultiplications, and the last stage requires 6 complex multiplications, resulting in a total of10 complex multiplications .# of complex mult. to implement = 10][k G# of complex mult. to recover and from = 0][k X ][k H e ][k G# of complex mult. to form = 16][k Y # of complex mult. to form the IDFT of = 10][k Y Hence, the total number of complex mult. = 36A direct implementation of a complex multiplication requires 4 real multiplications resultingin a total of 4 × 36 = 144 real multiplications for Method #3. However, if a complex multiplycan be implemented using 3 real multiplies (see Problem 11.13 ), in which case Method #3requires a total of 3 × 36 = 108 real multiplications.

11.29 Let y[n] denote the result of convolving a length- L sequence x [n] with a length- N sequenceh[n]. The length of y[n] is then L + N – 1. Here, L = 16 and N = 10, hence length of y[n] is25.

Method #1: Direct linear convolution - For a length- L sequence x [n] and a length- N sequenceh[n],

# of real mult. = .160)11016(102)1(210

11

=∑=n

N

n

n N L N n

Method # 2: Linear convolution via circular convolution - Since y[n] is of length 24, to getthe correct result we need to pad both sequences with zeros to increase their lengths to 24

before carrying out the circular convolution.# of real mult. = 25 ×25 = 625.

Method #3: Linear convolution via radix-2 FFT - The process involves computing the 16-

point FFT G of the length-16 complex sequence][k ],[][][ nh jn x ng e where isobtained by zero-padding h[n] to length 16. . Then, the 16-point DFTs, and of

][nhe][k X ],[k H e

][n x and , respectively, are recovered from . Finally, the IDFT of the product

yields y[n].

][nhe ][k G

][][][ k H k X k Y e

Now, the first stage of the 16-point radix-2 FFT requires 0 complex multiplications, thesecond stage requires 0 complex multiplications, the third stage requires 4 complex

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multiplications, and the last stage requires 6 complex multiplications resulting in a total of 10complex multiplications.# of complex mult. to implement = 10][k G# of complex mult. to recover and from = 0][k X ][k H e ][k G# of complex mult. to form = 16][k Y # of complex mult. to form the IDFT of = 10][k Y

Hence, the total number of complex mult. = 36

A direct implementation of a complex multiplication requires 4 real multiplications resultingin a total of 4 × 36 = 144 real multiplications for Method #3. However, if a complex multiplycan be implemented using 3 real multiplies (see Problem 11.13 ), in which case Method #3requires a total of 3 × 36 = 108 real multiplications.

11.30 (a) Since the impulse response of the filter is of length 72, the transform length N should begreater than 72. If L denotes the number of input samples used for convolution, then L = N – 71. So for every L samples of the input sequence, an N -point DFT is computed andmultiplied with an N -point DFT of the impulse response sequence h[n] (which needs to becomputed only once), and finally an N -point inverse of the product sequence is evaluated.Hence, the total number M R of complex multiplications required (assuming N is a power-

of-2) is given by .log)log( 222712048 N N N N

N

N M ⎥⎤

⎢⎡=

R

It should be noted that in developing the above expression, multiplications due to twiddlefactors of values ±1 and ± j have not been excluded. The values of M R for different valuesof N are as follows:

1) For N = 128, M R = 37312, 2) For N = 256, M R = 28672, 3) For N = 512, M R =27904, 4) For N = 1024, M R = 38912.

Hence, N = 512 is the appropriate choice for the transform length requiring 27904 complexmultiplications or equivalently, 27904 × 3 = 83712 real multiplications.Since the first stage of the FFT calculation process requires only multiplications by ±1, thetotal number of complex multiplications for N = 128 is actually

,27648log)log(222271

2048 =⎥⎤

⎢⎡=

N N

N M N N N N R or equivalently, 27648 × 3 = 82944

real multiplications.

(b) For direct convolution, # of real multiplications =

147456)1722048(722)1(272

11

=∑=n

N

n

n N L N n .

11.31 (a)

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,

0001000

0000100

00000100000001

0001000

0000100

0000010

0000001

78

68

58

48

38

28

18

08

8

⎥⎥⎥⎥⎥⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎢⎢⎢⎢⎢

⎣

⎡

=

W

W

W W

W

W

W

W

V ,

0100000

0010000

01000000010000

0000010

0000001

0000010

0000001

28

08

28

08

28

08

28

08

4

⎥⎥⎥⎥⎥⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎢⎢⎢⎢⎢

⎣

⎡

=

W

W

W W

W

W

W

W

V

,

1000000

1000000

00100000010000

0000100

0000100

0000001

0000001

48

08

48

08

48

08

48

08

2

⎥⎥⎥⎥⎥

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

W

W

W W

W

W

W

W

V .

1000000000001000 001000000000001001000000000001000001000000000001

⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

=E

As can be seen from the above, multiplication by each matrix ,3,2,1, =k k V requires at most8 complex multiplications.

(b) The transpose of the matrices given in Part (a) are as follows:

,

000000

000000

000000

00000010001000 01000100

0010001000010001

38

38

28

28

18

18

08

088

⎥⎥⎥⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎢

⎣

⎡

=

W W

W W

W W

W W t V ,

000000

0000001010000001010000000000000000

0000101000000101

28

28

08

08

28

28

0808

4

⎥⎥⎥⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎢

⎣

⎡

=

W W

W W

W W W W

t V

,

00000011000000000000 00110000

0000000000110000000000000011

48

08

48

08

48

08

48

08

2

⎥⎥⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

=

W W

W W

W W

W W

t V .

1000000000001000001000000000001001000000000001000001000000000001

⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

=EE t

It is easy to show that the flow-graph representation of is precisely the 8-

point DIF FFT algorithm of Figure 11.28 .

t t t t 8428 VVVED =

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11.32 .10,][][][]2[2

1

2 /

2

12

0

21

0

2 ∑

=

=

=

N N

N n

n N

N

n

n N

N

n

n N W n x W n x W n x X ll lll Replacing n by

2 N

n in the right-most sum we get

,][][][][]2[

12

02 / 2

12

0

22

12

0

2 ∑

=

=

=⎟⎠

⎞⎜⎝

⎛

N

n

n N

N

N

n

n N

N

N

n

n N W n x n x W n x W n x X llll .10

2 N

l

Likewise,

,][][][][]14[1

43

)14(

14

3

2

)14(

12

4

)14(

14

0

)14( ∑

=

=

=

=

N

N n

n N

N

N n

n N

N

N n

n N

N

n

n N W n x W n x W n x W n x X lllll

where .104

N l Replacing byn

4 N

n in the second sum, byn2

N n in the third

sum, and byn4

3 N n in the fourth sum, we get

4 /

14

0

4

14

0

4 ]4

[][]14[ N N

N N

N

n

n N

n N

N

n

n N

n N W W W W

N n x W W n x X llll ∑

=

=

.][][ 4 / 33

14

0

44

32 / 2

14

0

42

N N

N N

N

n

n N

n N

N N N

N N

N

n

n N

n N

N W W W W n x W W W W n x llll ∑

=

=

Now, and Therefore,,,1 4 / 32 jW W W W N N

N N

N N

N N lll ,12 / N

N W .4 / 3 jW N N

.10,][][][][]14[44 /

14

04

342

⎭⎩⎟⎠

⎞⎜⎝

⎛ ⎠

⎞⎜⎝

⎛ ∑

=

N n N

n N

N

n

N N N W W n x n x jn x n x X ll l

Similarly, 4 / )34(

14

0

)34(4

14

0

)34( ][][]34[ N N

N

n

n N

N

N

n

n N W W n x W n x X

=

=

∑ llll

4 / 3)34(14

0

)34(4

32 / )34(14

0

)34(2

][][ N N

N

n

n N

N N N

N

n

n N

N W W n x W W n x

=

=

∑ llll

4 / 3

14

0

344

14

0

34 ][][ N N

N N

N

n

n N

n N

N

N

n

n N

n N W W W W n x W W n x lll ∑

=

=

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4 / 93

14

0

344

4 / 62

14

0

342

][][ N N

N N

N

n

n N

n N

N N N

N N

N

n

n N

n N

N W W W W n x W W W W n x llll ∑

=

=

.10,][][][][44 /

3

1

4

04

342

⎭⎩⎟⎠⎞⎜

⎝⎛

⎠⎞⎜

⎝⎛ ∑

=

N n N

n N

N

n

N N N W W n x n x jn x n x ll

The butterfly here is as shown below which is seen to require two complex multiplications.

