II. SOLUTIONS TO HOMEWORK PROBLEMS Unit 1 Problem Solutions
|
|
- Kathlyn Garrett
- 7 years ago
- Views:
Transcription
1 II. SOLUTIONS TO HOMEWORK PROLEMS Unit Prolem Solutions (). () r r5= 6 (4). r = = r 6 r7 (2).72 6 ().52 6 (8).32. () r4 6 6 r6 (4).24 r 6 (3).84 6 (3).44 6 (7) = 64.E3 6 = E = = () r r2 (8). r = = () E.6 6 = E () = = /6 = () 2 E = = /8 = = /6 = = = /8 = = () = = r r5 () r2 6 r3 () () (2) () = = r r= 6 ().76 r5 6 (2) = () 5. 6 =. 2 = () 5. 6 = () E. 6 = E = =
2 .5 () () (Su) + (Multiply).5 (, ) See L p. 625 for solution..6,.7,.8 See L p. 625 for solution.. (). () r9 6 5 r (6) = = r3 6 r2 ().92 6 (4).72. () r5 = 6 6 (5).28 6 (4) = = () r2 6 6 r6 (4). 3.2 = 2.E 8 =. 2 2 E = 66.E 6 = E. (). 2 = = = /8 = (). 2 = = = /8 = = /6 = = = = /6 = () = /8 + 4/64 = r r (2) r r ().2 3 r 3 r ().63 3 () = () r r (2) r 4 4 r2 (3).84 r 4 (3) =
3 .2 ().4 ().4 () 52.4 = / + 4/2 = r r5 (8).46 9 r6 9 r (4) = = /64 = r r4 (7) r6 8 r5 (6) /64 = (or ) =. 2 (or. 2) 9 3/ r r5 (7) r5 8 r3 (6) = se 3 se () r5 8 8 r3 (5).6 r 8 (4) = =. 2.4 () r5 8 8 r5 (2).4 r 8 (3).2 9 3/32 = = = =. 2.5 () () (Sutrt) (Multiply).5() () (Su) (Mult) 7
4 .5() () (Su) (Mult).6 () () ().7 () Quotient ) Reminer.7() Quotient ) Reminer.7().8 ().8() Quotient ) Reminer Quotient ) Reminer Quotient ) Reminer.8().9 Quotient ) Reminer =
5 is possile, ut is not, euse there is no wy to represent 3 or 9. lternte Solutions: is not possile, euse there is no wy to represent 3 or is possile: () () () () = 83 lternte Solutions: () () () ().23 lternte.24 lternte Solutions: Solutions: = 94 () () () () () () 49 = lt.: = " " " r4 6 r3 (3).52 6 (8) = E.38 6 = E r7 6 r (2).96 6 (5) = 7. 6 = 7. 9
6 .26 () In 2 s omplement In s omplement ( ) + ( ) ( ) + ( ) () ( 2) ( 2).26 () In 2 s omplement In s omplement ( ) + ( 6) ( ) + ( 6) () ( 6) ( 6).26 () ( 8) + ( ) ( 8) + ( ) () ( 9) ( 9).26 () (2) (2).26 (e) ( ) + ( 4) ( ) + ( 4) () ( 5) ( 5).27 () In 2 s omplement In s omplement () In 2 s omplement In s omplement + + ().27 () + + () overflow overflow.27 () + + ().27 (e) In 2 s omplement In s omplement + + ().28 () In 2 s omplement In s omplement + + ().28 () () + + () overflow overflow.28 () + +
7 Unit 2 Prolem Solutions 2. See L p. 626 for solution. 2.2 () In oth ses, if =, the trnsmission is, n if =, the trnsmission is. 2.2 () In oth ses, if =, the trnsmission is YZ, n if =, the trnsmission is. Y Y Z Y Z 2.3 or the nswer to 2.3, refer to L p () = [( ) + ( )] + E + = + E () Y = (' + ( + )) + = (' + ) + = ( + ) + = + + = () ( + ) ( + ) (' + ) (' + E) = ( + ) (' + ) (' + E) y Th. 8 = (' + ) (' + E) y Th. 8 = ' + E y Th () (' + + ') (' + ' + ) (' + ') = (' + ' + ) (' + ') {y Th. 8 with = ' + '} = '' + '' + ' + '' + '' + ' = '' + '' + '' + '' 2.6 () + '' = ( + ') ( + ') 2.6 () W + WY' + ZY = (W + WY' + ZY) = ( + ') ( + ') ( + ') ( + ') = (W + ZY) {y Th. } = (W +Z) (W + Y) 2.6 () ' + E + E' = ' + E( +') = ' + E( +) = (' + E) (' + + ) = (' + E) ( + E) ( + E) (' + + ) ( + + ) ( + + ) 2.6 () YZ + W'Z + Q'Z = Z(Y + W' + Q') = Z[W' + (Y + Q')] = Z(W' + ) (W' + Y + Q') y Th (e) ' + '' + ' = ' ( + ') + ' = ' ( + ') + ' y Th. = (' + ') ( + ' + ') = (' + ') (' + ) ( + ' + ') y Th. = (' + ') ( + ') 2.6 (f) + + E = ( + + ) ( + + E) = ( + + ) ( + + ) ( + + E) ( + + E) 2.7 () ( ) ( E) ( ) = E pply seon istriutive lw (Th. 8) twie 2.7 () WYZ + VYZ + UYZ = YZ (W + V + U) y first istriutive lw (Th. 8) U V W E Y Z 2.8 () [()' + ']' = (')' = ( + ') 2.8 () [ + (' + )]' = '((' + ))' = + ' = '(' + (' + )') = '(' + ') = '' + '' 2.8 () (( + ') )' ( + ) ( + )' = (' + ') ( + )'' = (' + ')'' = ''
8 2.9 () = [( + )' + ( + ( + )')'] ( + ( + )')' 2.9 () G = {[(R + S + T)' PT(R + S)']' T}' = ( + ( + )')' y Th. with =(+(+)')' = (R + S + T)' PT(R + S)' + T' = '( + ) = ' = T' + (R'S'T') P(R'S')T = T' + PR'S'T'T = T' 2. () Y 2. () Y Y Y' Y' 2. () ' Y Z Y Z 2. () ' 2. (e) 2. (f) Y Z Y Z Y Y 2. () '' + ('')' = y Th () ( + ') + + ' = + ' y Th. 2. () + + '( + )' = + + ' y Th. 2. () (' + ')(' + E) = ' + 'E y Th (e) [' + ()' +E'] = ' + E' y Th (f) (' + )('E + )' + ('E + ) = 'E + + ' + y Th. 2.2 () ( + Y'Z)( + Y'Z)' = y Th () (W + ' + YZ)(W' + ' + YZ) = '+ YZ y Th () (V'W + )' ( + Y + Z + V'W) = (V'W + )' (Y + Z) y Th. 2.2 () (V' + W')(V' + W' + Y'Z) = V' + W' y Th. 2.2 (e) (W' + )YZ' + (W' + )'YZ' = YZ' y Th (f) (V' + U + W)(W + Y + UZ') + (W + UZ' + Y) = W + UZ' + Y y Th. 2.3 () = ' + + ( + ) = + + = 2.3 () 2 = '' + ' = ' + ' = ' + ' 2.3 () 3 = [( + )'][( + ) + ] = ( + )' ( + ) + ( + )' = ( + )' y Th. 5 & Th () Z = [( + )]' + ( + ) = [( + )]' + y Th. with Y = [( + ) ]' = '' + ' + ' 2.4 () ( + E + ) 2.4 () W + Y + Z + VU 2
9 2.5 () 2.5 () 2.5 () 2.5 () 2.5 (e) 2.5 (f) H'I' + JK = (H'I' + J)(H'I' + K) = (H' + J)(I' + J)(H' + K)(I' + K) + '' + ' = ( + '' + ') = [(' + )( + ') + '] = (' + + ')( + ' + ') ' + + E' = (' + + E') = [' + ( + E')] = (' + )(' + + E') ' + '' + E' = ' + '' + E' = ' ( + ') + E' = (' + E') ( + ' + E') = (' + E) (' + ') ( + E) ( + ' ) ( + ' + E) ( + ' + ') W'Y + W'' + W'Y' = '(WY + W') + W'Y' = '(W' + Y) + W'Y' = (' + W') (' + Y') (W' + Y + W') (W' + Y + Y') = (' + W') (' + Y') (W' + Y) ' + (' + E) = ' + ( + E)(' + E) = (' + + E)(' + ' + E) = ( + + E)(' + + E)( + ' + E)(' + ' + E) 2.6 () W + 'YZ = (W + ')(W + Y) (W + Z) 2.6 () VW + Y' + Z = (V++Z)(V+Y'+Z)(W++Z)(W+Y'+Z) 2.6 () '' + '' + 'E' = '(' + ' + E') 2.6 () = '[E' + (' + ')] = '(E' + )(E' + ' + ') + E' + ' = ( + E' + ') = [E' + ( + ')] = (E' + )(E' + + ') = ( + )( + E')( + ' + )( + ' + E') 2.7 () [(Y)' + (' + Y')'Z] = ' + Y' + (' + Y')'Z 2.7 () ( + (Y(Z + W)')')' = 'Y(Z + W)' = 'YZ'W' = ' + Y' + Z y Th. with Y = (' + Y') 2.7 () [(' + ')' + ('')' + '']' = (' + ')''( + ) = '' 2.7 () ( + ) ' + ( + )' = ' + ( + )' {y Th. with Y = ( + )'} = ' + '' 2.8 () = [(' + )']' + = [' + + '] + = () G = [()'( + )]' = ( + '') = 2.8 () H = [W''(Y' + Z')]' = W + + YZ 2.9 = (V + + W) (V + + Y) (V + Z) = (V + + WY)(V + Z) = V + Z ( + WY) y Th. 8 with = V W 2.2 () = + ' + ' + ' = + ' + ' (y Th. 9) = ( + ') + ' = (+ ) + ' (y Th. ) = + + ' = + ( + ') = + ( + ) = () Y + Z V eginning with the nswer to (): = ( + ) lternte solutions: = + ( + ) + 3 = + ( + )
10 2.2 () W Y Z W'Y WZ W'Y+WZ W'+Z W+Y (W'+Z)(W+Y) 2.2 () + +' (+)(+') ' +' 2.2 () Y Z +Y '+Z (+Y)('+Z) Z 'Y Z+'Y 2-2 () Y Z Y YZ 'Z Y+YZ+'Z Y+'Z 4
11 2.2 (e) Y Z +Y Y+Z '+Z (+Y)(Y+Z)('+Z) (+Y)('+Z) 2.22 ( + ) =, = [(+Y')Y] = Y' + Y, (Y) = + Y Unit 3 Prolem Solutions 3.6 () (W + ' + Z') (W' + Y') (W' + + Z') (W + ') (W + Y + Z) = (W + ') (W' + Y') (W' + + Z') (W + Y + Z) = (W + ') [W' + Y' ( + Z')] (W + Y + Z ) = [W + ' (Y + Z)] [W' + Y'( + Z')] = WY' ( + Z') + W'' (Y + Z) {Using ( + Y) (' + Z) = 'Y +Z with =W} = WY' + WY'Z' + W''Y + W''Z 3.6 () ( ) (' + ' + + ') (' + ) ( + ) ( + + ) = ( + + ) (' + ) ( + ) = ( + + ) (' + ) {Using ( + Y) (' + Z) = 'Y + Z with = } = ' + ' + ' = ' () + '' + '' + = + '(' + ') = (' + ) [ + (' + ')] {Using ( + Y) (' + Z) = 'Y + Z with =} = (' + ) [ + (' + ') (' + )] = (' + ) ( + ' + ') 3.7 () ''' + ' + ' + ' 3.8 = ' ('' + ) + (' + ') = ' [(' + ) ( + ')] + [(' + ') (' + )] {Using Y + 'Z = (' + Y) ( + Z) twie insie the rkets} = [ + (' + ) ( + ')] [' + (' + ') (' + )] {Using Y + 'Z = (' + Y) ( + Z) with = } = ( + ' + ) ( + + ') (' + ' + ') ( ' + ' + ) {Using the istriutive Lw} = [( ) + ] = ( + '' + ) = ('' + ) = (' + ) = ()' (' + ) + (' + )' = (' + ') (' + ) + (') = ' + ' + ' {Using ( + Y) ( + Z) = + YZ} = ' + ' + ' {Using + 'Y = + Y} 3.9 = ( Β) (Α ) is not vli istriutive lw. PROO: Let =, =, =. LHS: = = =. RHS: ( ) ( ) = ( ) ( ) = =. 5
