CARLETON UNIVERSITY Final EXAMINATION Wed, April 26, 2006, 14:00

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1 RLTO UIVRSITY Final ITIO We, pril 26, 26, 4: ame: umber: Signature: URTIO: 3 HOURS o. of Stuent: 29 epartment ame & oure umber: lectronic L 267, an oure Intructor() T.G.Ray an J. Knight UTHORIZ OR TUR OFF cell phone an peronal communication equipment an LV TH T TH FROT. ote, book, an non-communicating calculator are allowe. Stuent UST count the number of page in thi examination quetion paper before beginning to write, an report any icrepancy immeiately to a proctor. Thi quetion paper ha page. Thi examination paper ay ot be taken from the examination room. In aition to thi quetion paper, tuent require: an examination booklet ye no x may requet a Scantron heet ye no x leae anwer on the examination paper. The correct anwer houl eaily fit in the pace given. uewt You may ak for a booklet if you nee it. la vg oolean ( % for tating rule an map name.) If you ue a map, inicate in the pace unerneath it, for which function it i. % 73% If you ue algebra, inicate the rule ue at the right ie of each line. a) Simplify f = + ( + ).5% 94% = rule: (Simp; x + xy = x) 2% 59% b) Simplify g = + (+ ) If you can t o thi quickly, kip it. = ( + ) (eorgan an itbutive) = ( + ) (Reuction; x + xy = x + y ), (=) 2% 92% % 96% 2.5% 66% 2% 85% c) Simplify h = ( + ) + F G = ( + ) + ( + ) F G (eorgan) = ( + ) + F G (Reuction; x + xy = x + y) ) Fin the ual of k = + = () + () k ual = ( + )( + + ) (uality) e) Simplify m = ( + )( + + ) o ) firt m ual = + (From above) = ( + ) (itributive) m ual = ( + ) (Reuction; x + xy = x + y) m = + (Revere ual) lternate Solution m = (+) + Swap = + + (itributive) = + + = f) Simplify n = ( + ) + ( + )( + ) = ( + ) (Simp; x + xy = x) page T.G.Ray an J. Knight page of

2 arleton Univerity lectronic ngineering L267 ame 2% 65% 3% 55% 2 oolean a) Given q = ( + ) + fin q without uing overbar longer than one letter. q = [( + ()))] + ( ) q ual = [((+)) + + ]( + ) q = ((+) + + )( + ) arele peron, on t bracket all the arele q ual = ((+)) + (+)( + ) arele q = ((+)) + (+)( + ) b.i)fin any rouct-of-sum expreion for r = ( + )( + ) Remember Π of Σ expreion have no two-letter (or more) term r ual = (( + )) + (((+)) (ual) = (itrb) = + ( + ) + (itrib) = + + ( + =, =) r = ( + )( + ) (ual) b.ii) Fin the implet rouct-of-sum expreion for r = ( + )( + ) unle you have alreay one o. one above 3.5% 66% c) Fin a rouct-of-sum expreion for = (+) + = ( + + )( + + ) (itrb2 x+yz=(x+y)(x+z)) = ( + + )( + + )( + + ) (itrb2) Remember a Π of Σ expreion ha no two-letter (or more) term 4% 68% ) Simplify thi expreion to give the minimum number of gate. t = (+++)(+++)(+++F) Uually expreion with a lot of bracket are eaier to implify a ual. t ual = + + F (Take ual) t ual = ( + + F) (itb) t =(+ + + (F)) (Revere ual) 2% 42% e) rovie a circuit with input,,,,, an Y, an output F, uch that F = if Y =, F = if Y = F = if Y = F = if Y =. F= Y + Y + Y + Y F ither circuit will o 4-input U Y F T.G.Ray an J. Knight /4/7 page 2, of

3 arleton Univerity lectronic ngineering L267 ame 4% 68% 3 ux Logic Implement the function g uing only 2-input U an inverter. ecompoe the function in alphabetical orer. Removal of unneee U i expecte. If the expreion are the ame, mux can be hare g = g = = + + lway look for chance to ue reuction or implification + + If the expreion are the ame, mux can be hare eople mae mitake becaue they i not immeiately implify the logic expreion. It i very eay to ue x = x + ax, an x + a = x + xa 5% 83% 3% 8% 4 rogrammable Logic rogram the L to implement the logic efine by the K-map hown. The inverte line are mae thinner to help avoi uing the wrong line. write each term F G H ap of F here (Thi i the only ize of L available to you.) 5 Karnaugh ap a) Loop the map to give the minimum Sum-of-rouct logic. b) Write the equation any peron wrote a which i not minimum. ap of G ap of H T.G.Ray an J. Knight /4/7 page 3, of