11.33 From the flow-graph of the 8-point split-radix FFT algorithm given below it can be seenthat the total number of complex multiplications required is 2. On the other hand, the totalnumber of complex multiplications required for a standard DIF FFT algorithm is also 2.

11.34 If multiplications by are ignored, the flow-graph shown below requires 8 complexmultiplications = 24 real multiplications. A radix-2 DIF 16-point FFT algorithm, on theother hand requires 10 complex multiplications = 30 real multiplications.

1, ± j

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11.35 (a) N = 12. Choose and41 = N .32 = N Thus, and

The corresponding index mappings are indicated below:

⎩⎨⎧

≤≤≤≤+=

,,,

20304

21

21 nnnnn

⎩

⎨⎧

≤≤

≤≤+=.

,,20

3032

121

k

k k k k

0 1 2 3n1n2 x [0] x [1] x [2] x [3] x [4] x [5] x [6] x [7] x [8] x [9] x [10] x [11]

012

0 1 2 3k 1k 2 X [0] X [3] X [6] X [1] X [4] X [7] X [10] X [2] X [5] X [8] X [11]

012

X [9]

(b) N = 15. Choose and31 = N .52 = N Thus, and


⎩⎨⎧

≤≤≤≤+=

,,,

40203

21

21 n

nnnn

⎩⎨⎧

≤≤≤≤+=

.,,

40205

21

21 k

k k k k

0 1 2n1n2 x [0] x [1] x [2]

x [3] x [4] x [6]

x [5] x [7] x [8]

x [9] x [10] x [11]

0123

4 x [12] x [13] x [14]

0 1 2k 1k 2 X [0] X [5] X [10]

X [1] X [6] X [2]

X [11] X [7] X [12]

X [3] X [8] X [13]

0123

4 X [4] X [9] X [14]

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(c) N = 21. Choose and71 = N .32 = N Thus, and


⎩⎨⎧

≤≤≤≤+=

,,,

20607

21

21 nnnnn

⎩⎨⎧

≤≤≤≤+=

.,,

20603

21

21 k k k k k

0 1 2 3n1n2 x [0] x [1] x [2] x [3] x [4] x [5] x [6]

x [7] x [8] x [9] x [10] x [11]012

4 5 6

x [12] x [13] x [14] x [15] x [16] x [17] x [18] x [19] x [20]

0 1 2 3k 1k 2 X [0] X [3] X [6] X [9] X [12] X [15] X [18]

X [1] X [4] X [7] X [10] X [13]012

4 5 6

X [16] X [19] X [2] X [5] X [8] X [11] X [14] X [17] X [20]

(d) N = 35. Choose and71 = N .52 = N Thus, and


⎩⎨⎧

≤≤≤≤+=

,,,

40607

21

21 n

nnnn

⎩⎨⎧

≤≤≤≤+=

.,,

40605

21

21 k

k k k k

0 1 2 3n1n2 x [0] x [1] x [2] x [3] x [4] x [5] x [6]

x [7] x [8] x [9] x [10] x [11]012

4 5 6

x [12] x [13] x [14] x [15] x [16] x [17] x [18] x [19] x [20]

3 x [21] x [22] x [23] x [24] x [25] x [26] x [27]

4 x [28] x [29] x [30] x [31] x [32] x [33] x [34]

0 1 2 3k 1k 2 X [0] X [5] X [10] X [15] X [20] X [25] X [30]

X [1] X [6] X [11] X [16] X [21]012

4 5 6

X [26] X [31] X [2] X [7] X [12] X [17] X [22] X [27] X [32]

3 X [3] X [8] X [13] X [18] X [23] X [28] X [33]

4 X [4] X [9] X [14] X [19] X [24] X [29] X [34]

11.36 (a) N = 12. Choose and41 = N .32 = N

Thus,

,,, 93343 41 ==== −C B A

.444 31 == − D

⎩⎨⎧

≤≤≤≤+=

,,,

203043

21

1221 n

nnnn

⎩⎨⎧

≤≤≤≤+=

.,,

203049

21

1221 k

k k k k


0 1 2 3n1n2 x [0] x [3] x [6] x [9] x [4] x [7] x [10] x [1] x [8] x [11] x [2] x [5]

012

0 1 2 3k 1k 2 X [0] X [9] X [6] X [4] X [1] X [10] X [7] X [8] X [5] X [2] X [11]

012

X [3]

(b) N = 15. Choose and31 = N .52 = N ,,, 105535 31 ==== −C B A .633 5

1 == − D

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Thus,⎩⎨⎧

≤≤≤≤+=

,,,

402035

21

1521 nnnnn

⎩⎨⎧

≤≤≤≤+=

.,,

4020610

21

1521 k k k k k


0 1 2n1n2

x [0] x [5] x [10] x [3] x [8] x [6]

x [13] x [11] x [1]

x [9] x [14] x [4]

0123

4 x [12] x [2] x [7]

0 1 2k 1k 2

X [0] X [10] X [5] X [6] X [1] X [12]

X [11] X [7] X [2]

X [3] X [13] X [8]

0123

4 X [9] X [4] X [14] (c) N = 21. Choose and71 = N .32 = N

Thus,

The corresponding index mappings are indicated

below:

,,, 15533373 41 =×==== −C B A

.71777 31 =×== − D

⎩⎨⎧

≤≤≤≤+=

,,,

206073

21

2121 nnnnn

⎩

⎨⎧

≤≤

≤≤+=.

,,20

607152

12121

k

k k k k

0 1 2 3n1n2 x [0] x [3] x [6] x [9] x [12] x [15] x [18]

x [7] x [10] x [13] x [16] x [19]012

4 5 6

x [1] x [4] x [14] x [17] x [20] x [2] x [5] x [8] x [11]

0 1 2 3k 1k 2 X [0] X [15] X [9] X [3] X [18] X [12] X [6]

X [7] X [1] X [16] X [10] X [4]012

4 5 6

X [19] X [13] X [14] X [8] X [2] X [17] X [11] X [5] X [20]

(d) N = 35. Choose and71 = N .52 = N

Thus,

The corresponding index mappings are indicated

below:

,,, 15355575 71 =×==== −C B A

.213777 51 =×== − D

⎩⎨⎧

≤≤≤≤+=

,,,

406075

21

3521 n

nnnn

⎩⎨⎧ ≤≤ ≤≤+= .