12 3. () ( + W) (Y Z) + W' = ( + W) (YZ' + Y'Z) + W' = YZ' + Y'Z + WYZ' + WY'Z + W' 3. () ( ) + + = ' + ()' + + = ' + (' + ') + + = ' + ' + ' + + Using onsensus Theorem WYZ' + WY'Z + W' = ' + ' + ' ( onsensus term, eliminte ) = ' + ' + ' + (Remove onsensus term ) 3. () (' + ' + ') (' + + ') ( + + ) ( + + ) = (' + ' + ') ( + ' + ) (' + + ') ( + + ) ( + + ) onsensus term = (' + + ') ( + + ) = (' + ' + ') ( + ' + ) ( + + ) Removing onsensus terms 3. ( + ' + + E') ( + ' +' + E) (' + ' + ' + E') = [ + ' + ( + E') (' + E)] (' + ' + ' + E') = ( + ' + 'E' + E) (' + ' + ' + E') = ' + ( + 'E' + E) (' + ' + E') ' { onsensus term} = ' + ' + ' + E' + ''E' + 'E' + 'E' + 'E = ' + ' + ' + E' + ' +'E + 'E' = ' + ' + E' + ' + 'E' 3.2 ''E + ''' + E + = ''' + + 'E Proof: LHS: ''E + 'E + ''' + E + onsensus term to left-hn sie n use it to eliminte two onsensus terms = 'E + ''' + This yiels the right-hn sie. LHS = RHS 3.3 () (' + ' + ) (' + ) ( + ' + ') (' + + ) ( + ) = (' + + '') ( + ') = ( + '') + ('') {Using Y + 'Z = ( + Z)(' + Y) with = } = + '' + '' 3.3 () (' + ' + ') ( + + ') ( + ) (' + ) (' + + ) = [' + (' + ')] [ + ( + ')] = (' + ') + ' ( + ') = ' + ' + ' + '' 3.3 () (' + ' + ) ( + ') (' + + ') ( + ) ( + + ') = [' + (' + ) ( + ')] ( + ') = (' + + '') ( + ') {y Th. 4 with = } = ( + '') + '' {y Th. 4 with = } = + '' + '' 3.3 () 3.3 (e) ( + + ) (' + ' + ') (' + ' + ') ( + + ) = ( + + ) (' + ' + '') = ( ' + '') + '( + ) {y Th. 4 with = } = ' + '' + ' + ' ( + + ) ( + + ) (' + ' + ') (' + ' + ') = ( + + ) (' + ' + '') = (' + '') + '( + ) = ' + '' + ' + ' lt. soln's: ' + ' + '' + ' (or) ' + ' + ' + '' (or) ' + ' + '' + ' 3.4 () ' + '' + ' = '' + ' = ('' + ') = (' + '') {y Th. with Y = '} = ' + '' 6
13 3.4 () '' + ' + '' = '' + '( + ') = '' + '( + ) = '' + ' + ' 3.4 () 3.4 () 3.5 () 3.5 () 3.5 () ( + ') (' + ' + ) (' + + ') = ' + (' + ) ( + ') = ' + ( + ') = ' + (' + + ' + ) (' + ' + + E) (' +' + + E') = [' + ' + ( + ) ( + E) ( + E')] {y Th. 8 twie with = ' + '} = [' + ' + ( + )] = [' + ' + ] = '' + ' + + '' = ' ( + ') + ('' + ) = ' [( + ') ( + )] + [(' + ) ( + ')] {y Th. 4 twie with = n = } = [ + ( + ') ( + ) ] [' + (' + ) ( + ')] {y Th. 4 with = } = ( + + ') ( + + ) (' + ' + ) (' + + ') {y istriutive Lw} + '' + ''' + ' = ' (' + '') + ( + ') = ( + ' + '') (' + + ') {y Th. 4 with = } = ( + ' + ') ( + ' + ') (' + + ) (' + + ') + '' + '' + '' = ' [' + '] + [ + ''] = ' [( + ) (' + ')] + [( + ') ( + ')] = [ + ( + ) (' + ')][' + ( + ') ( + ')] = ( + + ) ( + ' + ') (' + + ') (' + + ') 3.5 () '' + '' + '' + = ('' + ) + ' (' + ') = ( + ') ( + ') + ' (' + ') ( + ) = [' + ( + ') ( + ')] [ + (' + ') ( + )] = (' + + ') (' + + ') ( + ' + ') ( + + ) 3.5 (e) WY + W'Y + WYZ + YZ' = WY ( + ' + Z) + YZ' = WY + YZ' = Y (W + Z') = Y (W + ) (W + Z') 3.6 () ( ) + '' = ()' + ' + '' = (' + ') + ' + '' = (' + ') + ' ( + ') = (' + ' + ') ( + ' + ) = (' + ' + ') ( + ' + ) ( + ' + ) 3.6 () ' ( ') + + ' = ' ['' + ] + + ' = ''' + ' + + ' = ''' + ( + ' + ') = ''' + ( + ' + ') = ''' + = + '' = (' + ) (' + ) 3.7 () 3.7 () ( Y) Z = (Y Z) Proof: LHS: Let Y =. Z = Z' + 'Z = ( Y) Z' + ( Y)' Z = ( Y ) Z' + ( Y) Z {y (3-8) on L p. 6) = ('Y + Y') Z' + (Y + 'Y') Z = 'YZ' + Y'Z' + YZ + 'Y'Z RHS: Let Y Z =. = ' + ' = (Y Z)' + ' (Y Z) = (Y Z) + ' (Y Z) = [YZ + Y'Z'] + ' [YZ' + Y'Z] = YZ + Y'Z' + 'YZ' + 'Y'Z LHS = RHS ( Y) Z = (Y Z) Proof: LHS: Let Y =. ( Z) = Z + 'Z' = ( Y) Z + ( Y)' Z' = ( Y ) Z + ( Y) Z' = (Y + 'Y') Z + (Y' + 'Y) Z' = YZ + 'Y'Z + Y'Z' + 'YZ' RHS: Let Y Z =. ( ) = + '' = (Y Z) + ' (Y Z)' = (Y Z) + ' (Y Z) = [YZ + Y'Z'] + ' [Y'Z + YZ'] = YZ + Y'Z' + 'Y'Z + 'YZ' LHS = RHS 3.8 () '' + ' + ' + ' + '' = '' + ' + ' + '' = ' + ' + '' 3.8 () W'Y' + WYZ + Y'Z + W'Y + WZ = W'Y' + WYZ + Y'Z + W'Y + WZ = W'Y' + WYZ + W'Y + WZ = W'Y' + W'Y + WZ 7
14 3.8 () ( + + ) ( + + ) (' + + ) (' + ' + ') = ( + + ) (' + + ) (' + ' + ') 3.8 () W'Y + WZ + WY'Z + W'Z' = W'Y + WZ + WY'Z +W'Z' + YZ = WY'Z + W'Z' + YZ YZ ( onsensus term) 3.8 (e) '' + '' + ' + ' + ' = '' + ' + ' 3.8 (f) ( + + ) ( + ' + ) ( + + ) (' + ' + ') = ( + + ) ( + ' + ) (' + ' + ') 3.9 Z = + E + + ' + 'E' = ( + + ' + 'E') + E = ( + E) (E ' + 'E') {y Th. 8 with = E} = ( + ) ( + E) ( + + ' + E + 'E') = ( + ) ( + E) (' + E + ' + + ) {Sine E + 'E' = E + '} = ( + ) ( + E) (' + E + ' + + ) {Sine ' + = ' + } = ( + ) ( + E) (' + E + ' +) {Sine + = } = ( + E) (' + E + ' + ) = ' + E + ' + + E + E' + E {eliminte onsensus term E; use + Y = where = E} 3.2 = ' + ' + + E = ' + + '' + E + = ( + ) (' + ) + ('' + E + ) = [( + ) (' + ) + ] [( + ) (' + ) + '' + E + ] = ( + ) (' + + ) ( + + '' + E + ) (' + + '' + E + ) + + ' = ( + ) (' + + ) ( + ) ( + + '' + E + ) (' + + ' + E + ) = ( + ) ( + ) (' + + ' + E + ) = ( + ) ( + ) (' + + ' + + E) = ( + ) ( + ) (' + + ' + ) = ( + ) (' + + ' + ) = (' + + ' ' + = ' + ' + + use onsensus, + Y = where = 3.2 'Y'Z' + YZ = ( + Y'Z') (' + YZ) = ( + Y') ( + Z') (' + Y) (' + Z) (Y + Z') = ( + Y') ( + Z') (' + Y) (' + Z) (Y + Z') = ( + Y') ( + Z') (' + Z) (Y + Z') = ( + Y') (' + Z) (Y + Z') lt.: (' + Y) (Y' + Z) ( + Z') y ing (Y' + Z) s onsensus in 3r step 3.22 () y + 'yz' + yz = y ( + 'z') + yz = y + yz' + yz = y + y = y lternte Solution: y + 'yz' + yz = y ( + 'z' + z) = y ( + z' + z) = y ( + ) = y 3.22 () 3.22 () y' + z + (' + y) z' = 'y + (' + y) {y Th. with Y = z} = y' + ' + y = + ' + y = + y = lt.: y' + z + (' + y) z' = (y' + z) + (y' + z)' = () (y' + z) ( + y') z = (y' + z + y'z) z = y'z + z + y'z = z + y'z lternte Solution: (y' + z) (+y') z = z ( + y') = z + zy' ' (' + ) + '' ( + ') +(' + ) ( + ') = '' + ' + '' + ''' + '' + = '' + '' + '' + ' Other Solutions: '' + + ''' + '' '' + + ''' + ' '' + + '' + '
15 3.22 (e) w'' + 'y' + yz + w'z' + 'z reunnt term = w'' + 'y' + yz + w'z' + 'z = 'y' + yz + w'z' + 'z Remove reunnt term = 'y' + yz + w'z' 3.22 (g) [(' + ' + ') ( + + ')]' + ''' + '' = ( + ') + '' (' + ) +''' + '' = + '+ ''' + '' + ''' + '' ' '' = + ''' + '' + ' = + '' + ' 3.22 (f) ' + ''+ 'E+ E'G+'E+''E = ' + 'E + E'G + 'E (onsensus) = ' + 'E + E'G 3.23 () ''' + ' + + '' + ' + '' 3.23 () = '' + ' + + ' onsensus = '' + ' WY' + (W'Y' ) + (Y WZ) 3.25 () = WY' + W'Y' + (W'Y')' ' + Y (WZ)' + Y'WZ = WY' + W'Y' + (W + Y) ' + Y (W' + Z') + Y'WZ = Y' + W' + 'Y + W'Y + YZ' + WY'Z + WY' = Y' + W' + 'Y + W'Y + YZ' + WY'Z + WY' = Y' + W' + W'Y + YZ' + WY' = + W' + W'Y + YZ' lternte Solutions: = W'Y + W' + WZ' + Y' = YZ' + W' + Y' + WY' = W' + 'Y + Z' + WY' = W' + Y' + WZ' + WY' ''' + + ' + '' + ' + '' = '' + + ' + ' = '' + + ' VLI: ' + ' + ' = ' ( + ') + ( + ') ' + ( + ') ' = ' + '' + ' + '' + ' + '' = ' + ' + ' lternte Solution: ' + ' + ' ll onsensus terms: ', ', ' We get = ' + ' + ' + ' + ' + ' = ' + ' + ' 3.25 () NOT VLI. ounteremple: =, =, =. LHS =, RHS =. This eqution is not lwys vli. In ft, the two sies of the eqution re omplements: [( + ) ( + ) ( + )]' = [( + ) ( + )]' = [ + + ]' = (' + ') (' + ') (' + ') 3.25 () VLI. Strting with the right sie, onsensus terms RHS = + '' + ' + ' + + ' = + '' + ' + ' + + ' = + '' + ' + ' + = LHS 3.25 () VLI: LHS = y' + 'z + yz' onsensus terms: y'z, z', 'y = y' + 'z + yz' + y'z + z' + 'y = y'z + z' + 'y = RHS 3.25 (e) NOT VLI. ounteremple: =, y =, z =, then LHS =, RHS =. This eqution is not lwys vli. In ft, the two sies of the equtions re omplements. LHS = ( + y) (y + z) ( + z) = [( + y)' + (y + z)' + ( + z)']' = ('y' + y'z' + 'z')' = [' (y' + z') + y'z']' =[(' + y'z') (y' + z' + y'z')]' = [(' + y') (' + z') (y' + z')]' (' + y') (y' + z') (' + z') 9
