0195136047.pdf

PROBLEMS 741

the initial value of the output be 1. Also, show the

schematic representation of the required decoder

in the receiver.

Input{bk}:0111010010

1100

15.3.10(a) In the so-calledfolded binary codeof 4 bits,
the leftmost digit represents the sign of an
analog signal’s quantized samples (with 0 for
negative and 1 for positive), and the next three
digits are the natural binary code words for the
magnitude of the quantized samples. Obtain a
table of binary code words.
(b) By letting the digit sequence in part (a) be
b 4 b 3 b 2 b 1 , which also represents a 4-bit natural
binary code for 16 levels labeled 0 through 15,
obtain the digit sequenceg 4 g 3 g 2 g 1 , where

gk=

{
b 4 , fork= 4
bk+ 1 ⊕bk, fork=1, 2, 3
which is known as theGray code. A unique
characteristic of the Gray code is that code
words change in only one digit between ad-
jacent levels. Check the same in your result.
[See Problem 15.3.9 for modulo-2 addition,
represented by⊕.]
15.3.11Consider the digit sequence

{bk}= 1001110010

11011000

(a) Sketch the polar and unipolar waveforms for

the sequence. Estimate the probability that a

binary 1 will occur in the next digit interval.

(b) Let{bk}be the input to the differential encoder

of Figure P15.3.9, and

(i) Find{ak}with the initial value 1 of the

output.

(ii) Find{ak}with the initial value 0 of the

output.

(iii) Check whether the original sequence

{bk}is recovered in both cases, when{ak}

of (i) and (ii) is put through a differential

decoder, as shown in Figure P15.3.11.

(c) For the first 10 digits of the sequence{bk}, il-

lustrate the Manchester-formatted waveform.

*15.3.12An audio message is band-limited to 15 kHz, sam-

pled at twice the Nyquist rate, and encoded by

a 12-bit natural binary code that corresponds to

Lb= 212 =4096 levels, all of which span the

message’s variations. Find the first-null bandwidth

(given byωb/ 2 π= 1 /Tb) required to support a

polar waveform format. Also, determine the best

possible signal-to-noise ratioS 0 /Nqwith a crest

factor of 3.8.

15.3.13In Equation (15.3.24), for polar and Manchester

PCM,Eb=A^2 Tb. Find the noise levelN 0 /2 at the

input to the polar PCM receiver that corresponds

toPe= 10 −^5 whenA=6 V andTb=0.5μs.

15.3.14For a unipolar PCM, the bit-error probability is

given by

Pe=

1

2

erfc





√√

√

√^1

8

(

Sˆi

Ni

)

PCM





whereas for polar and Manchester PCM,

Pe=

1

2

erfc





√√

√√ 1

2

(

Sˆi

Ni

)

PCM





Inputsequence {b +

k}

{ak − 1 }

Delay Tb

Modulo-2 adder

Output

sequence {ak}

Figure P15.3.9

Demodulated +

sequence {ak}

Delay Tb

Modulo-2 adder (see Problem 15.3.9)

Recovered

sequence {bk}

Figure P15.3.11