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1 7/31/ Boolen (Switching) Algebr Review Good Review book BeBop to the Boolen Boogie: An Unconventionl Guide to Electronics, 2 nd ed. by Clive Mxwell Hightext Publictions Inc. from Amzon.com for pprox. $27.14 ISBN Theorems nd Postultes + 0= 0 = 0 = ( ) ( ) + bc = + b + c + = 1 = 0 + = = + 1= 1 1 = ( ) + b = + b = + b = + b + b = ( ) b ( ) ( ) ( ) ( + ) = ( + ) ( + c) b + b = + b + b = b + bc = b + c + b + b c b + b= b b = + b b + c + bc = b + c + ( b) ( + c) ( b + c) = ( + b) ( + c) The bove tble is vilble for downlod in printble version from my homepge! While you re t it, print out the No. of 1's document which you will need for the Quine-McCluskey topics lter. In our review of Boolen Algebr, I ll be minly teching by exmple nd will be using the Theorems nd Postultes shown bove. A certin level of lgebr experience is ssumed, so the first six rows of rules re not going to be introduced.

2 7/31/ Boolen Exmple 1 (Truth tble): Problem Sttement: Use Truth tble to Prove : f (, ) b = + b = + b b b + b = + b = = = = 1 QE.. D.( Thus it is proven) Boolen Exmple 2 Problem Sttement Simplify the following switching function using: A + AB = A f (,b,c) b + = + b + c = + b + c = + c = = m - no gtes 1 level circuit - A WIR E! ( 4,5,6, 7 ) = M( 0,1,2, 3)

3 7/31/ This problem cn be solved nother wy. Boolen Exmple 3 Problem Sttement: Simplify the problem from the lst exmple by fctoring 1 st. f (,b,c ) = b+ + b+ c = x+ 1= 1 b 1 b c = ns Note tht the result for both exmples is the sme. Boolen Exmple 4 Problem Sttement: Simplify the following switching expression. (, b, c) Simplify f = c + bc fctor out the c = c ( + b ) ( b) = c + = c + bc ns

4 7/31/ MultiSIM Exmple 2-1 Problem Sttement: Use MultiSIMs Logic Converter to verify the results from the circuit shown. Once the circuit hs been built, connect the inputs nd output to the Logic Converter. Be creful to insure tht the inputs re connected with the MSB on the frthest input to the left. The input t the fr right end is where the circuit output is connected. This device is only used with circuits which only hve single output. Once the circuit is connected, select the Circuit to Tble button (see bove right). As cn be seen, the resulting tble mtches the tble on the previous pge.

5 7/31/ Boolen Exmple 5 Problem Sttement: Prove tht the following Theorem is true: ( ) ( ) ( ) ( ) ( ) f X,Y,Z = X + Y X + Z = X + YZ X + Y X + Z = X + YZ XX + XZ + XY + YZ = X + YZ X X + XZ + XY + YZ = X + YZ X X + XY X + YZ = X + YZ X + YZ = X + YZ Boolen Exmple 6 Problem Sttement: Simplify f( x, y,z) = m( 0, 3, 4,5, 7) ( x, y, z) f = x y z + xyz + xy z + xyz + xyz = xyz + x y z + xy z + xyz + xyz = xyz + (x + x) y z + x (y + y) z (+ = 1) (+ = 1) = xyz + y z + xz = yz + xz + xyz b b c (b+ bc = b+ c) = yz+ xz + yz reorder ns

6 7/31/ Note the second to lst line of the lst exmple. = yz+ x z + xyz b b c theorem (b + bc = b + c) = y z + xz + yz An ttempt hs been mde to show how tht prt of the expression mtches one of the theorems. Note how the (x) hs been ssigned to (b) in the theorem nd the ( X ) hs been ssigned to the ( b ) in the theorem. Let s tke look t how this ssignment might chnge if the problem chnged slightly: All tht hppened ws the (x) nd the ( X ) trded plces. Note tht the ssignment of (b) in the theorem is now mde to the ( X ) vice the (x) nd note the chnge in the nswer. The student might be tempted to ssign the ( b ) from the theorem to the ( X ), but this would be wrong. = y z + x z+ x yz b b c theorem ( b + bc = b + c ) = y z + x z + yz