16 3.25 (f) VLI: LHS = ' + ' + '' () VLI: onsensus terms: ', = ' + ' + '' + + ' + ' + ' '' = RHS LHS = (' + Y') ( Z) + ( + Y) ( Z) = (' + Y') ('Z' + Z) + ( + Y) ('Z + Z') = 'Z' + 'YZ' + Y'Z + 'YZ + Z' + YZ' = 'Z' + (Y' + 'Y)Z + Z' = Z' + Z( Y) = Z' + ( Y) = RHS 3.26 () LHS = (W' + + Y') (W + ' + Y) (W + Y' + Z) = (W' + + Y') (W + (' + Y) (Y' + Z)) = (W' + + Y') (W + ('Y' + YZ)) = (W' ('Y' + YZ) + W ( + Y ')) = W''Y' + W'YZ + W + WY' onsensus terms: 'Y' YZ = W''Y' + W'YZ + W + WY' + YZ + 'Y' = W''Y' + W''Z + W'YZ + YZ + W + WY' + 'Y' = W''Z + W'YZ + YZ + W + 'Y' = W'YZ + YZ + W + 'Y' 3.26 () LHS = + ''' + '' + = ( + ) + '' ( + ') = ( + ' ( + ')) (' + ( + )) = ( + ') ( + + ') (' + ) (' + + ) = ( + ') ( + + ') (' + ) (' + + ) ( + ' + ) onsensus: + ' + = ( + ') ( + + ') (' + ) ( + ' + ) = ( + ') (' + ) ( + ' + ) = RHS 3.27 () VLI. [ + = ] [' ( + ) = '()] [ + = ] [' + ' = '] 3.27 () NOT VLI. ounteremple: =, = = n = then LHS = + = RHS = = = LHS ut + = + = ; = + The sttement is flse () VLI. [ + = ] [( + ) + = () + ] [ + = ] [ + + = + ] 3.27 () NOT VLI. ounteremple: =, = = n = then LHS = + + = RHS = + = = LHS ut + = + = The sttement is flse () '' + + ' + ' + ''' + ' onsensus terms: () '' using '' + ' (2) ' using '' + (3) using ' + (4) '' using ''' + ' Using, 2, 3: '' + + ' + ' + ''' + ' + '' + ' + = '' + + ' (Using the onsensus theorem to remove the e terms sine the terms tht generte them re still present.) 3.28 () ''' + ' + '' + ' onsensus terms: () '' using ''' + ' (2) ' using '' + ' (3) ''' using ''' + '' (4) '' using ''' + ' (5) ' using ' + ' Using : ''' + ' + '' + ' + ', whih is the minimum solution. 2
17 Unit 4 Prolem Solutions See L p. 628 for solution. E y z 4.2 () Y = ''''E' + '''E' + ''E' (less thn gpm) + (t lest gpm) () Z = ''E' + 'E' + E' (t lest 2 gpm) + + (t lest 3 gpm) + (t lest 4 gpm) + (t lest 5 gpm) 4.3 = m(, 4, 5, 6); 2 = m(, 3, 4, 6, 7); + 2 = m(, 3, 4, 5, 6, 7) Generl rule: + 2 is the sum of ll minterms tht re present in either or 2. 2 n Proof: Let = i m i ; 2 = j m j ; + 2 = i m i + j m j = m + m + 2 m Σi = 2 n Σj = 2 n Σi = 2 n Σj = 2 n + m + m + 2 m = ( + ) m + ( + ) m + ( ) m = ( i + i ) m i Σi = 4.4 () 4.4 () 2 2n = 2 22 = 2 4 = 6 y z z z 2 z 3 z 4 z 5 z 6 z 7 z 8 z 9 z z z 2 z 3 z 4 z 5 'y' 'y ' y' y' 'y+y' '+y' y 'y'+y y '+y +y' +y 4.5 E Z lternte Solutions E Z 3 4 These truth tle entries were me on't res euse = n = n never our 2 These truth tle entries were me on't res euse when is, the output Z of the OR gte will e regrless of its other input. So hnging n E nnot ffet Z. 3 These truth tle entries were me on't res euse when n E re oth, the output Z of the OR gte will e regrless of the vlue of. 4 These truth tle entries were me on't res euse when one input of the N gte is, the output will e regrless of the vlue of its other input. 4.6 () Of the four possile omintions of & 5, = n 5 = gives the est solution: = ''' + '' + ' + = '' () y inspetion, G = when oth on t res re set to. 2
18 4.7 () Etly one vrile not omplemente: = '' + '' + '' = m(, 2, 4) 4.8 () (,,, ) = m(,, 2, 3, 4, 5, 6, 8, 9, 2) Refer to L for full term epnsion 4.7 () Remining terms re mterms: = M(, 3, 5, 6, 7) = ( + + ) ( + ' + ') (' + + ') (' + ' + ) (' + ' + ') 4.8 () (,,, ) = Π M(7,,, 3, 4, 5) Refer to L for full term epnsion 4.8 = 2 = 2 2 = 2 3 = 2 = 2 = 2 2 = = 3 > 2 2 = 2 2 = = 4 > = 6 > 2 3 = 2 3 = 3 > = 6 > = 9 > () 4.9 () 4.9 () 4.9 () = ' + ' ( + ') ( + ') = ' + ' + '' + '' + '''; = m(,, 4, 5, 6) Remining terms re mterms: = M(2, 3, 7) Mterms of re minterms of ': ' = m(2, 3, 7) Minterms of re mterms of ': ' = M(,, 4, 5, 6) 4. (,,, ) = ( + + ) (' + ) (' + ' + ') ( + + ' + ') = ( ) ( + + ' + ) (' + + ' + ') (' + ' + ' + ) (' + ' + ' + ') ( + + ' + ') = ( ) ( + + ' + ) (' ) (' ') (' + ' + + ) (' + ' + + ') (' + ' + ' + ) (' + ' + ' + ') ( + + ' + ') 4. () = m(, 4, 5, 6, 7,, ) 4. () = M(, 2, 3, 8, 9, 2, 3, 4, 5) 4. () ' = m(, 2, 3, 8, 9, 2, 3, 4, 5) 4. () ' = M(, 4, 5, 6, 7,, ) 4. () ifferene, i = i y i i ; i+ = i i ' + i 'y i + i y i 4. () i = s i ; i+ is the sme s i+ with i reple y i ' i y i i i+ i 4.2 See L p. 629 for solution. 22
19 4.3 Z Z = '''' + ''' + ''' + ' + + ' = ''' + + ''' + ' = ''' + + ''' + ' + + ''' (e onsensus terms) Z = ''' ''' ' ' ' ' ' ' Z 4.4 Z Z = '' + '' + ' + ''' + '' + '' + ' = ' + '' + ' + '' = ' + ' + '' + ' (e onsensus terms) Z = ' + ' + ' ' ' ' Z 4.5 () The uzzer will soun if the key is in the ignition swith n the r oor is open, or the set elts re not fstene. K S' The two possile interprettions re: = K + S' n = K( + S') 4.5 () You will gin weight if you et too muh, or you o not eerise enough n your metolism rte is too low. W E' M The two possile interprettions re: W = ( + E') M n W = + E'M 4.5 () The speker will e mge if the volume is set too high n lou musi is plye or the stereo is too powerful. V M S The two possile interprettions re: = VM + S n = V (M + S) 4.5 () The ros will e very slippery if it snows or it rins n there is oil on the ro. V S R O The two possile interprettions re: V = (S + R) O n V = S + RO 4.6 Z = Z = (E + ''''E')'; Y = '''E 4.8 () 3 = 6 = ; = '''E'G 4.8 () = ; Y = '''E'G' 4.8 () = 2 ; 64 = 2 ; 3 = 2 ; 27 = 2 ; 32 = 2 ; Z = ('')' = + 23
20 4.9 2 = M(, 3, 4, 5, 6, 7). Generl rule: 2 is the prout of ll mterms tht re present in either or 2. Proof: 2 n Let = ( i + M i ); 2 = ( j + M j ); 2 = ( i + M i ) ( j + M j ) Πi = 2 n Πj = 2 n Πi = 2 n Πj = = ( + M ) ( + M ) ( + M ) ( + M ) ( 2 + M 2 ) ( 2 + M 2 )... = ( + M ) ( + M ) ( M 2 )... 3 n = ( i i + M i ) Πi = Mterm M i is present in 2 iff i i =. Mterm M i is present in iff i =. Mterm M i is present in 2 iff j =. Therefore, mterm M i is present in 2 iff it is present in or G H J () (,,, ) = m(5, 6, 7,,, 3, 4, 5) = M(,, 2, 3, 4, 8, 9, 2) () G (,,, ) = m(, 2, 4, 6) = M(, 3, 5, 7, 8, 9,,, 2, 3, 4, 5) () H (,,, ) = m(7,, 3, 4, 5) = M(,, 2, 3, 4, 5, 6, 8, 9,, 2) () J (,,, ) = m(4, 8, 2, 3, 4) = M(,, 2, 3, 5, 6, 7, 9,,, 5) 4.22 f () f = m(, 2, 4, 5, 6,,, 2, 4, 5) () f = M(, 3, 7, 8, 9, 3) () f ' = m(, 3, 7, 8, 9, 3) () f '= M(, 2, 4, 5, 6,,, 2, 4, 5) You n lso work this prolem lgerilly, s in prolem You n lso work this prolem using truth tle, s in prolem f(,, ) = ' ( + ') = ' + '' = ' ( + ') + ' ( + ') ' = ' + '' + '' + ''' m 3 m 2 m 2 m 4.2 () f = m(, 2, 3) 4.2 () f = M(, 4, 5, 6, 7) 4.2 () f ' = m(, 4, 5, 6, 7) 4.2 () f ' = M(, 2, 3) 4.23 () (,,, ) = m(3, 4, 5, 8, 9,,, 2, 4) 4.23 () = '' + ''' + '' + ''' + '' + '' + ' + '' + ' (,,, ) = M(,, 2, 6, 7, 3, 5) = ( ) ( ') ( + + ' + ) ( + ' + ' + ) ( + ' + ' + ') (' + ' + + ') (' + ' + ' + ') 4.24 () (,,, ) = m(, 3, 4, 7, 8, 9,, 2, 3, 4) = '''' + '' + ''' + ' + ''' + '' m m 3 m 4 m 7 m 8 m 9 + ' + '' + ' + ' m m 2 m 3 m () (,,, ) = M(, 2, 5, 6,, 5) = ( ') ( + + ' + ) ( + ' + + ') ( + ' + ' + ) M M 2 M 5 M 6 (' + + ' + ) (' + ' + ' + ') M M 5 24
21 4.25 () If on't res re hnge to (, ), respetively, = ''' + + '' + ' = '' +, 4.25 () If on't res re hnge to (, ), respetively 3 = ( + + ) ( + + ') = () If on't res re hnge to (, ), respetively 2 = '''+ '' + '' + ' = ' 4.25 () If on't res re hnge to (, ), respetively 4 = ''' + ' + '' + = '' E Z These truth tle entries were me on't res 2 euse = n 2 = n never 2 our. 2 These truth tle entries were me on't res 2 euse when one input 2 of the OR gte is, the output will e regrless of the vlue of its other 2 input () G (,, ) = m(, 7) = M(, 2, 3, 4, 5, 6) 4.27 () G 2 (,, ) = m(,, 6, 7) = M(2, 3, 4, 5) 4.28 's Y Z () = Y = '' + '' + '' + ' + '' + '' + ' + '' + ' + ' Z = ''' + ''' + ''' + ' + ''' + ' + ' + ' 4.29 W Y Z () = ''' + ''' + '' + ''' + '' + '' + ' + ''' + '' + '' + ' + '' + ' + ' + Y = '''' + ' + ' + ' + Z = '''' + '' + '' + '' + '' + '' + ' + '' () Y = ( ) ( ') 4.29 () Y = ( ') ( + + ' + ) ( + + ' + ) ( + ' + + ) ( + + ' + ') ( + ' + + ) (' ) (' + ' + ' + ') ( + ' + + ') ( + ' + ' + ) (' ) (' ') (' + + ' + ) (' + + ' + ') (' + ' + + ) Z = ( ) ( + ' + + ') ( + ' + ' + ) (' ') (' + + ' + ) (' + ' + + ) (' + ' + ' + ') Z = ( ') ( + + ' + ) ( + ' + + ) ( + ' + ' + ) (' ) (' + ' + + ') (' + ' + ' + ) 25
22 4.3 S T U V W Y Z 4.3 S T U V W Y Z 5 = 4 + = 5 = = = = = = = = = = = = = = = = = =37 Note: Rows through hve on't re outputs. Note: Rows through hve on't re outputs. S =, T =, U =, V =, W =, =, Y =, Z = S =, T =, U = + +, V = ' + '' +, W = '' +, = '' + ', Y = ' + '' +, Z = 4.32 Notie tht the sign it 3 of the 4-it numer is etene to the leftmost full er s well. S 4 S 3 S 2 S S Y 4 Y 3 Y Y Y 4.33 Y Sum out Y Sum out S 3 S 2 S S H.. 2 H.. H.. H
