1

Solution 1: (7.7-2)

Baseband bandwidth: ( ) ( )M

RRBW bS

BB

2log21

21 αα +=+=

Passband bandwidth: BBPB BWBW 2=

Part a)

HzBWBB 15008

12000

16log2

12000)01(0

2

==+=⇒=α HzBWPB 30001500*2 ==

Part b)

HzBWBB 18008

120002.1

16log2

12000)2.01(2.0

2

=×=+=⇒=α HzBWPB 36001800*2 ==

Solution 2: (7.7-3)

Part a)

Quantization error: L

mv

p2=∆

825622200005.02

2005.0

2

8=⇒===⇒≥⇒≤⇒≤

∆nLLm

L

mm

v n

p

p

p

Therefore, each sample requires 8=n bits.

Since each 4-ary pulse represents 2 bits, each sample will require four 4-ary pulses.

Part b)

Nyquist rate is kHz4.622.3 =× .

The sampling rate is 25% above Nyquist rate which kHzf s 825.14.6 =×=

Each sample requires four 4-ary pulses and therefore the pulse rate will be

32000800044 =×== sS fR pulses/sec

Minimum Bandwidth: HzR

BW S

BB 160002

32000

2=== ,

HzBWBW BBPB 320001600022 =×==

Part c)

( ) ( ) HzR

BW S

BB 200002

3200025.01

21 =+=+= α , HzBWBW BBPB 400002000022 =×==

2

Solution 3: (7.7-4)

In this question we use polar full-width rectangular pulses.

Part a)

For binary signaling: 2

b

BB

RBW =

For 16-ary signaling: 816log22 2

bbs

BB

RRRBW ===

Reduction in bandwidth: 482

=÷ bb RR

Part b)

For binary case, we use 2/A± and the energy of signal would be

442

22

0

2A

T

EPT

Adt

AE

b

b

bb

T

b

b

==⇒=

= ∫

For 16-PAM, we use:

2/15,2/13,2/11,2/9,2/7,2/5,2/3,2/ AAAAAAAA ±±±±±±±±

SPAM TAAAAAAAA

E

+++++++=

4

225

4

169

4

121

4

81

4

49

4

25

4

9

416

2 22222222

16

b

S

PAM

PAMSPAM PT

EPT

AE 85

485 16

16

2

16 ==⇒=

Solution 4: (7.7-5)

Part a)

bPB

b

BB RBWR

BW =⇒=2

For binary case, we use 2/A± and the energy of signal would be

88242

2222/

0

2A

T

EP

TATAdt

AE

b

b

b

bb

T

b

b

==⇒=×=

= ∫

Part b)

M

RBW

M

RRBW b

PB

bS

BB

22 loglog22=⇒==

3

( ) ( ) [ ]SSMPAM T

AM

MT

AMAAA

ME

8)1(...531

2

8

)1(...

8

5

8

3

8

2 2222

22222

−++++=

−++++=

83

1

86

22223

S

SMPAM

TAMT

AMM

ME

−=

−=

( )24

1 22AM

T

EP

S

MPAM

MPAM

−==

Proof of [ ]6

)1(...75313

2222 MMM

−=−++++

[ ] ( )2

4412)1(...75312/)2(

0

2/)2(

0

22/)2(

0

22222 MkkkM

Mm

k

Mm

k

Mm

k

++=+=−++++ ∑∑∑−=

=

−=

=

−=

=

23

)1(6)12)(1(2

22

)1(4

6

)12)(1(4

MmmmmmMmmmmm+

++++=+

+×+

++×=

23

8124

23

66264 23223MmmmMmmmmm

+++

=+++++

=

23

2

28

2

212

2

24

23

M

MMM

+

−+

−+

−

=23

2

28

2

212

2

24

23

M

MMM

+

−+

−+

−

=

( )( )

623

)2(42

6

2 32

3MMMM

MM −

=+−

+−+−

=

Solution 5: (7.7-6)

⇒=n2256 =n Number of bits per sample 8=

sec/8.3521.448 kbitsfnR Sb =×=×=

( ) ⇒≤=+=+=+= 30log2

441

log2

8,352)25.01(

log2)1(

21

222 MMM

RRBW bS

BB αα

835.760

441

2 22260

441log =≥⇒≥⇒≥

MMMM

Note that we have taken smallest possible M satisfying the bandwidth requirement to achieve the best

possible performance.

4

Solution 6: (7.8-1)

Part a)

MHzHzRBW bPB 1106===

Part b)

kHzff

f CC 502

100

2

21 ==−

=∆

kHzR

B b 5002

10

2

6

===

( ) ( ) kHzBfBWPB 11005005022 =+=+∆=

Solution 7: (7.8-2)

Part a)

MHzHzRBW bPB 2.110)2.01()1( 6=×+=+= α

Part b)

kHzR

B b 6002

10)2.01(

2)1(

6

=+=+= α

( ) ( ) kHzBfBWPB 13006005022 =+=+∆=

Solution 8:

Part a)

710−=

=

n

P

e

AQP

σ

From Q-function table: 20.5=n

PA

σ

VAmV Pn

31020.51 −×=⇒=σ

Power of the signal for a polar full width pulse is: ( ) WattsAP Preceived

6232 1004.271020.5 −−×=×==

Part b)

1000010log1040 4==⇒= XXdB

Transmitted power WattsPreceived 2704.01004.271000010000 6=××==

−

Part c)

Average number of errors in one hour 36101000003600 7=××=

−