,,40602115

213521 k

k k k k

0 1 2 3n1n2 x [0] x [5] x [10] x [15] x [20] x [25] x [30]

x [7] x [12] x [17] x [22] x [27]012

4 5 6

x [32] x [2] x [14] x [19] x [24] x [29] x [34] x [4] x [9]

3 x [21] x [26] x [31] x [1] x [6] x [11] x [16]

4 x [28] x [33] x [3] x [8] x [13] x [18] x [23]

0 1 2 3k 1k 2 X [0] X [15] X [30] X [10] X [25] X [5] X [20]

X [21] X [1] X [16] X [31] X [11]012

4 5 6

X [26] X [6] X [7] X [22] X [2] X [17] X [32] X [12] X [27]

3 X [28] X [8] X [23] X [3] X [18] X [33] X [13]

4 X [14] X [29] X [9] X [24] X [4] X [19] X [34]

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11.37 N = 12. Choose and41 = N .32 = N

Thus,

,,, 93343 41 ==== −C B A .444 3

1 == − D

⎩

⎨⎧

≤≤≤≤+=

,,,

203043

2

11221 n

nnnn

⎩

⎨⎧

≤≤≤≤+=

.,,

203049

2

11221 k

k k k k


0 1 2 3n1n2 x [0] x [3] x [6] x [9] x [4] x [7] x [10] x [1] x [8] x [11] x [2] x [5]

012

0 1 2 3k 1k 2 X [0] X [9] X [6] X [4] X [1] X [10] X [7] X [8] X [5] X [2] X [11]

012

X [3]

Alternately,⎩⎨⎧

≤≤≤≤+=

,,,

203049

21

1221 nnnnk

⎩⎨⎧

≤≤≤≤+=

.,,

203043

21

1221 k k k k k


0 1 2 3k 1k 2 X [0] X [3] X [6] X [9] X [4] X [7] X [10] X [1] X [8] X [11] X [2] X [5]

012

0 1 2 3n1n2 x [0] x [9] x [6] x [4] x [1] x [10] x [7] x [8] x [5] x [2] x [11]

012

x [3]

Hence, and],[][ k Y k X 22 = .,,,],)([][ 51012612 12 K=++=+ k k Y k X

11.38 (a) N = 6. Choose and21 = N .32 = N

Thus,

,,, 33323 21 ==== −C B A .422 3

1 == − D

⎩⎨⎧ ≤≤≤≤

+= ,,

, 2010

23 21621 n

nnnn ⎩⎨⎧ ≤≤

≤≤+= .

,, 20

1043 2

1621 k k

k k k


0 1n1n2 x [0] x [3]

x [2] x [5]

x [4] x [1]

012

0 1n1k 2G[0,0] G[1,0]

G[0,1] G[1,1]

G[0,2] G[1,2]

012

0 1k 1k 2 X [0] X [3]

X [4] X [1]

X [2] X [5]

012

3-ptDFT

3-ptDFT

2-ptDFT

2-ptDFT

2-ptDFT

G[0,0]

G[1,0]

G[0,1]

G[1,1]

G[0,2]

G[1,2]

x [0]

x [1]

x [2]

x [3]

x [4]

x [5]

X [0]

X [1]

X [2]

X [3]

X [4]

X [5]

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(b) N = 10. Choose and21 = N .52 = N

Thus,

,,, 55525 21 ==== −C B A .622 5

1 == − D

⎩⎨

⎧

≤≤≤≤

+= ,,

, 2040

25 21

1021 nn

nnn ⎩⎨

⎧

≤≤≤≤

+= .,

, 2040

65 21

1021 k k

k k k The corresponding index mappings are indicated below:

0 1n1n2 x [0] x [5] x [2] x [7]

01

x [4] x [9]2

0 1n1k 2G[0,0] G[1,0]G[0,1] G[1,1]G[0,2] G[1,2]

012

0 1k 1k 2 X [0] X [5] X [6] X [1] X [2] X [7]

012

x [6] x [1]3 x [8] x [3]4

G[0,3] G[1,3]3G[0,4] G[1,4]4

X [8] X [3]3 X [4] X [9]4

2-ptDFT

2-ptDFT

2-ptDFT

G[0,0]G[1,0]

x [0]

x [1]

x [2]

x [3]

x [4]

x [5]

X [0]

X [1]

X [2]

X [3]

X [4]

X [5]

2-ptDFT

2-ptDFT

5-ptDFT

5-ptDFT

x [7]

x [9]

x [6]

x [8]

X [6]

X [8]

X [7]

X [9]

G[0,1]

G[1,1]

G[0,2]

G[1,2]

G[1,3]

G[1,4]

G[0,3]

G[0,4]

(c) N = 12. Choose and31 = N .42 = N

Thus,

,,, 44434 31 ==== −C B A .333 4

1 == − D

⎩⎨⎧

≤≤≤≤+=

,,,

302034

21

1221 nnnnn

⎩⎨⎧

≤≤≤≤+=

.,,

302034

21

1221 k k k k k


0 1 2k 1n2G[0,0] G [1,0] G [2,0]G[0,1] G [1,1] G [2,1]G[0,2] G [1,2] G [2,2]

012

0 1 2n1n2 x [0] x [4] x [8] x [3] x [7] x [11] x [6] x [10] x [2]

012

x [9] x [1] x [5]3 G [0,3] G [1,3] G [2,3]3

0 1 2k 1k 2 X [0] X [4] X [8] X [3] X [7] X [11] X [6] X [10] X [2]

012

X [9] X [1] X [5]3

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G[0,0]

G[1,0] x [0]

x [1]

x [2]

x [3]

x [4]

x [5]

X [0]

X [1]

X [2]

X [3]

X [4]

X [5]

4-ptDFT

x [7]

x [9]

x [6]

x [8] X [6]

X [8]

X [7]

X [9]

3-ptDFT

3-ptDFT

3-ptDFT

4-pt

DFT

4-ptDFT

x [10]

x [11]

X [10]

X [11]

3-ptDFT

G[0,1]

G[1,1]

G[0,2]

G[1,2]

G[1,3]

G[2,1]

G[0,3]

G[2,0]

G[2,2]

G[2,3]

(d) N = 15. Choose and51 = N .32 = N

Thus,

,,, 103353 51 ==== −C B A

.655 31 == − D

⎩⎨⎧

≤≤≤≤+=

,,,

204035

21

1021 nnnnn

⎩⎨⎧

≤≤≤≤+=

.,,

2040610

21

1021 k k k k k


0 1n1n2

x [0] x [5] x [2]

x [7]

01

x [4]

x [9]

2

0 1n1k 2

G[0,0] G[1,0]G[0,1] G[1,1]G[0,2] G[1,2]

012

0 1k 1k 2 X [0]

X [5]

X [6] X [1]

X [2] X [7]

012

x [6]

x [1]

3

x [8] x [3]

G[3,1]G[2,1]

3

G[2,2] G[3,2]

4

X [8]

X [3]

3

X [4] X [9]

2

x [12] x [11] x [14]

x [10] x [13]

2

G[2,0] G[3,0]

2 4

X [10] X [12]

X [11] X [14] X [13]

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G[0,0]

G[1,0]

x [0]

x [1]

x [2]

x [3]

x [4]

x [5]

X [0]

X [1]

X [2]

X [3]

X [4]

X [5]

x [7]

x [9]

x [6]

x [8]

X [6]

X [8]

X [7]

X [9]

G[0,1]

G[1,1]

G[0,2]

G[1,2]

G[1,3]

G[1,4]

G[0,3]

G[0,4]

