ECE 2030 -- Introduction to Computer Engineering EXAM #4 SOLUTION April 20, 2007 Page 1 of 2 Problem 1. (8 points) 0/01/11/00/00/01/0C A BA. (4 points) 00 110 D 1 1A 0 B0C001 B. (4 points) Problem 2. (15 points) D2 D1 S2 S1 S0 X S2 S1 S0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 1 0 0 1 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 00 01 11 10 S0X S2S1 00 0 0 0 1 01 0 1 x x 11 0 1 x x 10 1 1 0 1 S2S1 S0X 00 0 1 0 0 01 0 1 x x 11 1 0 x x 10 100000 01 11 10 00 01 11 10 S0X S2S1 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 XECE 2030 -- Introduction to Computer Engineering EXAM #4 SOLUTION April 20, 2007 Page 2 of 2 Problem 3. (13 points) Memory Configuration Total bits in memory # of memory address lines # of memory data lines Size of individual memory ICs Number of memory ICs needed 4G x 8 32G = 235 32 8 1G x 1 4 x 8 = 32 2M x 4 223 21 4 512K x 1 4 x 4 = 16 64M x 16 1G = 230 26 16 1M x 4 64 x 4 = 256 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) X Y Z rwe imm en imm va auena/s luen LF (3210)suenSTstenlden r/w msel Description ex) 15 7 10 1 0 x 1 1 0 x 0 x 0 0 x 0 R10 = R15 – R7 a) 30 13 19 0 0 0 0 0 0 0 0 0 1 0 0 1 M[R30] = R13 4 x 22 1 0 x 0 x 0 x 0 x 0 1 1 1 R22 = M[R4] b) 22 16 22 1 0 x 0 x 1 0010 0 x 0 0 x 0 R22 = R22 R16 12 x 18 1 1 -8 0 x 0 x 1 00 0 0 x 0 R18 = R12 · 256 12 x 19 1 1 -5 0 x 0 x 1 00 0 0 x 0 R19 = R12 · 32 c) 18 19 18 1 0 x 1 0 0 x 0 x 0 0 x 0 R18 = R18+R19 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. X Y X Y 0 0 0 = LF0 1 0 1 = LF1 0 1 0 = LF2 1 1 0 = LF3 For c1) and c2), multiplication by 2n is equivalent to shifting left by n places. Either logical shift (ST=00) or arithmetic shift (ST=01) may be used for shifting
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