fratstock.eu · 2017-05-08 · Full file at 2.7 (a) ya ( )tx 12 1 0yt ayt bxt ( ) ( ) ( ) ( ) initial condi bt tions (b) The poles are at 2 sa a 11 0 a 4 (

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2.1

(a) 2

0

E T

P

(b) 1 210 2

0

E T

P

T

(c)

1

2

E

P

(d) 2

2

E

AP

(e)

3

8

E

P

(f)

0

E T

P

(g) 1

40

Ea

P

(h) 2 3

40

A TE

P

(i)

1/3 1/33 32/3

3

3 3 3 3 31lim lim 3 lim 0

Tt

T T T

E

dT T

P tT T T

t


2.2

(a) 8

(b) 0

(c) 2

4

(d) 1


2.3

(a)

1(

5)X s

s

(b) 6ln(5 ) ln(5)( ) 5

(1

ln)

(5)

t tx

s

t e e

X s

(c) 5

1s

(d) 5

1s


2.4

(a) 2 2

2

3 2 3

2 ( ) 4 4( ) ( )

2 4 4 2 4 4( )

t u t t t u t

s sX s

s s s s

x t

(b)

2 3 2 3

2

3 2 3

( )

(

2 ( ) 4 4 ( )

2 4 4 26 20 4

33 3 3)

t tx t t u t t t ee

X s

u t

s s

ss s s

(c)

0

0 0

0 0

0 02 2 2 2 2 2

0 0 0

cos ( )4

cos cos sin sin ( )4 4

1 1cos sin ( )

2 2

1 1 1)

(

(2

)

2 2

t u t

t t u t

t t u t

sX s

s s s

x t

s

(d)

30

30 0

30 0

0 02 2 22 2 2

0 0 0

( ) cos ( )

cos cos sin sin ( )

cos sin ( )

3(

3

3 3

1 3

2 2

3 31 3 1)

2 2 23 3 3

t

t

t

x t e u t

e u t

e

t

t t

t t

s

s

u t

X ss

s

s


2.5

(a)

(2 ) (2 )

2

2

2

( )

( )

cos( ) 2sin( ) ( )

c

1 12 2

2 2

1 1( ) ( )

2 2

1 1

2 2

5 63.4 (s )o

j t j t

t jt jt jt jt

t

t

X s

je je u t

t t u

j j

s j s j

x t j e j e u t

e e e

e

e u t

t

t

(b)

2

2 2

( )

( ) 4 1 (

4 4 4

2 1 1

( ) 4 4 )4t t t t t

X s

u t t e u t

s s s

x t e e te e

(c)

2

2 2 2

4 4 4

2 2

( ) 4 4 4

( )1

( ) 4 1 ( )t t t t t

s s

x t e te e e

X ss

u t t e u t


2.6

(a)

3 2

3 2 1

3 2 1

( ) 3 ( )

( )

2t t t

s s s

x t e e e

X s

u t

(b)

2 3 2

2 3 2

229 54 216 108

3 2 13 3 2

(

135 13( )

229 135 26 216 108 13) ( )t t t

X s

t t

s s ss s s

t e ux e et t

(c)

2 3 2

2 3 2

31 1( )

32 31 30 21

32 30 21 20

3 2 13 3 2

( ) 20 ( )t t t

s s ss s s

x

X s

t t e t e e u tt


2.7

(a) 1 2 1 0( ) ( ) ( ) ( ) initial condi( ) tionsy t a y t b x tt xy a b t

(b) The poles are at 21 1 04as a a

(i) For real and distinct poles, we require 21 04a a . In this case, the poles

are 21 1 04as a a .

(ii) For real and repeated poles, we require 21 04a a . In this case, the poles

are 1 1,a as s .

(iii) For complex conjugate poles, we require 21 04a a . In this case, the poles are

21 0 14s a j a a

(c) (i) 1

0

02n

aa

a

In general, the poles are at 2 1n ns

(ii) For real and distinct poles, we require 1 . In this case, the poles

are 2 1n ns

(iii) For real and repeated poles, we require 1 . In this case, the poles

are ,n nss .

(iv) For complex conjugate poles, we require0 1 . In this case, the poles

are 21n njs

(v) This form is preferred because the nature of the poles is determined by a single parameter.


(d) (i)

2 2

2 2

0 0

2 20

2 2 2 2

1 10

2

1 10

2

20

2

2 1 2 1

2 1 1

( ) (

( )

sinh (

)2 1

( )2 1

11

)

n n

nn n

n

n n

n n n n

t t

n

t t

n

tn

t

n

b b

H ss

b

b

s s s

h t e e u t

ee e u t

e t

b

bu t

(ii)

0

2

0( ) ( )

( )

n

n

t

b

s

h t b u

H

te

s

t

(iii)

2 2

0 0

2 2

2 2

1 10

2

20

2

2 1 2 1

1 1

( ) ( )2 1

sin 1 )

(

(1

)

n nn n

n

n n

n n

j t j tt t

n

t

n

n

b b

j j

s j s j

bh t e e u t

j

b et u

s

e

t

H

e

(e) An overdamped system does not display any oscillations in its impulse response, whereas the underdamped system does display oscillations. Hence, the term damped refers to oscillations: if oscillations are present, the oscillations have not been damped; if there are no oscillations, the oscillations have been damped.


2.8

(a)

6

6 6 5 5 5( )

6 6 6

( ) 5 1 ( )

( )

t

X ss s s s s

y t

Y s

e u t

(b)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

t

x(t)

y(t)

(c)

6

6 6 5 30( )

6 6 6

( ) 3

)

( )

(

0 t

s sX s

s s s s

v t e u

V s

t

(d)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

t

v(t)


2.9

(a)

2 2

6

6 6 5 5 5 / 6 5 / 6( )

6 6 65

( ) 5

(

1 ( )6

)

t

X ss s s s s s

y

s

t

Y

t e u t

(b)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

t

x(t)

y(t)

(c) 2

6

6 6 5 30 5 5( )

6 6 ( 6) 6

( ) 5

(

( )

)

1 t

s sX s

s s s s s s sV s

ev t u t

(d)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

3

4

5

t

v(t)


2.10

(a) 2 2

2 3

6 6 5 5 15 10( )

5 6 5 6 2 3

( ) 5 1 3

(

2 ( )

)

t t

X ss s s s s s s s

y t e e u

Y

t

s

(b)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

t

x(t)

y(t)

(c) 2 2

2 3

6 6 5 30 30( )

5 6 5 6 2 3

( ) 30 ( )

( )

t t

s sX s

s s s s s s s

v t u t

V s

e e

(d)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

t

v(t)


2.11

(a)

2 2 2 2

2 3

25 10156 6 5 5 6 32( )5 6 5 6 2 35 3 2

( ) 5 ( )6 2

)

3

(

t t

X ss s s s s s s s

y t t e e u

Y

t

ss

(b)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

5

10

15

20

25

t

x(t)

y(t)

(c) 2 2 2

2 3

6 6 5 5 15 10( )

5 6 5 6 2 3

( )

( )

5 1 )3 (2t t

s sX s

s s s s s s s sV s

e ev t u t

(d)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

t

v(t)


2.12

(a)

2 2

5 10 10

5 5

10 5 10 5125 5 5 4 4( )

10 125 10 125 5 10 5 10

10 5 10 5( ) 5 ( )

4 4

1 55 1 2

125

cos 10 sin 10 ( ) 5 1 10 26.6 (

(

o )2

)

2c s

t j t j t

t t

j j

X ss s s s s s s j s j

j jy t e e e

Y

u t

e t t u t t

s

e u t

(b)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1

2

3

4

5

6

t

x(t)

y(t)

(c)

2 2

5 10 10 5

125 125125 125 5 4 4

( )10 125 10 125 5 10 5 10

125 125( ) ( ) 1

( )

sin2

0 ( )4

t j t j t t

s s j jX s

s s s s s s j s j

v t u t

V s

e e t u tj

e e

(d)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-10

0

10

20

30

40

t

v(t)


2.13

(a)

2 2 2 2

5 10 10

5 5

2 4 3 4 3125 5 55( )

10 125 10 125 5 10 5 10

2 4 3 4 3

125 20 20( )

20 2( ) 5 ( )

5

4 3 15 cos 10 sin 10 ( ) 5 10 36

0

2 2cos .

5 10 10 5 2

t j t j t

t t

j j

X ss s s s s s s s j s j

j jy t t e e e u t

t e t t u t t e t

Y s

9 ( )u t

(b)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

2

4

6

8

10

t

x(t)

y(t)

(c)

2 2 2

5 10 10

5 5

10 5 10 5125 5 5 4 4( )

10 125 10 125 5 10 5 10

10 5 10 5( ) 5 ( )

4 4

1 55 1 2cos 10 sin 10 ( ) 5 1 10 26.6 ( )

2 2

125( )

cos

t j t j t

t t

sj j

sX s

s s s s s s s j s j

j jv t e e e u t

e t t u

V

t e t u t

s

(d)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

2

4

6

8

t

v(t)


2.14

The transfer function is ( )

( )( )

Y s k

XH

s s ks

. This system has one pole at s k . The system is

stable if the pole is in the left-half plane. The pole is in the left-half plane when 0k .