23 Unit 5 Prolem Solutions 5.3 () f 5.3 () 5.3 () f2 f3 r e f s t 5.3 () f4 y z f = '' + ' + ' f2 = 'e' + 'f + e'f f3 = r' + t' f4 = 'z + y + z' 5.4 () 5.4 () 5.4 () = ' + ' + + ' + '' = ' + ' + = ( + '+ ') ( + + ') 5.5 () 5.5 () See L p. 63 for solution. 2 2 lt: Z = ' ' 2 + ' 2 ' + 2 ' 2 ' ' 2 ' 2 Z = ' ' 2 + ' 2 ' + 2 ' 2 ' ' 2 ' Z = ' ' 2 + ' 2 ' + 2 ' 2 ' ' () * * * * 5.6 () * * * lt: = ''' + ' + ' + ' + ' = ''' + ' + ' + ' + ' (*) inites minterm tht mkes the orresponing prime implint essentil. lt: = ' + '' + + '' = ' + '' + + ' (*) Inites minterm tht mkes the orresponing prime implint essentil. ' m 5 ; ''' m ; ' m ;' m 2 m 3 or m 5 ; ' m 3 ; '' m 8 or m 27
24 5.6 () * 5.7 () * * * = '' + ' + '' (*) Inites minterm tht mkes the orresponing prime implint essentil. lt: = ''' + ' + ''' + ' + ''' = ''' + ' + ''' + ' + '' '' m 2 ; '' m 6 ; ' m or m 5.7 () 5.7 () 5.7 () = '' + ''' + = ''' + ' + ' + ' + ' = () = ('+ ') ('+ ') ( + + ) ('+ + ) = ' ' + ' + ' ' 5.8 () lt: = ('+ ) ('+ ') ( + ) ('+ ) = ('+ ) ('+ ') ( + ) ('+ ') = '' + '' + 28
25 5.9 () E = (' + ' + + E) (' + + ' + ') ( + ' + ' + E) (' + + E) ( + ' + ) (' E') (' + ' + ' + E') E lt: = ''E + '' + ' E + ' ' + ' E + E' + ''E' + '' = ''E + '' + ' E + ' ' + ' E + E' + ''E' + ''E' 5.9 () E E = (' + ' + E) (' + ' + + E) ( + ' + E') ( + + ' + E) ( + + ) (' + + E') lt: = ' ' + ' E' + E + ''' + ' E' + ' E = ' ' + ' E' + E + ''E' + ' + ''E = ' ' + ' E' + E + ''' + ' E' + ''E = ' ' + ' E' + E + ''E' + ' E' + ''E 29
26 5. () e 5. () e Essentil prime implints: ''E' (m 6, m 24 ), 'E' (m 4 ), E (m 3 ), ''E (m 3 ) Prime implints: ''E, ''E', 'E, 'E', E, '', 'E, ''E', '' 5. E = ('+ + ') ('+ '+ E ) ( + '+ E') ( + + E') ( ) ('+'+ + ) ( + + E') 5.2 () = '' + ' + ' + = M(,, 9, 2, 3, 4) = ( ) ( ') (' + ' + + ) (' + ' + + ') (' + ' + ' + ) (' ') lt: = ('+ + ') ('+ '+ E ) ( + '+ E') ( + + E') ( ) ('+'+ + E ) ( + + E') ' 5.2 () () ' = ''' + ' + ' = ' + ' + ' + '' Minterms m, m, m 2, m 3, m 4, m, n m n e me on t res, iniviully, without hnging the given epression. However, if m 3 or m 4 is me on t re, the term ' or the term ' (respetively) is not neee in the epression. = ('+ '+ ) ( + + ) ('+ + ') 3
27 5.4 () 5.4 () () R E S T 5.4 () 5.4 (e) N P Q = ' + ' 2 = E' + E' + = T' + R = + ' = N'P + N Q 5.4 (f) 5.5 () f Y Z 5.5 () G E 5.5 () p q r 5.5 () s t u = Y' + 'Z' + Z f = '' + + ' f = '' + + G = E ' + 'E' G = E ' + ' G = E ' + E' = p'r + q'r' + p q = p'q' + p r' + q r = s' 5.5 (e) 5.5 (f) e f 5.6 () 5.6 () = '' + ' + = '' + ' + g = 'e' + f ' 5.6 () = '' + ' + = ('+ ) ( + '+ ') ('+ ) ('+ + ') 5.7 (), () 5.7 () = ' + ' + ' ' lt: = ('+ + ) ('+ '+ ') ('+ '+ ) = ('+ + ) ('+ '+ ') ('+ '+ ') 3
28 5.8 () 5.8 () 2 2 Z 2 2 = ( + 2+ ) ( + + 2) ( + 2'+ '+ 2') ( '+ 2'+ + 2') ( '+ '+ 2 ) ( ) 5.9 () 5.9 () = ' + ' + ' + ''' + ' = ''' + ''' + ''' lt: = ' + ' + ' + ''' + ' lt: = ''' + ''' + '' 5.9 () 5.9 () W Y Z 5.9 (e) = '' + ' + ' lt: = 'Y' + W'Z + Y'Z + W Z' = 'Y' + W Y' + W'Z + W Z' = 'Y' + W Y' + W'Z + W ' = '' + ' + ' () 5.2 () 5.2 () = ' + ' ' + ' + = '' + ' + ' + + = '' + ' + ' 32
29 5.2 () 5.2 () 5.2 () = ' ' + '' + = ('+ ) ('+ + ') ( + '+ ) ( + ) = ('+ ) ('+ + ') ( + ) ('+ ) 5.22 () lt: = ('+ ) ('+ + ') ( + ) ('+ ) 5.22 () = ('+ '+ ) ('+ ') ( + + ) w y z = ' + '' + ' + ' w y z = 'y' + w'z + y'z + w z' = (w + '+ z ) (w + y'+ z ) (w'+ y'+ z') lt: = 'y' + w y' + w'z + w z' = 'y' + w y' + w'z + w ' lt: = (w + '+ z ) (w + y'+ z ) (w'+ '+ y') () = '' + ' + ' Notie tht = n never our, so minterms 5 n 5 re on t res. = '' + ' + ' + = M(,, 9, 2, 3, 4) = ( ) ( ') (' ' ) (' + ' + + ) (' + ' + + ') (' + ' + ' + ) 33
30 5.24 () 5.24 () ' = ' + ''' + ' = ('+ '+ ) ( + + ) ('+ + ') Prime implints for f ': 'e, '', 'e', 'e, ''e', ''e, ''e Prime implints for f: ''e', e, 'e', e',, e', ''e, ''e, '', 'e 5.27 or : ''e', 'e, 'e', '', 'e, ''e, ''e 5.28 () = + E * * * (*) Inites minterm tht mkes the orresponing prime implint essentil. ''' m ; 'e' m 28 ; ''e m 25 ; '' m 2 * or G: 'e, ', 'e', e, 'e, ''', ''e' 5.28 () E ''', 'e', ''e, '', 'e', 'e', 'e', '''e, ''e, '', 'e', 'e', 'e' 5.29 () E = '' + ' ' + '' + ''E' + ' + E + ''E lt: = '' + ' ' + '' + ''E' + ' + E + ''E 34
31 5.29 () E lt: = '''E + ' '' + E + ''E' + E + 'E' + '''E + ' E = '''E + ' '' + E + ''E' + E + 'E' + ' 'E + ' E = '''E + ' '' + E + ''E' + E + 'E' + ' 'E + E 5.3 E = '''' + 'E + ' + 'E' + ' + ''E + 'E 5.3 E = ''E' + '' + E' + '''E' + ''E + ''E 5.32 () w y z v = v' y'z' + 'y'z + v z + w 'y z' + v w 5.32 () w 5.33 () y z E v = ( + y + z) (v + y' + z') (v + ' + z') (v + ' + y') (v' + w + z) lt: = ( + + E ) ('+ ) ( + ') ( + '+ '+ E') = ( + + E ) ('+ ) ( + ') ( + '+ '+ E') 35
32 5.33 () E lt: = ( + ') ( + '+ E') ('+ + E') ('+ '+ '+ ) ( + + E ) ( + '+ E ) = ( + ') ( + '+ E') ( + '+ ) ('+ + E') ('+ '+ '+ ) ( + '+ E ) 5.34 () w y z v lt: = (v'+ w'+ '+ y + z') (w + y'+ z') (v + y') (w + + y ) (v'+ + y + z ) (w'+ + y') = (v'+ w'+ '+ y + z') (w + y'+ z') (v + y') (w + + y ) (v'+ w'+ + z ) (w'+ + y') = (v'+ w'+ '+ y + z') (w + y'+ z') (v + y') (w + + y ) (v'+ w'+ + z ) ( + y'+ z') 5.34 () 5.35 () E = ( + + E) (' + ' + ') (' ) ( + ' + ) ( + ' + E) ( + + E) = ' + + ' hnging m to on't re removes ' from the solution () 5.36 () w y z v m 4 = v' y' + v'w z' + y z + v w''y' + v w'y z' + w' z m 8 = ' + ' + ' + '' m 4, m 3, or m4 hnge the minimum sum of prouts, removing '', ', or ', respetively. m 3 = v' y' + v'w z' + y z + v w''z' + v w' y + w'y'z = v' y' + v'w z' + y z + v w''y' + v w'y z' + w'y'z = v' y' + v'w z' + y z + v w''z' + v w'y z' + w'y'z 5.36 () 36 v'wz' m 8 ; yz m 3 ; v'y' m 4
33 Unit 6 Prolem Solutions 6.2 () ü, 5 - '' 5 ü, 9 - '' 9 ü 5, 7 - ' 2 ü 9, - ' 7 ü 2, 4 - ' ü 7, 5-4 ü, 5-5 ü 4, 5 - Prime implints: '', '', ', ', ',,, 6.2 () ü, - ''', 3, 5, 7 -- ' ü, 8 - ''', 5, 3, ü, 3 -ü 6, 7, 4, ü, 5 -ü 6, 4, 7, ü 8, - '' 6 ü 3, 7 -ü ü 5, 7 -ü 7 ü 6, 7 -ü 4 ü 6, 4 -ü 5 ü, 4 - ' 7, 5 -ü 4, 5 -ü Prime implints: ''', ''', '', ', ', 6.3 () , 5 '', 9 '' 5, 7 ' 9, ' 2, 4 ' 7, 5, 5 4, 5 f = ' + '' + ' + f = ' + '' + ' () , 3, 5, 7 ' 6, 7, 4, 5, ''', 8 ''' 8, '', 4 ' f = ' + + ''' + '' f = ' + + ''' + '' f = ' + + ''' + ' 37
34 6.4 ü, 3 -ü, 3, 5, 7 -- ' 2 ü, 5 -ü, 5, 3, ü, 9 -ü, 5, 9, 3 -- ' 3 ü 2, 3 -ü, 9, 5, ü 2, 6 -ü 2, 3, 6, 7 -- ' 6 ü 2, - '' 2, 6, 3, ü 4, 5 -ü 4, 5, 6, 7 -- ' ü 4, 6 -ü 4, 5, 2, 3 -- ' 2 ü 4, 2 -ü 4, 6, 5, ü 3, 7 -ü 4, 2, 5, ü 5, 7 -ü 5, 7, 3, ü 5, 3 -ü 5, 3, 7, , 7 -ü 9, 3 -ü 2, 3 -ü 7, 5 -ü 3, 5 -ü 3, 5 -ü Prime implints: '', ', ', ', ', ', , 3, 5, 7 ', 5, 9, 3 ' 2, 3, 6, 7 ' 4, 5, 6, 7 ' 4, 5, 2, 3 ' 5, 7, 3, 5 2, '' f = ' + '' + ' + ' f = ' + '' + ' + ' f = ' + '' + ' + ' 6.5 ü, 5 -ü, 5, 9, 3 -- ' 4 ü, 9 -ü, 9, 5, ü 4, 5 -ü 4, 5, 2, 3 -- ' 5 ü 4, 2 -ü 4, 2, 5, ü 8, 9 -ü 5, 7, 3, ü 8, 2 -ü 5, 3, 7, ü 5, 7 -ü 8, 9, 2, 3 -- ' ü 5, 3 -ü 8, 2, 9, ü 9, -ü 9,, 3, ü 9, 3 -ü 9, 3,, ü 2, 3 -ü 2, 3, 4, , 4 -ü 2, 4, 3, , 5 -ü, 5 -ü 3, 5 -ü 4, 5 -ü Prime implints: ', ',, ',, 38
35 6.5 (ont) P (, 5, 9, 3) ' P2 (4, 5, 2, 3) ' P3 (5, 7, 3, 5) P4 (8, 9, 2, 3) ' P5 (9,, 3, 5) P6 (2, 3, 4, 5) (P + P4 + P5) (P2 + P4 + P6) (P + P2 + P3 + P4 + P5 + P6) (P3 + P5 + P6) = (P4 + PP2 + PP6 + P2P5 + P5P6) (P3 + P5 + P6) = P3P4 + P4P5 + P4P6 + PP2P3 + PP2P5 + PP2P6 + PP3P6 + PP5P6 + PP6 + P2P3P5 + P2P5 + P2P5P6 + P3P5P6 + P5P6 = = (' + ) or ( + ') or ( + ') or ( + ) or ( + ') or ( + ') P4 P3 P5 P2 P5 P4 P6 P5 P6 P4 P6 P 6.6 () E = E = E E = MS + EMS = ' + ''' + ' + E ('' + ) or E ('' + ) MS = ''' + ' + ' 6.6 () E = = G = E = ; = G = E MS = '' + MS = '' + G E MS = '' = ; E = G = G = ; E = = MS = '' + ' MS = '' + Z = '' + + E ('' + ') + () + G (') MS = 2 MS 3 = ' or ' or 39
