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Unformatted text preview: 1. (10 pts) Proof of Hamming.
State and prove the Hamming Bound n _ '3;
M g N ‘ _ W (2;? mm 1“ :: GQC$C§IAEI 6“ [Q J
R
1/0 {WM 6:: y 57% 6: ~E
E 4 w~w~m~w w. ﬁn!“ 3 "m mi Wetsmﬂ {a} g‘ {a
:5»: “K: @ 2. (10 pts) GF(11).
Recall that GF(11) may be described as the integers 0,  ‘ v ; 10 with modulmll addition
and meltipiieetion. Find an inseger that corresponds to a, primitive element and express
all of the nonzero elements of GF(11) as powers (module 11) of an integer less than 11
(Le. a single element. of GF(11)). (3W golw‘l¥€M F .5 M. C an {kg} kiwi/g 3. (12 p135) Plowing a Galois Field.
This problem investigates GF(64). (a) (4 pts) What are the possible multipiicative orders of elements in GF(64)? HOW
many elements have each order? 055:; Me: Mu ﬁg}? F3“ gave: 0 “cfﬁ m”. I “jgdwms 66f“: 6 ﬁve I" '4’Lw5‘e 6W5; Ligfﬁicﬁgxhé‘yﬁ ﬂ
wmigw 43 Wait?!” a it 0+1 awe? w {* (b) (4 pts) Consider factoring (1:63 w 1 over It just so happens that the
irreducible factors are .Lfﬁeet+w+1 W "i" 5'35 r 3:2 + :2: +1
‘ $3 + $2 + 1
~» ééé: s6 + as" + 1:3 if a; + 1
r ’ $4 + :62 + 1
1
‘ $4 + x + 1 6 $5 + 1
“Wrmw’m‘mwm“4,..ivkwm—ImMéw““'
"E m + 9: + i E
For each. irreducible polynomiai above, identify the multiplicative order of the ele—
ments that are the conjugates that are the poiynomial’s roots. You can write this
information beside each poiynomial. You will ﬁnd the following multiplications
useful: (w+1)(m2+m+1)mm3 ——1
($3+$2+1)(m3+$+1){$+1)291:7 ——1
(m2+m+1)(m5+m4+m2+m+1)(m6+m5+m4+m2+1)(m7+1) 2222144 C>Wla CW“? efewwl and: mfcger‘ .11. :57 $1M mad ﬁg (Kiwi?) chi he “lies
eipwn'b m” {be/“Clef a: Moral?" piewewl“; can ape“de ‘3 13$ 42%;; weevil} a"? (Kaye!) wwi“ 431%.. i ‘R\
w mam "3%; mam»; (em wear grewum xﬁxrixriife‘ewlm .. . \ \ “.2” m, é {)LDMQJ»; a” (BUhing" and “pf; (agwﬁiiﬁ’ﬁmexil) m3 X+Mi~l wag xzv’rE‘rl raw—Wain gnaw woe“ erf awaitsv” w?
M We em “ﬁeﬁmw 257% (aliiliﬂgexgel} “n+3 sari" ﬁle3:63“ a“? “f” 5 Mmj‘ﬁﬂﬁ d"? oweiev' (a Q ‘4' \ﬁ i . ‘ Lli‘twa “@tlﬂfﬁmﬁli‘eon 5;» WE:le MN? gaff ‘5 3’3 iyfl‘w*'l #3 ﬁnal? (kgL Vary {LI3‘53" E.» moi. a vim Names; We: {.2 me am 9wa “Me
a me We ram any? f9 A mm} m i} Emqu q “ mgrm E <23 é 3 MV id" be: (c) (4 ptS) Now suppose that $6+$+1 is a primitive polynomial for GF(64). Let or be
a root of this primitive polynomial. Express 0:24 as a. poiynomial with coefﬁcients
in GF(2) with degree no more than 5. 2%} as? “f’é’uén 6x m X ME 4. (10 pts) BCH Code Design. The following fable lists the conjugates arid their corresponding minimal polynomials
for GF(64). Table 1: Cyclotomie cosets (conjugates) and their minimal polynomials for GF(64). Conjugacy Class Minimal Poiynomial 1 :17 + 1 a,a2,a“,(x8,oclb,a32 a: +zr+l 01311 (16,0!12! @211 053.35 Q48 $6 +34 +1.2 + 1L. +1 0552043030437, Q20, Q34? arm $6 + $5 + $2 +2: + 1 mm: of? 87 O; 5} 1.6 + IT} ‘1. (19:5118:04‘56 + .132 +1 ali’a22ia255a373a443a50 mh+m5+ I3+$2 +1 or ,oz‘ ,0: or or or xii—Hr +2: +x—l—1
/ 0615,0530]QBQ’G51)Q57?(360 ' $5+$5+$4+m2+1 ‘ 0:21, 0412 m? + 2: +1 "We’d Jaeﬁaajassja‘ﬁ $6+I5+m4+$+1
a27,a45,oc54 233+23+1
0531, @473 (15530659, @613 aw _ its + $5 + 1 Design a binary BCI—I code with length n 2 63 for design distance 6 ﬂ 9. You may
express the generator polynomial as a product of minimal polynomials. To get full.
