ABCFFpuFpd 000001 001float00 010110 011floaECE 2030 -- Introduction to Computer Engineering EXAM #1 SOLUTION February 2, 2007 Page 1 of 3 Problem 1. (16 points) A. (7 points) (i) 1001011012 = 28 + 25 + 23 + 22+ 20 = 256 + 32 + 8 + 4 +1 = 30110 (ii) 18210 = 128 + 32 + 16 + 4 +2 = 27 + 25 + 24 + 22 + 21 = 101101102 (iii) 100111012 = 010 011 1012 = 2358 (iv) 6378 = 6x82 + 3x81 + 7x80 = 6x64 + 3x8 + 7x1 = 384 + 24 + 7 = 41510 (v) 6D916 = 0110 1101 10012 = 110110110012 (vi) 70310 = 2x256 + 11x16 + 15 = 2x162 + 11x161 + 15x160 = 2BF16 (vii) B5D916 = 1011 0101 1101 10012 = 001 011 010 111 011 0012 = 1327318 B. (9 points) (i) (A + B) (A + C) (B + C) = (A A + A C + B A + B C) (B + C) = A C B + A C C + B A B + B A C + B C B + B C C = A B C + A B + A B C + B C = A B (C + 1 + C) + B C = A B + B C (U) (ii) (A + C + D) (B + C D) = (A + C D) (B + C + D) = A B + A C + A D + C D B + C D C + C D D = A B + A C + A D + C D (B + 1) = A B + A C + A D + C D (Q) (iii) A + C + B + (D + A) = A C + B (D + A) = A C + B (D A) = A C + A B D (T) Problem 2. (10 points) A. (4 points) For each function listed below, complete the corresponding truth table. F(A, B, C) = A C + C(A + B) G(A, B, C) = A B C + B(C + A) A B C G 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 A B C F 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0ECE 2030 -- Introduction to Computer Engineering EXAM #1 SOLUTION February 2, 2007 Page 2 of 3 B. (6 points) F = ( A B + C ) ( B + C + A D ) + B D + ( A + B ) = ( A B + C ) + ( B + C + A D ) + ( B D ) ( A + B ) = ( A B ) ( C ) + ( B + C ) ( A D ) + ( B + D ) ( A + B ) = ( A B ) ( C ) + ( B + C ) ( A + D ) + ( B + D ) ( A + B ) = ( A B C ) + ( B + C ) ( A + D ) + ( B + D ) ( A + B ) = A B C + B A + B D + C A + C D + B A + B B + D A + D B = A (B C + B + C + B + D) + B (D + D) + C D = A (1 + B C + C + D) + B (1) + C D = A + B + C D Problem 3. (14 points) A. (6 points) FPU = B C + A C FPD = B (A + C) A B C F Fpu Fpd 0 0 0 0 0 1 0 0 1 float 0 0 0 1 0 1 1 0 0 1 1 float 0 0 1 0 0 0 0 1 1 0 1 short 1 1 1 1 0 1 1 0 1 1 1 1 1 0 FB C A B A C CECE 2030 -- Introduction to Computer Engineering EXAM #1 SOLUTION February 2, 2007 Page 3 of 3 B. (8 points) X Y Y Z X Y Y Z G = Y(Z + X Y) = Y + Z(X + Y) J = (R + T) (S + T) = R T + S T T T T J S T S R GR Problem 4. (10 points) A. (4 points) F = (A + B) (B + B C) (D E) A BCD E F D E C B A B. (6 points) F # of gates: _5__ NOR _6__
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