CARLETON UNIVERSITY. 1.0 Problems and Most Solutions, Sect B, 2005

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1 RLETON UNIVERSIT eprtment of Eletronis ELE 2607 Swithing iruits erury 28, 05; 0 pm.0 Prolems n Most Solutions, Set, 2005 Jn. 2, #8 n #0; Simplify, Prove Prolem. #8 Simplify Reue to four letters (literls). rw gte level iruit Solution. #8. Using si Priniples Using the Simplifition Theorems (sso) (Simp +z=) = + ( +) + (+=) = + + (sso) = + + (sso) = + + = + + ((+)=) ((+)=) = + ( + ) + = + ( + ) + (ist ) (ist ) = = (ist ) (Simp +z=) = ( + ) + ( + ) = + (+=) = () + ()) (=) = + Prolem. #0 Prove ( + )( + )( + ) = + + Solution. #0 Using for the first step ( + )( + )( + ) (ist ) = ( + + +)( + ) = ( + + +)( + ) = ( +)( + ) = = = + + (=) (Simp +z=) (ist ) (=) (+=) Using 2 for the first step ( + )( + )( + ) = ( + )( + ) ontinue s efore (ist 2). Reuing to 5 letters is esy. Reuing to 4 letters is esy if you look he in these notes n fin the onsensus theorem. rleton University SITHING IRUITS J.Knight, 2/28/05

2 rleton University Prolems n Most Solutions, Set, 2005 Prove #0 y trying ll 8 omintions (+) ( + ) ( + ) = LHS + + = RHS = = = = 0 = = 0 0 = = = = = = 0 = = 0 0 = = 0 Left Hn Sie equls Right Hn Sie for ll omintions, thus epressions re equl Jn. 7, #27 n #32, onsensus OR, Prolem. #27 ) Simplify Use onsensus Solution. #27 (+E) + (+)(+E) + (+) (+E) + (+)(+E) + (+) = (+E) + (+E)(+) + (+) = (+E) + (+) (sso) (onsensus zu +uv +vz = zu + vz) Prolem. #27 ) Simplify (+E) + (+)(+E) + (+) Solution. #27 (+E) + (+)(+E) + (+) = (+E) + (+) + (+)(+E) = (+E) + (+)(+E) Prolem. #32 (sso) (onsensus zu +uv +vz = zu + vz) Prove tht: = = Systems n omputer Engineering erury 28,

3 rleton University Prolems n Most Solutions, Set, 2005 Solution. #32 = + = + = lternte proof using truth tles (efinition of ) (efinition of ) = + ( = ) ( = ) = + (efinition of NOR) (efinition of NOR) = Hene ll three epressions re equl Jn. 9, #38 n #45, Prity n NOR -> S of P# Prolem. #38 (or 39 on the newer notes) ht hppens if two its re flippe uring trnsmission? ill the prity error etetion iruit shown ientify the errors? p z Error Solution. #38 The prity it P is generte y n OR gte. P= 0 if n even numer of inputs re. P = if n o numer of inputs re. If one it flips, the numer of its must hnge y. However if two its flip, the numer of its must hnge y 0 or 2. In either se P stys the sme, n the OR t the reeiver will not inite n error. Prolem. #37 Prolem on Prity esign n even-prity iruit whih gives output if n even numer of inputs re one. Solution. #37 p Jn. 24, #47 n #5 Systems n omputer Engineering erury 28,

4 rleton University Prolems n Most Solutions, Set, 2005 Prolem. #47 or the iruit on the right, fin the NN-NOR iruit when the onnetion on the levels shown re selete. G Solution. #47 Put inverting irles k-to-k Remove irles whih go to gtes whih whih re lrey NN or NOR G G e on t nee this irle either This gte nees 2 input inverters. Put the inversion on the output, n only one inverter is neee G G hnge the inverte-input form to the inverte-output form. G G Note rel gtes nnot hve input irle. Prolem. #5 Use emorgn s generl lw to remove ll ut one of the inverting rs from the inosur iruit on Slie 34. The epression is f = (+) ( ) Systems n omputer Engineering erury 28,

