Exercise_Midterm1_Solutions

# Exercise_Midterm1_Solutions - BCD’ Y1...

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SOLUTIONS: 1. a. F (A,B,C) = A’ B’ C’ + A’ B’ C + A B’ C’ + A B’ C + A B C’ + A B C [Distributive] = A’B’ (C’ + C) + AB’ (C’ + C) + AB (C’ + C) [ Complementarity] = A’B’ + AB’ + AB [Idempotency] = A’B’ + AB’ + AB’ +AB [ Distributive] = B’ (A’ + A) + A (B’ + B) [ Complementarity] = B’ + A b. F(A,B) = (A’+B) ( A+B) [Distributive] = AA’ + A’B+ AB+ BB [Complementarity ] = A’B+AB+BB [Distributive] = B ( A’+A+1) [Complementarity] = B . 1 [ identity ] = B 2. a. a’b’ + a’c’ + bc’ [Complementarity & Distributive] = a’b’ + (a’c’b’+a’c’b) + bc’ [combining] = (a’b’ + a’b’c’) + (bc’a’ + bc’) [ commutativity & associativity] = a’b’ + bc’ [covering] b. ( a’+ b)(a’+ c’)(b’+ c’) [Associative] = ( a’+ b)(b’+ c’)(a’+ c’)[commutativity] = (a’+ b)(b’+ c’)[consensus]

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3. a) Truth Table b) SOP Y0 = (A ’B’C’D)+(A’B’CD’)+(A’BC’D’)+(A’BCD)+(AB’C’D’)+(AB’CD)+(ABC’D)+(A

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Unformatted text preview: BCD’) Y1 = (A’B’CD)+(A’BC’D)+(A’BCD’)+(A’BCD)+(AB’C’D)+(AB’CD’)+(AB’CD)+(AB C’D’)+(ABC’D)+(ABCD’) Y2 = ABCD c) POS Y0 = (A+B+C+D)(A+B+C’+D’)(A+B’+C+D’)(A+B’+C’+D)(A’+B+C+D’)(A’+B+C’+ D) (A’+B’+C+D)(A’+B’+C’+D’) Y1 = ( A+B+C+D)(A+B+C+D’)(A+B+C’+D)(A+B’+C+D)(A’+B+C+D)(A’+B’+C’+D’ ) Y2 = πM (0,1,2,3,4,5,6,7,8,9,10,11,12,13,14) 4. a. The simplified function is: F = B ’ D + A ’ B ’ C ’ + ACD + ABD ’ b. 5. a. F(A,B,C,D) = (A + B ’ + C ’ )(A + B + C) b. F(A,B,C,D) = D (A ’ + C ’ ) 6. a. Since the universal gates {AND, OR, NOT can be constructed from the NAND gate, it is universal. b. The given function is universal. 7. a. Hence, the result is proved. The same result will be obtained for Expansion for y and Expansion for z (Try it!!!) b. Thus, proved. Same result will be obtained for Expansion for y. 8. f ( a , b , c , d ) = S m (0,1, 2, 3, 4, 5, 7,8,12) + S d (10,11)...
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Exercise_Midterm1_Solutions - BCD’ Y1...

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