5-ptDFT

5-ptDFT

5-ptDFT

3-ptDFT

3-ptDFT

3-ptDFT

3-ptDFT

3-ptDFT x [10]

x [13]

x [11]

x [14]

x [12]

X [14]

X [11]

X [12]

X [13]

X [10]

G[2,4]

G[2,3]

G[2,2]

G[2,1]G[2,0]

11.39 Note that 1536 = 256 × 6. Now an -point DFT , with divisible by 6, can be computed

as follows:

where

N N

][][][][][ 6 / 22

6 / 1

1

06 / 0 N

k N N

k N

N

n N

nk N k X W k X W k X W n x k X ∑

=],[][][ 6 / 5

56 / 4

46 / 3

3 N

k N N

k N N

k N k X W k X W k X W

.50,]6[][

16

06 / 6 / ≤∑

=lll

N

r

rk N N W r x k X For ,1536 N we thus get

where

][][][][][ 25633

153625622

1536256115362560 k X W k X W k X W k X k X k k k

],[][ 25655

153625644

1536 k X W k X W k k .50,]6[][5111

0256256 ≤∑

=lll

r

rk W r x k X

X [k ]

768k W

6 z

x [n] DFT256-point

z

z

z

6

6

6

6

6

z

DFT256-point

DFT256-point

DFT256-point

DFT256-point

DFT256-point

x [n]000

x [n]001

x [n]010

x [n]011

x [n]100

x [n]101

768k W

768k W

768k W

1536k W

k ] X [ 256000

k ] X [ 256001

k ] X [ 256010

k ] X [ 256100

k ] X [ 256101

k ] X [ 256011

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Now an N -point FFT algorithm requires N N

22log complex multiplications and

complex additions. Hence, an

N N 2log

6 N -point FFT algorithm requires )6 / (log 212

N N complex

multiplications and )6 / (log 26

N N complex additions. In addition, we need complex

multiplications and complex additions to compute the N -point DFT X[k]. Hence, for N =

1536, the evaluation of using six 256-point FFT modules requires

N 5

N 5

][k X N N N 5)6 / (log 212

complex multiplications and870415365)256(log128 2 = N N N 5)6 / (log 26

972815365)256(log256 2 = complex additions. It should be noted that a directcomputation of the 3072-point DFT would require 6431296 complex multiplications and2357760 complex additions.

11.40 (a) # of zero-valued samples to be added is 512 – 498 = 14.

(b) Direct computation of a 512-point DFT of a length-498 sequence requires (498) 2 =248004 complex multiplications and 497 ×498 = 247506 complex additions .(c) A 512-point Cooley-Tukey type FFT algorithm requires 2304)512(log256 2 = complexmultiplications and 46308)512(log512 2 = complex additions.

11.41 Hence, Since.ll α= z .lll α=φ−θ− oo j j

oo eeV A α is real, we have ,/, α== 11 oo V A

., 00 =φ=θ oo 11.42 (a) or)()()( z X z H zY = ).]1[]0[)(]1[]0[(]2[]1[]0[ 1121 z x x zhh z y z y y Now,

]),1[]0[])(1[]0[()1()1(]2[]1[]0[)1()( 0 x x hh X H y y yY zY ],0[]0[)()(]0[)()( 1 x h X H yY zY =

}).1{]0[])(1[]0[{)1()1(]2[]1[]0[)1()( 2 x x hh X H y y yY zY

From Eqs. (6.114) and (6.115), we arrive at

),()(

)()(

)()(

)()(

)()( 2

22

21

11

10

00

0 zY z I z I

zY z I z I

zY z I

z I zY where ),1)(1()( 1

21

10 z z z z z I

),1)(1()( 12

101

z z z z z I ).1)(1()( 11

102

z z z z z I Therefore,

),1()1)(1(

)1)(1()(

)(112

11

021

01

1

2

1

100

0 = z z z z z z

z z z z z I

z I

),1()1)(1(

)1)(1()()( 2

112

110

12

10

11

1 = z z z z z

z z z z

z I z I

and

).1()1)(1(

)1)(1()(

)( 1121

121

120

11

10

02

2 = z z z z z z

z z z z

z I z I

Hence,

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)()1()()1()()1()( 211

21

12

011

21

zY z z zY z zY z z zY

222

1102

1122

102

11 )()()()()()(

⎟⎠

⎞⎜⎝

⎛ ⎠

⎞⎜⎝

⎛ z zY zY zY z zY zY zY

1

2

1

2

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

⎠

⎞⎜

⎝

⎛ z x x hh x x hh x h

221

21 ])1[]0[])(1[]0[(]0[]0[])1[]0[])(1[]0[(

⎟⎠

⎞⎜⎝

⎛ z x x hh x h x x hh

.]1[]1[])0[]1[]1[]0[(]0[]0[ 21 z x h z x h x h x h

Ignoring the multiplications by ,21 computation of the coefficients of require the values

of and , which can be evaluated using only 3 multiplications.

)( zY

)( 2 zY ),(),( 10 zY zY

(b) or)()()( z X z H zY = ]2[]1[]0[]2[]1[]0[]4[]3[]2[]1[]0[ 21214321 z x z x x zh zhh z y z y z y z y y

Now,

]),2[4]1[2]0[])(2[4]1[2]0[()()(21

0 x x x hhhY zY

]),2[]1[]0[])(2[]1[]0[()1()( 1 x x x hhhY zY ],0[]0[)()( 2 x hY zY = ]),2[]1[]0[])(2[]1[]0[()1()( 3 x x x hhhY zY

]).2[4]1[2]0[])(2[4]1[2]0[()()(21

4 x x x hhhY zY

From Eqs. (6.114) and (6.115), we arrive at

),()(

)()(

)()(

)()(

)()(

)()(

)()(

)()( 4

44

43

33

32

22

21

11

10

00

0 zY z I

z I zY

z I

z I zY

z I

z I zY

z I

z I zY

z I

z I zY where

),1)(1)(1)(1()( 141312110 z z z z z z z z z I

),1)(1)(1)(1()( 14

13

12

101

z z z z z z z z z I

),1)(1)(1)(1()( 14

13

11

102

z z z z z z z z z I

),1)(1)(1)(1()( 14

12

11

103

z z z z z z z z z I

).1)(1)(1)(1()( 13

12

11

104

z z z z z z z z z I Therefore,

),1)(1()1)(1)(1)(1(

)1)(1)(1)(1()(

)( 21211

121

104

103

102

101

14

13

12

11

00

0 = z z z z z z z z z z z

z z z z z z z z z I z I

),1)(1()1)(1)(1)(1()1)(1)(1)(1(

)()( 2

4111

321

141

131

121

10

14

13

12

10

111 = z z z

z z z z z z z z z z z z z z z z

z I z I

),1)(1()1)(1)(1)(1(

)1)(1)(1)(1()(

)( 2412

124

123

121

120

14

13

11

10

22

2 = z z z z z z z z z z

z z z z z z z z

z I z I

),1)(1()1)(1)(1)(1(

)1)(1)(1)(1()(

)( 24111

32

134

132

131

130

14

12

11

10

33


z z z z z z z z

z I

z I

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).1)(1()1)(1)(1)(1(

)1)(1)(1)(1()(

)( 21211

121

143

142

141

140

13

12

11

10

44


z z z z z z z z

z I z I

Hence,

)())())( 14

413

4121(

32

04

4132

211(

121 zY z z z z zY z z z z zY

)())()1( 3441

341

21(32

2441

245

zY z z z z zY z z

)() 44

2131

211(

121 zY z z z z

1412

133

213

2012

12 )()()()()(

⎟⎠

⎞⎜⎝

⎛ z zY zY zY zY zY

2424

133

224

513

2024

1 )()()()()( ⎟⎠

⎞⎜⎝


3412

136

116

1012

1 )()()()( ⎟⎠

⎞⎜⎝

⎛ z zY zY zY zY

.)()()()()( 4424

1

36

1

24

1

16

1

024

1 ⎟

⎠

⎞⎜

⎝


Substituting the expressions for and in the aboveequation, we then arrive at the expressions for the coefficients in terms of thecoefficients and Thus,