5 rleton University Prolems n Most Solutions, Set, 2005 Solution. #5 f = ((+) ( )) ( ) (Use the generl emorgn s lw first) f ul = (( )+(+)) (+) => f = (( )+(+)) (+) f = f = (( )+(+)) (+) f = (( ) (+)) (+) f = ((+) (+)) ( ) ( + ) = ( + ) = Jn. 26 #52 n #52 Prolem. #52 in the simplest Σ of Π epression for the mp ) on the right. Solution. #52 John s poor solution Tom s solution Prolem. #52 rite the yer in inry oe eiml. In 2005, you shoul hve 6 its. Solution. #52 000, 0000, 0000, 00 Jn. 3 #59 n #60 K-mp Prolem. #59 = ( + + )e + ( + + )e = ( + )e + ( + )e + Plot on the 5 vrile mp on the right. Plot on the split-squre mp on the right. Prolem. #59 (se on the lst prolem) irle the 5 vrile mp n reue to 2 letters. irle the split-squre mp to give 4 terms of 3 letters eh. e=0 e= e=0 e= Split-squre mp Systems n omputer Engineering erury 28,

6 rleton University Prolems n Most Solutions, Set, 2005 Solution. #59 n 59 e=0 e=0 e=0 e= e=0 e= e= e e e e Split-squre mp e e= e e e Split-squre mp = e + + e + e e 2 #6 or #62 Mult output mps. Prolem. #6 in the Σ of Π epressions with miniml gtes for the two-output iruit E,. Soln hs 5 gtes. If it is not pure Σ of Π, it n e one in 5 two-input gtes, or with ftoring, 4 gtes Solution. $ E E E E E irle hlf mps (none) Lote s tht pper on one mp only. Epn irles roun these s to mimum. size. Use your he if the irles n e epne in severl wys irle ny squres not previously irle. E= + = + = ( + ) Systems n omputer Engineering erury 28,

7 rleton University Prolems n Most Solutions, Set, 2005 Prolem. #62 in the minimum numer of gtes for the iruit to implement E, n G. This is hr prolem. ou shoul re over the emple for the 7-segment isply rivers efore ttempting it. Solution.# E= = G= irle the hlf mps Ientify the s tht pper on only one mp. irle them, n epn the irles s muh s possile. Look he if there re two wys to irle Ientify s tht n no longer e prtilly shre euse their position is lrey irle on the other mps. irle them n epn the irle L= letters, 27 gte inputs, gtes E= +L =+L G= Prolem. # ) in the Σ of Π equtions using the minimum totl numer of gtes 0 if no gtes re shre etween outputs. 0 ) in the wy to reue this to 9 gtes if gtes re shre etween E= mps. One solution hs: two 5-input gtes, one 4-input gte, five 3-input gtes, n one 2-input gte = G= Systems n omputer Engineering erury 28,

8 rleton University Prolems n Most Solutions, Set, 2005 Solution.# irle hlf mps irle isolte s irle newly isolte s L= M= N= P= E= +Q+N =+L+M+N+P 7 gtes 27 letters G=L+M+N+P+Q R = E= +Q+N G=R+Q =+R 8 gtes, 2 letters e 7 # 3 two ff, #67 (or #69) Prolem. #3 Solution. #3 Q 2 Q 2 omplete the wveforms for 2 n Q 2. ssume the flip-flop outputs re initilly zero s shown. Q just fter lok ege, tkes on the vlue of just efore the ege. Q 2 just fter lok ege, tkes on the vlue of 2 =Q just efore the ege. Step through eh lok ege. LK Q = 2 Q 2 LK Q = 2 Q 2 Systems n omputer Engineering erury 28,

9 rleton University Prolems n Most Solutions, Set, 2005 Prolem. #67 Multiply out to get four terms of three letters eh. The nswer shoul e very symmetri on n \ Krnugh mp. ( + )( + + )( + + )( + + )( + + ) Solution. #67 ( + )( + + )( + + )( + + )( + + ) = ( + )( + + )( + + )( + + )( + + ) = ( + )( + )( + + )( + + ) = ()( + + )( + + ) = ()[ ( + ) + ( + )] = [ + ) + + ] = Simplify first (sso) ( + )( + ) = ( + )( +) = swp ( + )( + y) = y + () () lterntely, if you like simplifition rule + = etter thn ( + )( + ) =, use the ul. = ( + )( + + )( + + )( + + )( + + ) tke ul ul = ( )+( )+( )+( )+( ) (sso) = ( )+( )+( )+( )+( ) + = = ( )+ +( )+( ) = + ( ) + ( ) + = tke ul k = ( + + )( + + ) Some people might use 2 s elow. This is true ut it oesn t simplify s well here. ( + )( + + )( + + )( + + )( + + ) (2) ( + )( + + )( + + )( + + ) e 9 # or #3 or #4 Prolem. # ht logi funtion is performe y the iruit on the right? Solution. # K 0V 0 0 5V 0 5V 0 0 5V This is NN +5V K=? Pro. Systems n omputer Engineering erury 28,

