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COSC 3330/6308 First Review Session

COSC 3330/6308 First Review Session. Fall 2012. First Question. Simplify the following Boolean expression. w (v x y + y) +w' y + v w x + y. Answer. w (v x y + y) +w' y + v w x + y = v w x y + w y + w' y + v w x + y. v w x y + w y + w' y + v w x + y. Answer.

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COSC 3330/6308 First Review Session

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  1. COSC 3330/6308First Review Session Fall 2012

  2. First Question • Simplify the following Boolean expression. • w (v x y + y) +w' y + v w x + y

  3. Answer • w (v x y + y) +w' y + v w x + y =v w x y + w y + w' y + v w x + y

  4. v w x y +w y + w' y + v w x + y Answer

  5. v w x y +w y + w' y + v w x + y Answer

  6. v w x y +w y + w' y + v w x + y Answer

  7. Answer • By Karnaugh maps: • v w x + y • By algebra: • w (v x y + y) +w' y + v w x + y =v w x y + w y + w' y + v w x + y =y (v w x + w + w' + 1) + v w x =y + v w x

  8. Second Question • Give a simplified implementation for the following expression using only NAND gates. • w' (x  y) + w x y' + x' y'

  9. Answer • w' (x  y) + w x y' + x' y' =w' (x' y + x y') + w x y' + x' y' =w' x' y + w' x y' + w x y' + x' y'

  10. Answer • w' x' y + w x y' + w' x y' + x' y' • w' x' + y'

  11. Answer • Using algebra: • w' (x  y) + w x y' + x' y' =w' (x' y + x y') + w x y' + x' y' =w' x' y + w' x y' + w x y' + x' y' =w' x' y + x y' (w' + w) + x' y' =w' x' y + x y' + x' y' =w' x' y + y' =w' x' y + w' x' y' + y' =w' x'( y + y') + y' = w'x' + y'

  12. Reminder • A NAND B = (A B)' = A' + B' • (NOT A) NAND(NOT B) ( A' B')' = A+ B • NOT(A NAND B) = (A B)'' = A B • (A NAND B) NAND (C NAND D) =((AB)' (CD)')' = AB + CD

  13. w' x' y Answer (w'x')' = w + x ((w + x)y)' = w'x' + y'

  14. Third Question (I) • What is the main reason for using a base plus displacement representation of memory addresses in an instructions set?

  15. Answer • What is the main reason for using a base plus displacement representation of memory addresses in an instructions set? • To be able to access a very large address space with as few bits as possible in order to keep instructions as short as possible • MIPS IS uses 5 bits to specify which register and 16 bits for the displacement

  16. Third Question (II) • What is the main reason for requiring all instructions to have the same length? • To allow the CPU to fetch the next instruction before the current instruction is decoded • Would not work if we had 16-bit and 32-bit instructions

  17. Fourth Question • Assume we have a very basic microprocessor doing 4-bit arithmetic. • How would you represent the decimal value – 8 in signed arithmetic?

  18. Answer • How would you represent the decimalvalue – 8 in signed arithmetic? • 1000

  19. Fourth question (II) • What would be its result of adding 0100 to 0101 assuming that the numbers being added were • Unsigned integers? • Signed integers?

  20. Answer • What would be its result of adding 0100 to 0101 assuming that the numbers being added were • Unsigned integers? • Signed integers? • 0100 + 0101 = 1001 • 1001 represents 9 in unsigned arithmetic • 1001 represents -7 in signed arithmetic

  21. Explanation • We use two complement's arithmetic • To change the sign of a positive value • Negate all its bits then add 1 • To change the sign of a positive value • Negate all its bits then add 1 • For 1001 we do • not(1001) + 1 = 0110 + 1 = 0111 = 7

  22. Fifth question • Implement a two-bit counter going to the cycle 00, 01, 10, 11, 00, … when its input is on. You may use the flip-flops and the gates of your choice.

  23. Answer • Implement a two-bit counter going to the cycle 00, 01, 10, 11, 00, … when its input is on. You may use the flip-flops and the gates of your choice. • How many flip-flops do we need?

  24. Answer • Implement a two-bit counter going to the cycle 00, 01, 10, 11, 00, … when its input is on. You may use the flip-flops and the gates of your choice. • How many flip-flops do we need? • Two because we have 22 =4 states

  25. Answer • Implement a two-bit counter going to the cycle 00, 01, 10, 11, 00, … when its input is on. You may use the flip-flops and the gates of your choice. • How many flip-flops do we need? • Two because we have 22 =4 states

  26. Answer • Draw the Karnaugh maps for these flip-flops: • Least significant bit y

  27. Answer • Draw the Karnaugh maps for these flip-flops: • Least significant bit y

  28. Answer • Draw the Karnaugh maps for these flip-flops: • Most significant bit x

  29. Answer • Draw the Karnaugh maps for these flip-flops: • Most significant bit x

  30. Answer • The system equations are • y = y  i • x = x  iy = x(i' + y') + x' iy

  31. Answer • We will use T flip-flops for x and y • y will be triggered by input i • x will be triggered by input iy

  32. Answer

  33. Sixth Question (I) • Which decimal values are stored in the following single precision floating point numbers?

  34. Sixth Question (II) • Sign is negative • Exponent is 129 -127 = 2 • Coefficient is 1.01two = 1 + ¼ • - (1 + ¼) 22 = -5

  35. Sixth Question (III) • Sign is positive • Exponent is 124 – 127 = -3 • Coefficient is 1.1two = 1+ ½ = 1.5 • 1.5 2-3 =1.5/8 = 3/16

  36. Seventh Question • Multiply the two following floating-point numbers:

  37. Answer • Multiply the two following floating-point numbers: • Compute the new sign • Add the exponents • Multiply the coefficients

  38. Answer • Multiply the two following floating-point numbers: • Compute the new sign: Negative • Add exponents: (129 – 127) + (129- 127) • Multiply the coefficients: 11

  39. Answer Result is -1 24 Sign bit is 1 Biased exponent is 127 + 4 =131 Coefficient is 1

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