2.15

The system transfer function is 2

(

(( )

) 1

) 1

Y s

X s sH

ss

a

. The poles are

2 4

2

a as

.

(a) The system is stable when the poles are in the left-half plane. The poles are in the left-half plane for 0a .

(b) The system impulse response exhibits oscillations when the poles have a non-zero imaginary part. The poles have a non-zero imaginary part when 2 4 0a , which implies

2 2a . (Note that the system response is oscillatory and stable only for 0 2a .)

These results are summarized by the plot below. This plot, known as a “root-locus” plot, plots the location of the poles as a function of a.

-10 -5 0 5 10-1

-0.5

0

0.5

1

Real(s)

Imag

(s)

p1

a = 0

a -> infinity a -> -infinity

-10 -5 0 5 10-1

-0.5

0

0.5

1

Real(s)

Imag

(s)

p2

a = 0

a -> infinity

a -> -infinity


2.16

The system transfer function is 2

)( )

(

( )

Y s a

X s s as aH s

. The poles are

4

2

a a as

.

(a) The system is stable when the poles are in the left-half plane. The poles are in the left-half plane for 0a .

(b) The system impulse response exhibits oscillations when the poles have a non-zero imaginary part. The poles have a non-zero imaginary part when 4 0a and 0a , which implies 0 4a . (Note that the system response is oscillatory and stable for these values of a.)

These results are summarized by the plot below. This plot, known as a “root-locus” plot, plots the location of the poles as a function of a.

-10 -5 0 5 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real(s)

Imag

(s)

p1

a = 0a -> infinity a -> -infinity

-10 -5 0 5 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real(s)

Imag

(s)

p2

a = 0

a -> infinity

a -> -infinity


2.17

0 0 0 00cos

2 2 2 2j t j t j t j tj jA A A A

x e et A t e e e e

(a)

0 0

0 0

2 22 2

j j

j j

A Aj e e

A e e

X

A

(b) 0 02 2j jX

A Af e f f e f f


2.18

0

2 2

0

2 2 2

1 1

2 2

2

4

at j ft at j ftf e e dt e e dt

j f a j f a

a

a f

X

0

0

2 2

1 1

2

at j t at j te e dt e e dt

j a j

X

a

a

a


2.19 The easiest way to compute the Fourier transform is to recognize that x t , plotted below

T T can be produced by convolving the function plotted below with itself.

2

T

2

T

sin fTT

fT

Fourier transform

1

T

Thus, the Fourier transform may be computed as follows:

2

2 2 2

sin sin sinfT fT fTf T T T

fT T TX

f f

Similarly,

2

2

2 2 2

2

sin sin

2

sin

2

T T T

j T TT T

X TT


2.20 The easiest way to compute the Fourier transform is to recognize that x t is produced by the

convolution of the two functions shown below.

2 1

2

T T

2 1

2

T T 2 1

2

T T2 1

2

T T

2 1

2 12 1

sin f T TT T

f T T

Fourier transform

2 1

2 12 1

sin f T TT T

f T T

Fourier transform

2 1

A

T T

Multiplying the two Fourier transforms produces

2 1 2 1

2 12 1 2 1

sin sinf T T f T Tf A T T

f T T T TX

f

Similarly

2 1 2 1

2 12 1 2 1

2 2si

2 2

n sinT T T

X

T

j A T TT T T T


2.21

(a) 21 j feX f

(b) Let je . Then 21 j fX f e

2 22 2

21 2

1

cos 2

j f j fX f X f X f e e

f

(c)

f

2X f

21

21

1

0


2.22

(a) 531

jYj

ej

j

(b)

3 5 5

1 5j j

ej

Y j

(c) The symmetry properties of X j show that x t is real. Hence,

1

jY

jj

(d)

2

1Y j

j


2.23

(a)

2 1000 2 60 2 60 2 10006 6

2 60 2 602 1000 2 10006 6

5 3 16 3 5

16 3 5

16 6 2 60 10sin 2 10006

cos

j jj t j t j t j t

j t j tj t j t

x t j e e e e e j e

e e j e

j t

e

t

(b) Yes, x t is periodic. The period is

1 1 1LCM ,

60 1000 10

.


2.24

Using the identity 21( )

2 2te u t

j f

and the property 02

0j fte X f x t t for the last

term, we have

2 2 2 2 2( 2)

2( 2)

1 12

2 2

cos 2 2 2

j t j t t

t

x e

e

t e e u t

t u t


2.25

(a)

2

2 2 22 23

3

1 1 1 1

2 22 2

2

f X fa j f aa f a

B

XB

a

(b) 2 22 2

3

3

1 2

22 2

22

2 2

1

a af

aa f a

a

XB

B

(c) 22 23

2 22

3

1

2

ln(2)

2

Bf ff e X f e e

B

X


2.26 Using the transform pair

1sin

0 otherwise

22

2

T t Tf

fTT

T

,

set 1

2T and apply Parseval’s theorem:

1

2 22

21

2

sin1 1

fdf dt

f


2.27

(a)

90

2 2

2

22

1 90

220

9

0

0

1( )

2

1 1

2 2

2tan

0.92

2 2

0.9tan

2

at

B

t dt e dta

X f X fa j f a f

Bdf a

a f

aB

E x

a

aa

(b)

90

2 2 2

22

2 22 22

1 9

0

02

902 22220

0

90

1( )

2 4

2 2

22 tan

0.9 42

22

4

at at

B

t dt e dt e dta

a aX f X f

a f a f

Ba

E x

B

a

adf

a aBa f

(c)

2

2 2

90 902 2 2 2

90

2

2 2

2 2 2 2

0 0 0

( ) 1

0.9 2 0.45

t

f f

B Bf f f f

B

t dt e dt

X f e X

E x

e

e df e df e df e

f

df

Now, using 2

0

22

2

u

due

and

2

21

2

u

z

z e duQ

the two intergrals are

2

2

2 2

2

90 90 90

2 2

0 0

2 2 290

4 4

1 1 2 1

24 4 2 2

1 1 1 14

4 2 2 2

uf

u uf

B B B

e df e du

e df e du e d Qu B

Putting this together produces, we have that 90B is determined from

902 0.5 40.45 Q B


2.28

(a)

2 32 3

0 0 0

2 3 2 3

0 0 0

cos cos cos2 3

cos cos 2 cos 34 4 4 12

A A

A A A

y t A t t t

AtA t t

(b)

3

2

2 3

4 1000 10004 2

2000 2000 3000 3024

008

AAA

f f f f

A

Y

Af f f f

0 1000 2000 3000−1000−2000−3000

4

16

A

23

4

4

AA

23

4

4

AA

4

64

A4

64

A6

576

A6

576

A

(c) 6 6 4 4

4 6

2 2 2 4 63 3

1 9576 576 64 64THD 100 1009 8

4

4 4

6

4

1

A A A AA A

A AAA

AAA

A plot of the THD is shown below. Note that the THD increases as the amplitude of the sinusoid increases.


0 1 2 3 410

-4

10-2

100

102

A

TH

D (

%)


2.29

(a)

0 1

2 2 2 2 22

0 0 0

3 3 3 3

0 1 0

3 3

0 0

3 3

1 1 1

31

1 1

1 10 0

cos cos4

17cos 2 cos 2 cos cos

32 32 32

99 27cos cos cos 3 cos 3

256 32 256 4

3 3cos cos

64 64

3 3cos 2 cos

4 4

2 2

216 16

At t A t

A A A A Ax t t t t t

A A A Ax t t t t t

A At t

A At t

y t

x

1 1

2 3 3 2 2

0 1 0 1

3 3 2 2

0 0 0

3

1

1 1

1

3

0 01

3

0 0

3

17 99 3cos cos cos 2 cos 2

64 4 768 32 64 4

cos 3 co8 8

2

s 3 cos cos768 12

cos cos64 64

cos 2 cos 216

2

16

A A A A A At A t t t

A A A At t t t

A At t

A At t

(b)

2 3

3

2 2

3 3

2

17 9960 60

64 8 1536

37000 7000

2 64

120 120 14000 14000128 8

180 180 21000 210001538 24

6740 674016

A A AY f f f

A Af f

A Af f f f

A Af f f f

Af f

f

2

3 3

3 3

7060 7060

6880 6880 7120 7120128 128

13940 13940 14060 140

16

6032 32

Af f

A Af f f f

A Af f f f


23

128

A

23

1538

A

22

16

A

23

128

A

2217

64

A

223

642

AA

3 299

15368

AA

22

128

A

22

8

A

23

1538

A

23

24

A

22

16

A

23

128

A 23

32

A

23

32

A

060

120

1806740

70006880 7060

7120

14000

13940 14060

21000

6740

700068807060

7120

14000

1394014060

2100060

120

180

3 299

15368

AA

22

128

A

23

128

A

22

8

A 23

32

A

23

32

A

23

24

A

f

(c) 2 2 2 2 2 2 2 2 22 3 3 2 2 3 3 2 3

2 22 2

2 2 22 3 3

2 2 2 2 2 2 2 2 2128 1538 128 16 16 128 32 8 32

ID99 3

2 28 2 6153 4

2128 1

6

538 128

A A A A A A A A A

A A A A

A A A

2 2 22 3 2

2 22 2

2 216 32 8

99 38 2 641536

A A A

A A A A

A plot of the ID is shown below. Note that the ID is a function of A.