4 arleton Univerity lectronic ngineering L267 ame 4% 82% 6 rouct-of-sum Karnaugh ap On thi map, (a) Loop the zeroe, to prouce the minimum rouct-of-sum logic. (b) Write the equation. ircling the zero give F = Uing eorgan F = (+++)(++)(++)(+++) 4% 7% 7 Five- Karnaugh ap a) Loop the map to fin the minimum logic. b) Write the equation. F = map of map of 5% 7% 8 Synchronou ircuit For the tate table an timing iagram hown, with R a the initial tate: i) omplete the waveform an ii) fill in the tate for the machine hown, with boxe to how the uration of the tate. ext = = Output Z = = R output z x x x x lock R Output Z ot people fin it eaier to write the output a hown in the extra column. Then after the tate i calculate, the output i either x, x, or. T.G.Ray an J. Knight /4/7 page 4, of

5 arleton Univerity lectronic ngineering L267 ame 9 Reuction 7% (i) Fin the equivalent tate in the table below. 8% (ii) ake a new tate table w. Revie Table ith the minimum number of tate. ext = = Output YZ ext = = Output YZ F J H F J H H J H F F J H H H J F H J F H J H F J J H J H J H H H F fixe thank to atalin J F H lternately one can ort the table by tate with the ame output. H F F F J H J J H J H J H J H F T.G.Ray an J. Knight /4/7 page 5, of

6 arleton Univerity lectronic ngineering L267 ame Synchronou ircuit (like II controller) 4.5% 6% The tate in the graph below are the ecimal value correponing to the output of four flip flop,, an. Thu for tate 5;,,, =,,,. Output Z= in tate where it i not hown. omplete the waveform for,,, an Z. Their initial value are all a hown. 9 z= z= Z tate lock ynch Reet The main problem wa with aynchronou reet. ll the flip flop go to zero immeiately when the reet ignal i applie, an they remain zero until the next riing clock ege. Since the next tate after i, the tate goe to on the next riing clock ege after the reet ignal i remove. The inary umber omparator Lab Reviite 6% 52% If 3 2 >Y 3 Y 2 then 32 =. 3 If Y 3 Y 2 > 3 2 then Y 32 =. Since both cannot happen at once 32, Y 32 =, i impoible. onier O 32. If 32, Y 32 =, never happen, then for 32, Y 32 =, one can chooe the output of O 32 to be anything, ince it never happen. lo, Y can never be,. ever, ow you know about Karnaugh map, eign the ba ( 32 ) output of block O 32 to give whenever 32 >Y 32 Y Ue the implet circuit you can fin. Y Ue abbreviation b = 32, Y a =Y, ba = 32. Y ba =Y 32, etc. b Y a b a a) Fill in the map for ba = b a >Y b Y a ( 32 >Y 32 Y ). b) Loop the map. c) Write the equation. ) raw the circuit. b Y 3 2 Y 2 Y Y O 32 O 32 Y 32 Y ever, b Y b a Y a O 32 ba Y ba 32 Y 32 Y a a on t care are the loop Y a a +Y b b They overrie the, which are quare where the binary number b a i larger than the number Y b Y a (hae quare). ba = a Y b + b Y b a b b a >Y b Y a Y b ap of ba = b a >Y b Y a Y a b Y b a T.G.Ray an J. Knight /4/7 page 6, of

7 arleton Univerity lectronic ngineering L267 ame 8% 72% 2 Synchronou Graph raw the tate graph for a machine to meet all the following requirement: Ha one input, an two output Y an Z. Y= only in the cycle after the machine receive the complete equence =. Z= only in the cycle after the machine receive the complete equence =. The leftmot bit of i receive firt. Overlappe equence are to be etecte incluing a equence overlapping itelf. Y an Z are except for the ingle clock perio after the appropriate equence i complete. The machine tart at the RST tate. raw only the tate graph. Ue the tate provie below an o OT change anything alreay given. Reet YZ= ake ure you have two exit from every tate. YZ= YZ= YZ= YZ= YZ= YZ= YZ= YZ= T.G.Ray an J. Knight /4/7 page 7, of

