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Unformatted text preview: Problems and Solutions (Chapter 4) 1. Explain why channel coding reduce bandwidth efﬁciency of the link in
high signaltonoise ratio conditions. [Solution] Channel coding is a technique used to overcome the deﬁciencies of the
noisy transmission medium. It basically adds redundancy to the trans—
mitted bits. Due to redundant bits the cli'ective bandwidth available for
useful data transmission is loss. . Can channel coding be considered as a post detection technique? [Solution] Since we have redundant bits we can use them to detect/correct errors.
Thus it can be considered as a post detection technique. What is the main idea behind channel coding? Does it improve the per
formance of mobile communication? {Solution} Channel coding is a technique used to overcome the deﬁciencies of the
noisy transmission medium. It basically adds redundancy to the trans
mitted bits so that the receiver is able to detect and/ or correct the errors
in the received information using the additional bits. It improves the
robustness but decreases the eEective bandwidth. . If the code generator polynomial is 9(3) = 1H:2 for a (5, 3) code. Find the linear block code generator matrix G.
[Solution] 'g(a:)=1+23 n=ak=an—k=a
p‘=rem(1i;{§‘—‘1) fort: 1m 1:,
p1=remtni:Zi;)= 14101. p2 =rem(zé:—l’q)=m—> [011. p3 ”emulate 1—» [1 0]. thus OOH
GHQ
HOG
Her G=[I,,P}=[ OHO
‘L—l 24
5. The following matrix represents a generator matrix for a (7, 4) block code.
1 0 0 0 1 1 0
G _ 0 l 0 0 0 1 1
F' 0 0 1 0 1 1 1
0 0 0 1 1 0 1 What is the corresponding parity check matrix H? [Solution]
n=7,k=4.
Given
1000110
_ _0100011
G—{I"P]“' 0010111
0001101
Thus
110
_011
P‘111
101
110
011
111
HT=[IP]=101
“"" 100
010
001
Therefore, parity check matrix
1011100
H={HT]T= 1110010
0111001 6. Find the linear block code generator matrix G, if the code generator poly
nomial is 9(3) = 1 + 2:2 + 2:3 for a. (7, 4) code. [Solution]
9(50) 2 1 +3:2 +13
11:7, 10:4, n—k=3, p‘ = rem(1i;:5_—‘l for i = 1 to k, p1.=rem(m§2;n)=1+x2~+ [1 o 1], p2=rem((ﬁ§§3;q)=1+z+x2—+ [1 1 1],
p3=rem(?ﬁlp};;7)=l+z> [1 10],
p4 =fem(rﬁ§lm) =$+$2—9 [0 1 1], .thus 
1 0 0 0 1 0 1
_ _ 0 1 0 0 1 1 1
G"[I"P]‘ 0010110
I 0 0 0 1 0 1 1
7. Repeat Problem 4.6 if g(z)=1+z3 fora.(7, 4) code.
[Solution] '
g(2:)=1+2:3
n2ﬁk=Ln—k=&
pi=rem( Err*1 fori=1to 1:, p1: T6m(ﬁl:7235} = 1 —» [1 o 01.
p2 =rem(ﬁ‘:—237) =36 {0 1 0],
p3 mama—3;) =.2 + [o 0'11,
p‘ = rem(ﬁ(_f_—?,7) = 1 —+ [l 0 0], thus
10 00 10 0
_ 0 10 00 10
Gﬁmm‘ 001 0001
0 00 11 00 8. Consider the rate r = 1/2 in the convolutional encoder shown: below. Find
the encoder output (Y1 Y5} produced by the message sequence 10111. . . . Figure for problem 4.8. [Solution] If we consider the input to be X and the ﬂip ﬂop to be D then from the
ﬁgure we can derive the following transformation table. 26 10. Y2=X
Dn=Xn1
D1: Therefore for the input string 10111.. . .
The output string is 1110110101. . . . . Find the state diagram for Problem 4.8. [Solution]
The state diagram for the above problem is as follows Figure for problem 4.9. The states represent the value in the ﬂip ﬂop and the arrows represent the
transitions from one state to another. ix): represents that if X is the input then YY is the output. Initially the state machine is in 0 state. The following ﬁgure shows the encoder for a. 1/2 rate convolutional code.