36 6.7 () ü, 4 - ''' 4 ü 4, 5 - '' 3 ü 3, 7 - ' 5 ü 3, - ' 9 ü 5, 7 - ' 7 ü 5, 3 - ' ü 9, - ' 3 ü 9, 3 - ' Prime implints: ''', '', ', ', ', ', ', ' 6.7 () 2 ü 2, 6 - '' 4, 5, 2, 3 -- ' 4 ü 2, - '' 4, 2, 5, ü 4, 5 -ü 9,, 3, ü 4, 6 - '' 9, 3,, ü 4, 2 -ü ü 5, 3 -ü 2 ü 9, -ü ü 9, 3 -ü 3 ü, - ' 5 ü 2, 3 -ü, 5 -ü 3, 5 -ü Prime implints:, ', '', '', '', ' 6.8 () , 4 ''' 4, 5 '' 3, 7 ' 3, ' 5, 7 ' 5, 3 ' 9, ' 9, 3 ' f = ''' + ' + ' + ' f = ''' + ' + ' + ' 6.8 () , 6 '' 2, '' 4, 6 '', ' 4, 5, 2, 3 ' 9,, 3, 5 f = ' + + '' + '' f = ' + + '' + ' f = ' + + '' + '' 6.9 () ü, 3 -ü, 3, 9, -- ' 2 ü, 9 -ü, 9, 3, -- 4 ü 2, 3 -ü 2, 3,, -- ' 3 ü 2, -ü 2,, 3, -- 9 ü 4, 2 - '' 3, 7,, 5 -- ü 3, 7 -ü 3,, 7, ü 3, -ü 9,, 3, ü 9, -ü 9, 3,, 5 -- ü 9, 3 -ü,, 4, ü, -ü, 4,, ü, 4 -ü 2, 3, 4, ü 2, 3 -ü 2, 4, 3, , 4 -ü 7, 5 -ü, 5 -ü 3, 5 -ü 4, 5 -ü 4 Prime implints: '', ', ',,,,
37 6.9 () (ont) , 2 '', 3, 9, ' 2, 3,, ' 3, 7,, 5 9,, 3, 5,, 4, 5 2, 3, 4, 5 f = ' + '' + + ' + f = ' + '' f = ' + '' () ü, -ü,, 8, 9 -- '' ü, 8 -ü, 8,, ü, 5 -ü, 5, 9, 3 -- ' 5 ü, 9 -ü, 9, 5, ü 8, 9 -ü 8, 9,, -- ' 9 ü 8, -ü 8,, 9, -- ü 8, 2 -ü 8, 9, 2, 3 -- ' 2 ü 5, 7 - ' 8, 2, 9, ü 5, 3 -ü ü 6, 7 - ' 3 ü 9, -ü 9, 3 -ü, -ü 2, 3 -ü Prime implints: ', ', '', ', ', ' , 7 ' 6, 7 ',, 8, 9 '', 5, 9, 3 ' 8, 9,, ' 8, 9, 2, 3 ' f = ' + '' + ' + ' 6.9 () f = ' + + '' + + f = ' + + '' + + f = ' + + '' + + ' 6. Prime implints: ', ', ', ', ', ''' f = ' + ' + ' + ''' + ' f = ' + ' + ' + ''' + ' 4
38 6. ü, 2 -ü, 2, 4, 6 --ü, 2, 4, 6, 8,, 2, 'E' 2 ü, 4 -ü, 2, 8, --ü, 2, 8,, 4, 6, 2, ü, 8 -ü, 2, 6, 8 -- ''E', 4, 8, 2, 2, 6,, ü, 6 -ü, 4, 2, ü 2, 6 -ü, 4, 8, 2 --ü 6 ü 2, -ü, 8, 2, -- 9 ü 2, 8 -ü, 8, 4, 2 -- ü 4, 6 -ü, 6, 2, ü 4, 2 -ü 2, 6,, 4 --ü 8 ü 8, 9 -ü 2,, 6, ü 8, -ü 4, 6, 2, 4 --ü ü 8, 2 -ü 4, 2, 6, ü 6, 8 -ü 8, 9,, -- '' 4 ü 6, 7 - '' 8, 9, 2, 3 -- '' 9 ü 6, 4 -ü 8,, 9, -- 2 ü 9, -ü 8,, 2, 4 --ü 29 ü 9, 3 -ü 8, 2, 9, ü, -ü 8, 2,, 4 --, 4 -ü 2, 3 -ü 2, 4 -ü 8, 9 - '' 3, 29 - 'E 4, 3 - E' 2, 29 - 'E , 7 '' 8, 9 '' 3, 29 'E 4, 3 E' 2, 29 'E, 2, 6, 8 ''E' 8, 9,, '' 8, 9, 2, 3 '', 2, 4, 6, 8,, 2, 4 'E' = E' + '' + ''E' + '' + '' + 'E + 'E' 42
39 6.2 () ü, -ü,, 2, 3 --ü,, 2, 3, 8, 9,, --- '' ü, 2 -ü,, 8, 9 --ü,, 8, 9, 2, 3,, ü, 8 -ü, 2,, 3 --, 2, 8,,, 3, 9, ü, 3 -ü, 2, 8, --ü 3 ü, 9 -ü, 8,, ü, 7 -ü, 8, 2, -- 9 ü 2, 3 -ü, 3, 9, --ü ü 2, 6 - ''E', 9, 3, -- 7 ü 2, -ü, 9, 7, ''E 2 ü 8, 9 -ü, 7, 9, ü 8, -ü 2, 3,, --ü 2 ü 3, -ü 2,, 3, ü 9, -ü 8, 9,, --ü 28 ü 9, 25 -ü 8,, 9, ü, -ü 3 ü 7, 2 - ''E 3 ü 7, 25 -ü 2, 2 - '' 2, 28 - 'E' 2, 23 - 'E 28, 3 - E' 23, 3 - E 3, , 6 ''E' 7, 2 ''E 2, 2 '' 2, 28 'E' 2, 23 'E 28, 3 E' 23, 3 E 3, 3, 9, 7, 25 ''E,, 2, 3, 8, 9,, '' f = '' + ''E + ''E' + E + E' + '' f = '' + ''E + ''E' + 'E' + 'E () f = + E' + ''E + ''' +''E + '''E' f = + E' + ''E + ''E + ''' + '''E' 6.3 = ' + ' + ' + '' + ' = ' + ' + ' + '' + ' = ' + ' + ' + ' + '' 6.4 Prime implints: e, 'e, 'e, 'e', '', '', ''e, ''e', ''e' Essentil prime implints re unerline: = e + ''e' + 'e' + ''e + '' = e + ''e' + 'e' + ''e + 'e 43
40 6.5 ü, 3 -ü, 3, 7, 9 -- ''' 2 ü, 7 -ü, 7, 3, ü 2, 3 -ü 2, 3, 8, 9 -- '''E 32 ü 2, 8 -ü 2, 8, 3, ü 6, 7 -ü 6, 7, 8, 9 -- ''' 7 ü 6, 8 -ü 6, 8, 7, ü 6, 48 - ''E'' 48 ü 32, 48 - ''E'' 9 ü 3, 9 -ü 26 ü 7, 9 -ü 28 ü 8, 9 -ü 5 ''E 8, 26 - ''E' 29 ü 26, 3 - 'E' 3 ü 28, 29 - 'E' 39 ''E 28, 3 - '' 63 E ''E 39 ''E 63 E 6, 48 ''E'' 32, 48 ''E'' 8, 26 ''E' 26, 3 'E' 28, 29 'E' 28, 3 '', 3, 7, 9 ''' 2, 3, 8, 9 '''E 6, 7, 8, 9 ''' 6.5 () 6.5 () 6.5 () G = ''E + E + ''' + '''E +''E'' +''' +''E' G = ''E + E + ''' + '''E +''E'' +''' + 'E' Essentil prime implints re unerline in 6.5 (). If there were no on't res, prime implints 5, (26, 3), (28, 29), n (28, 3) re omitte. There is only one minimum solution. Sme s (), eept elete the seon eqution. 6.6 () Prime implints: 'E, '', 'E'', 'E, 'E'', E, 'E, ''E', ''E, ''E', ''E', ''E' G = E + 'E'' + 'E'' + 'E + ''E' + 'E + 'E G = E + 'E'' + 'E'' + 'E + ''E' + 'E + ''E G = E + 'E'' + 'E'' + 'E + ''E' + '' + 'E G = E + 'E'' + 'E'' + 'E + ''E' + '' + ''E 44
41 6.6 () Essentil prime implints re unerline in 6.6(). 6.7 Prime implints:, ',,,, ' Minimum solutions: (' + ); (' + ); ( + ); ( + ); ( + ') 6.6 () If there re no on t res, the prime implints re: 'E, '', 'E'', 'E, 'E'', E, 'E, ''E' G = E + 'E'' + 'E + 'E'' +'E + ''E' + '' G = E + 'E'' + 'E + 'E'' +'E + ''E' + 'E 6.8 () E 6.8 () G E E E = ' + ' + '' + E (''' + ') MS MS Z = ' + '' + E (' + ') + (') + G ('') MS MS MS 2 MS () Eh minterm of the four vriles,,, epns to two minterms of the five vriles,,,, E. or emple, m 4 (,,,) = ''' = '''E' + '''E = m 8 (,,,,E) + m 9 (,,,,E) 6.9 () Prime implints: ''', ', ', ''E, E, E, ''E = ''' + ' + ' + ''E + E = ''' + ' + ' + ''E + E E = ''' + ' + ' + ''E + E = ''' + ' + ' + ''E + E 6.2 E * * This squre ontins +, whih reues to. G = 'E' + E + ('') + () MS MS MS 2 45
42 7. () f Unit 7 Prolem Solutions f f = '' + '' + ' + ' ' Sum of prouts solution requires 5 gtes, 6 inputs f = ('+ ') ( + ) ( + + ') ( + '+ ') f = ('+ ') ( + ) ( + '+ ') ('+ + ') f = ('+ ') ( + ) ( + + ') ('+ '+ ') f = ('+ ') ( + ) ('+ + ') ('+ '+ ') Prout of sums solution requires 5 gtes, 4 inputs, so prout of sums solution is minimum. 7. () eginning with the minimum sum of prouts solution, we n get f = ' ( + ') + ' (' + ') 5 gtes, 2 inputs So sum of prouts solution is minimum. 2 eginning with minimum prout of sums solution, we n get f = ( + ) (' + ') (' + ' + ')) gtes, 4 inputs () ' + E' + E' + ' + ''E' = E' ( + ) + ''E' + ' ( + ) = ( + ) (E' + ') + ''E' levels, 6 gtes, 3 inputs 7.2 () E + E + E + G + G + G = E + G + E ( + ) + G ( + ) = (E + G) [ + ( + )] levels, 6 gtes, 2 inputs
43 7.3 (,,, )n = ' + ' or (' + ') = ( + ) (' + ') You n otin this eqution in the prout of sums form using Krnugh mp, s shown elow: ' ' N-OR ' ' NN-NN OR-NN ' ' ' ' NOR-OR ='+' (')'=[(')'(')']' (')'=[(+'+')('++')]' (')'=(+'+')'+('++')' (')'=[('+')']' ' ' ' ' ' ' OR-N ' ' NOR-NOR ' ' =(+)('+') (')'=['+(+)'+('+')']' (')'=['+''+]' (')'=('')'()' (')'=[[(+)('+'))']' ' ' N-NOR ' ' NN-N = ' + ' = ( + ) ( ) ('+ ') 7.4 (,,, ) = m(5,,, 2, 3) = ' + ' + ' = ' ( + ) + ' = ' ( + ) + ' gtes, inputs = ' + ' + ' ' ' ' ' 47
44 7.5 Z = ( + + ) ('+ ') ('+ ') ('+ ') Z = ( + + ) (' + ''') gtes, inputs ' 2 3 Z 7.6 Z = + + '' = ( + ) + '' ' ' ' Z 7.7 Z = E + E + E = E ( + + ) = E [ + ( + )] ' ' ' E' Z 7.8 or the solution to 7.8, see L P = ' + + '' 6 gtes 2 = '' + '' + ' 7. f (,,, ) = m(3, 4, 6, 9, ) f 2 (,,, ) = m(2, 4, 8,,, 2) f 3 (,,, ) = m(3, 6, 7,, ) = ' + ' + ' ' 2 = ' + ' ' + '' + '' 2 = ' + ' ' + '' + '' gtes 48 3 = ' + ' + '
45 7. = ( + ) ( + ') ('+'+)('++') 8 gtes 2 2 = ('++)('+'+)(+')('++') 2 = (+'+)('+'+)(+')('++') = (++)('+) 2 = (++)('++)('+) 9 gtes 3 = ('++)(+)(+') 7.3 () Using = (')' from Equtions (7-23()), p. 94: f = [(')' ()' ('')' (')']'; f 2 = [' (')']'; f 3 = [()' ('')' ()']' ' ' ' ' f ' f 2 ' ' ' f () Using = (')' from Equtions erive in prolem 7.2: f = [( + + )' + (' + )']' f 2 = [( + + )' + (' + + )' + (' + )']' f 3 = [(' + + )' + ( + )' + ( + ')']' ' f ' ' f 2 ' ' f 3 49
46 7.4 () f = ( + + ) ( + + ') ('+ '+ ') ('+ '+ ') 5 gtes, 6 inputs 7.4 () eginning with the sum of prouts solution, we get f = ' + ' + ' (' + ') = ' + ' + ' (' + ') ( + ) 6 gtes, 4 inputs ut, eginning with the prout of sums solution ove, we get f = ( + + ') (' + ' + '') 5 gtes, 2 inputs, whih is minimum n f = ' + ' + '' + '' f = ' + ' + '' + '' (two other minimun solutions) 5 gtes, 4 inputs miniml ' ' ' ' ' ' ' ' ' ' ' 7.5 () rom K-mps: = ' + ' + ' 4 gtes, inputs = ( + + ) ( + ) (' + ') 4 gtes, inputs, miniml 7.5 () rom K-mps: = + + '' 4 gtes, 9 inputs = (' + ) ( + ' + ) 3 gtes, 7 inputs, miniml ' ' ' ' 7.5 () rom K-mps: = + '' + = + '' + ' 4 gtes, inputs = ( + ) ( ' + ) ( + + ') 4 gtes, inputs, miniml ' ' 7.5 () rom K-mps: = ' + + ' 4 gtes, 9 inputs, miniml = ( + ) (' + + ') (' + + ) = ( + ) (' + + ') ( + + ) 4 gtes, inputs ' ' 5