8 arleton Univerity lectronic ngineering L267 ame 6% 33% 3 Hazar a) Fin any ingle-variable change hazar in the circuit repreente by the equation below.. f = ( + ) + + ( + )( + ) o hazar ( + ) + + ( + )( + ) hooe firt = = for no goo reaon. + + ( + ) + + (+) Fixe thank to atalin an anon) xpan on or firt ince they have no hazar. = = + = = + Static ynamic in in b) oify the equation repreenting the circuit o that it prouce the ame logic function but without hazar. (Hint Σ of Π) Recall Σ of Π ha only tatic- hazar that can foun on a map. c) Write the new hazar free equation 4 Race an ycle + + = = + + o hazar f = ( + ) + + ( + )( + ) = ( + ) + + +) wap = itb f Hazar Free = o hazar hooe on the right becaue ha no hazar hooe in the mile ha no hazar. hooe on the left becaue ha no hazar 7% There are everal circumtance, mot of them unpleaant, that can be aociate with tate-tranition in aynchronou tate-table. Four almot imilar table are hown below. are a an b. 38% a) ircle the table tate. In each tate table, look for the circumtance tate at the top of the table. b) Lightly hae, with pencil, the circle for the table tarting tate for the circumtance. c) Lightly cro-hatch, with pencil, the circle for LL poible final table tate for the circumtance, if any. e ure the marker can tell your cro-hatch from your haing! ) raw arrow howing the path taken through the next-tate part of the table. Only conier thing that might happen uring normal operation of the circuit. ritical Race nother ritical Race i) ii) tate next tate p + q + tate next tate p + q + ab ab ab ab ab ab ab ab ritical race mut alway en in column with at leat two table tate = + = = o hazar = + Static in The wap an itb law remove all but tatic- ( ) hazar. lot on map an get one hazar term to mak hazar. iii) ycle tate next tate p + q + ab ab ab ab The cycle here i alway in the ab= column but it can be entere by a tranition from any of 4 table tate. iv) oncritical Race tate next tate p + q + ab ab ab ab T.G.Ray an J. Knight /4/7 page 8, of

9 arleton Univerity lectronic ngineering L267 ame % 8% 5 o either a) or b) but not both. a) ynchronou ignment. For the following tate table, revie the table an elect a critical-race-free tate aignment. o not a any tate or change the table tate. Keep a tate. o not increae the number of row in the table. -- mean on t care. Y= Y= ext Y= F F -- F -- F Y= table tate raw arrow connecting allowe tranition. Show tranition on a map-like graph F F Sen racy tranition through intermeiate tate. Write your revie tate table below. Fill in the ext column uing letter only.you nee only enter your change. Write your tate aignment in the table/map on the right. Y= Y= ext Y= Y= F F F F F F F Write your tate aignment here Only put letter here Further ote (for partial mark): If you cannot get a race free anwer, then preent your bet anwer above, an inicate here the number of race in your olution. umber of race If you cannot get tate to be, then preent your tate aignment above changing a require. T.G.Ray an J. Knight /4/7 page 9, of

10 arleton Univerity lectronic ngineering L267 ame % 6% b) ultiple Output (ii Lab). (If you o thi quetion, kip the previou one) The ii ontroller count the SL pule (thee enable it flip-flop) an generate two ignal from that count: The UIT ignal tell the erial-to-parallel hift-regiter that all the ata-bit are rea, an it houl top hifting for the next two count. ote the top bit i till to come. The atavali ignal turn off TIV an mut be jut after the tart of the STRT-IT in SI to initiate raiing the TIV ignal. Thi circuit nee to count ample pule: tart, 8 ata bit an top. In the ii ontroller eign below: a) omplete filling in the Karnaugh map for the next-tate logic an the output. b) Loop the map to give the minimum number of gate, an thee houl be minimum ize. c) Write own the equation you get.. Table ext output ata- Vali output uit = = = 2= 3= 4= 2= 3= 4= 5= 5= 6= 7= 8= 6= 7= 8= 9= 9= = 9 UIT UIT atavali 5 ap of uit = SL ap of atavali ap of + ap of + ap of + ap of + uplicate map K =, + = +, + = + +, + = +, + = uit = K, atavali = K + on t forget to o half-map firt. on t forget to hare loop when poible. T.G.Ray an J. Knight /4/7 page, of