ee321ch13

ee321ch13 - \3.\ 5 ' ' £6955 ‘5 lfiZfikrlflo ,...

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Unformatted text preview: \3.\ 5 ' ' £6955 ‘5 lfiZfikrlflo , u," 1:: IL” Cm uh = :44 C03 L{01 3-7 1 q ‘ xxx — - - - ' ' him ‘/ ®.U3f;ldrs W Max fighter a. ’Qow gfnfixa Anumflnw K a 61vth Arc cmw-N§' QVDM \Gll ewe, 0001100le ~= l+igtk23+1ce = Llolv” ‘0’) *® “(L-elm $0.“ mace; ‘ (3,“ “he: “300100012 (Q (,(cw‘M gaexfl +09le = 10?. u, on M LA Acg tuxst an med Me. ' KM H C. ‘ - c? L‘JOIO‘OI; - heK Gvou?‘\m> qIN Sbm‘vua fijflt'k-Wj mot-{k 49% \SB 9.4L flxék w€3civ~g em '. is a gut-J»; weMML 'Jto CO‘AUQJJC QYGM , .xb'mwg Jro' \Ne-K I. ' ' _ ' k ' 10.1%M C’s—’5 (OS Rex = Lam) filmyb J: \o\ m iofl owl Mu LQ Oxvfiwm 5W kw Mao in 3YDU‘35 e4. B143 30w 5'ch WW? 9% a“ km dmuiqr ' G. Rizzoni, Pdnciples and Applications of Electrical Engineering Problem solotions. Chapter 13 'Sectlon.13.2:' The Binary Number System Problem 13.1 Solution: Know quantifies: . The base 10 representation of five numbers: 401m, 273m. 1510, 38m, 56m. . Find: _. The hex and the binary representation for these numbm. Analysis: Using the methodologies introduced in paragraph 13.2; a) 19115.1100100012 b) 111,5,1000100012 c} P15. 11112 d) 2615.1001101 6) 33m 1119002 Problem 13.2 Solution: Knowu quantities: ' I The hex representation of five numbers: A15. 6615. 4715, 2115, 1315. Find: The base 10 and the binary repmsentation for these numbers. Analysis: Using the memodoiogies introduced in paragraph 13.2: a) 1010, 1010; b) 1021m31001101 c) 71;". 10001112 d) 3310. 1000012 «3) 19m. 1001!; emblem 13.4 Solution: I Known quantifies: . r The binary representation of six numbers: 11112, 10011012, 11001013, 10111002, 11101;, 1010002. find.- Thehexandthe baSe lflrepresentafion for time dumbera. Analysis: Using the methodologies inlroduced in paragraph 13.2: a) F16» I510 b) 41315.77") 0) 6516. 10110 d) 5C“: 921.3 6) 1133;, 29m fl 2315,4010 13.3 J j +BO\C>HH' I,_fl________.___._._———— Hmmo' I \+\ '-‘- 0 BC‘\"\,\ 0. twig cg: k JR: kw (D‘Qmm. ML l-HH ='\ u “ v " ax (1A Mamas' Wx Me. 0.99s! '_x:a;.'7§ ® “$3 mum. k 1‘ g ‘ “ “ = o H WC‘LQV—L * \5 (h. “DRMQ-l, “JG-RUE. - “a {fix La I @332! 059:“ Maul-- 1"\\ooc>\o ’I‘ . ' M1315 :x = letgcckcfl vafiug \ \ (Sam 0 = Lina H'L’rl\= 018 _q%- ._., .- #5. 'ch 4, m mavens M mad 90.38,. rm 3' (q T0005 CEMTXQWVQW/E a4 \DXHOQ ' 009th 3‘0 0134mm ' 115% wp‘a ~ h\ “a”: \bfi-tm cop? ‘1 _ . . ehdq SULCQSSWL- ‘na Ck. Q9] ‘WL ‘e‘C‘d‘qs Fa.c\,\._ I (FM QML mom/$5M o.\c>blc3>0 I ' ' _ ' . - K/‘Wf’t ca\-,LM\)L'¢L) Musis _' g, W“‘“§+P P kg 1‘1 '1 n“ P0152, 03cm my}. G. Rizzoni. Principles and Applications of Electrical Engineering Problem 13.5 Solution: ’ Known quantities: Three couples of binary numbers. Find: ' The addition for each couple. Analysis: Using the meflrodologies introduced in paragraph 13.2: at) 11111010 13) 100010100 c) 110000100 7 m Problem 13.6 Solution: Knuwn quantifies: ' Three couples of binary numbers. Find: The subduction for each couple. Analysis: Using the methodologies introduced in paragraph 13.2: a) 11100 101101110 0) 1000 I -—-—————-————.—__.