ECE 2030 -- Introduction to Computer Engineering EXAM #1 SOLUTION February 14, 2012 Page 1 of 3 Problem 1. (15 points total; 2 pts each, parts i-vi; 3 pts, part vii) (i) 1010110102 = 28 + 26 + 24 + 23+ 21 = 256 + 64 + 16 + 8 +2 = 34610 (ii) 84.2510 = 64 + 16 + 4 + 0.25 = 26 + 24 + 22 + 2-2 = 1010100.012 (iii) 101010112 = 010 101 0112 = 2538 (iv) 437.48 = 4x82 + 3x81 + 7x80 + 4x8-1 = 256 + 24 + 7 + 0.5 = 287.510 (v) 5CE16 = 0101 1100 11102 = 101110011102 (vi) 60110 = 2x256 + 5x16 + 9 = 2x162 + 5x161 + 9x160 = 25916 (vii) A19D16 = 1010 0001 1001 11012 = 001 010 000 110 011 1012 = 1206358 Problem 2. (20 points) A. (15 points) (i) (S) ΠM (0, 1, 2, 5, 6) = Σm (3, 4, 7) = A B C + A B C + A B C (ii) (U) Σm (0, 1, 2, 5) = A B C + A B C + A B C + A B C = A C + A B + B C (iii) (T) (A + B) (A + B + C) (B + C) = ( A B + A C + B A + B C ) (B + C) = A B C + A C B + B A + B A C + B C = A B + B C + A B C (iv) (Q) (B + C + D) + (A + B D) + A B D = B C D + A B D + A B D = B C D + A B D + A B D (v) (P) A + C + B + A C = A C + B ( A C ) = A C + B A + B C = A B + A C + B C B. (5 points) NOTE: Other solutions possible F = ( B + D + A C ) + B + C + ( A + D ) (A C + D ) = ( B + D ) ( A + C ) + B C ( A + D ) ( ( A C ) D ) = A B + A D + B C + C D + B C ( A + D ) (A + C ) D = A B + A D + B C + C D + B C D ( A + C D ) = A B + A D + B C + C D + A B C DECE 2030 -- Introduction to Computer Engineering EXAM #1 February 14, 2012 Page 2 of 3 A B C J 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 A B C K 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 Problem 3. (12 points) A. (6 points) J(A, B, C) = A C + B(A + C) K(A, B, C) = A B C + A(C + B) J(A, B, C) = Σm ( 0, 2, 3, 4, 6 ) K(A, B, C) = ΠM ( 0, 3, 5, 7 ) B. (6 points) F = (A B) (B + B C) (C + D) PHYSICAL gates: _1_ AND _1_ OR _3_ NAND _0_ NOR _4_ INVERTER Problem 4. (14 points) A. (6 points) A B C Fpu Fpd F 0 0 0 0 0 float 0 0 1 0 0 float 0 1 0 1 0 1 0 1 1 1 1 short 1 0 0 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 1 1 1 0 1 0 B. (8 points) G = (X Y + Z) (X + Y) = X Z + Y Z = Z (X + Y) + X Y Y X Z X X G Y Z Y X Y A B C D FECE 2030 -- Introduction to Computer Engineering EXAM #1 February 14, 2012 Page 3 of 3 00 01 11 10 CD AB 00 0 0 1 1 01 0 0 1 1 11 1 1 0 1 10 0 1 1 1 00 01 11 10 CD AB 00 0 0 1 1 01 0 0 1 1 11 1 1 0 1 10 0 1 1 1 00 01 11 10 CD AB 00 0 0 1 1 01 0 0 1 1 11 1 1 0 1 10 0 1 1 1 Problem 5. (14 points) A. (6 points) Prime implicants: A C ess A C D ess B C ess ess C D ess ess A B D ess ess A B D ess ess B. (8 points) Final SOP – version 1 Final SOP – version 2 Final POS F(A, B, C, D) = A C + C D + A B C + A B D minimal sum-of-products (SOP – version 1) F(A, B, C, D) = A C + B C + A B D + A C D minimal sum-of-products (SOP – version 2) F(A, B, C, D) = ( A + C ) ( B + C + D ) ( A + B + C + D ) minimal product-of-sums (POS) 00 01 11 10 CD AB 00 0 0 1 1 01 1 0 1 1 11 0 1 0 1 10 0 1 1
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