1. Perform the following additions in 6 bit twos complement . Circle the 6 bit a
ID: 669379 • Letter: 1
Question
1. Perform the following additions in 6 bit twos complement. Circle the 6 bit answer for each. Show your work (carries and sums). Additionally, indicate for each: 1) whether there was a carry out of one and 2) whether there was overflow.
a. 111001 b. 010001
+110111 +010011
------- -------
c. 010011 d. 101101
+101011 +100011
------- -------
2. Give the base 10 values of the two numbers being added and the sum from 1a.
3. a. Use the bit shifting technique to multiply the following positive (not twos complement) binary number by 16.
1101
b. Explain how you found the answer.
c. What is the base 10 value of the original number?
d. What is the base 10 value of the product?
4. a. Use the bit shifting technique to divide the following positive (not twos complement) binary number by 8.
10110
b. Explain how you found the answer.
c. What is the base 10 value of the original number?
d. What is the base 10 value of the quotient?
5. Consider a 13 bit floating point number representation with the following fields:
1 bit sign
4 bit exponent (excess 8)
8 bit fraction
Convert the following floating point number using the given representation into its base 10 equivalent. Show your work.
Convert the following floating point number using the given representation into its base 10 equivalent. Show your work.
c. Do a binary addition of the two numbers in a. and b. above. Show your work.
d. Take the resulting sum from c. and place it back into the floating point format. Is there a problem with the result? Explain.
e. Convert the following base 10 number into the floating point representation (normalized) described above. Show your work.
-9.75
f. Show the computer representation of the largest number that can be represented using this format.
g. What is the base 10 equivalent for your answer to f. Show your work.
Explanation / Answer
this question has multiple subparts. Please post 1 more question for the same.
1
a)
111001 b. 010001
+110111 +010011
------- -------
1110000 100100
there will be overflow and carry bit set for part a
there will be no overflow and carry bit set for part b
c. 010011 d. 101101
+101011 +100011
------- -------
111110 1010000
there will be no overflow and carry bit set for part c
there will be overflow and carry bit set for part d
2)
it is 57 and 55. Result is 112.
3)
multiply by 16 means right shift by 4.
so 1101 will become 11010000
base 10 will be 208
base 10 of the product will be 208
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