10 rleton University Prolems n Most Solutions, Set, 2005 Prolem. #2 To implement P=( + ): () Into wht form oes one hve to hnge the epression for P? Solution.2 Sum of Prouts. + Prolem. #2 () Progrm P=( + ) on the rry on the right? Solution. 2 P Note tht one nnot just leve ll the fuses unlown s this puts into the output. Lter on we will isonnet the vertil trnsistors P Prolem. #4 Using the mps on Slie 9 (repete elow), efine new epression U = + M + N = + + ) ou will nee U, n some other help, to reefine, n g to use only three prout terms inste of the four terms shown on the net pge. This llows them to fit in the PL shown on pge 24 whih hs only three terms for eh output. J = N = K = P = K K L = R = N J L J M = = J + + K + N = J + L + = + + K M N P K = N + M + P + K K L M P R R P K e = K + P f = L + + g = + M + P + R Systems n omputer Engineering erury 28,

11 rleton University Prolems n Most Solutions, Set, 2005 J = N = K = P = K K L = R = N J L J M = U= + N + M = J + K + + N + M H = N + M = J + L + = + + = J + K + U K M N P K = N + M + P + K = e + N + M ) inish progrmming the PL on pge 24 K L M P R N P K e = K + P f = L + + g = + N + M + P + R g = U + P + R U e M= N= L= L= J= U K= J= U f U P= R= g P= K= e e M= N= Systems n omputer Engineering erury 28, 2005

12 rleton University Prolems n Most Solutions, Set, 2005 e 4, #7 Mu logi, #9 flip-flops Prolem. #7 in MU implementtion of = + e + e + + e Orer the MUs -e--- Solution. #7 = + e + e + + e + e + e + + e e e e (use S) e + e (use S & S2) e e Prolem. #9 esign the stte grph for ounter tht ounts own 0->7->6->5->4->3->2->->7... in inry. Note tht the zero sttes omes up on reset, ut nnot e entere gin. Solution. # Stte Grph Systems n omputer Engineering erury 28,

13 rleton University Solve Prolems 2.0 Solve Prolems Prolem. lgeri Simplifition Hint: onsensus shoul e use on one of the following prolems. The figure shows wy to rememer onsensus. Use oolen lger to simplify: Solutions: ).... ( ) ns: very smll ) E +... E +.. E ns: + ) M. N. L+ Q. P. N + P. R. M + Q. L. M. P + M. R ns: + + ) ns: three lettes ).... ( ) = = = ) E +... E +.. E = E +... E +.. E =. +.. E ) M. N. L+ Q. P. N + P. R. M + Q. L. M. P + M. R = M. N. L+ Q. P. N + Q. L. M. P + P. R. M + M. R = N. (Q. P) + (Q. P). (L. M) + (L. M). N + M. R = N. (Q. P) + (L. M). N + M. R ) = +. + = + + Use Repetely use. = Use. =0 Repetely use +. = Look for +. = Look for onsensus Use onsensus Use +. = + Use +. = + Prolem. e) Use one OR n one N to implement: + E + Solution.: + E + = ( + E + ) Use E Prolem. emorgn s Theorem ) kerly 4.5. kerly s ook mkes the following sttement in prolem. emorgn s theorem sttes the omplement of +. is. +. However oth funtions re for,, =,,0. i) How n funtion n its omplement oth e for the sme input omintion? ii)is kerly leing you stry? iii) Eplin. Systems n omputer Engineering erury 28,

14 rleton University Solve Prolems Solution. i) funtion n is omplement must lwys hve opposite vlues. The tle shows the funtions re not omplements ii) es, kerly is leing you stry. He forgot to put the rkets roun the N epressions efore using emorgn s Theorem. iii) The right nswer is erive y pply the generl emorgn s lw: Let = + = + () (put rkets roun the Ne letters) ul = ( + ) (tke the ul) = ( + ) omplement the single letters in the ul to get Prolem () Implement the following using (i) only NN gtes (ii) only NOR gtes. (Input inverters my e use) = Solution. This n e one lgerilly or grphilly lgeri metho, prt (i); hnge to NNs: = Use generl emorgn s Lw (,,..+, ) = (,,,...,+) = (++)(+)(+)(++) = ( ) ( ) ( ) ( ) Trnsform NORs to NNs using + = Interpret the result: Re ots n rs re top level NNs lue ots n rs re ottom level NNs lk rs re input inverters Grphil Metho, prt (i); hnge to NNs: = Systems n omputer Engineering erury 28,