0 1 2 3 4 510

-6

10-4

10-2

100

A

ID


2.30

0

0

0

2

222

0 0

sin 21 2

even2

2 odd

( )

0

0 ,

,

kj t

Tk

Tk kj t j

k

Tk

k

x t c

k

c d

e

Ak

AAe e

kTA

k

t

j

k

k

0

0

kk

kk

X c f

X c f

kf

T

kf

T

00

1

T 0

3

T 0

5

T 0

7

T0

1

T

0

3

T

0

5

T

0

7

T

f

X f2

A

A

3

A

5

A

7

A

A

3

A

5

A

7

A


2.31

0

0

0

0 0

0

2

2 222

0 002

0sin1 1 2

1 even

2

( )

0

0 ,

,2 odd

kj t

Tk

Tk kT kj t j t jkT T

kT

k

k

e

k

x t c

k

c dt dt k

k

Ae Ae A ekT T

Aj

k

0

0

kk

kk

X c f

X c f

kf

T

kf

T

00

1

T 0

3

T 0

5

T 0

7

T0

1

T

0

3

T

0

5

T

0

7

T

f

X f

2A

2

3

A

2

5

A

2

7

A

2A

2

3

A

2

5

A

2

7

A


2.32

0

11

0 0

2

12

01

10 00

0

sin21

2

( )k

j tT

k

TkT j t j kT T

k

k

x t e

Tk

TTAe A e

c

TkT TT

c dt

0

0

kk

kk

X c f

X c f

kf

T

kf

T

00

1

T 0

3

T 0

5

T 0

7

T0

1

T

0

3

T

0

5

T

0

7

T

f

X f1

0

TA

T1

1

10

sin2

2

fT

TA

fTT

0

2

T 0

4

T 0

6

T 0

8

T0

2

T

0

4

T

0

6

T

0

8

T


2.33

0

0

0

0 0

0

2

2 22

2 20 0 0 00

22 2

02

1 2 1 22 1 1 0 , even

2, od

( )

d

kj t

Tk

Tk kTj t j t kT

kT

k

T

e

Ak

A A Ate A t e

x t c

c d kT T T T k

Ak

k

t dt

0

0

kk

kk

X c f

X c f

kf

T

kf

T

00

1

T 0

3

T 0

5

T 0

7

T0

1

T

0

3

T

0

5

T

0

7

T

f

X f2

A

2

2A

2

2

9

A

2

2

25

A

2

2

49

A

2

2

9

A

2

2

25

A

2

2

49

A

2

2A


2.34

0

0 0

0 0

0 0

2

32 2

2 20 0

4

0 0

4

2 2

4

4

1 4 1 4 42 sin

2

0 0

0 , even

4sin , odd

2

( )k

j tT

k

T Tk k

j t j

k

tT T

kT T

x t c

c dt d

e

A A A kte A t e j

T T T T kt

k

k

A kj k

k

0

0

kk

kk

X c f

X c f

kf

T

kf

T

00

1

T 0

3

T 0

5

T 0

7

T0

1

T

0

3

T

0

5

T

0

7

T

f

X f

2

4A

2

4

9

A

2

4

25

A

2

4

49

A

2

4

9

A

2

4

25

A

2

4

49

A

2

4A


2.35

0

0

0

2

2

0 00

01 2

02

( )k

j tT

k

kT j tT

k

k

x t e

Ak

Ate

AT T

c

k

dt

j k

c

0

0

kk

kk

X c f

X c f

kf

T

kf

T

00

1

T 0

3

T 0

5

T 0

7

T0

1

T

0

3

T

0

5

T

0

7

T

f

X f2

A

2

A

6

A

10

A

14

A

4

A

8

A

12

A

0

2

T 0

4

T 0

6

T0

2

T

0

4

T

0

6

T

14

A

12

A

10

A

8

A

6

A

4

A

2

A


2.36

0

0

0

2

2 2

0 00

2

1 1 111 2

sin

14

0

14

0 , odd, 1

2,

(

e n

)

ve1

kj t

Tk

k

kT j tT

k

k

x t c

kc d

e

Ak

A t eT T A

j k

Ak

Aj k

k

k

t

k

k

A

0

0

kk

kk

X c f

X c f

kf

T

kf

T

00

1

T0

1

T

f

X f2A

4

A

2

3

A

2

15

A

2

35

A

0

2

T 0

4

T 0

6

T0

2

T

0

4

T

0

6

T

2

3

A

4

A

2

15

A

2

35

A


2.37

0

0

0

4

4

20 00

2 2sin

1

( )

2

4

kj t

Tk

kT j tT

k

k

e

AA t e

T T

x t c

c dtk

0

0

2

2

kk

kk

X c f

X c f

kf

T

kf

T

0

f

X f2A

2

3

A

2

15

A

2

35

A

0

2

T 0

4

T 0

6

T0

2

T

0

4

T

0

6

T

2

3

A

2

15

A

2

35

A


2.38

n

t t nTp

is periodic with period T. Hence it can be represented by a Fourier series

2 k

j tT

kk

t cp e

where the Fourier series coefficients are

2 22 2

2 2

1 1 1T T

k kj t j t

T Tk

nT T

c t e dtnT t e dtT T T

Thus, 21 k

j tT

k

t epT

Now using the transform pair 2 2

2k

j tT

k

Te

we have

22

k

kj

T TP

.


2.39 The key to solving this problem is to understand what H f does to 1cos 2 f t . The Fourier

transform of 1cos 2 f t is 1 1

1 1

2 2f f f f . When this is the input, the Fourier

transform of the output of H f is 1 12 2f f f f

j j whose inverse Fourier

transform is 1 12 21si 2

2n

2j f t j f tj j

e e f t . Armed with this relationship, we may now

compute y t :

1 2 10 0

0 1 0 1 0 0

0 1 0 1 0 0

2

0 1 0 2

2 2

2 2

cos 2 cos 2 cos 2 sin 2 sin 2 sin 2

1 1 1 1cos 2 cos 2 cos 2 cos 2

2 2 2 21 1 1 1

cos 2 cos 2 cos 2 cos 22 2 2 2

cos 2 cos 2

t f t f t f t f t f t f t

f f t f f t f f t f f t


f t f f t

y

f

1161

1162

−11

61

−11

62

f (kHz)

1

2

1

2

1

2

1

2

Y f


2.40 The Fourier transform of the upper input to the adder is

1160

1165

−11

60

−11

65

f (kHz)

1155

−11

55

The Fourier transform of the output of H f is

0−5 5f (kHz)

j

j Now, if z t Z f are a Fourier transform pair, then

0 0 0

1 1sin 2

2 2t f t Z f f Z f f

j jz

Applying this relationship to the output of H f produces the Fourier transform of the lower

input to the adder:

1160

1165

−11

60

−11

65

f (kHz)

1155

−11

55

The result is

1160

1165

−11

60

−11

65

f (kHz)

Y f


2.41 The key to solving this problem is to understand what H f does to 1cos 2 f t . The Fourier

transform of 1cos 2 f t is 1 1

1 1

2 2f f f f . When this is the input, the Fourier

transform of the output of H f is 1 12 2f f f f

j j whose inverse Fourier

transform is 1 12 21si 2

2n

2j f t j f tj j

e e f t . Armed with this relationship, we may now

compute y t :

1 2 10 0

0 1 0 1 0 0

0 1 0 1 0 0

2

0 1 0 2

2 2

2 2

cos 2 cos 2 cos 2 sin 2 sin 2 sin 2

1 1 1 1cos 2 cos 2 cos 2 cos 2

2 2 2 21 1 1 1

cos 2 cos 2 cos 2 cos 22 2 2 2

cos 2 cos 2

t f t f t f t f t f t f t



f t f f t

y

f

11

58

1159

−11

58

−11

59

f (kHz)

