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COSC 3330/6308 First Review Session Fall 2012
COSC 3330/6308First 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
Answer
v w x y +w y + w' y + v w x + y
v'w' v'w vw vw'
x'y'
x'y
xy
xy'
Answer
v w x y +w y + w' y + v w x + y
v'w' v'w vw vw'
x'y' 0 0 0 0
x'y 1 1 1 1
xy 1 1 1 1
xy' 0 0 1 0
Answer
v w x y +w y + w' y + v w x + y
v'w' v'w vw vw'
x'y' 0 0 0 0
x'y 1 1 1 1
xy 1 1 1 1
xy' 0 0 1 0
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
Second Question
Give a simplified implementation for the following expression using only NAND gates.
– w' (x y) + w x y' + x' y'
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'
Answer
w' x' y + w x y' + w' x y' + x' y'
w' x' + y'
x'y' x'y xy xy'
w' 1 1 0 1
w 1 0 0 1
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'
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
Answer
w'
x'
y
(w'x')' = w + x
((w + x)y)' = w'x' + y'
Third Question (I)
What is the main reason for using a base plus displacement representation of memory addresses in an instructions set?
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
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
Fourth Question
Assume we have a very basic microprocessor doing 4-bit arithmetic.
How would you represent the decimal value – 8 in signed arithmetic?
Answer
How would you represent the decimalvalue – 8 in signed arithmetic?
– 1000
Fourth question (II)
What would be its result of adding 0100 to 0101 assuming that the numbers being added were
Unsigned integers? Signed integers?
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
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
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.
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?
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
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
Answer
Draw the Karnaugh maps for these flip-flops:
– Least significant bit y
x'y' x'y xy xy'
i
i'
Answer
Draw the Karnaugh maps for these flip-flops:
– Least significant bit y
x'y' x'y xy xy'
i' 0 1 1 0
i 1 0 0 1
Answer
Draw the Karnaugh maps for these flip-flops:
– Most significant bit x
x'y' x'y xy xy'
i'
i
Answer
Draw the Karnaugh maps for these flip-flops:
– Most significant bit x
x'y' x'y xy xy'
i' 0 0 1 1
i 0 1 0 1
Answer
The system equations are – y = y i– x = x iy = x(i' + y') + x' iy
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
Answer
Sixth Question (I)
Which decimal values are stored in the following single precision floating point numbers?
1 129 010000000000000000000000000…
0 124 10000000000000000000000000…
Sixth Question (II)
Sign is negative Exponent is 129 -127 = 2 Coefficient is 1.01two = 1 + ¼ - (1 + ¼) 22 = -5
1 129 0100000000000000000000000000…
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
0 124 10000000000000000000000000…
Seventh Question
Multiply the two following floating-point numbers:
1 129ten 000000000000000000000 …
0 129ten 000000000000000000000 …
Answer
Multiply the two following floating-point numbers:
– Compute the new sign– Add the exponents– Multiply the coefficients
1 129ten 000000000000000000000 …
0 129ten 000000000000000000000 …
Answer
Multiply the two following floating-point numbers:
– Compute the new sign: Negative– Add exponents: (129 – 127) + (129- 127)– Multiply the coefficients: 11
1 129ten 000000000000000000000 …
0 129ten 000000000000000000000 …
Answer
Result is -1 24
– Sign bit is 1– Biased exponent is 127 + 4 =131– Coefficient is 1
1 131ten 0000000000000000000000…
