midterm_solns - 1. (10 pts) Proof of Hamming. State and...

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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—Im-Méw““' "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 $1M mad fig (Kiwi?) chi- he “lie-s eipwn'b m” {be/“Clef a: Moral?" piewewl“; can ape“de ‘3 13$ 42%;; weevil} a"? (Kaye!) wwi“ 431%.. i ‘R\ w mam "3%; mam»; (em wear grew-um xfixrixriife‘ewlm .. . \ \ “.2” m, é {)LDMQJ»; a” (BU-hing" and “pf; (agwfiiifi’fimexil) m3 X+Mi~l wag xzv’rE-‘rl raw—Wain gnaw woe“ erf awaits-v” w? M We em “fiefimw 257% (aliiliflge-xgel} “n+3 sari" file-3:63“ 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 {LI-3‘53" E.» moi. a vim Names; We: {.2 me am 9wa “Me a me We ram any? f9 A mm} m i} Emqu q “ mgr-m 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. (19:51-18:04‘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 Jae-fiaajassja‘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 “ti-23, 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 ~E-m 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 Add-Comparc-Select 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 tam-cg“. a: N335 (‘0 413‘ I w J; ‘g-W‘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? FM@,§@W‘W {e13 M e) z a kak (b) (3 p133) Is this encoder degree-preserving? 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 thgfifiufiW-Pi‘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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This note was uploaded on 03/29/2011 for the course EECS 100 taught by Professor Dun during the Spring '11 term at CSU Channel Islands.

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midterm_solns - 1. (10 pts) Proof of Hamming. State and...

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