1

2

1

2

1

2

1

2

Y f


2.42 The Fourier transform of the upper input to the adder is

1160

1165

−11

60

−11

65

f (kHz)

1155

−11

55

The Fourier transform of the output of H f is

0−5 5f (kHz)

j

j Now, if z t Z f are a Fourier transform pair, then

0 0 0

1 1sin 2

2 2t f t Z f f Z f f

j jz

Applying this relationship to the output of H f produces the Fourier transform of the lower

input to the adder:

1160

1165

−11

60

−11

65

f (kHz)

1155

−11

55

The result is

1160

1155

−11

60

f (kHz)

Y f

−11

55


2.43

(a)

3

2

333

2

1000 2000 3000−3000 −2000 −1000

1000 2000 3000−3000 −2000 −1000

1000 2000 3000−3000 −2000 −1000

f

f

f

X f

H f

Y f

1 1 1 1

2cos 1000 2cos 20002 2f tY t

(b)

2

1000 2000 3000−3000 −2000 −1000

1000 2000 3000−3000 −2000 −1000

1000 2000 3000−3000 −2000 −1000

f

f

f

X f

H f

Y f

2

3

2

3

2


2.44

(a)

1

2

111

2

440 2000 3000−3000 −2000 −440

1000 2000 3000−3000 −2000 −1000

440 2000 3000−3000 −2000 −440

f

f

f

X f

H f

Y f

1 1 1 1

2

1

2cos 2 440 2cos 2 2000t ty t

(b)

1

2

1

2

440 880 3000−3000 −880 −440

1000 2000 3000−3000 −2000 −1000

440 3000−3000 −440

f

f

f

X f

H f

Y f

1 1

2

1

1

4

1

4

1

2

1

2

−880 880 2cos 2 440 cos 2 880y t t t


(c)

1

2

111

2

600 2600−2600 −600

1000 2000 3000−3000 −2000 −1000

f

f

f

X f

H f

Y f

2

1

11

600−600 2600−2600

11

2cos 2 600 2cos 2 2600t ty t

(d)

8

A2

A

8

A

60 7000−7000 −60

1000 2000 3000−3000 −2000 −1000

3000−3000

f

f

f

X f

H f

Y f

2

1

2

A

4

A

4

A

60−60

cos 2 602

At ty


(e) 3

4

1800 3600−3600 −1800

1000 2000 3000−3000 −2000 −1000

f

f

f

X f

H f

Y f

2

1

3

4

3

4

3

4

3

4

1800−1800

3

4

3cos 2 1800

2t ty


2.45

fH fH

t702cos t702cos

tx ty

139 140 141−141 −140 −139 −1 0 1

−1 0 1

4

1

2

1

4

1

4

−1 0 1

2

11

69 70 71−71 −70 −69

f

f

f

f

H f

Y f


2.46

11930−11930 20 70 120

2

1

2

f

f

f

H f

fH fH

t59302cos t39302cos

tx ty fG fG

1

2

1

2

1

2

−120 −70 −20

40 70 100−100 −70 −40

2

40 70 100

1

−100 −70 −40

1

f3970 4000 40303860

1

2

1

2

1

2

1

2

−4030 −4000 −3970 −3860

2

f

G f

3950 4000 4050−4050 −4000 −3950

2

f3970 4000 4030

1

−4030 −4000 −3970

Y f1


2.47

11930−11930 20 70 120

2

1

2

f

f

f

H f

fH fH

t59302cos t39302cos

tx ty fG fG

1

2

1

2

1

2

−120 −70 −20

2

11

f3950 4000 40503860

1

21

2

1

2

1

2

−4050 −4000 −3950 −3860

2

f

G f

3950 4000 4050−4050 −4000 −3950

2

f

1 Y f

20 70 120−120 −70 −20

20 70 120−120 −70 −20

3950 4000 4050

1

−4050 −4000 −3950


2.48

(a)

70070

5

Y f

4

716971 69 f (MHz)

(b)

70070

5

Y f

4

716971 69 f (MHz)

(c)

0 1 1 0

1 0 1 0

1 0 1 0

1 0 1 0

cos 2 cos 2 sin 2 cos 2

cos 2 cos 22 2

sin 2 sin 22 2

cos 2 sin 22 2

t f t t f t Q t f t f t

I t I tf t f t

Q t Q tf t f t

I t Q ty t

r I

f f

f f

f tff t f

In part (a), 0 930f so that

cos 2 70 sin 2 702 2

I t Q ty t t t

In part (b), 0 1070f so that

cos 2 70 sin 2 702 2

I t Q ty t t t

The difference is the sign of the “quadrature term” (the term involving sin 2 ft ).


2.49 (a)

f0.10.1

X f

9

(b)

f0.10.1

Y f

1


2.50 (a)

f

Fourier transform of input to H(f)

H(f)

f

Fourier transform of input to G(f)

G(f)

sum of “house” and “half-ellipse”

f455455 2f1f

1

2

450

46 1 5

0

0 8 0

f

f

2f 1f

H f

f

G f

55


(b)

f


Note the overlapping spectra here. This is bad because there is not way to use a filter to isolate the desired spectrum from the combination.

It is not possible to design an H(f) and G(f) to produce the desired output. The signal centered at 250 Hz must be removed before the mixer preceding the filter H(f). The typical solution is to add an additional filter as follows:

fH fH tx

tf02cos kHz 7050 f

tr

tf12cos

fG fG

kHz 4551 f

imagerejection

BPF

This filter needs to remove the spectral content that interferes with the desired signal. In tunable systems (i.e., systems that need to isolate different channels), this BPF may need to be tunable.


2.51 (a)

f


H(f)

f

Fourier transform of input to G(f)

G(f)

sum of “house” and “half-ellipse”

f455455 2f1f

1

2

450

46 1 5

0

0 8 0

f

f

2f 1f

H f

f

G f

55


(b)

f


Note the overlapping spectra here. This is bad because there is not way to use a filter to isolate the desired spectrum from the combination.

It is not possible to design an H(f) and G(f) to produce the desired output. The signal centered at 250 Hz must be removed before the mixer preceding the filter H(f). The typical solution is to add an additional filter as follows:

fH fH tx

tf02cos kHz 7050 f

tr

tf12cos

fG fG

kHz 4551 f

imagerejection

BPF

This filter needs to remove the spectral content that interferes with the desired signal. In tunable systems (i.e., systems that need to isolate different channels), this BPF may need to be tunable.


2.52 (a)

f15 19 23 38 530

X f

(b)

LPF 1

BPF 1

BPF 2 ×2

LPF 2 x t

l t r t

l t r t

2l t

2r t

0cos 4 f t

f

1

15−15

LPF 1

f

2

15−15

LPF 2

f

1

5323

BPF 1

38−53 −23−38

f

1

2117

BPF 2

−21 −17 (c) l(t) + r(t) is available at the output of LPF 1.


2.53

(a) Using 020

j f tt e X fx f , we have

0 0

0 0

2 20

2 2

0 0

1 1cos 2

2 2

1 1

2 21 1

2 2

j f t j f t

j f t j f t

t f t t e e

t e t e

X f

x x

x x

f X f f

(b) 2

cA

f

2cA

cfcf B cf Bcfc Bf c Bf

(c) cf B

(d) The minimum theoretical approach would be to space the carriers 2B Hz apart. But, this would require ideal low-pass filters. Real filters require a transition band thus necessitating a larger spacing. In commercial broadcast AM, for example, the channel assignments are every 2B = 10 kHz, but broadcast stations are not assigned to adjacent channels in any given geographical area. Instead, the closest spacing is every other channel slot.


2.54

(a)

f

2mA

2mA

mf mf

M f

(b)

cos 2 cos 2

cos 2 cos 22 2

4

m m c c

m c m cc m c

c m

m

m cc m c m c m

t A f t A f t

A A A Af f t f f t

A AS f f f f f f f f f f f f

s

f

(c)

f

4c mA A

cfc mf f c mf fcfc mff c mff

4c mA A

4c mA A

4c mA A

S f


2.55

(a) i.

2cA

f

2cA


R f

2

2cA

f

2

4cA

2 cf2 cf B 2 cf B2 cf

2 c Bf 2 c Bf

2

4cA

B B

X f

ii. The filter must satisfy the conditions illustrated below.

B B

2

2

cA

H f

f

(b) i.

2 2

cos 42 2 c

A At m t tx f

ii. The filter removes the double frequency term and scales the baseband term by

2

2

A.

Thus the filter output is y t m t

(c) i.

2 2

cos cos 42 2

cos

c cc

A At m t m t f t

y t m t

x


ii. The phase offset scales the output by the constant cos . This does not cause

any distortion, especially when is small. As 2

, the result is disastrous.

There are only two solutions: either must be known (or estimated from the received signal) or the modulation format should be altered to allow non-coherent detection.