15 rleton University Solve Prolems lgeri Metho, prt (ii); hnge to NORs: irst tor The NN gte iruit uses Σ of Π. = ul = (++)(+)(+)(++) ul = [++)(+)][(+)(++)] ul = [+(+)][+(+)] ul = [++][++)] ul = (++) + + (++) + (++) = ++ + The NOR gte iruit uses Π of Σ, i.e. one must ftor To ftor, we tke the ul, multiply out, n tke the ul gin. efore multiplying, fin ommon terms, n use 2. Then multiply out (using ) n ontinue multiplying out using e oul hve looke for omplemente letters n use Swp Note is not equl Tke the ul gin, to get the originl to ul = (+)(++)(++)(+++) Now pply the Generl emorgn Lw = Generl emorgn s Lw (,,..+, ) = (,,,...,+) = (+)+(++)+(++)+(+++) Trnsform NNs to NORs using = + Interpret the result: Re + n rs re top level NORs = (+)+(++)+(++)+(+++) lue + n rs re ottom level NORs lk rs re input inverters Grphil Metho, prt (ii); hnge to NORs: Use the K-mp of to fin on K-mp. = = Use Generl emorgn on to get s Π of Σ = (+)(++)(++)(+++) Then use Grphil emorgn to hnge to NORs Mp of Mp of Prolem. in n epression for G, (G inverse) whih hs inversion rs only over single symols. Tht is terms like + re ll right, ut terms like + or re not llowe. G = ( + E) + ( + E) Systems n omputer Engineering erury 28,

16 rleton University Solve Prolems Solution. pply the Generl emorgn s Lw Put rkets roun ne epressions G =[( + [E])] + [([] + [E])] Tke ul G ul = [++( [+E])] [++([+] [+E])] Invert ll single letters. The result is G using the G =[++( [+E])] [++([+] [+E])] generl emorgn s Lw (,,..+, ) = (,,,...,+) One oul lso o this y pplying + =, n = +, mny mny times. Tht woul e muh longer. Prolem. Sum-of-Prouts n Prout-of-Sums () in the minimum Σ of Π epression for. = Solution. ) This woul e very hr to o lgerilly, use mp = Now irle the sme squres so s to minimize the iruit. = ns: Mp of Mp of () Using the mp for, write out the mp for. Then fin the minimum Σ of Π epression for Solution. ) = rom the mp of, fin the mp for Then irle the squres to fin the minimum for = Mp of ns: = Mp of Solution. () rom Σ of Π epression for, fin the minimum Π of Σ epression for. ns: ( + + )( + + )( + + )( + + ) ) Mp of Then irle the mp of to fin the minimum for = Trnsform k to using Generl emorgn = (++)(++)(++)(++) Generl emorgn s Lw (,,..+, ) = (,,,...,+) Systems n omputer Engineering erury 28,

17 rleton University Solve Prolems 6.3 rz, ill, liff n onn ought r on shres. rz owns 0%, ill owns 20%, liff owns 30% n onn 40%. To voi rguing out where they shoul go, they instlle 4 swithes in the r. They ll rie together, n they eh hve their own set. hen estintion is suggeste they ll vote on it y pushing, or not pushing their swith. 66% mjority of the shres lights light n then they go there. Less thn 66% n the light stys out. Then they wit for new suggestion. () erive truth tle for the output () to the light. () in the miniml Σ of Π epression for. n you mke it simpler using ftroring? () rw the iruit using gtes n inluing the swithes, showing where the swith wires onnet. o not leve gte inputs floting (isonnete) for more thn n instnt. () If you put gtes in your iruit in (), o you relly nee them? Solution V 0% 20% 30% 40% Mp of '' in % Mp of 66% 40% = = ( + ) % % % ll others 50 () iruit without ny gtes. V V is the ommon nme for power supply onnetion in eletroni iruits. The other en of the power supply is the groun symol.. Push-utton Swith 6.4 in the minimum Σ of Π for the mp on the right. is on t re. Solution: \ = The following sttement will hnge two squres to on t res. Given the funtion = ut the inputs re not ompletely inepenent. e re given tht never equls when ==. Minimize with this onitions on the inputs.- Systems n omputer Engineering erury 28,

18 rleton University Solve Prolems Solution here == Prolem. Sine ertin inputs never hppen, the outputs for those inputs never hppen. Thus we n mke those outputs to simplify the logi. here The minimum s Σ of Π epression for is given elow. minimize the logi to three 2-input gtes in wy to implement this with only three 2-input gtes. here n == = + + +, onsiering impossile inputs = + + Solution = = + pply the generl emorgn Lw = ( + )( + ) Systems n omputer Engineering erury 28,