HW #5
SOLUTIONS
1/7
ECSE2610 Computer Components and Operations
SOLUTIONS HW # 5
Due March 3, 2010
1. (10 points) Number Systems, Codes, Computer Arithmetic
a.
(1 point each, 6 points total)
i.
Convert 0010.1111
2
to a decimal number.
____2.9375
10
__________
10.11110000
ii.
Convert 01101010
2
to a hexadecimal number.
____006A
16
______________
01101010
iii.
Convert AB
16
to a binary number.
____10101011
2
_________
AB
iv.
Convert 83.76
10
to a binary number.
____01010011.11000010
2
_
83.7578
v.
Convert 83
10
to a BCD number
.
____1000 0011
BCD
_______
8
3
vi.
Convert 01010101
2
to its 2scomplement negative.
____10101011
2scomp
______
81
81
b.
(3 points) Add using 2scomplement arithmetic 01000111
2
scomp
 10101110
2
scomp
.
Does this result in an overflow? Explain your answer.
Result =
________10011001
2 scomp
________________________________
Overflow =
Yes. Subtracting a negative from a positive should not result in negative.
c.
(1 point) Convert the numbers of part (b) to decimal and add them. What answer do you
get?
S
u
m
=
________71 – (82) = 153
Overflow =
because 153 > 127, the largest 8bit 2’s comp number/
.
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SOLUTIONS
2/7
2.
(20 points)
a.
(10 points)
Simplify the following Boolean function to sumofproducts form using the laws
of Boolean algebra. Show each step.
You do not need to identify the name of the property
you use.
i
= X’Z + Y’(X’Z)’
= X’Z + Y’
by Rule 11D (Jan 28)
A more pedestrian approach is to find the SOP canonical form and recombine minterms (as with Kmap):
X’Z + Y’Z’ + XY’
= X’Y’Z + X’YZ + XY’Z’ +X’Y’Z’ + XY’Z + XY’Z
X’Y’Z + X’Y’Z’ = X’Y’,
XY’Z’ + XY’Z = XY’,
X’Y’Z + X’YZ = X’Z
and X’Y + XY’ = Y’
Therefore G = X’Z + Y’
TT#1 has columns for each of the ORed terms, TT#2 has columns for each AND term.
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 Spring '08
 Ji
 Boolean Algebra, Binary numeral system, Decimal, Karnaugh map, Canonical form, prime implicants

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