2.56

(a)

cos 2 cos 2 cos 2

envelope cos 2

c c c m m c

c m m

m t f t A A f t f t

A A f t

A

(b)

t

envelope output

(c) The envelope detector does not preserve sign information. The solution is to reformat the signal so that the envelope is never negative.


2.57

(a)

2cA

f

2cA


S f2

cAA

2cAA

(b) Write the modulated signal as

cos 2 cos 2 cos 2c c c c c ct A m t A f t AA f t A m ts f t

The first term is a replica of the unmodulated carrier. Because this replica is part of the transmitted signal, the term “transmitted carrier” is used.

(c) The envelope detector output is cA A m t . Assuming 0A m t for all t, the output

may be written as c cA A A m t . The term cA A needs to be subtracted from the envelope

detector output to produce an appropriately scaled version of the desired signal. (In other words, the envelope detector needs to be AC-coupled.)

The condition that guarantees correct output is min 0A m t .


2.58

(a) mA A

(b)

2 2c m c m

c m m

A A A Af A f f f f fA A

(c)

1 1

2 2

2 2 2

2 2 2

c c

c c m c mc c m c m

c c m c mc c m c m

f A f f A f f

A A A A A Af f f f f f

A A

S

f

A A A Af f f f f

f

f ff

(d)

f

4c mA A

cfc mf f c mf fcfc mff c mff

4c mA A

4c mA A

4c mA A

S f2cA A

2cA A


2.59

(a)

2 2 2 22

2 2 2 2

2 2 2 2

2 22

0

tot

tot

cos 2 cos 2

cos 2 cos 22 2

cos 48 8

cos 4 cos 44 4

cos 48 8

1

4

power ratio 1

m

c m m c

c m c mc m c m

c m c mc m

c m c mc

c m c mc m

Tc m

mm

m

m

m

t A A f t f t

A A A Af f t f f t

A A A As t f f t

A A A Af t f t

A A A Af f t

A A

s

P

P s t dtT

P

P

P

(b)

2 2 2 22

2 2 2 2 2 2 2 2

2 2

2

cos 2 cos 2 cos 2

cos 2 cos 2 cos 22 2

cos 4

cos 4 cos 48 8 8 8

cos 2 cos2

2 2

2 42

c c c m m c

c m c mc c c m c m

c cc

c m c m c m c mc m c m

c m c mc m m

c

t A A f t A A f t f t

A A A AA A f t f f t f f t

A As t f t

A A A A A A A Af f t f f t

A AA

s

A A

fA AA

f t f t

A

2

2 2 2 2

cos 2 cos 42 2

cos 4 cos 44 4

2m c mc m m

c m c mc m

AA A AAf t f t

A A A

f

Af t f t


2 2 2 2

2tot

2 2

2 2 2 2 2tot

0

(from part (a))4

4

14power ratio

14

1pow

1

2

22

er ratio3

m

c mm

c c m

m

c m

m

c c m

m

m

T

A AP

A A AP t dt

A AP

P A A A A

A

A

A

T

A

A

s


2.60 (a)

0 0.5 1 1.5 2 2.5 3 3.5 4-1

-0.5

0

0.5

1

t

s(t)

(b)

0 0.5 1 1.5 2 2.5 3 3.5 4-1

-0.5

0

0.5

1

t

s(t)


2.61

(a)

0

( ) 2 cos 2 sin 2mm m

m

t

m

A ft f f dA f

ft

Let m

m

A f

f

, then ( ) sin 2 mt f t and cos 2 sin( ) 2c c mfs t A t f t .

(b) Using the identity cos( ) sin( )jX X j Xe , we see that cos( ) Re jXX e . Thus,

( ) ( )2 22 ( ) Re Recos c cj f t t j f tj tc c c cf t t A e A e eA . Substitute

( ) sin 2 mt f t to obtain the desired result.

(c) 0 0

si

0

n2

0 0

22

0

1 1( ) mm mj f tj kf t j kf t

k

T T

s t e dt e e dtT T

c

Using the substitution 2 mx f t , the integral becomes

n2

0

si1

2j kx

k kxe dxc J

Thus c( o) 2sc k c mk

s t A J f kf t

(d)

2c

k c m c mk

Af J kf kfS f f f f

The bandwidth is infinite.

(e) The power at the carrier frequency is determined by the 0k in the answer to part (d). The 0k term is

02c

c cfJ f f fA

The condition of zero power at the carrier frequency occurs when 0 0J . Thus the

curious and interesting situation occurs at the values of corresponding to the zeros of

0 ·J . The first five zeros are 2.4048, 5.5201, 8.6537,11.7915,14.9309 .


2.62 (a)

22 2

22 2

2 22 22

4

4 2

2

4

ck c m c m

k

ck c m c m

k

ck c m c m

k

c ck k

k k

S f f f f

S f f f f

S df

Af J kf kf

Af J kf kf

AP f J kf kf f f f f df

A AJ J

(b) The table of the Bessel functions is | k=0 k=1 k=2 k=3 k=4 k=5 k=6 k=7 k=8 ------+-------------------------------------------------------------------------------- B = 1 | 0.58553 0.19364 0.01320 0.00038 0.00001 0.00000 0.00000 0.00000 0.00000 ------+-------------------------------------------------------------------------------- B = 2 | 0.05013 0.33261 0.12449 0.01663 0.00116 0.00005 0.00000 0.00000 0.00000 ------+-------------------------------------------------------------------------------- B = 5 | 0.03154 0.10731 0.00217 0.13310 0.15306 0.06819 0.01717 0.00285 0.00034 ------+--------------------------------------------------------------------------------

The table of the power ratio R for different values of K is

2

1

20 2

K

kk

J JR

| K=0 K=1 K=2 K=3 K=4 K=5 K=6 K=7 K=8 ------+-------------------------------------------------------------------------------- B = 1 | 0.58553 0.97282 0.99922 0.99999 1.00000 1.00000 1.00000 1.00000 1.00000 ------+-------------------------------------------------------------------------------- B = 2 | 0.05013 0.71535 0.96433 0.99759 0.99990 1.00000 1.00000 1.00000 1.00000 ------+-------------------------------------------------------------------------------- B = 5 | 0.03154 0.24616 0.25049 0.51670 0.82282 0.95921 0.99356 0.99926 0.99993 ------+--------------------------------------------------------------------------------

The bold numbers indicate the smallest value of K which captures 98% of the power. The values are

98

98

98

2 for 1

3 for 2

6 for 5

K

K

K

from which we deduce that 98 1K . Thus we have

98BW 2 12m mK f f


2.63

(a)

2 2

2

cos ( ) 2 ( ) sin ( )

2 2 ( ) sin ( )

c c c c c

c c c

dA f t t A f t f t t

dtA f fm t f t t

(b) The envelope detector output is 2 2 ( )c cA f fm t . To produce the desired output, we

need ( )c ff m t for all t.

(c)

t

2c c mf fA A

2c cA f

2c c mf fA A

envelopedetectoroutput

t

2c mA fA

y t

2c mfA A


2.64

(a) FM

0

( )( )

( )p

p

F ss

k

skH

k F ss

(b) FM ( ) ( ) ( )) ( ) (( )V

ds s s s v tH t

ds

t which is the desired result.

(c)

0 0

( )( )

( ) 1p

p p

sk F s ss F s

s k k F s k k

The filter is a differentiator.

0 1k should be avoided.


2.65

(a)

1 11 1 1 1

( ) (( ) ( 11) ( ) 12 2 2

)2

n n nn

h n uu n u n nn

(b) 11 1 1

( ) ( 1)2 3

( )2

n n

u n u nh n

(c) 21

( 2)2

( )n

h n u n

(d) ( 1) 2 ( 2) 3 ( 4)) ) 4 (( 3n nh n nn


2.66

(a) 11 1

( 1) ( 1)2 2

( ) 2n n

u n nh n u

(b)

1

2

1

1 1

1 1

2

3 4 3 4

3 4 ( 1) 3 4 (

1

1)

53 4 5

1 1 11 1

3 4 3 4

1 1 1 1(

12

53 4 5

)3 4 3 4

1 15

3 4

15

3 4 12

1

n n

jj j

jj j

j z j zz

j z j z

j

H

h n

j

j u n j j u n

j e j e

j e j

e

j ee

1

1

3

4

5

1

tan

2( ) 10 sin ( 1)

n

h n n u n

(c)

1 1

1 1

1

2

1

1 13 3

4 4

3 31 ( 1) 1 ( 1)

4 4

5

1 1 1 11 1

4 3 4 3

1 1 1 1( )

4 3 4 3

1 1

4 3

1 1

4 3

3 5 51

12 4 4 4

5

12

n n

jj j

z z

z

j z j z

j j u n j j u n

j e j e

j

j j

H

h n

j e

2

1

1

4

3 5 51

4 4 4

tan

5( ) sin ( 1)

2

3

5

12

jj j

n

e j e

h n n u n

j e


(d)

2 1

211

1 1

11

2

2 21142

1 14 1 1

2 4

n n

z zz

zz

h n n n

H

u


2.67

(a) 1 2 1( ) () )( Y z z YY z z zz

(b) 1

1 2( )

1

z

zY z

z

(c) 2

1 1

1 1

1 1

1 1 5 1 5

2 2

5 5 5 5

10 101 5 1 5

2 2

5 5 5 5

10 101 5 1 5

1 12 2

5 5 1 5 5 5 1 5( )

10 2 10 2

1 1 5 1 5

2 25

( )

( 1)

n n

n n

z z

z zz z

z z

z z

z

z

n u

z

y

Y

n

( 1)u n


2.68

(a)

−3 −2 −1 0 1 2 3 4 5 6 7

1

(b)

65

10

1

1n

n

Xz

z zz

(c)

−3 −2 −1 0 1 2 3 4 5 6 7

1

−1

(d) 61G z z

(e)

61 1 1 6

1

11 1 1

1

zz X zG z X z X z z z

zz

This answer is the same as that from part (d).