Determine the encoder output produced by the message sequence 1011... I: ﬂip Hopshmresista)
e Modulol Edda 11. 27 Figure for problem 4.10. [Solution] . From the ﬁgure given we can derive the following equations
Y1 =X$D1$Da®Ds Y2 = X 9 D1 63 De D? = X"‘1
a = D1“
m=m* Using these equations we can construct the following table Thus for the input sequence 1011. . . , the output sequence is 11110111... . Consider a. SAW ARQ system between 2 nodes A (transmitting node )
and B (receiving node). Assume that data frames are all of the same
length and require 1" seoonds for transmission. Acknowledgement frames
require R seconds for transmission and there is a propagation delay P on 12. the link (in both directions). One in every 3 frames that A sends is in
error at B. B responds to this with a NAK and this erroneous frame is
received correctly in the (ﬁrst) retransmission by A. Assume that nodes
send new data. packets and acknowledgements as fast as possible, subject
to the rules of stop and wait. What is the rate of information transfer in
frames/second from A to B? [Solution]
The time taken to transmit a data frame from one node to the other is Transmission Time + Propagation Delay = T + P The time taken to transmit an ACK frame from one node to the other is Transmission Time + Propagation Delay = R + P Time required for successful transmission of a frame is T+P+‘R+P=T+R+2P (time for data frame from A to B and positive ACK from B to A). Time required for successful transmission of a frame after receiving a neg
ative ACK is ' 2(r+ R + 2P) (Additional T + R + 2P required for successful retransmission) Given that one in every 3 frames is in error, the time required to transmit
3 consecutive frames is T+R+2P+T+R+2P+2(T+R+2P)=4(T+R+2P) This cycle repeats for every 3 frames so the information transfer rate is 3
4(T+R+2P) frames/second. Compare and gamma GBN ARQ and SR ARQ schemes.
[Solution]  GEN ARQ SR ARQ Provides better channel utilization as Provides better channel utilization as
compared to SW ARQ compared to SW ARQ
All frames transmitted after a frame Only the frame for which an NAK is for which an NAK is received must be received is retransmitted
retransmitted. Better suited when burst errors are Better suited when single packet errors
The receiver does not maintain a buﬂ'er The receiver has to maintain a butler
for the frames received as packets are to order the packets explicitly 13. Consider the block diagram of a typical digital transmission system. Spec 14. ulate where one would use source coding and channel coding. Differentiate
between them. Would they increase or decrease the original message size?
(Hint: we want to transmit most efficiently, i.e., message size should be
the smallest possible but enough redundancy should be added to correct
small errors so that retransmission is avoided as far as possible). Infonnanon Message Prince . EC‘ﬂhzrunel . M 0d ulster l I
Figure for problem 4.13.
[Solution] Source coding is a technique used to ensure that the information to be
transmitted is coded using as few bits as possible. The basic purpose of
source coding is to better utilize the available bandwidth by transmitting
more information using fewer bits. Channel coding is a technique used to overcome the deﬁciencies of the
noisy transmission medium. It basically adds redundancy to the trans
mitted bits so that the receiver is able to detect and/or correct the errors
in the received information using the additional bits. Source codingdecreases the original message size by using compression
techniques whereas channel coding adds redundant information to over
come transmission errors thus increasing the transmitted message size. Can you interleave an interleaved signal? What do you gain with such a
system? '
[Solution] Yes, we can interleave an interleaved signal. It can increase the robust ness and avoid burst errors. However there may be possibility of de
interleaving. 30 15. 16. 17. 18. Why do you need both error correction capability and ARQ in a cellular
system? Explain clearly. [Solution] In a cellular system, the last communication link is between the BS and
the MS. The wireless link is exposed to loss and interference in the open
air medium and hence the error rate is much higher than the wired link. Error correction techniques and ARQ sches can provide an elegant way
to implement endto—end errorfree transmission. Error correction capa
bility can improve the link quality. It will decrease the probability of
retransmission. However, channel coding can not correct all the errors
in the transmission. Hence, the ARQ is also necessary if we need the
errorfree transmission. In a twostage coding system, the ﬁrst stage provides (7, 4) coding while
the second stage supports (11, 7) coding. Is it better to have such a
twostage coding scheme as compared to a singlestage (11, 4) complex
coding? Explain your answer in terms of the algorithmic complexity and
error correcting capabilities. [Solution] The twostage coding system is equivalent to the singlestage coding in
terms of algorithm complexity and errorcorrecting capabilities. For al
gorithm complexity, there are 3 XOR operations in the (4, 7) coding and
4 XOR operation in the ('7, 11) coding. Hence, there are total 7 XOR
operations in the twostage coding system which is same as singlestage
coding system. The errorcorrecting capabilities of twostage and single
stage coding should be same, since only single bit error can be corrected
in these two systems. Under what scenarios would cyclic codes be preferred over interleaving
and vice versa? [Solution] Interleaving is preferred for trafﬁc where burst errors are more probable.
