Semester One Week3. Workshop - Data Representation
1. Find the decimal value represented by the binary sequence 101011012
in each of the following notations:
(a) Unsigned notation
(b) Signed magnitude notation
(c) Excess notation
(d) Two’s complement natation
Model Answer:
(a) Unsigned notation
101011012 = 128 + 32 + 8 + 4 + 1 = 173
(b) Signed magnitude notation
101011012 = −(32 + 8 + 4 + 1) = −45
(c) Excess notation
101011012 = unsignedvalue − 128 = 173 − 128 = 45
(d) Two’s complement natation
101011012 = (−128) + 32 + 8 + 4 + 1 = −83
2. Represent -85 in each of the following notations.
(a) Signed magnitude notation
(b) Excess notation
(c) Two’s complement nation
Model Answer:
(a) Signed magnitude notation
negative number implies the sign bit is 1.
11010101
85 = 01010101, hence -85=
(b) Excess notation
-85 + 128= 43 and 43 = 00101011. hence -85 = 00101011.
(c) Two’s complement nation we first find the two’ complement for 85 = 85 =
01010101. hence -85 = 10101011
3. What is the range of of number represented in 8-bit Tow’s complement
notation?
Model Answer:
positive range 0 to 27 − 1 and negative range - 27 to 0
1
4. State the advantages of Two’s complement over signed notation.
Model Answer:
signed notation have two representation of 0 and difficult to do arithmetic operation.
Two’s complement have one representation of 0 and all arithmetic operations can
be achieved by addition only.
5. Find the decimal value represented by the following single precision
floating point number.
1
10000101
1100 0000 0000 0000 0000 000
Model Answer:
Sign bit = 1 hence it is a negative number.
Biased exponent = 10000101 = 128 + 3 + 5 = 136.
Real exponent = 136 - 127= 9
Normalized mantissa = 1100 0000 0000 0000 0000 000.
Real mantissa = 1.11
hence the number represented is (−)1.11 × 29 = (−)1110000000 = (−)(29 + 28 + 27 )
6. Represent the following decimal numbers in single precision floating
point format.
(a)
−1
64
(b) 125.875
Model Answer:
(a)
−1
64
−1
is a negative number hence, the
64
1
=
2−5 = 0.00001 = 1.0 × 2− 5
64
sign bit is 1
Real exponent = -5
Biased exponent = -5 +127 = 122 = 01111010
Real mantissa = 1.0
Normalized mantissa = 0000 0000 0000 0000 000
hence the floating point representation of −1
is
64
1011110100000000000000000000
(b) 125.875
125.875 is positive, hence the sign bit is 1
125.875 = 01111101.111 = 1.111101111 × 26
Real exponent = 6
Biased exponent = 6 + 127 = 133 = 10000101
Real mantissa = 1.111101111
Normalized mantissa = 1111 0111 1000 0000 000
hence the floating point representation is:
0 10000101 111 0111 1000 0000 000
2
7. What is the difference between positive overflow and positive underflow
overflow in single precision floating point representation.
Model Answer:
Positive overflow occur if the number represented if greater then largest number
represented in this representation(2127 (2 − 2−23 ))
Positive underflow occur if the number represented if greater then largest number
represented is between 0 and the smallest positive number (2−127 )
8. (a) Represent 0.4 in the following floating point representation: sign(bit),
exponent(4-bits, bias 7) and a 6-bits normalised mantissa.
(b) convert the result back to decimal
(c) find the relative error of representing 0.4 in the floating point
representation.
Model Answer:
(a) It is a positive number implies sign bit =1.
convert 0.4 to binary the usual way by multiplying by 2 and extract the
integer part. We will keep doing that till we have a 6 bit sequence required
for the mantissa.
.
0.4 × 2
0 0.8 × 2
1 1.6 × 2
1 1.2 × 2
0 0.4 × 2
0.4 = 0.01100110 = 1.100110 × 2−2
0 0.8 × 2
1 1.6 × 2
1 1.2
0 0.4 (stop here)
real exponent = - 2 and the biased exponent = −2 + 7 = 5 = 0101
real mantissa = 1.100110
hence the normalized mantissa = 100110
the representation of 0.4 in this representation is then
00101100110
(b) we now convert back the floating point 0 0101 100110 back to decimal
which is 1.100110 × 2−2 = (0.3987)notchecked
(c) relative error =
0.4−0.3987
0.4
3