(f)

6

1 1

1 1

1 1

zz G zX

z z

This answer is the same as that from part (b).


2.69

(a) 1

11

13

H zz

(b)

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

Real Part

Imag

inar

y P

art

(c)

2

1

3

1 2

1 2

0 0

1 0

41

3

13 12

9 3

n

n

n u n

y n h n x n

h n n n n

h n h n h n

h

n

n

n

n


2.70

(a) 1

11

14

H zz

(b)

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

Real Part

Imag

inar

y P

art

(c)

2

1

1 2

1 2

0 0

1 0

1

1

4

3 1

3

4

16 42

n

n

n u n

y n h n x n

h n n n n

h

h n h n h n

n

n

n

n


2.71

(a) IIR

(b) FIR

(c)

1 2 3

1 21 2 3

76

31 1

16

8

1

6

z z zz

zH z z

zH

zH

(d) IIR

(e) 11 1 7( 1) ( 2) ( 3) 6 ( 1) 8 ( 2) ( 3)

6 6)

3( y n y n y n x n x n x ny n

(f)

1 1 1

1 1 1

2 2 2 2

3 6 3 6

1 1

2

1 1

1 1 13

z z z

zz z z

H

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

Real Part

Imag

inar

y P

art


(g)

1 1 1

11 1

1 1

2 31 11 1 12 3

1 1(

23 11 2

3)

n n

zH

z zz

z z z

h n u n

(h)

4 4

( )

( ) ( ) ( 1) ( 2)

( ) ( 1) ( 2)

0 0

6 1

9 2

7 1 13 13 3

4 2

(

9 3

( ) )

n n

x n

h n n n n

h n h n h n

n

n

n

y n n

n

h

0 5 10 15 200

2

4

6

8

10

12

n

y(n)


2.72

(a) 81z zH

H z has 8 zeros equally spaced on a circle of radius 1

8 and 8 poles at the origin as

shown below.

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

8

Real Part

Imag

inar

y P

art

The ROC is the entire z-plane except for z = 0.

(b) 8

1 1

1H z zG z

G z has 8 poles equally spaced on a circle of radius 1

8 and 8 zeros at the origin as

shown below. The ROC for a causal stable system is 1

8z .


-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

8

Real Part

Imag

inar

y P

art

(c) a multiple of 8

0 otherwise

n

gn

n


2.73

(a) 1 2 0 1 21 2 1 2n a n a n b x n b x ny by x ny

(b) The poles are 1 2

21

42

1 1a

za

a

i. For real, distinct poles we require

22

2 4a

a . The poles are 1 2

21

42

1 1a

za

a

.

1 0 1 02 2 20 0 2

1 1 1 1

221

0

11 221

1 0 1 02 2 20 0 2

1 1 1 1

221

11 221

0

1 0 1 02 20 0

1 1

1 4

1 4

1

22 2

2

22 2

2

2

1 1 4

1 4

1 4

1 1 1 4

2 2

a b a ba ab b

a a

z ba

z

a b a ba ab b

a a

az

h n b n

a b a ba ab b

a

a

az

a

aH

a

a

a

az

a

a

a

a

a

21

21 1 2

22 121

1 0 1 02 2 210 0 2

1 1 1 1 22

2 121

1 4

1 1 41 4

1 4

1 1 41

1

22

12

4

2

2

n

n

au n

a b a ba ab b

a a

a

a aa aa

a

a an

a a

au

a

ii. For real, repeated poles, we require 2

1 24aa . The poles are 1 1,2 2

za a


1 2 2 1 11 10 1 0 1

110 0 0 02 2 2

1 1 11 1 1

1

1 1 10 0 0

2 22

1 1 1

1

42

2 2 2

2 1 12 2 2

n n

a aa z b a z za

z b b b za a a

z z z

a a ah n b n b n n b

b zH b

nu un

iii. For complex conjugate poles, we require 21 24 0a a . Let 2 2

1 24a a , then the

poles are 1

2 2

ajz

.

2 21 11 0 1 0 2 0 1 0 1 0 2 0

01 11 1

121 0 1 0 2 0 1

0

21 0 1 0 2 0

2 2 2 2

2 2 2 2

12 2 2 2

2 2

1 1

n

a b a b a b a b a b a bj jz

z ba a

j z j z

a b a b a b ah n b n j j u n

a b a b a b

H

j

z

1

1 12 2

na

j u n

(c) i.

The poles are 211 2

14

2 2jz a

aj a re from which we obtain

2

1 221

tan 1 4

r a

a

a

ii. 1 2 21 2 cosr rB z z z


iii.

cosr

r

The location of the pole intersects the real axis at 1

2 times the coefficient of 1z .

The distance from the origin is the square root of the coefficient of 2z .

iv.

2 21 1

0 0 01 1

0 0

2sin 2sin

1 1

sin 1 1sin

j j

j j

n

z z

z bre z r

jre j

e z

rh n

re

H b

b n b n u n

b

(d) The case of real, repeated poles corresponds to 0 . In this case,

21 1

2 4 2

a aar .

Also note that 0

silim

n 11

sin

nn

. Now, the answer to part (c)-iv becomes

0 0

0 0

0 0

10 0 0

sin 1 1sin

1 1

1 2 1

2 1 1

n

n

n

n n

rn b n b n u n

b n b n r u n

b n b n r u n

b n b r u n b n r n

h

ur

Using 1

2r

a gives the answer from part (c)-iv.


2.74 (a) ( ) 0.5F z

1 1 1 1

1 1 1 1 1

1

1( )

1 1

( ) 1 (

0.5 0.5( )

1 0.5 1 0.5 1 0.

0.5) ( 1

5

)n

zz z zY z

z zX z

z z

y u n

z

n

0 2 4 6 8 10-1

0

1

2

n

u(n)

y(n)

(b) ( ) 1F z

11

1( )

1(

( ) ( )

)

1

Yz

z z X zz

y n u n

0 2 4 6 8 10-1

0

1

2

n

u(n)

y(n)


(c) ( ) 1.5F z 1 1 1 1

1 1 1 1 1

1

1( )

1 1

1.5 1.5 0.5( )

1 0.5 1 0.5 1

( ) 1 0.5( 0.5) (

.5

1

0

)n

z z zY z

z z

zX z

z

n

zz

y u n

0 2 4 6 8 10-1

0

1

2

n

u(n)

y(n)

(d) Unlike the case with continuous-time systems, a first-order system can show an oscillatory transient reponse.


2.75

(a)

1

111

1 1

2 211 1122

1 1

2

11

11

2

n

zV

zz

zz

v n nu

0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

n

u(n)

y(n)

(b)

1

1 1

1 1

1

11

1 1 1z

z

v n

zV

z

n

0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

n

u(n)

y(n)


(c)

1

111

2 21

1122

1

3 31

11

2

1

3

3

2

n

zV

zz

zz

v n nu

0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

n

u(n)

y(n)

(d) Even though this is a first-order system, it can still exhibit oscillations.


2.76

(a)

1

1 2

1

1

1

41 1

4 8

zz

zH

z

(b)

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

Real Part

Imag

inar

y P

art

(c)

1 11

11 1

1

1

9 7 9 7 6

1117 7

9 7 7

9

7 7

308 308 308 308

11 11 1

8 8 8 8

7 1

308 308 8 8

7 1

308 308

7 7(

6

)8

11

8

n

n

j z j z zY z

j z j z

j j

n j j

z

y n

1u n


2.77 The closed loop transfer function is

1

1 2

2

0.

( ) 1 0.25

( ) 1 0.75

0.25

0.