),(),(),(),( 3210 zY zY zY zY ),( 4 zY ]}[{ n y

]}[{ nh ]}.[{ n x ],0[]0[)(]0[ 2 x h zY y =

],0[]1[]1[]0[)()()()(]1[ 4121

332

132

0121

x h x h zY zY zY zY y

],0[]2[]1[]1[]2[]0[)()()()()(]2[ 4241

332

245

132

0241

x h x h x h zY zY zY zY zY y

],1[]2[]2[]1[)()()()(]3[ 4121

361

161

0121

x h x h zY zY zY zY y

].2[]2[)()()()()(]4[ 424136

124116

10241 x h zY zY zY zY zY y =

Hence, ignoring the multiplications by ,,,,,61

41

45

32

121 and ,

241 computation of the

coefficients of require the values of and which can be evaluated using only 5 multiplications.

)( zY ),(),(),(),( 3210 zY zY zY zY ),( 4 zY

11.43 or)()()( z X z H zY = 1121 ]1[]0[]1[]0[]2[]1[]0[ z x x zhh z y z y y

.]2[[]2[])0[]1[]1[]0[(]0[]0[ 21 z x h z x h x h x h Hence, ],0[]0[]0[ x h y = ].1[]1[]2[],0[]1[]1[]0[]1[ x h y x h x h y = Now,

].1[]0[]1[]1[]0[]1[]1[]0[]0[])1[]0[])(1[]0[( y x h x h x h x h x x hh = As a result,evaluation of requires the computation of 3 products, , and

. In addition, it requires 4 additions,)()( z X z H ]1[]1[],0[]0[ x h x h

])1[]0[])(1[]0[( x x hh ],1[]0[],1[]0[ x x hh and]1[]1[]0[]0[])1[]0[])(1[]0[( x h x h x x hh .

11.43 Let the two length- sequences be denoted by and Denote the sequence]}[{ nh ]}.[{ n x N

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generated by the linear convolution of and][nh ][n x as Let and].[n y ∑ =

= 10

][)( N

nn znh z H

.][)( 1

0

=

= N n

n zn x z X Rewrite and in the form and

where

)( z H )( z X ),()()( 12 /

0 z H z z H z H N

),()()( 12 /

0 z X z z X z X N ,][)( 1)2 / (

00 ∑ =

= N n

n znh z H

,][)( 1)2 / (

0 21 ∑ =

N n

n N znh z H ,][)(

1)2 / (00 ∑

= = N

nn zn x z X

.][)( 1)2 / (

0 21 ∑ =

N n

n N zn x z X Therefore, we can write

)()()()()( 12 /

012 /

0 z X z z X z H z z H zY N N

)()()()()()()()( 1101102 /

00 z X z H z z X z H z X z H z z X z H N N

where),()()( 212 /

0 zY z zY z zY N N ),()()( 000 z X z H zY =

and),()()()()( 01101 z X z H z X z H zY ).()()( 112 z X z H zY = Note that and are

products of two polynomials of degree

)(0 zY )(1 zY

2 N

each, and hence, require2

2 ⎟⎠⎞⎜⎝⎛ N

multiplications each.

Now, we can write ).()()()()()()( 1010101 zY zY z X z X z H z H zY Since,

is a product of two polynomials of degree)()()()( 1010 z X z X z H z H 2

N each, it can be

computed using2

2⎟⎠

⎞⎜⎝

⎛ N multiplications. As a result, )()()( z X z H zY = can be computed using

2

23 ⎟⎠

⎞⎜⎝

⎛ N multiplications instead of multiplications. If is a power-of-2,2 N N 2

N is even, and

the same procedure can be applied to compute and reducing further the

number of multiplications. This process can be continued until the sequences beingconvolved are of length 1 each.

),(),( 10 zY zY ),(2 zY

Let )( N R denote the total number of multiplications required to compute the linearconvolution of two –length sequences. Then, in the method outlined above, we have N

)2 / (3)( N N R R ⋅ with .1)1( =R A solution of this equation is given by .3)( 2log N N =R

11.45 The dynamic range of a signed –bit integer is given by B η )12()12( )1()1( B B η

which for is given by32 B ).12()12( 3131 η

(a) For and the value of a 32-bit floating-point number is given by

Hence, the value of the largest number is and the value of the

smallest number is The dynamic range is therefore

6 E ,25 M

).(2)1( 31 M E s η ,232≈.232 .22 32×

(b) For 7 E and the value of a 32-bit floating-point number is given by,24 M

).(2)1( 63 M E s η Hence, the value of the largest number is and the value of the,264≈

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smallest number is The dynamic range is therefore.264 .22 64× (c) For and the value of a 32-bit floating-point number is given by

Hence, the value of the largest number is and the value of the

smallest number is The dynamic range is therefore

8 E ,23 M

).(2)1( 127 M E s η ,2128≈

.2128

.22 128

× Hence, the dynamic range in a floating-point representation is much larger than that in a fixed-

point representation with the same wordlength.

11.46 A 32-bit floating-point number in the IEEE Format has E = 8 and M = 23. Also, theexponent E is coded in a biased form as 127 E with certain conventions for special casessuch as E = 0, 255, and M = 0 (See page 637 of text).

Now, a positive 32-bit floating-point number represented in the “normalized” form have anexponent in the range 0 < E < 255, and is of the form ).1(2)1( 127 M E s

∆ η Hence, the

smallest positive number that can be represented will have E = 1, and

and has therefore a value given by

,000022

4 4 34 4 21 Kbits

M =

.1018.12 38126 × For the largest positive number, E = 254, and Thus, here.1111

224 4 34 4 21 K

bits

M = 22)11111(2)1( 127

22

1270 ×4 4 34 4 21 K

bits

∆η

.104.3 38× Note: For representing numbers less than ,2 126 IEEE format uses the “de-normalized”

form where E = 0, and ).0(2)1( 126 M s∆

η In this case, the smallest positive number

that can be represented is given by 149

22

1260 2)10000(2)1( ≈4 4 34 4 21

K

bits

∆η

.104013.1 45× 11.47 For a two’s-complement binary fraction the decimal equivalent for

is simply For

,21 baaas K∆

0s .21

ibi ia = ,1s the decimal equivalent is given by

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

bb

i

iia 22)1(

1

=

=

∑ )21(22211

bbb

i

ii

b

i

i a bib

iia

= 22

1

Hence, the decimal equivalent of is given by.211

ib

iia

= baaas K21∆

.21

ib

iias

=

11.48 For a ones’-complement binary fraction the decimal equivalent for,21 baaas K∆

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0s is simply For.21

ibi ia = ,1s the decimal equivalent is given by

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ∑

=

b

i

iia

1

2)1( =

=

∑ )21(2211

bb

i

ii

b

i

i a ib

iia

= 2

1

Hence, the decimal equivalent of is given by.211

ib

iia

= baaas K21∆

.2)21(1

ib

ii

b as

= ∑

11.49 .01111153125.0,10110068750.0,11001078125.0 103102101 ∆∆∆ =ηηη

.01111110110011001021 ∆∆∆ =ηη

Dropping thee overflow bit and adding )( 21 ηη to :3η

,9375.0111100011111011110)( 10321 =∆∆∆nηη where we have dropped the carry bit in the MSB location. Note that the final sum is correctinspite of the overflow .