Cyclic codes are used for noisy channels where individual bit errors are
much more frequent. Polynomial 1 + 27 can be factorized into three irreducible polynomials
(1 +x)(l +3 +33)(1 +z2 +x3) with (1 +m+a:3) and (1 +x2 +29) as
primitive polynomials. Using 1 +x+ 9:3 as generator polynomial calculate
the (7, 4) cyclic code word for given message sequence 1010. [Solution] Let d(:r) represents the polynomial for the message “1010"I that we are . trying to encode, then, d(:c) =2+$3 19. 20. 21. 31
Let g(:r) represent the generating function, then 9(5): 1+0:+.1.3 Let the code be represented by its corresponding polynomial C(x).
From the theory of cyclic codes, we have 0(5) =9(5)d(3)
= (1+z+23) (2+?)
=11.'+a:2+1t3+x6 Therefore, the cyclic code is = (1 0 0 1 1 1 0). Repeat Problem 18 with 1 +32 +273 as generator polynomial and compare
the results. [Solution] M3?) = [:30 + 32)]mod[1 + 2:2 + 13] = :1: + 3:2 Therefore, C(z) = 2, + z“ + 33 +25 or 0111010, with 011 as parity and
1010 as message bits respectively. Develop the encoder and syndrome calculator with 1+1:2 +33 as generator
polynomial in Problem 18. [Solution] ‘
Encoder for the (7, 4) cyclic code generated by 9(15) 2 1 + x2 + 11:3: Figure of Answer 1 of Problem 20. Syndrome calculator for the (7, 4) cyclic generated by 9(5) = 1 + :52 +29: What is an RSC code? Why are these codes called be systnatic?
[Soiution] RSC code is Recursive Systematic Convolutional (RSC) codes. Since the
parity sequence is appended at the end of the information sequence in the
encoded sequence, it is called systematic. 32 Module2 Flip—top
adder 22. Describe brieﬂy syndrome decoding and incomplete decoding? [Solution]
Syndrome decoding: In linear blodr encoding, n—k parity bits are added to the original message
vector In which is 1: bits long. Hence, the transmitted vector c and the
received vector x at the decoder are it hits sequences. Syndrome decoding
consists of calculating the 1 x (n— k) vector 3 which would either uniquely
map the received vector x to a possible match in c or if it is not possible to
correctly decode the received vector x, a nonnull vector value of s would
indicate errors in transmission. In a linear block coding transmission, the
syndrome vector 8 can be calculated by s=xHT, where H = [PTIn—k] and H is called the parity check matrix and is given by concatenating the
transpose of the parity matrix P (k by n — A: matrix) and the identity
matrix LP]: (11  k by n— 1:: matrix). Incomplete decoding: Incomplete decoding is not an independent decoding method. Actually,
it is combined with eidsting decoding methods to improve their error cor
rection capability. For example, the target of the maximum likelihood
(ML) decoding is to ﬁnd the likely codeword after a codeword is received.
However, it is possible to have more than one such codeword. If using
the existing ML decoding, the decoder will arbitrarily select one of them.
However, it will request a retransmiseion in the incomplete ML decod
ing. It can improve the error correction capability since the probability
of continuous decoding failure is much smaller than probability of single
decoding failure. Furthermore, incomplete decoding can be used in the
other probability decoding method. 23. 24. 25. 33 Prove that the average transmission time in tenns'of block duration, T53,
for SelectiveRepeat ARQ is given by: T511 = 1PACK + 2oPACK (1 — PACK) + 3.13m]: (1 — PACK )2 +   ° . where PACK is the probability to return 8. ACK in the transceiver side.
Also, solve the above equation for PACK = 0.5. ' {Solution} .
The probability of receiving an ACK with no losses = PACK The probability of receiving an ACK after one packet loss = PACK (1 
PACK) The probability of receiving an ACK after two packet losses 2 PACKG 
Plies)2 so. Thus, total time is given by Tss = 1PACK + 2«PACK (1  PACK) + 3PACK (1  PACK)2 +    . For PACK = 0.5, T511 = 2. In Stopand—Wait ARQ, let the probability of the transmitting side receiv
ing 5 ACK after exactly one loss of ACK be P = 0.021. Find the Average
transmission time in terms of a. block duration if : D = (Round trip propagation delay) R5 = bit rate, 11 = number of bits in a block.
(Hint: The probability for the considered case = PACK (1  PACK ))
[Solution] notes).
9 11 Compare a. block with s. convolutional interleaver. [Solution] A block interleaver accepts a set of symbols and rearranges them, with
out repeating er omitting any of the symbols in the set. The number of
symbols in each set is ﬁxed for a. given interleaver. The interleaver‘s oper . ation on a set of symbols is independent of its operation on all other sets of symbols. Whereas a convolutional interleayer consists of a. set of shift
registers, each with a. ﬁxed delay. Each new symbol from the input signal
feeds into the next shift register and the oldst symbol in that register
becomes part of the output signal. ...
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