25

0.12575

K

Y z z

X z K z

z z

z

z

KK z

This system has two poles at

2

1 2

0.75 0.75

2,

K Kp

Kp

The goal is to find the values of K for which both poles are inside the unit circle. That is, find the values of K for which 1| | 1p and 2| | 1p . The following Matlab script plots the poles for

0 3K : %% ex2_77 K = 0:0.001:3; p1 = 0.5*(K-0.75 + sqrt((0.75-K).^2-K)); p2 = 0.5*(K-0.75 - sqrt((0.75-K).^2-K)); plot(real(p1),imag(p1),'b-',real(p2),imag(p2),'r.-',... real(p1(1)),imag(p1(1)),'bo',... real(p1(end)),imag(p1(end)),'bs',... real(p2(1)),imag(p2(1)),'ro',... real(p2(end)),imag(p2(end)),'rs','LineWidth',2); legend('p_1','p_2'); grid on; %% plot unit circle for reference hold on; plot(exp(j*2*pi*[0:0.01:1]),'k:'); hold off;


This script produces the following plot

-1 -0.5 0 0.5 1 1.5 2-1

-0.5

0

0.5

1

p1

p2

From this plot, we see that 2p is always inside the unit circle and that 1p is outside the unit circle

when K is too large. Using the data vector p1 produced by the script, we see that 1p is outside the

unit circle for 2.334K . Repeating the same script for 0K produces the following plot

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

p

1

p2

Form this plot, we see that 1p is always inside the unit circle and that 2p is outside the unit

circle when K gets too negative. Using the data vector p2 produced by the script we see that 2p

is outside the unit circle for 0.2K Putting this all together, the system is stable for 2.3340.2 K


2.78

(a)

1

1 2

1

21 1

21

4

zz

Hz z

(b) There is one zero at 0z and two poles at

4

1 3

4z j

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

Real Part

Imag

inar

y P

art

(c)

11 1

11 1

1 1

1

21 3 1 3

312 1211 3 1 3

4 4 4 4

1 3 1 3 1 3 1 3 21

12 4 4 12 4 4 3

1 1 2cos 1 1

3 2 3

tan

1 1

n n

n

j jzz z

zz

j z j z

j jy n j j u n

n u n

Y

1

1

3

tan 3


2.79

1

1 212

1

z KzYH

KX zz

K z z

The poles are at 2

1 1 2

2

K Kz

K .

The position of the poles in the z-plane (as a function of K) is illustrated below.

-1.5 -1 -0.5 0 0.5 1 1.5 2-1

-0.5

0

0.5

1pole 1

real part

imag

. pa

rt

K -K

-2 -1.5 -1 -0.5 0 0.5 1-1

-0.5

0

0.5

1pole 2

real part

imag

. pa

rt

K -K

Several observations are in order

The poles are real and distinct for 2 3K and 2 3K


The poles are real and repeated for 2 3K and 2 3K

The poles are complex conjugates for 2 3 2 3K .

The pole at 2

2

2

1 1K Kz

K is inside the unit circle for 0 2K and 4K .

The pole at 2

2

2

1 1K Kz

K is inside the unit circle for 2K .

The system is stable for 0 2K


2.80

(a)

1

1

2

3

12

3

z

zH z

(b)

-2 -1 0 1 2-2

-1

0

1

2

Real Part

Imag

inar

y P

art

(c)

2

1

32

3

j

j

j

ee

He

(d)

2

2 2 4 2 21

3 3 9 3 3 12 2 2 2 4

13 3 3 3 9

1 1

j j j j

j

j j j j

e eH e

e e

e e e e

The system is an all pass filter!


2.81

(a) j je eX

(b) 4

1

j jj

j

e ee

eX

(c)

8 84 48 84 4

4 4

28 8

1 1

2 2 2 2 2 2

1 2cos

4 2 82 21

cos8 4

2

c s

11 1

2

o

n nj jj jn

j n j nj j

j jj

j

j j j jj j

e ee en e e

e

e

e ee ee e

x e e

e e

X

(d)

9

sin2

sin2

jeX

(e) 1 2 1 2j

l

e lX l

(f)

5 5

2 23 3

7 72 2

3 3

j

l

l l

e

j l j l

X


2.82

(a)

3sin

3 838

8

x

n

nn

(b)

1sin

2

1

2

n

n

x n

(c) 10 1 2 2 3 3n n n n nx


2.83 The DTFT of the sequence x n is

jX e

2j 2j

0.5 0.5

(a)

jY e

1

2

j

0.5 0.5

1

2

1

2

1

2

j j j

(b)

0.5 0.5

1 1

2j 2j jY e

(c)

0.5 0.5

jY e

1

2

1

2

1

2

1

2

j j jj

(d)

0.5 0.5

jY e

1 1

2j 2j


(e)

0.5 0.5

jY e

1

2

j1

2

j

1

2

j1

2

j

(f)

0.5 0.5

jY e

1

2j

1

2j


2.84

(a)

jY e

0.5 0.5

1

2

(b)

0.5 0.5

jY e

1

(c)

0.5 0.5

jY e

1

2

1

2

(d)

0.5 0.5

jY e

1

(e)

0.5 0.5

jY e

1

2

(f)

0.5 0.5

jY e

1


2.85 From the input/output relation we have

212 7 12 5j j j j j j j je e e e e X e e eY Y Y X

from which we obtain

2

12 5

12 7

j j

j jj

Y e e

e eX e

From the block diagram we have

1 2

j

j j

j

Y eH e H e

X e

and we are given 1

1 3

311

3

jj

j

H eee

.

Putting this all together gives,

1 2 22

2 2

2

12 5

12 7

12 5 8 2

3

3

1

313 4 14

12

4

2 7

jj j j

j j j

jj

j j j jj

n

eH e H e H e

e e e

eH e

e e e e e

h n u n


2.86

1

1

1

11 1

1 11 1

1 1

11

131

141

1

211

311 11 1 1 11 133 4

2 1 3 11 1

23

2

4

3 3 4 4

4

n n

Y

Y

u u

zz

zX z

z

zz zzH z

X z z zz z

h n n n n

The frequency response is

2

11

1 11 1

3 4

5cos

497 91 1

cos cos 272 72

2

6

j

j

j j

j

eH e

e e

H e


2.87

(a)

a

1

a

1 1j jera

a re

222

2 2

1

1

1 1 1 1 2co1 1

11

1

1

s1

2 cos

1

j j

j

j j

e ea a rrH e

ra

zaH z

a

ra e ae

az

(b)

1

a

a

1 1j jera

a re

2 22

2 2

1

1

2 cos1

1 1 1 1 2cos

11

1

1

1

1

1

j jj

j j

r raH e

e ea ra r

zH z

za

a e ae

a

a


(c)

a

1

a

1 1j jera

a re

2 22

22 2

1

1

11

1 1 1

2 cos1

2 cos

1

| |11

1

1

j j

j

j j

era

a

ea a r

H era e ra

zaH

ae

azz a

(d)

a

1

a

1 1j jera

a re

222

2 2

1

1

1 2 cos1

2 c

| | 1

11

1

os

j jj

j j

ra a e ae

ra e a

rH e

ra

a

e

az

zH z

Both the systems of parts (c) and (d) are all-pass systems. However, only the system in part (d) has a z-transform in the standard form.


2.88

(a) The pole-zero plot (below) shows a pole at

8

9z which means the frequency response

has a large amplitude in the vicinity of . Hence this is a high-pass filter as shown.

-2 -1 0 1 2-2

-1

0

1

2

Real Part

Imag

inar

y P

art

-0.5 0 0.50

20

40

60

80

100

normalized frequency (cycles/sample)

|H(e

j |2

(b) The pole-zero plot (below) shows two poles at

8

9z and a zero at

8

9z which means

the frequency response has a large amplitude in the vicinity of 0 and a small amplitude at . Hence this is a low-pass filter as shown.

-2 -1 0 1 2-2

-1

0

1

2

2

Real Part

Imag

inar

y P

art

-0.5 0 0.50

0.5

1

1.5

2

2.5x 10

4


|H(e

j |2


(c) The pole-zero plot (below) shows complex-conjugate poles at

8

9z j and a zero at the

origin, which means the frequency response has a large amplitude in the vicinity of

2

. Hence this is a band-pass filter as shown.

-2 -1 0 1 2-2

-1

0

1

2

2

Real Part

Imag

inar

y P

art

-0.5 0 0.50

5

10

15

20

25


|H(e

j |2


2.89

(a)

-4 -2 0 2 4 6 80

0.5

1

1.5

2

n

x(n)

(b)

3

2

6 4cos cos c

2

2 2 o 3s4

j j j

j

X e

X

e e

e

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50

5

10

15

20

frequency (cycles/sample)

|X(e

j |2

(c)

-4 -2 0 2 4 6 80

0.5

1

1.5

2

n

x(n)

(d) /2 3 /2

3

3 /2 9 /2

[0] 2 1 1 4

[1] 2 2

[2] 2 0

[3] 2

3

2

j j

j j

j j

X

X e

X

e

e

e

X e


(e)

0

/2 /2 3 /2

3

/2 / 9 23 3 2 /

2 1 1 4

2 2

2 3 0

2 2

j j

j j

j

j

j j j

X e

X e e

X e e

X e e

e

e

These numbers are the same as those obtained in part (d).

