1. COSC 3330/6308 First 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. Answer v'w' v'w vw vw' x'y' x'y xy xy'
v w x y + w y + w' y + v w x + y

5. Answer v'w' v'w vw vw' x'y' x'y 1 xy xy'
x'y 1 xy
xy'
v w x y + w y + w' y + v w x + y

6. Answer v'w' v'w vw vw' x'y' x'y 1 xy xy'
x'y 1 xy
xy'
v w x y + w y + w' y + v w x + y

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 x'y' x'y xy xy' w' 1 w w' x' y + w x y' + w' x y' + x' y'w

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. Answer w' x' y
(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 decimal value – 8 in signed arithmetic?
1000

19. Fourth question (II)
What would be its result of adding 0100 to 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?
= 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
For 1001 we do
not(1001) + 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
x'y'
x'y xy
xy' i i'

27. Answer
Draw the Karnaugh maps for these flip-flops:
Least significant bit y
x'y'
x'y xy
xy' i' 1 i

28. Answer
Draw the Karnaugh maps for these flip-flops:
Most significant bit x
x'y'
x'y xy
xy' i' i

29. Answer
Draw the Karnaugh maps for these flip-flops:
Most significant bit x
x'y'
x'y xy
xy' i' 1 i

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? 1
129 …
124 …

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

35. Sixth Question (III) 124 10000000000000000000000000… Sign is positive
124 …
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: 1
129ten …

37. Answer Multiply the two following floating-point numbers:
Compute the new sign
Add the exponents
Multiply the coefficients 1
129ten …

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

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