__.._.._________m__ ‘ Problem' 13.7 Solution: Known quantifies: . Three eight-bit binary numbers in sign-magifimde form. find: The decimal value of these numbers. Using the methodologies introduced in paragraph 13.2: 59-120 b) -31 c) 121 W Problem 13.8 Solution: . Known quantities: Three decimal timbers. Find: . ' The sign-magnimde form binary representation. _ Analysis: ‘ Using the methodoiogies introduced in paragraph 13.2: _ a)01111110 b)11111110 (901101100 _ 611100010 W 13.4 Problem solutions. Chapter 13 ' G. Rizaoni, Principles and Applications of Elecm'ca! Engmoering Problem solutions. Chapter 13 Problem 13.9 Solution: Known qunnflfia: - l ' Pour binary numbers. ' - Find: ‘ Tho two's complement of those four numbers. Analysis: . Using the memodologiss introdpcefi in paragraph 13.2: a) 2‘-1111=10000-1111=0001 b) 27-1001101=0110011 . c) 27-10111oo=0100100 cl) 25-11101=100000-11101=ocou - ' E M _ i i i | i i E i 13.5 G. Rizzoni, Principles and Applications of Eiectrical Engineering Problem solutions, Chapter 13 Section 13.3: Boolean Afgebra‘ Problem 13.1! (— eleven. Solution: Known quantities: Theexpression B =AB+AB. . Find: The 11th table that proves that the expression is true. Analysis: Using a truth gable as explained in paragraph 13.3: we prove limit the expression is true. \/C.DM?&R‘\UC9 CDUJMNQ 1?) RA MB ‘\ R B Problem 13.11 Solution: Known quantities: -_ . Theexpression BC+BC +§A = A+B. Find: . The truth table that proves that the expression is true. Analysis: ‘ Using a Iruth table as explained in paragraph 13.3: "EL-4' BE +EA = AH?» \v/ a [Infill- , Chapter 13 ....._.--._fi_.....mmm.- Problem 13.1g Solution: Known quantities: The expression (X +Y)v(j{_+X -Y)=Y. Find: ' The proof that the expression is true using the perfect induction method. Analysis: Using a truth table as explained in paragraph 13.3: we prove that the expression is true. ISJLI F4: REE. +Kac, +ABC+AEL :F -; LE+A§RZ +L§C+fl13cw F: KZ‘i‘EC F 2 TS (El-MA (are) (2% L222) (2.5) . _.____——wmm—_.__mmmw______n__wnnu. Uh. b.” h“ ‘+. L Eu A .IMI Q B A.. LI a C «a» R. A A + +. den. mm. B ALA 3 . F BA? tsrrllillll]!lil:.l|:, . .. Q‘pqu‘f—LA \\ “Dammgak’g "V ecu {M ch-JQL lrélflfl 59w figs-PPQ-‘g I c5\'€_J(€,\/M{n{fi3 1U; X311Xtt'O‘O 0:; & MAMA suck me thb‘e‘k "J: 3H, a MOT 3J6. \3.2(. BIL? B30 \3.'5S ' Solution: Known quanfifles: The multiplexer circuit. Find: a) The truth table for the multiplexer b) The binary function performad by the multiplexer. h) , BinaryAdditiOmSisthesumandCis them. \332 -- , _ Solution: Known quantities: ‘ n _ r h The circuit of Figure P11”. ' 5 Find: z - | The ability ofthc circuit of performmg a conversion from 4—bit binary numbers to 4-bit Gray code. Analysis: ‘ We canstuct the truth table for this circuit as shown below: g The output is clearly a Gray code since each num‘oer only changes by one bit relative to the previous number. I i w I 1 3 i i I i | 13.33 ...
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This note was uploaded on 05/20/2008 for the course EE 321 taught by Professor Perks during the Spring '08 term at Cal Poly.

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ee321ch13 - \3.\ 5 ' ' £6955 ‘5 lfiZfikrlflo ,...

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