11.49 The transformation ω β αω ˆcoscos is equivalent to ,22

ˆˆ

⎟⎠

⎞

⎜⎝

⎛ ωωωω β α

j j j j eeee

which by analytic continuation can be expressed as .2ˆˆ

2

11

⎟⎠

⎞

⎜⎝

⎛ z z z z β α Now, let

be a Type 1 linear-phase FIR transfer function of degree As indicated in Eq. (8.128),

can be expressed as

)( z H

.2 M

)( z H ∑ = ⎟⎠

⎞⎜⎝

⎛ = M n

n M z zna z z H

01

2][)( with a frequency response

given by with,)](cos[)(0=

= M n

n jM j naee H ωωω ∑ = M n

nna H 0

)](cos[)( ωω(

denoting

the amplitude function or the zero-phase frequency response. The amplitude function or thezero-phase frequency response of the transformed filter obtained by applying the mapping

is therefore given byω β αω ˆcoscos .)ˆcos]([)ˆ(0= M

nnna H ω β αω

( Or, equivalently,

the transfer function of the transformed filter is given by

∑ =

⎟⎠

⎞

⎜⎝

⎛ M n

n M z z

na z z H 0

1

2ˆˆ

][)ˆ( β α A convenient way to realize is to consider the

realization of the parent transfer function in the form of a Taylor structure as outlined inProblem 8.17 which is obtained by expressing in the form

)ˆ( z H

)( z H )( z H

.2

1][)(

0

2∑ =

⎟⎠

⎞

⎜⎝

⎛ = M n

nn M z

zna z H Similarly, the transformed filter can be realized by

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replacing each block2

1 2 z in the Taylor structure by the block .

2

ˆ1ˆ

21

z z β α

Now, for a lowpass-to-lowpass transformation we can impose the condition.)()ˆ(

00ˆ ==ωω

ωω H H ((

This condition is met if 1 β α and .10 <α In this case, the

transformation reduces to .ˆcos)1(cos ωααω From the plot of the mapping given below, itfollows that as is varied between 0 and 1,α .ˆ cc ωω <

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

cos ω

c o s ω

^

On the other hand if is desired along with a lowpass-to-lowpass transformation, we

can impose the condition

.ˆ cc ωω >.)()ˆ(

ˆ π ωπ ωωω == H H

(( This condition is met if and

The corresponding transformation is now given by Fromthe plot of the mapping given below, it follows that as is varied between 0 and 1,

α β 1

.11 ≤α .ˆcos)1(cos ωααω α .ˆ cc ωω >

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

cos ω

c o s ω

^

11.51 (a) .][)( ∑−

=

−=1

0

N

n

n z n x z X ,)()(

][)()( z D z P

z

z n x z X z X

N

n

n

z

z z (

(

(

((

(

( =⎟⎟

⎠

⎞⎜⎜

⎝

⎛

α−+α−== ∑

−

= −

−

α−+α−= −

−−1

01

1

1 11

11

where ,)()]([][)( nn N N

n

N

n

n z z n x z n p z P 111

0

11

01 −−−−

=

−−

=

− +α−α−== ∑∑ (((( and

.)(][)( 111

01 −−−−

=α−== ∑ N n

N

n z z nd z D (((

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(b) ,][][

)()(

)(][/

/k Dk P

z D z P

z X k X N k j

N k j

e z e z (

(

((

((

(( ===

ππ

== 2

2 where N k je z z P k P /)(][ π== 2(((

is

the –point DFT of the sequence and N ][n p N k je z z Dk D /)(][ π== 2(((

is the –point DFT

of the sequence

N

].[nd

(c) Let and[ ]t N p p p ][][][ 110 −= LP [ ].][][][ t N x x x 110 −= LX Without anyloss of generality, assume in which case4= N

( )][][][][][)( 3210 323

0 x x x x z n p z P

n

n α−α+α−== ∑=

−((

( 1222 332212103 −α+α+α−α++α−+ z x x x x (][][)(][)(][

( ) 2222 33211203 −α−α+−α+α−α+ z x x x x (][][)(][)(][

( ) .][][][][ 323 3210 −α+α−α+α−+ z x x x x ( Equating like powers of 1− z ( we can write

or It can be shown

that the elements

XQP ⋅= .

][][][][

)(][][][][

⎢⎢⎢⎢

⎣

⎡

αα−α+α−α

α+α−α−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

23213

1

3210

23

22

2

p p p p

)(

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

α−α−α+

αα+α−α−α

3210

1321

322

2232

x x x x

,,,, 30 ≤≤ sr q sr of the 4 ×4 matrix Q can be determined as follows:

(i) The first row is given by ,)(, s

sq α−=0

(ii) The first column is given by ,)()!(!

!)(,

r r r r r r

C q α−−

=α−=333

0 and

(iii)

The remaining elements can be obtained using the recurrence relation.,,,, sr sr sr sr qqqq 1111 −−−− α+α−=

In the general case, we only change the computation of the elements of the first column

using the relation .)()!(!

)!()(,

r r r

N r r N r

N C q α−

−−−=α−= −111

0

11.52 The highpass transfer function can be expressed as [ ],)()()( 1021 z A z A z H herew

1

1

03038.01

3038.0)( =

z

z z A and

21

21

1639.04012.01

4012.0639.0)( =

z z

z z z A . The tunable highpass

transfer function is obtained by applying the lowpass-to-lowpass transformation of Eq.

(11.115) where the tuning parameter is given by .]2 / )ˆ6.0sin[(]2 / )ˆ6.0sin[(

p

p

ωπ ωπ α

= The tunable

highpass transfer function is then given by [ ],)(ˆ)(ˆ),( 1021 z A z A z H α where from Eqn.

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(11.119)1

1

0]9077.03038.0[1

]9077.03038.0[)(ˆ

= z

z z A

α

α and from Eqn. (11.123)

.]1448.0639.0[]439.34012.0[1

]439.34012.0[]1448.0639.0[)(ˆ

21

211 =

z z

z z z A

αα

αα

11.53 The lowpass transfer function can be expressed as [ ],)()()( 1021

z A z A z H where

1

1

01584.01

1584.0)( =

z

z z A and .

3554.04191.01

4191.03554.0)(

21

21

1 = z z

z z z A The tunable bandpass

transfer function is obtained by applying the lowpass-to-bandpass transformation of Eq.(11.124) where the tuning parameter is given by oω β cos with denoting the centerfrequency of the bandpass filter. The tunable bandpass transfer function is then given by

oω

[ ],)(ˆ)(ˆ),( 1021

z A z A z H α where⎟⎠

⎞

⎜⎝

⎛

⎟⎠

⎞

⎜⎝

⎛

= 1

1

1

1

11

11

0

1584.01

1584.0

)(ˆ

z

z

z

z

z

z

z A

β

β

β

β

and

.