47 7.6 () In this se, multi-level iruits o not improve the solution. rom K-mps: = ' + + ' + '' 5 gtes, 6 inputs, miniml = (' + + ) ( + + ) (' + ' + ) ( + + ') 5 gtes, 6 inputs, lso miniml Either nswer is orret. ' ' ' ' 7.6 () Too mny vriles to use K-mp; use lger. E y onsensus, then use + Y = E + E + ' + EG + E + E G E ' = E + ' + EG + E = E ( + G) + (' + E) gtes, 3 inputs, miniml E () = M(,, 2, 4, 8) () = ( + + ) ( + + ) ( + + ) ( + + ) = ( + + )( + + ) or = ( + + )( + + ) or = ( + + )( + + ) This solution hs 5 gtes, 2 inputs. eginning with the sum of prouts requires 6 gtes. 5
48 7.8 () (w,, y, z) = ( + y' + z) (' + y + z) w y' z ' y z OR-N w y' z ' y z NOR-NOR w' ' y z' y' z' N-NOR w' ' y z' y' z' NN-N w rom Krnugh mp: = wy + w'y' + wz w y w ' y' w z N-OR w y w ' y' w z NN-NN w' ' y' w' y w' z' OR-NN w' ' y' w' y w' z' NOR-OR 7.8 () (,,, ) = m(4, 5, 8, 9, 3) rom Kmp: = '' + '' + ' = '' + '' + ' = ' ( + ) (' + ' + ) ' ' ' ' ' N-OR ' ' ' ' ' NN-NN ' ' ' ' OR-NN ' ' ' ' NOR-OR ' ' OR-N ' ' ' NOR-NOR ' ' ' N-NOR ' ' ' NN-N ' 52
49 7.9 () y z = (y'+ z ) ('+ y + z') rom Kmp: = (y' + z) (' + y + z') y z' y' z f y z' y' z f 7.9 () y z ' z ( ) or ' y' y' z' f ' z ( ) or ' y' y' z' f = y z + y'z' + 'y' y z y z = y z + y'z' + 'z 7.2 () Using OR n NOR gtes: ' ' f ' ' f = '' () Using NOR gtes only: ' ' ' f ' ' ' f = ('+ ) ('+ ) ('+ ) ('+ ) ' ' 53
50 7.2 () NN gtes: = ' + ' + (Refer to prolem 5.4 for K-mp) NOR gtes: = ( + ' + ') ( + + ') 7.2 () NN gtes: = '' + '' + (Refer to prolem 5.8() for K-mp) NOR gtes: = ( + ) (' + ') (' + ) (' + ') = ( + ) (' + ') (' + ) (' + ) 7.2 (e) NN gtes: =''+'E'+E+'''+''E+'E' =''+'E'+E+''E'+'+''E =''+'E'+E+'E'+''E'+''E =''+'E'+E+'E'+''E+'E (Refer to prolem 5.9() for K-mp) NOR gtes: = (' + ' + + E) ( + ' + E') (' + ' + E) ( + + ' + E) ( + + ) (' + + E') 7.2 (g) NN gtes: f = 'y' + wy' + w'z+ wz' f = 'y' + wy' + w' + w'z f = 'y' + wy' + y'z+ wz' (Refer to prolem 5.22() for K-mp) NOR gtes: f = (w + ' + z) (w + y' + z) (w' + y' + z') f = (w + ' + z) (w + y' + z) (w' + ' + y') 7.2 () NN gtes: = '' + ' + '' (Refer to prolem 5.8() for K-mp) NOR gtes: = (' + ') (' + ') ( + + ) (' + + ) 7.2 () NN gtes: = '' + ''E + 'E + 'E + E' + '' + ''E' + '' = '' + ''E + 'E + 'E + E' + '' + ''E' + ''E' (Refer to prolem 5.9() for K-mp) NOR gtes: = (' + + E) ( + ' + ) (' + + ' + ') (' + ' + + E ) ( + ' + ' + E) (' E') (' + ' + ' + E') 7.2 (f) NN gtes: = ' + ' + ' + '' (Refer to prolem 5.22() for K-mp) NOR gtes: = ( + + ) (' + ' + ) (' + ') 7.22 () 7.22 () f = (' + ' + e) ( + ' + e') ( + + ) E ' e' e ' ' ' ' f = ( + ' + ) ( + ' + e') (' + ' + e) ( + + ) ( + ' + ') 54
51 7.23 ' ' ' ' ' ' f f = (' + ) (' + + ) ( + ') = ( + ') [' + ( + )] = ( + ') (' + + ) 7.24 () Z = e'f + 'e'f + 'e'f + gh = e'f ( + ' + ') + gh 7.24 () Z= (' + +e + f)(' + ' + )(' + ' + )(g+h) = [' + + '' (e + f)] (g + h) g h e' f z e f g h ' z 7.25 = e' + '' + = ( + ') (' + e') + = ( + ' + ) (' + + e') ' ' ' e ' lternte: = (' + + ) (' + + e') 7.26 f = 'yz + vy'w' + vy'z' = 'yz + vy' (z' + w') w z ' y z v y' f 7.27 () 7.27 () = ' + ' + ''' + ' = ' + ' + ''' + ' rw N-OR iruit n reple ll gtes with NNs. = ( + + ') ('+ ) ('+ ) ('+ '+ ') rw OR-N iruit n reple ll gtes with NORs. 55
52 7.27 () = (' + ') + ' ('' + ) ' ' ' ' ' lterntive: = ' ('' + ) + (' + ') = (' + ') + ' ('' + ) = ' ('' + ) + (' + ') = (' + ') + ' ('' + ) 7.28 () 7.28 () = + '' + '' + '' + '' + '''' = ( + + ') ( + ' + ) ( + ' + ) ( + ' + ') (' + + ) (' + + ') (' + + ') (' + ' + ) 7.28 () Mny solutions eist. Here is one, rwn with lternte gte symols. = ' (''' + ' + ') + ('' + '' + ) = ' ('('' + ) + ') + (('' + ) + '') ' ' ' ' ' ' ' 56
53 7.29 () = '' '' = ( + '') + ( + '') Mny NOR solutions eist. Here is one. = ( + ) (' + + ) ( + + ') ( + ' + ' + ) = ( + ) [ + ( + ') (' + ' + )] (' + + ) = ( + ) [ ( + ) + ' ( + ') (' + ' + )] = ( + ) [ ( + ) + ' ((' + ) + '')] ' ' ' 7.29 () = ' + + '' + ' = ( + ') + ('' + ') = ( + ') + [(' + ') ( + ')] = ( + ) ( + ) ('+ + ') ('+ + ') ' ' ' ' ' ' ' ' ' ' = (' + + ') (' + + ') ( + ) ( + ) = ( + (' + ')) ( + (' + ')) 7.3 = m(,, 2, 3, 4, 5, 7, 9,, 3, 4, 5) = + '' + '' + = + ' (' + ') + lternte solution: = + (' + ) ( + ' + ') = '' + ''
54 7.3 () 7.3 () ' ' ' ' ' ' 7.3 () ' ' ' ' ' ' ' ' ' ' 7.3 Z = [' + + E(' + GH)] G H ' E ' G' H' ' E' ' ' 58
55 7.32 f f2 f3 f = '' + f 2 = '' + '' 8 gtes f 3 = + + '' 7.33 = ' + ' + '' 6 gtes = '' + ' + '' 7.34 f y z f2 y z f3 y z f = 'y z + ' y z' + y' f 2 = y' z + 'y z + y z' 8 gtes f 3 = y' + y'z + 'y z' + y z' 7.35 () f f2 f = (' + + ) (' + ' + ') (' + ) 6 gtes f 2 = (' + + ) (' + ' + ') ( + ') 59
56 7.35 () f irle 's to get sum-of-prouts epressions: f = ' + ''' + ' 6 gtes f 2 = ' + ''' + ' Then onvert iretly to NN gtes. f () irle s 2 f = ( + + ) ('+ ') ('+ ) 7 gtes f 2 = ( + + ) (' + ) (' + ' + ') 7.36 () irle 's to get sum-of-prouts epressions: 2 Then onvert iretly to NN gtes f = ' + ' + ' 7 gtes f 2 = ' + ' + ' 6
57 7.37 () f = + ' + ' 2 f 2 = ' + ' + ' ' ' ' f f () f = ( + ) ( + ) ( + ) ( + + ') 2 f 2 = ( + ) ( + ) ( + + ') ('+'+') ' f f 2 ' ' ' 7.38 () f f = '' + ' + ' f 2 = '' + ' + ' 7.38 () f2 f = ('+ ) ('+ ') ( + ') ( + '+ ') f 2 = ('+ ) ('+ ') ( + '+ ') ( + ' + ) 6
58 7.39 () The iruit onsisting of levels 2, 3, n 4 hs OR gte outputs. onvert this iruit to NN gtes in the usul wy, leving the N gtes t level unhnge. The result is: ' ' e f ' g h () One solution woul e to reple the two N gtes in () with NN gtes, n then inverters t the output. However, the following solution vois ing inverters t the outputs: = [( + ') + ] (e' + f) = e' + 'e' + e' + f + 'f + f = e' ( + ') + (e' + f) + f ( + ') 2 = [( + ') + g'] (e' + f) h = h (e' + 'e' + f + 'f) + g'h (e' + f) = h [e' ( + ') + f ( + ')] + g'h (e' + f) ' e' e f ' f g' h h 2 62
59 Unit 8 Prolem Solutions 8. W Y V Z t (ns) 8.2 () = ''' + + ' 8.2 () (ont) Stti -hzrs re: n = ( + ') ('+ + ) ( + + ') Stti -hzrs re: n 8.2 () 8.2 () t = ( + ') (' + + ) ( + + ') (' + + ) ( + + ') 8.3 () t = ''' + + ' + '' + E G t (ns) Glith (stti '' hzr) 8.3 () Moifie iruit (to voi hzrs) 8.4 G = '' + + ' = ; = Z; = Z = ; = + Z = ; E = ' = ; = ' = ; G = = ; H = + = See L Tle 8-, P = =, = = So = ' + '' + = ut in the figure, gte 4 outputs =, initing something is wrong. or the lst NN gte, = only when ll its inputs re. ut the output of gte 3 is. Therefore, gte 4 is working properly, ut gte 3 is onnete inorretly or is mlfuntioning. 63
60 W Y V Z t (ns) = Z; = ; = Z' = ; = Z = ; E = Z; = + + = ; G = ( Z)' = ' = ; H = ( + )' = ' = = = =, so = ( + ' + ') (' + + ') (' + ' + ) = ut, in the figure, gte 4 outputs =, initing something is wrong. or the lst NOR gte, = only when ll its inputs re. ut the output of gte is. Therefore, gte 4 is working properly, ut gte is onnete inorretly or is mlfuntioning. 8.7 Z = ' + '' + ' ' Stti -hzrs lie etween n Without hzrs: Z t = '' + ' + '' + '' + '' 8. () (,,, ) = m(, 2, 5, 6, 7, 8, 9, 2, 3, 5) There re 3 ifferent minimum N-OR solutions to this prolem. The prolem sks for ny two of these. 8. () = + ' + ' ' + ''' Solution : -hzrs re etween n = ( + + ') ( + ' + + ) (' + ' + ) (' + + ') -hzr is etween Either wy, without hzr: t = ( + + ') ( + ' + + ) (' + ' + ) ( + ' + ') (' + + ') = + ' + ''' + ' Solution 2: -hzrs re etween n 64 = + ' + ''' + ' ' Solution 3: -hzrs re etween n Without hzrs: t = + ' + ''' + '' + ''' + ' = ( + + ') ( + ' + + ) (' + ' + ) ( + ' + ') -hzr is etween