credit you should ﬁnd 'the highest rate code possible. E Lei“ l“ 345 avail‘3 i; V 3cm MMMM we“? K" WCMQ : Mela??? MEG?) Mmgé’fwl Mime Cw??? 5. (12 pts) Shortened RS Codes. Recall that Reed Solomon (RS) cocles over CF97”) are
cyclic codes that have the form gar) = (w+a)(m+a2)(sv+a5‘1) (I) (4 pts) Prove that RS codes are maximum distance separable (MDS) (Le. that
they achieve the Singleton bound). You may use Without proof the Singleton
bound and the BCH bound that relates dmin and 6. (lea (.3 e: k 6 w i
“:3? 7Q» k—H '3" 5 (b) (4 pts) Let C (5) be the “shortened code” obtained by taking all codewords in the
original RS code C that have 0 in the last 3 positions and deleting those last .5
positions. Deleting those last 8 positions creates a code of length n — s, Where n is
the length of the original Reed Solomon code C. Find the dimension is (the length
of the message word or the number of coefficients in the message polynomial) for
the shortened code (3(3), Hintﬂonsider the generator polynomiai inﬁerpretation
of RS codes, which message polynomials cause codeword polynomials With Zeros
in the s highest—degree coeﬁicients? (c) (4pts) Show that the shortened code C (3) is MDS‘ To get full credit your proof
must be complete Faw‘i‘ {6%.}; éwfmuw é \aﬁ SQ; W? wwé “ti23, stew & «3: (£63
x1
Aﬁécwﬂm») 2:: Wm dRCCtQM’JﬂmQQQ} 5 CE
C, ,czﬁfé} gifoaé é" ng CngFCcr ;C2_‘3 Qt? CMvgﬁ (f3: “
C.E {é mﬁtgwg (JCS) Wfﬁm kfﬁﬂ %» €§€§3 A685“: mﬁs‘amiafﬂﬂ 1:3? M533 10 6. (10 pts) Finite Tmcebock Viterbi with Hard Decoding. Congider the rate—U4 encoder G(D) m [1 + D ~ D2 D 1 ~Em D 1 + D23 with the
leftmost polynomiai corresponding to the most signiﬁcant bit. (a) (1 points) Draw the canonical feedforward encoder circuit. (b) (2 points) Draw one stage in the trellis, labeiing the branches with the correspond—
ing binary output symbois. You will use this diagram for the Viterbi decoding of
the next part. (c) (4 pts) Suppose that the decoder has been operating for a While implementing
hard decoding, and the states contain the path metrics shown in Table 1. rTable 2: Path metrics gust before the next four points are received. 1 State Path Metric 00 _ 3 01 7 10 5 11 3 The next four received three—bit symbols are 1118, 1800, 1110, and 0100. Use
the standard Viterbi AddComparcSelect operation to til} out the four stages of
the trellis associated with these four received points. Clearly identify the survivor
paths and associated path metrics. Note that you CANNOT assume that the receiver begins in state zero, since the
decoder is already in steady state with the metrics in the above table. You should
use the above table to initiaiize your trellis path metrics. «i
e ‘ cf:
Wciww @
"mm, 3 
ma at “NM “A 12 (d) (3 pts) Use finite—traceback Viterbi decoding with a decision depth. of three trellis
branches to decode the two input bits associated with the ﬁrst two received points. "(at %,.g%s~ MN w W %me wag. {39% ﬁat (133 M mum a! m l f ge tamcg“. a: N335 (‘0 413‘ I w J; ‘gW‘ﬁ: Q'Q a“: £2. wwﬁﬁux 34m (11“; {W MW 5: a; w» :3; 13 7. (17 pts) Minimality of convolutionel codes. Consider the following generator matrix GU?) for a oonvolutional code. 132 + 1310 1 + D2 1
G{D) m [1+D2+DEOD D2+D10+D100 I] (a) (3 pte) Is this encoder deiay—preserving? Show your "work. {LN
ﬂ} (jam? [email protected],§@W‘W {e13 M e) z a kak (b) (3 p133) Is this encoder degreepreserving? Show your work. E €> <3
£)_ :3 {5}” ﬁt a? Q’ we? E} \ fywﬁv \ss‘ :? 14 . D2? D39 (0) (4 pts) Is this encoder non—catastrophic? Show your work. E+Dl \{3 M {ms a, : Ciﬁﬁrﬁﬂ e“ e?"
E) #3) H.) g $6 ﬁi‘ b‘. on :2 E + ﬁw “E Ema GQwrﬁﬁwwee thgﬁﬁuﬁWPi‘m W 12%,. thvmcﬁef b (d) (4 pts) At least one of the previous parts has an answer of ’7110.” Pick a previous
part where the answer was "110” ané replace a row of G (D) with a new row that
lowers the degree of the encoder but maintains the renge~space (row—space) of the
code. eﬁe” seel 3 j T [1 1 a] We?” (Jr Day” mew _ NW“ E 01;? 102:: ND: ‘1
Ar“. .1 1‘ e E (e) (3 pts) Is your new encoder minimal or not? Show your work. [32% b\ 1’3 G mu 3 "3‘ £3.33.  l W m“ '““ (1.636?“ :2 ‘gwwa ﬁlm» _W.W..Mﬂ=w.mm.....m 16 ...
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 Spring '11
 Dun
 pts, Coding theory, Reed Solomon, Polynomial code

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