3554.04191.01

4191.03554.0

)(ˆ2

12

11

2

12

11

1

1

1

1

1

1

1

1

1

⎟⎠

⎞

⎜⎝

⎛ ⎟⎠

⎞

⎜⎝

⎛

⎟⎠

⎞

⎜⎝

⎛ ⎟⎠

⎞

⎜⎝

⎛ =

z

z

z

z

z

z

z

z

z z

z z

z A

β

β

β

β

β

β

β

β

M11.1 (a) Numerator coefficients =[0.0528 0.0797 0.1295 0.1295 0.0797 0.0528] Denominator coefficients =[1.0000 -1.8107 2.4947 -1.8801 0.9537 -0.2336]

0 5 10 15 20 25 30-0.2

0

0.2

0.4

0.6

Time index n

m p

u e

Impulse response samples

(b) Numerator coefficients =[0.0084 -0.0335 0.0502 -0.0335 0.0084]Denominator coefficients =

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[1.0000 2.3741 2.7057 1.5917 0.4103]

0 5 10 15 20 25 30-0.4

-0.2

0

0.2

0.4

Time index n

m p u

e


(c) Numerator coefficients = [0.0003 0 -0.0019 0 0.0057 0-0.0095 0 0.0095 0 -0.0057 0 0.0019 0 -0.0003]Denominator coefficients = [1.0000 1.7451 4.9282 6.11959.8134 9.2245 10.4323 7.5154 6.4091 3.4595

2.2601 0.8470 0.4167 0.0856 0.0299]

0 5 10 15 20 25 30

-0.4

-0.2

0

0.2

0.4

Time index n

m p

u e


M11.2 The modified MATLAB program is given below:

n = 0:60; w = input('Normalized angular frequency vector = ');

num = input('Numerator coefficients = ');den = input('Denominator coefficients = ');x1 = cos(w(1)*pi*n); x2 = cos(w(2)*pi*n);x = x1+x2;% Generate the output sequence by filtering the inputy = filter(num,den,x);% Plot the input and the output sequencesfigure(1)stem(n,x);xlabel('Time index n'); ylabel('Amplitude');title('Input sequence');figure(2)stem(n,y);xlabel('Time index n'); ylabel('Amplitude');

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title('Output sequence');

The plots generated by the above program for the filter of Example 9.14 for an inputcomposed of a sum of two sinusoidal sequences of angular frequencies, 0.3 π and 0.6 π, aregiven below:

0 10 20 30 40 50 60-2

-1

0

1

2

Time index n

A m p

l i t u d e

Input sequence

0 10 20 30 40 50 60

-1

-0.5

0

0.5

1

Time index n

m p

u e

Output sequence

The blocking of the high-frequency signal by the lowpass filter can be demonstrated byreplacing the statement stem(n,x); with stem(n,x2); and the statementy=filter(num,den,x); with y=filter(num,den,x2); . The plots of the high-frequency input signal and the corresponding output are shown below:

0 10 20 30 40 50 60-1

-0.5

0

0.5

1

Time index n

m p

u e

Input sequence

0 10 20 30 40 50 60

-0.2

-0.1

0

0.1

0.2

0.3

Time index n

m p

u e

Output sequence

M11.3 The plots generated by the program of Exercise M11.2 for the filter of Example 9.15for an input composed of a sum of two sinusoidal sequences of angular frequencies, 0.3 π and0.6π, are given below:

0 10 20 30 40 50 60-2

-1

0

1

2

Time index n

A m p

l i t u d e

Input sequence

0 10 20 30 40 50 60

-0.6

-0.4

-0.2

0

0.2

0.4

Time index n

m p

u e

Output sequence

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The blocking of the low-frequency signal by the highpass filter can be demonstrated byreplacing the statement stem(n,x); with stem(n,x1); and the statementy=filter(num,den,x); with y=filter(num,den,x1); . The plots of the low-frequency input signal and the corresponding output are shown below:

0 10 20 30 40 50 60-1

-0.5

0

0.5

1

Time index n

m p

u e

Input sequence

0 10 20 30 40 50 60

-0.2

-0.1

0

0.1

0.2

Time index n

m p

u e

Output sequence

M11.4 % The factors for the transfer function of% the lowpass filter of Example 9.14 are% num1 = [0.0528 0.0528 0];% den1 = [1.0000 -0.4909 0];% num2 = [1.0000 0.6040 1.0000];% den2 = [1.0000 -0.7624 0.5390];% num3 = [1.0000 -0.0949 1.0000];% den3 = [1.0000 -0.5574 0.8828];N = input( ‘Total number of sections = ‘ );for k = 1:N;

num(k,:) = input( 'Numerator factor = ' );den(k,:) = input( 'Denominator factor = ' );

endn = 0:60;

w = input( 'Normalized angular frequency vector = ' );x1 = cos*w(1)*pi*n); x2 = cos*w(2)*pi*n);x = x1 + x2;% Plot the input sequencefigure(1)stem(n,x);xlabel( 'Time index n' ); ylabel( 'Amplitude' );title( 'Input sequence' );si = [0 0];for k = 1:N;

y(k,:) = filter(num(k,:),den(k,:),x,si);x = y(k,:);

end% Plot the output sequencefigure(2)stem(n,y(N,:));xlabel( 'Time index n' ); ylabel( 'Amplitude' );title( 'Output sequence' );

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The plots generated by the above program for the filter of Example 9.14 for an inputcomposed of a sum of two sinusoidal sequences of angular frequencies, 0.3 π and 0.6 π, aregiven below:

0 10 20 30 40 50 60-2

-1

0

1

2

Time index n

A m p

l i t u d e

Input sequence

0 10 20 30 40 50 60

-1

-0.5

0

0.5

1

Time index n

m p

u e

Output sequence

M11.5 % The factors for the transfer function of

% the highpass filter of Example 9.15 are% num1 = [0.0084 -0.0167 0.0084];% den1 = [1.0000 1.3101 0.5151];% num2 = [1.0000 -2.0000 1.0000];% den2 = [1.0000 1.0640 0.7966];

The plots generated by the program of Exercise M11.4 for the filter of Example 9.15 for aninput composed of a sum of two sinusoidal sequences of angular frequencies, 0.3 π and 0.6 π,are given below:

0 10 20 30 40 50 60-2

-1

0

1

2

Time index n

A m p

l i t u d e

Input sequence

0 10 20 30 40 50 60

-0.6

-0.4

-0.2

0

0.2

0.4

Time index n

m p

u e

Output sequence

M11.6 To apply the function direct2 to filter a sum of two sinusoidal sequences, we replacethe statement y = filter(num,den,x,si); in the MATLAB program given in thesolution of Exercise M11.2 with the statement y = direct2(num,den,x,si); . The

plots generated by running the modified program for the data given in this problem are given below:

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0 10 20 30 40 50 60

-2

-1

0

1

2

Time index n

A m p

l i t u d e

Input sequence

0 10 20 30 40 50 60

-1

-0.5

0

0.5

1

Time index n

m p

u e

Output sequence

M11.7 The MATLAB program that can be used to compute all DFT samples using the functiongoertzel and the function fft is as follows:

clearN = input( 'Desired DFT length = ' ) ;x = input( 'Input sequence = ' );for j = 1:N

Y(j)=goertzel(x,j);enddisp( 'DFT samples computed using goertzel are ' ) ;disp(Y)disp( 'DFT samples computed using fft are ' ) ;X = fft(x); disp(X);

Results obtained for two input sequences of lengths 8 and 12, respectively, are given below:

Desired DFT length = 8Input sequence = [1 2 3 4 4 3 2 1]