61 Unit 9 Prolem Solutions 9. See L p. 636 for solution. 9.2 See L p. 636 for solution. 9.3 See L p. 637 for solution. 9.4 See L p. 637 n igure 4-4 on L p y y y 2 y 3 y y y y 2 3 y y y y 2 3 y y y y 2 3 = y 3 + y 2 = y 3 + y 2' y = y 3 + y 2+ y + y 9.6 See L p. 638 for solution. 9.7 See L p. 638 for solution. 9.8 See L p for solution. 9.9 The equtions erive from Tle 4-6 on L p. re: 9. Note: 6 = 4 ' n 5 = 4. Equtions for 4 through n e foun using Krnugh mps. See L p for nswers. = 'y' in + 'y in ' + y' in ' + y in out = ' in + 'y + y in See L p. 639 for PL igrm. 9. () = '' + ' + ' Use 3 N gtes ' = ['' + ' + ']' = [' ( +') + ']' = [( + + ') (' + ')]' = '' + Use 2 N gtes 9. () = '' + '' Use 2 N gtes ' = ('' + '')' = [(' +') (' + ') (' + ') (' + ')]' = Use 4 N gtes 9.2 () See L p. 64, use the nswer for 9.2 (), ut leve off ll onnetions to n '. 9.2 () See L p. 64 for solution. 9.3 Using Shnnon s epnsion theorem: = 'e' + ''e + ''e + 'e' = ' (e' + ''e + 'e') + (''e + ''e + 'e') = ' [e' ( + ') + ''e] + [(' + ') 'e + 'e'] = ' (e' + ''e) + (''e + ''e + 'e') The sme result n e otine y splitting Krnugh mp, s shown to the right. E = = 65
Words Symbols Diagram. abcde. a + b + c + d + e
Logi Gtes nd Properties We will e using logil opertions to uild mhines tht n do rithmeti lultions. It s useful to think of these opertions s si omponents tht n e hooked together into omplex networks. To
More informationAngles 2.1. Exercise 2.1... Find the size of the lettered angles. Give reasons for your answers. a) b) c) Example
2.1 Angles Reognise lternte n orresponing ngles Key wors prllel lternte orresponing vertilly opposite Rememer, prllel lines re stright lines whih never meet or ross. The rrows show tht the lines re prllel
More informationLesson 2.1 Inductive Reasoning
Lesson.1 Inutive Resoning Nme Perio Dte For Eerises 1 7, use inutive resoning to fin the net two terms in eh sequene. 1. 4, 8, 1, 16,,. 400, 00, 100, 0,,,. 1 8, 7, 1, 4,, 4.,,, 1, 1, 0,,. 60, 180, 10,
More informationCS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001
CS99S Lortory 2 Preprtion Copyright W. J. Dlly 2 Octoer, 2 Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes to oserve logic
More informationMaximum area of polygon
Mimum re of polygon Suppose I give you n stiks. They might e of ifferent lengths, or the sme length, or some the sme s others, et. Now there re lots of polygons you n form with those stiks. Your jo is
More information1 Fractions from an advanced point of view
1 Frtions from n vne point of view We re going to stuy frtions from the viewpoint of moern lger, or strt lger. Our gol is to evelop eeper unerstning of wht n men. One onsequene of our eeper unerstning
More informationThe remaining two sides of the right triangle are called the legs of the right triangle.
10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right
More informationChapter. Contents: A Constructing decimal numbers
Chpter 9 Deimls Contents: A Construting deiml numers B Representing deiml numers C Deiml urreny D Using numer line E Ordering deimls F Rounding deiml numers G Converting deimls to frtions H Converting
More informationCS 316: Gates and Logic
CS 36: Gtes nd Logi Kvit Bl Fll 27 Computer Siene Cornell University Announements Clss newsgroup reted Posted on we-pge Use it for prtner finding First ssignment is to find prtners P nd N Trnsistors PNP
More informationMA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!
MA 5800 Lesson 6 otes Summer 06 Rememer: A logrithm is n eponent! It ehves like n eponent! In the lst lesson, we discussed four properties of logrithms. ) log 0 ) log ) log log 4) This lesson covers more
More informationMATH PLACEMENT REVIEW GUIDE
MATH PLACEMENT REVIEW GUIDE This guie is intene s fous for your review efore tking the plement test. The questions presente here my not e on the plement test. Although si skills lultor is provie for your
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More information50 MATHCOUNTS LECTURES (10) RATIOS, RATES, AND PROPORTIONS
0 MATHCOUNTS LECTURES (0) RATIOS, RATES, AND PROPORTIONS BASIC KNOWLEDGE () RATIOS: Rtios re use to ompre two or more numers For n two numers n ( 0), the rtio is written s : = / Emple : If 4 stuents in
More informationwww.mohandesyar.com SOLUTIONS MANUAL DIGITAL DESIGN FOURTH EDITION M. MORRIS MANO California State University, Los Angeles MICHAEL D.
27 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This publication is protected by opyright and written permission should be obtained or likewise. For information regarding permission(s),
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More informationLec 2: Gates and Logic
Lec 2: Gtes nd Logic Kvit Bl CS 34, Fll 28 Computer Science Cornell University Announcements Clss newsgroup creted Posted on we-pge Use it for prtner finding First ssignment is to find prtners Due this
More informationVolumes by Cylindrical Shells: the Shell Method
olumes Clinril Shells: the Shell Metho Another metho of fin the volumes of solis of revolution is the shell metho. It n usull fin volumes tht re otherwise iffiult to evlute using the Dis / Wsher metho.
More information1. Definition, Basic concepts, Types 2. Addition and Subtraction of Matrices 3. Scalar Multiplication 4. Assignment and answer key 5.
. Definition, Bsi onepts, Types. Addition nd Sutrtion of Mtries. Slr Multiplition. Assignment nd nswer key. Mtrix Multiplition. Assignment nd nswer key. Determinnt x x (digonl, minors, properties) summry
More information5.6 POSITIVE INTEGRAL EXPONENTS
54 (5 ) Chpter 5 Polynoils nd Eponents 5.6 POSITIVE INTEGRAL EXPONENTS In this section The product rule for positive integrl eponents ws presented in Section 5., nd the quotient rule ws presented in Section
More informationIf two triangles are perspective from a point, then they are also perspective from a line.
Mth 487 hter 4 Prtie Prolem Solutions 1. Give the definition of eh of the following terms: () omlete qudrngle omlete qudrngle is set of four oints, no three of whih re olliner, nd the six lines inident
More informationInterior and exterior angles add up to 180. Level 5 exterior angle
22 ngles n proof Ientify interior n exterior ngles in tringles n qurilterls lulte interior n exterior ngles of tringles n qurilterls Unerstn the ie of proof Reognise the ifferene etween onventions, efinitions
More informationModule 5. Three-phase AC Circuits. Version 2 EE IIT, Kharagpur
Module 5 Three-hse A iruits Version EE IIT, Khrgur esson 8 Three-hse Blned Suly Version EE IIT, Khrgur In the module, ontining six lessons (-7), the study of iruits, onsisting of the liner elements resistne,
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 249-25 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationPractice Test 2. a. 12 kn b. 17 kn c. 13 kn d. 5.0 kn e. 49 kn
Prtie Test 2 1. A highwy urve hs rdius of 0.14 km nd is unnked. A r weighing 12 kn goes round the urve t speed of 24 m/s without slipping. Wht is the mgnitude of the horizontl fore of the rod on the r?
More informationVectors Summary. Projection vector AC = ( Shortest distance from B to line A C D [OR = where m1. and m
. Slr prout (ot prout): = osθ Vetors Summry Lws of ot prout: (i) = (ii) ( ) = = (iii) = (ngle etween two ientil vetors is egrees) (iv) = n re perpeniulr Applitions: (i) Projetion vetor: B Length of projetion
More informationAngles and Triangles
nges nd Tringes n nge is formed when two rys hve ommon strting point or vertex. The mesure of n nge is given in degrees, with ompete revoution representing 360 degrees. Some fmiir nges inude nother fmiir
More informationCHAPTER 31 CAPACITOR
. Given tht Numer of eletron HPTER PITOR Net hrge Q.6 9.6 7 The net potentil ifferene L..6 pitne v 7.6 8 F.. r 5 m. m 8.854 5.4 6.95 5 F... Let the rius of the is R re R D mm m 8.85 r r 8.85 4. 5 m.5 m
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationBayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the
More informationEnd of term: TEST A. Year 4. Name Class Date. Complete the missing numbers in the sequences below.
End of term: TEST A You will need penil nd ruler. Yer Nme Clss Dte Complete the missing numers in the sequenes elow. 8 30 3 28 2 9 25 00 75 25 2 Put irle round ll of the following shpes whih hve 3 shded.
