ECE 2030 Introduction to Computer Engineering April 20 2007 Problem 1 8 points EXAM 4 SOLUTION Page 1 of 2 0 0 1 0 1 0 A 4 points A B 1 1 C 0 0 1 0 B 4 points 0 0 1 1 0 A 0 D 1 C 0 B 0 0 1 0 Problem 2 15 points S0X S0X S2S1 00 01 11 10 00 0 0 0 1 01 0 1 x 11 0 1 10 1 1 S2S1 X S2 S1 S0 00 01 11 10 0 0 0 0 0 0 1 00 0 1 0 0 0 0 0 1 0 1 0 x 01 0 1 x x 0 0 1 0 1 0 1 x x 11 1 0 x x 0 0 1 1 0 0 0 0 1 10 1 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 1 1 0 x x x 0 1 1 1 x x x 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 0 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 x x x 1 1 1 1 x x x D2 D1 S0X S2S1 S2 S1 S0 00 01 11 10 00 1 0 0 1 01 0 0 x x 11 0 0 x x 10 0 1 1 0 D0 D2 S1 X S0 X S2 S1 S0 D1 S2 S0 X S2 S0 X D0 S2 S1 X S2 S1 X ECE 2030 Introduction to Computer Engineering April 20 2007 EXAM 4 SOLUTION Page 2 of 2 Problem 3 13 points Memory Configuration Total bits in memory 4G x 8 32G 2 2M x 4 2 64M x 16 1G 2 of memory address lines of memory data lines Size of individual memory ICs Number of memory ICs needed 32 8 1G x 1 4 x 8 32 21 4 512K x 1 4 x 4 16 26 16 1M x 4 64 x 4 256 35 23 30 Problem 4 14 points ex R10 R15 R7 solution given below as an example a Fill in Description with the operation defined by the listed control signal values b R22 M R4 AND NOT R16 c R18 288 R12 multiplication hint 288 256 32 st en ld r w msel en x 0 0 x 0 R10 R15 R7 0 0 1 0 0 1 M R30 R13 x 0 x 0 1 1 1 R22 M R4 1 0010 0 x 0 0 x 0 R22 R22 R16 x 0 x 1 00 0 0 x 0 R18 R12 256 0 x 0 x 1 00 0 0 x 0 R19 R12 32 1 0 0 x 0 x 0 0 x 0 R18 R18 R19 imm imm au a s lu LF su ST en va en en 3210 en X Y Z rwe ex 15 7 10 1 0 x 1 1 0 x 0 a 30 13 19 0 0 0 0 0 0 0 4 x 22 1 0 x 0 x 0 22 16 22 1 0 x 0 x 12 x 18 1 1 8 0 12 x 19 1 1 5 18 19 18 1 0 x Description b c NOTES Don t care values may be assigned Other register numbers may be used for intermediate results For b2 the function X AND NOT Y X Y is defined by the truth table to the right For c note that 288 R12 256 32 R25 256 R12 32 R12 where multiplication n X 0 1 0 1 Y 0 0 1 1 XY 0 1 0 0 For c1 and c2 multiplication by 2 is equivalent to shifting left by n places Either logical shift ST 00 or arithmetic shift ST 01 may be used for shifting left LF0 LF1 LF2 LF3
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