7 7/31/ The following is n exmple of how n SOP (Sum of Products) expression cn be converted into n expression which would only require NAND gtes to implement by using De Morgn s Theorem. Boolen Exmple 7 (NAND ONLY LOGIC) Problem Sttement: Implement f( x, y,z) = y z + xz + yz in NAND only logic. st 1, double negte both sides of the eqution. We know tht = so if we double negte both sides, the expression will remin UNCHANGED! f(x, y,z) = y z + xz + yz Once this hs been done, perform DeMorgn's on the bottom negtion, only pplying the theorem to the signs of the expresssion ( y z) ( xz) ( yz) = the resulting expression cn be inplemented by using ONLY NAND gtes. ANSWER

8 7/31/ The designer cn use the sme technique to implement POS (Product of Sums) expression s NOR only circuit. Boolen Exmple 8 (NOR ONLY LOGIC) Problem Sttement: Implement the following expression in NOR only logic. f( x, y,z) (x y + z)( y ( x, y, z) = + + = f (x + y + z)(y + z) As before, only pply De'Morgn's to the bottom negtion nd then only to the signs in the expression. z) = (x + y + z) + (y + z) As noted erlier, the min-term list cn esily be converted into n lgebric SOP expression. If POS expression is desired, express the min-term list s mx-term list. Then the simplifiction cn be crried out. Boolen Exmple 9 Problem Sttement: Implement nd simplify the following expression in POS formt. f ( x, y, z) = m(0,3, 4,5,7) = M(1, 2, 6) = ( x + y + z)( x + y + z)( x + y + z) b b ( + b)( + b) = = ( x + y + z)( y + z) Answer This exmple lso shows simplifiction technique which cn be very helpful. Note tht the exmple equtes + y z to the in the theorem ( ) ( ) + b + b =.

9 7/31/ MultiSIM Exmple 2-2 Problem Sttement: Use MultiSIMs Logic Converter to verify the proof: Prove : f (, b) = + b = + b Open the Logic Converter nd select the two bits to the left s shown bove. Then type the eqution in the entry re t the bottom of the device. Use ( ) when negtion is required (see bove right). Once the eqution is complete, select the A B Tble button. The resulting tble is shown below:

10 7/31/ Boolen Exmple 1 0 Problem Sttement: Find the miniml POS Boolen Expression f ( xyz,, ) = ( x + z) ( x + y) ( x + y + z) ( x + y + z) reorder nd pply two different theorems x y x y z b = ( + b) = x + z x y z + + b b c ( + b) + b+ c = ( + b)( + c) = ( x + y) ( x + z) ( y + z )

11 7/31/ MultiSIM Exmple 2-3 Problem Sttement: Use MultiSIM s Logic Converter to verify the simplifiction from the previous exmple. In order to progrm this exmple into MultiSIM you first need to convert the (x, y, z) switching list into n (A, B, C) switching list.? ( xyz,, ) = ( + ) ( + ) ( + + ) ( + + ) = ( x + y) ( x + z) ( y + z) f x z x y x y z x y z? ( ABC,, ) ( ) ( ) + = ( ) f = A+ C A+ B A+ B + C A+ B C A+ B A+ C B + C Next, open the Logic Converter, select the A, B, nd C columns in the tble nd progrm the eqution into the entry box. Don t forget to use the ( ) s the negtion symbol. Once you progrm the eqution into the converter, select the A B Tble button. The result cn be seen in the figure bove. simple Finlly, select the " Tble A B " button. The simplified eqution will pper in the eqution entry re s shown below. Note tht the result is s predicted.

12 7/31/ Boolen Exmple 1 1 Problem Sttement: Simplify the following expression: (,,, ) = ( ) ( ) + ( ) ( ) f A B C D AC AD CD BD A C A D C D B D C A D B = ( A C) ( A D) ( C D) ( B D) = = AA + A D + AC + CD + BC + C D + BD + DD 0 0 = AD+ AC + CD + C D + BC + BD remember D ( C + C) 1 tht ny vrible AND'd with "1" is equl to tht vrible, so D = A D + AC + B C + B D + D b + b = + b B + D = A D + AC + BC + B + D = + + AC + + ( 1 ) reorder.... A D D B C B b b + b = = + b = = = D D B = D + B + AC nswer Implement s NAND only expression = D + B + AC = ( D) ( B) ( AC) ( D) ( B) ( AC) = nswer

13 7/31/ MultSIM Exmple 2-4 Problem Sttement: Verify the simplifiction in the previous exmple using Multisim.