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More informationBoğaziçi University Department of Economics Spring 2016 EC 102 PRINCIPLES of MACROECONOMICS Problem Set 5 Answer Key
Boğziçi University Deprtment of Eonomis Spring 2016 EC 102 PRINCIPLES of MACROECONOMICS Prolem Set 5 Answer Key 1. One yer ountry hs negtive net exports. The next yer it still hs negtive net exports n
More informationSECTION 7-2 Law of Cosines
516 7 Additionl Topis in Trigonometry h d sin s () tn h h d 50. Surveying. The lyout in the figure t right is used to determine n inessile height h when seline d in plne perpendiulr to h n e estlished
More information2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration
Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls -: The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N): - counting numers. {,,,,, } Whole Numers (W): - counting numers with 0. {0,,,,,, } Integers (I): -
More informationOrthopoles and the Pappus Theorem
Forum Geometriorum Volume 4 (2004) 53 59. FORUM GEOM ISSN 1534-1178 Orthopoles n the Pppus Theorem tul Dixit n Drij Grinerg strt. If the verties of tringle re projete onto given line, the perpeniulrs from
More informationNational Firefighter Ability Tests And the National Firefighter Questionnaire
Ntionl Firefighter Aility Tests An the Ntionl Firefighter Questionnire PREPARATION AND PRACTICE BOOKLET Setion One: Introution There re three tests n questionnire tht mke up the NFA Tests session, these
More informationThe art of Paperarchitecture (PA). MANUAL
The rt of Pperrhiteture (PA). MANUAL Introution Pperrhiteture (PA) is the rt of reting three-imensionl (3D) ojets out of plin piee of pper or ror. At first, esign is rwn (mnully or printe (using grphil
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationc b 5.00 10 5 N/m 2 (0.120 m 3 0.200 m 3 ), = 4.00 10 4 J. W total = W a b + W b c 2.00
Chter 19, exmle rolems: (19.06) A gs undergoes two roesses. First: onstnt volume @ 0.200 m 3, isohori. Pressure inreses from 2.00 10 5 P to 5.00 10 5 P. Seond: Constnt ressure @ 5.00 10 5 P, isori. olume
More information1.2 The Integers and Rational Numbers
.2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl
More informationRatio and Proportion
Rtio nd Proportion Rtio: The onept of rtio ours frequently nd in wide vriety of wys For exmple: A newspper reports tht the rtio of Repulins to Demorts on ertin Congressionl ommittee is 3 to The student/fulty
More informationDiaGen: A Generator for Diagram Editors Based on a Hypergraph Model
DiGen: A Genertor for Digrm Eitors Bse on Hypergrph Moel G. Viehstet M. Mins Lehrstuhl für Progrmmiersprhen Universität Erlngen-Nürnerg Mrtensstr. 3, 91058 Erlngen, Germny Emil: fviehste,minsg@informtik.uni-erlngen.e
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationCOMPLEX FRACTIONS. section. Simplifying Complex Fractions
58 (6-6) Chpter 6 Rtionl Epressions undles tht they cn ttch while working together for 0 hours. 00 600 6 FIGURE FOR EXERCISE 9 95. Selling. George sells one gzine suscription every 0 inutes, wheres Theres
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More informationMath Review for Algebra and Precalculus
Copyrigt Jnury 00 y Stnley Oken. No prt of tis doument my e opied or reprodued in ny form wtsoever witout epress permission of te utor. Mt Review for Alger nd Prelulus Stnley Oken Deprtment of Mtemtis
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationSection 5-4 Trigonometric Functions
5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationTwo hours UNIVERSITY OF MANCHESTER SCHOOL OF COMPUTER SCIENCE. Date: Friday 16 th May 2008. Time: 14:00 16:00
COMP20212 Two hours UNIVERSITY OF MANCHESTER SCHOOL OF COMPUTER SCIENCE Digitl Design Techniques Dte: Fridy 16 th My 2008 Time: 14:00 16:00 Plese nswer ny THREE Questions from the FOUR questions provided
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More information19. The Fermat-Euler Prime Number Theorem
19. The Fermt-Euler Prime Number Theorem Every prime number of the form 4n 1 cn be written s sum of two squres in only one wy (side from the order of the summnds). This fmous theorem ws discovered bout
More informationOn Equivalence Between Network Topologies
On Equivlene Between Network Topologies Tre Ho Deprtment of Eletril Engineering Cliforni Institute of Tehnolog tho@lteh.eu; Mihelle Effros Deprtments of Eletril Engineering Cliforni Institute of Tehnolog
More informationMultiplication and Division - Left to Right. Addition and Subtraction - Left to Right.
Order of Opertions r of Opertions Alger P lese Prenthesis - Do ll grouped opertions first. E cuse Eponents - Second M D er Multipliction nd Division - Left to Right. A unt S hniqu Addition nd Sutrction
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationSOLVING EQUATIONS BY FACTORING
316 (5-60) Chpter 5 Exponents nd Polynomils 5.9 SOLVING EQUATIONS BY FACTORING In this setion The Zero Ftor Property Applitions helpful hint Note tht the zero ftor property is our seond exmple of getting
More informationLISTENING COMPREHENSION
PORG, přijímí zkoušky 2015 Angličtin B Reg. číslo: Inluded prts: Points (per prt) Points (totl) 1) Listening omprehension 2) Reding 3) Use of English 4) Writing 1 5) Writing 2 There re no extr nswersheets
More informationLearning Outcomes. Computer Systems - Architecture Lecture 4 - Boolean Logic. What is Logic? Boolean Logic 10/28/2010
/28/2 Lerning Outcomes At the end of this lecture you should: Computer Systems - Architecture Lecture 4 - Boolen Logic Eddie Edwrds eedwrds@doc.ic.c.uk http://www.doc.ic.c.uk/~eedwrds/compsys (Hevily sed
More informationChapter. Fractions. Contents: A Representing fractions
Chpter Frtions Contents: A Representing rtions B Frtions o regulr shpes C Equl rtions D Simpliying rtions E Frtions o quntities F Compring rtion sizes G Improper rtions nd mixed numers 08 FRACTIONS (Chpter
More informationMULTIPLYING OUT & FACTORING
igitl ircuit Engineering MULTIPLYING OUT & FTORING I IGITL SIGN Except for #$&@ fctoring st istributive X + X = X( + ) 2nd istributive (X + )(X + ) = X + (X + )(X + )(X + ) = X + Swp (X + )(X + ) = X +
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationPentominoes. Pentominoes. Bruce Baguley Cascade Math Systems, LLC. The pentominoes are a simple-looking set of objects through which some powerful
Pentominoes Bruce Bguley Cscde Mth Systems, LLC Astrct. Pentominoes nd their reltives the polyominoes, polycues, nd polyhypercues will e used to explore nd pply vrious importnt mthemticl concepts. In this
More informationFurther applications of area and volume
2 Further pplitions of re n volume 2A Are of prts of the irle 2B Are of omposite shpes 2C Simpson s rule 2D Surfe re of yliners n spheres 2E Volume of omposite solis 2F Error in mesurement Syllus referene
More informationDATABASDESIGN FÖR INGENJÖRER - 1056F
DATABASDESIGN FÖR INGENJÖRER - 06F Sommr 00 En introuktionskurs i tssystem http://user.it.uu.se/~ul/t-sommr0/ lt. http://www.it.uu.se/eu/course/homepge/esign/st0/ Kjell Orsorn (Rusln Fomkin) Uppsl Dtse
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More information0.1 Basic Set Theory and Interval Notation
0.1 Bsic Set Theory nd Intervl Nottion 3 0.1 Bsic Set Theory nd Intervl Nottion 0.1.1 Some Bsic Set Theory Notions Like ll good Mth ooks, we egin with definition. Definition 0.1. A set is well-defined
More informationFirm Objectives. The Theory of the Firm II. Cost Minimization Mathematical Approach. First order conditions. Cost Minimization Graphical Approach
Pro. Jy Bhttchry Spring 200 The Theory o the Firm II st lecture we covered: production unctions Tody: Cost minimiztion Firm s supply under cost minimiztion Short vs. long run cost curves Firm Ojectives
More informationForensic Engineering Techniques for VLSI CAD Tools
Forensi Engineering Tehniques for VLSI CAD Tools Jennifer L. Wong, Drko Kirovski, Dvi Liu, Miorg Potkonjk UCLA Computer Siene Deprtment University of Cliforni, Los Angeles June 8, 2000 Computtionl Forensi
More informationQuick Guide to Lisp Implementation
isp Implementtion Hndout Pge 1 o 10 Quik Guide to isp Implementtion Representtion o si dt strutures isp dt strutures re lled S-epressions. The representtion o n S-epression n e roken into two piees, the
More informationPure C4. Revision Notes
Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd
More informationVectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a.
Vectors mesurement which onl descries the mgnitude (i.e. size) of the oject is clled sclr quntit, e.g. Glsgow is 11 miles from irdrie. vector is quntit with mgnitude nd direction, e.g. Glsgow is 11 miles
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint
More informationH SERIES. Area and Perimeter. Curriculum Ready. www.mathletics.com
Are n Perimeter Curriulum Rey www.mthletis.om Copyright 00 3P Lerning. All rights reserve. First eition printe 00 in Austrli. A tlogue reor for this ook is ville from 3P Lerning Lt. ISBN 78--86-30-7 Ownership
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationClause Trees: a Tool for Understanding and Implementing Resolution in Automated Reasoning
Cluse Trees: Tool for Understnding nd Implementing Resolution in Automted Resoning J. D. Horton nd Brue Spener University of New Brunswik, Frederiton, New Brunswik, Cnd E3B 5A3 emil : jdh@un. nd spener@un.
More informationEquivalence Checking. Sean Weaver
Equivlene Cheking Sen Wever Equivlene Cheking Given two Boolen funtions, prove whether or not two they re funtionlly equivlent This tlk fouses speifilly on the mehnis of heking the equivlene of pirs of
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationWHAT HAPPENS WHEN YOU MIX COMPLEX NUMBERS WITH PRIME NUMBERS?
WHAT HAPPES WHE YOU MIX COMPLEX UMBERS WITH PRIME UMBERS? There s n ol syng, you n t pples n ornges. Mthemtns hte n t; they love to throw pples n ornges nto foo proessor n see wht hppens. Sometmes they
More informationGeometry 7-1 Geometric Mean and the Pythagorean Theorem
Geometry 7-1 Geometric Men nd the Pythgoren Theorem. Geometric Men 1. Def: The geometric men etween two positive numers nd is the positive numer x where: = x. x Ex 1: Find the geometric men etween the
More informationGENERAL OPERATING PRINCIPLES
KEYSECUREPC USER MANUAL N.B.: PRIOR TO READING THIS MANUAL, YOU ARE ADVISED TO READ THE FOLLOWING MANUAL: GENERAL OPERATING PRINCIPLES Der Customer, KeySeurePC is n innovtive prout tht uses ptente tehnology:
More information84 cm 30 cm. 12 in. 7 in. Proof. Proof of Theorem 7-4. Given: #QXY with 6 Prove: * RS * XY
-. Pln Ojetives o use the ie-plitter heorem o use the ringle-ngle- isetor heorem Emples Using the ie-plitter heorem el-worl onnetion Using the ringle-ngle- isetor heorem Mth kgroun - Wht ou ll Lern o use
More informationRotating DC Motors Part II
Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors
More informationStyleView SV32 Change Power System Batteries
F the ltest User Instlltion Guie n StyleLink Softwre Downlo plese visit: Enontrrá l versión más reiente el mnul e instlión el usurio y el softwre e StyleLink en: Si vous souhitez téléhrger le ernier mnuel
More informationSolution to Problem Set 1
CSE 5: Introduction to the Theory o Computtion, Winter A. Hevi nd J. Mo Solution to Prolem Set Jnury, Solution to Prolem Set.4 ). L = {w w egin with nd end with }. q q q q, d). L = {w w h length t let
More informationOxCORT v4 Quick Guide Revision Class Reports
OxCORT v4 Quik Guie Revision Clss Reports This quik guie is suitble for the following roles: Tutor This quik guie reltes to the following menu options: Crete Revision Clss Reports pg 1 Crete Revision Clss
More informationAnswer, Key Homework 10 David McIntyre 1
Answer, Key Homework 10 Dvid McIntyre 1 This print-out should hve 22 questions, check tht it is complete. Multiple-choice questions my continue on the next column or pge: find ll choices efore mking your
More information2 DIODE CLIPPING and CLAMPING CIRCUITS
2 DIODE CLIPPING nd CLAMPING CIRCUITS 2.1 Ojectives Understnding the operting principle of diode clipping circuit Understnding the operting principle of clmping circuit Understnding the wveform chnge of
More informationPrinter Disk. Modem. Computer. Mouse. Tape. Display. I/O Devices. Keyboard
CS224 COMPUTER ARCHITECTURE & ORGANIZATION SPRING 204 LAYERED COMPUTER DESIGN. Introdution CS224 fouses on omputer design. It uses the top-down, lyered, pproh to design nd lso to improve omputers. A omputer
More information32. The Tangency Problem of Apollonius.
. The Tngeny olem of Apollonius. Constut ll iles tngent to thee given iles. This eleted polem ws posed y Apollinius of eg (. 60-70 BC), the getest mthemtiin of ntiquity fte Eulid nd Ahimedes. His mjo wok
More informationOverview of IEEE Standard 91-1984
Overview of IEEE tnr 9-984 Explntion of Logi ymols emionutor Group DYZA IMPOTANT NOTICE Texs Instruments (TI) reserves the right to mke hnges to its prouts or to isontinue ny semionutor prout or servie
More informationUnit 5 Section 1. Mortgage Payment Methods & Products (20%)
Unit 5 Setion 1 Mortgge Pyment Methos & Prouts (20%) There re tully only 2 mortgge repyment methos ville CAPITAL REPAYMENT n INTEREST ONLY. Cpitl Repyment Mortgge Also lle Cpitl & Interest mortgge or repyment
More informationHow To Balance Power In A Distribution System
NTERNATONA JOURNA OF ENERG, ssue 3, ol., 7 A dynmilly S bsed ompt ontrol lgorithm for lod blning in distribution systems A. Kzemi, A. Mordi Koohi nd R. Rezeipour Abstrt An lgorithm for pplying fixed pitor-thyristorontrolled
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More informationRIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS
RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is
More information