Computing and Information Systems/Creative Computing

Running head: COMPUTING AND INFORMATION SYSTEMS/CREATIVE
COMPUTING
COMPUTING AND INFORMATION SYSTEMS/CREATIVE COMPUTING
Name of the Student
Name of the University
Author Note

1COMPUTING AND INFORMATION SYSTEMS/CREATIVE COMPUTING
Question 1:
a. The representation of the positive decimal number 1025.25 in IEEE 754 32-
bit format is given by, 0 1000 1001 0000 0000 0101 0000 0000 000
Now, the steps to transform the decimal number into binary format is shown below.
1025.2510 = 10000000001.012 = 10000000001.01*2(^0) = 1.000000000101*2(^10).
a. Hence, the exponent is 10 and biasing it by 127 gives 10+127 = 137. So, normalizing
the biased exponent gives 137 = 10001001. Now, the mantissa is 1.000000000101.
The 1 in the left of radix point is not stored in IEEE 754 32-bit format (Tang,
Schneider and Tsen 2015). Hence, the number in the IEEE 754 32-bit format is given
below.
Sign Exponent Mantissa
0 1000 1001 0000 0000 0101 0000 0000
000
b. The IEEE 754 number is given by,
1.10111 x 2^(−130)
In IEEE 754 32-bit denormalised form the number is given as follows. The exponent
is given in decimal form.
Sign Exponent Mantissa
0 -13010 10111 000 0000 0000 0000
000
Question 2:

2COMPUTING AND INFORMATION SYSTEMS/CREATIVE COMPUTING
a. The two’s complement addition is given by
1001 + 1010 = 10011. Now, in decimal equivalent 1001 = -1 and 1010 = -2. Now, the
addition of two negative numbers gives a negative number. Now, in 2’s complement
addition the extra bit is discarded and the sign bit of result is 0 which is wrong
(Nannarelli 2017). Hence, there is an overflow which means the magnitude of result is
too large to represent in 4 bit format.
b. After sign extending the two numbers in 6 bits the numbers will be
111001 and 111010. The sign bit which is 1 is moved to the leftmost position and
the extra two bits are filled with copy of the sign bit which is 1.
Now, addition gives,
111001
+111010
=1110011
Now, discarding the extra bit in result and representing in 6 bit format the result is
110011. Here, the sign bit is the left most bit which is 1 or the result is negative
(Nannarelli 2017). Now, removing the sign bit and extra two bits from left which
were added to the 4 bit signed binary number, the magnitude bits are 011 = 3.
Hence, the result is -3 and there is no overflow.
Question 3:
a. A decoder with 4 inputs and 16 outputs is shown below. A 4 to 16 decoder is
constructed using two 3 to 8 decoder (Kirichenko et al. 2015).