DFT samples computed using goertzel areColumns 1 through 420.0000 -5.8284-2.4142i -0.0000-0.0000i -0.1716-0.4142iColumns 5 through 80-0.0000i -0.1716+0.4142i 0.0000-0.0000i -5.8284+2.4142i

DFT samples computed using fft areColumns 1 through 420.0000 -5.8284-2.4142i 0 -0.1716-0.4142iColumns 5 through 80 -0.1716+0.4142i 0 -5.828 +2.4142i

Desired DFT length = 12Input sequence = [1 2 4 8 10 12 7 3 -4 5 0 6]

DFT samples computed using goertzel areColumns 1 through 454.0000 -13.0622-21.0885i 1.5000+19.9186i -4.0000-2.0000iColumns 5 through 8

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4.5000+2.5981i -0.9378+10.0885i -18.0000+0.0000i-0.9378-10.0885i

Columns 9 through 124.5000-2.5981i -4.0000+2.0000i 1.5000-19.9186i -

13.0622 +21.0885i

DFT samples computed using fft are54.0000 -13.0622-21.0885i 1.5000+19.9186i -4.0000-2.0000iColumns 5 through 8

4.5000+2.5981i -0.9378+10.0885i -18.0000+0.0000i-0.9378-10.0885i

Columns 9 through 124.5000-2.5981i -4.0000+2.0000i 1.5000-19.9186i -13.0622 +21.0885i

M11.8 The MATLAB program that can be used to verify the plots of Figure 11.43 is given below:

%Program_11_8.m[z,p,k] = ellip(5,0.5,40,0.4);a = conv([1 -p(1)],[1 -p(2)]);b =[1 -p(5)];c = conv([1 -p(3)],[1 -p(4)]);

w = 0:pi/255:pi;alpha = 0;an1 = a(2) + (a(2)*a(2) - 2*(1 + a(3)))*alpha;an2 = a(3) + (a(3) -1)*a(2)*alpha;g = b(2) - (1 - b(2)*b(2))*alpha;cn1 = c(2) + (c(2)*c(2) - 2*(1 + c(3)))*alpha;cn2 = c(3) + (c(3) -1)*c(2)*alpha;a = [1 an1 an2];b = [1 g]; c = [1 cn1 cn2];h1 = freqz(fliplr(a),a,w); h2 = freqz(fliplr(b),b,w);h3 = freqz(fliplr(c),c,w);ha = 0.5*(h1.*h2 + h3);ma = 20*log10(abs(ha));alpha = 0.1;an1 = a(2) + (a(2)*a(2) - 2*(1 + a(3)))*alpha;an2 = a(3) + (a(3) -1)*a(2)*alpha;g = b(2) - (1 - b(2)*b(2))*alpha;cn1 = c(2) + (c(2)*c(2) - 2*(1 + c(3)))*alpha;cn2 = c(3) + (c(3) -1)*c(2)*alpha;a = [1 an1 an2];b = [1 g]; c = [1 cn1 cn2];h1 = freqz(fliplr(a),a,w); h2 = freqz(fliplr(b),b,w);h3 = freqz(fliplr(c),c,w);hb = 0.5*(h1.*h2 + h3);mb = 20*log10(abs(hb));alpha = -0.25;an1 = a(2) + (a(2)*a(2) - 2*(1 + a(3)))*alpha;an2 = a(3) + (a(3) -1)*a(2)*alpha;g = b(2) - (1 - b(2)*b(2))*alpha;cn1 = c(2) + (c(2)*c(2) - 2*(1 + c(3)))*alpha;cn2 = c(3) + (c(3) -1)*c(2)*alpha;

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a = [1 an1 an2];b = [1 g]; c = [1 cn1 cn2];h1 = freqz(fliplr(a),a,w); h2 = freqz(fliplr(b),b,w);h3 = freqz(fliplr(c),c,w);hc = 0.5*(h1.*h2 + h3);mc = 20*log10(abs(hc));plot(w/pi,ma,'r-',w/pi,mb,'b--',w/pi,mc,'g-.');axis([0 1 -80 5]);xlabel( 'Normalized frequency' );ylabel ('Gain, dB' );legend( 'b--' , '\alpha = 0.1 ' , 'r-' , '\alpha = 0 ' , 'g-.' , '\alpha = -0.25' );

M11.9 The MATLAB program that can be used to verify the plots of Figure 11.44 is given below:%Program_11_9.m

w = 0:pi/255:pi; wc2 = 0.31*pi;

f = [0 0.36 0.46 1];m = [1 1 0 0];b1 = remez(50, f, m);h1 = freqz(b1,1,w);

m1 = 20*log10(abs(h1));n = -25:-1;c = b1(1:25)./sin(0.41*pi*n);d = c.*sin(wc2*n);q = (b1(26)*wc2)/(0.4*pi);b2 = [d q fliplr(d)];h2 = freqz(b2,1,w);

m2 = 20*log10(abs(h2)); wc3 = 0.51*pi;

d = c.*sin(wc3*n);q = (b1(26)*wc3)/(0.4*pi);b3 = [d q fliplr(d)];h3 = freqz(b3,1,w);

m3 = 20*log10(abs(h3));plot(w/pi,m1,'r-',w/pi,m2,'b--',w/pi,m3,'g-.');axis([0 1 -80 5]);xlabel( 'Normalized frequency' );ylabel( 'Gain, dB' );legend('b--','\omega_c = 0.3\pi','r-','\omega_c = 0.41\pi','g-.','\omega_c = 0.51\pi')

M11.10 The MATLAB program to evaluate Eqs. (11.155) is given below:%Program_11_10.mx = 0:0.001:0.5;y = 3.140625*x + 0.0202636*x.^2 - 5.325196*x.^3 +0.5446778*x.^4 + 1.800293*x.^5;x1 = pi*x;z = sin(x1);plot(x,y);xlabel( 'Normalized angle, radians' );ylabel( 'Amplitude' );title( 'Approximate sine values' );grid;axis([0 0.5 0 1]);pauseplot(x,y-z);xlabel( 'Normalized angle, radians' );ylabel( 'Amplitude' );title( 'Error of approximation' );grid;

The plots generated by the above program are as indicated below:

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0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

Normalized angle, radians

A m p

l i t u d e

Approximate sine values

0 0.1 0.2 0.3 0.4 0.5

-2

-1

01

2

3

4x 10 -5

Normalized angle, radians

A m p

l i t u d e

Error of approximation

M11.11 The MATLAB program to evaluate Eqs. (11.156) and (11.135) is given below:%Program_11_11.mk = 1;for x = 0:.01:1

op1 = 0.318253*x+0.00331*x^2-0.130908*x^3+0.068524*x^4-0.009159*x^5;op2 = 0.999866*x-0.3302995*x^3+0.180141*x^5-

0.085133*x^7+0.0208351*x^9;arctan1(k) = op1*180/pi;arctan2(k) = 180*op2/pi;actual(k) = atan(x)*180/pi;k = k+1;

endsubplot(211)x = 0:.01:1;plot(x,arctan2);ylabel( 'Angle, degrees' );xlabel( 'Tangent Values' );subplot(212)plot(x,actual-arctan2,'--');ylabel( 'Tangent Values' );xlabel( 'error, radians' );

The plots generated by the above program are as indicated below:

0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

A n g

l e ,

d e g r e e s

Tangent Values 0 0.2 0.4 0.6 0.8 1

-1

-0.5

0

0.5

1x 10 -3

a n g e n

a u e s

error, radians

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Chapter11 SM[1]