(f)

-10 -5 0 5 10 150

0.5

1

1.5

2

n

x(n)

(g) /4 3 /4

/2 3 /2

3 /4 9 /4

3

5 /4 15 /4

3 /2 9 /2

7 /4 21 /4

[0] 2 1 1 4

[1] 2 2

[2] 2 2

[3] 2 2

[4] 2 0

[5] 2 2

[6] 2 2

[7] 2 2

2

2

3

2

2

j j

j j

j j

j j

j j

j j

j j

X

X e j

X e

e

X e j

X e

X e j

X e

X e

e

e

e

e

e j


(h)

0

/4 /4 3 /4

2 /4 /2 3 /2

3 /4 3 /4 9 /4

4 /4 3

5 /4 5 /4 15 /4

6 /4 3 /2 9 /2

7 /4 7 /4 21

2

2

2 1 1 4

3

2

2 2

2 2

2 2

2 0

2 2

2 2

2

j

j j j

j j j

j j j

j j j

j j j

j j j

j j j

e

e e

e e

X e e

X e

X e e

X e

X

X e j

X e

e j

e

X e e

e

e j

e

e

/4 22 j

These numbers are the same as those obtained in part (g).

(i) The values obtained in part (d) are a subset of those obtained in part (g). This is because the DFT length of part (g) is divisible by the length of the DFT length of part (d).


2.90 Start with the sequence 1x n

2

40 1 2 95

nx1

n3 6 7 8

2

1

1

Draw the length-4 periodic extension of 1x n – call it 1,4x n – and delay it by 3. This

produces the sequence

1

2

40 1 2 95

1,4 3x n

n3 6 7 8

2

1

Note that the samples at 0,1,2,3n are the same as those of 2x n .

Now the DTFT of 1,4 3x n is 2

34

je

times the DTFT of 1x n . Thus we have 3

21,4 1[ ] [ ]

j kX Xk e k

and

3

21 2[ ] [ ]

j ke kX X k

Thus, 3

22 1 1[ ] [ ] [ ]

j kXX k e k kX


2.91 (a)

decimal numbers

3-bit binary equivalent

3-bit binary equivalent index of input

0 000 000

1 001 100

2 010 010

3 011 110

4 100 001

5 101 101

6 110 011

7 111 111

(b) The 3-binary equivalent of the indexes of the inputs are reversed (bit reversed) versions of the

indexes associated with the natural (temporal) order.


2.92

(a) jdX e

8

10

9

10

8

10

9

10

100

(b) The negative frequency component of cX f shows up as the positive frequency

component of jdX e , and the positive frequency component of cX f shows up as the

negative frequency component of jdX e .


2.93

(a) jdX e

42

51

51

51

1

42

5

1

51

5

102

(b) The negative and positive frequency components of cX f are preserved on the negative

and positive frequency components of jdX e , but there is aliasing.


2.94

(a) jdX e

4

13

9

13

9

13

4

13

2

2

52

(b) The negative frequency component of cX f shows up as the positive frequency

component of jdX e , and the positive frequency component of cX f shows up as the

negative frequency component of jdX e .


2.95

(a) Write cos

4n w n nx

where 1 0 7

0 otherwise

nw n

.

Then

7

2

7

2 4

1

2 4 4

sin1

2 4 4sin

2

sin sin4 41 1

2 21 1sin sin

2 4

4

4

4

4

2

j

j

j je WX e

e

e

7

2 4j

e

-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.50

1

2

3

4

5

/2 (cycles/sample)

|X(e

j )|


(b)

-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.50

1

2

3

4

5

/2 (cycles/sample)

|X(e

j )|

(c)

-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.50

1

2

3

4

5

/2 (cycles/sample)

|X(e

j )|


2.96

(a)

-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.50

0.5

1

1.5

2

2.5

3

3.5

4

/2 (cycles/sample)

|X(e

j k)|

(b)

-0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.50

0.5

1

1.5

2

2.5

3

3.5

4

/2 (cycles/sample)

|Y(e

j k)|

(c) The DTFT of x n is

7 7

2 4 2 4

sin sin4 41 1

2 21 1si

4

n sin4

4

2 2 4

j jjX e e e

which has zero-crossings at 1l for 0l . The length-8 DFT samples jX e at

4

k which corresponds to the zero-crossings. On the other hand, the DTFT of y n is


3 3

7 7

2 2

sin sin1 1

2 21

4 43 3

1sin si

2 3n

32

j jjY e e e

which has zero-crossings at 3 4

12

l for 0l . The length-8 DFT samples jX e at

4

k which does not correspond to any of the zero-crossings.


2.97

3.3

43.3

4

jX e

80,000


2.98 (a)

jdS e

5

12 1

2

5

12

1

2

normalized frequency(cycles/sample)

96000

(b)

jdS e

200

441

1

2

1

2


88200

200

441

(c) The DTFT in part (b) is wider than the DTFT in part (a). This is due to the fact that the sample rate used in part (b) is closer to the minimum sample rate defined in the sampling theorem than the sample rate used in part (a).


2.99

jdS e

3

10

1

2

1

2


500

4

10

2

10

2

10

3

10

4

10

400


2.100 (a)

jdS e

1

2

1

2


280

224

4

56 14

56

24

56

4

56

14

56

24

56

(b)

jdS e

1

2

1

2


200

160

1

4

1

4

(c) There are two key differences between the spectra in parts (a) and (b).The first difference is the bandwidth: the spectrum in part (a) is the spectrum of an oversampled signal whereas the spectrum in part (b) is the spectrum of a critically sampled signal (that is, it is sampled at the minimum sample rate defined by the sampling theorem). The second difference is the fact that the two spectra are “flipped” relative to each other. In part (a) the positive frequency component of the continuous-time signal shows up on the positive frequency axis in the DTFT domain. In part (b) the negative frequency component of the continuous-time signals shows up on the positive frequency axis in the DTFT domain.


2.101 (a)

jY e

1

2

1

2


120

1

10

1

10

100

Y f

12

11

10

f

(b)

jZ e

1

2

1

2


120

1

10

1

10

100

pZ f

120

11

100

f1020 10 20


(c)

jZ e

1

2

1

2


120

1

10

1

10

100

pZ f

120

11

100

f1020 10 20

positive- and negative-frequency components of the continuous-time signal overlap when sampled

The fundamental problem with this approach is that the positive- and negative-frequency components of the z(t) overlap when sampled. This was not a problem when using the complex-valued version of the signal. This problem illustrates one of the advantages of signal-processing using complex-valued signals: one does not have to worry about negative- and positive-frequency components aliasing on top of each other. The disadvantage is complexity: a complex-valued signal is a “two-dimensional” signal. Consequently, addition and multiplication require more resources.


2.102 The DTFT of the sampled sequence is

jS e

0.4 0.6

500400

0.2 0.80.20.40.60.8 Neglecting scaling constants, the sampled sequence may be written as (see Exericse 2.48)

cos 0.6 sin 0.6nT nT n nTs Q nI .

To produce I nT and Q nT from s nT , we need 0 0.6 .

As it is written the system produces

1

21

2

nT I nT

y

x

nT Q nT

Hence the system must be modified either by changing the sign on y nT or by changing the

sign on 0sin n -- that is, use 0sin n in place of 0sin n on the lower mixer.


2.103

(a)

f

1

5−5

H f

f

1

H j

10 10

(b)

/ 2

cycles/sample

1

5T−5T

jH e

rads/sample

1

10 T 10 T

jH e

(c) Construct the diagram of the impulse sampled waveform shown below

pX f H f

01

T120

T

201

T 20 5 5

1f

TX

1f

TX

X f

From this diagram, we see that the minimum sampling rate is determined from relation 1

20 5T from which we obtain

125

T .


2.104

(a) j Tc j eH

(b) In the interval we have

0 otherwise

j TT

dj e WH We

(c) sin

1( )

2

Wj T

j nTd

W

Tn W

Wn e

Th

W

e

T

Td

n

(d) sin

( )d

Tn

nT

TW h

Tn

(e) 1sin

1, ( )

12

2

2

d

nT

nW hT

n

Documents

fratstock.eu · 2017-05-08 · Full file at 2.7 (a) ya ( )tx 12 1 0yt ayt bxt ( ) ( ) ( ) ( ) initial condi bt tions (b) The poles are at 2 sa a 11 0 a 4 (