Question

Convert the following 32-bit IEEE floating point numbers to decimal:

0100 1100 1110 0110 1111 1000 0000 0000
1011 0101 1110 0110 1010 0110 0000 0000

Determine whether or not the following pairs are equivalent by constructing truth tables:

[(wx'+y')(w'y+z)] and [(wx'z+y'z)]
[(wz'+xy)] and [(wxz'+xy+x'z')]

Using DeMorgan’s Law and Boolean algebra, convert the following expressions into simplest form:

(a'd)'
(w+y')'
((bd)(a + c'))'
((wy'+z)+(xz)')'

Draw the circuit that implements each of the following equations:

AB'+(C'+AD')+D
XY'+WZ+Y'
(AD'+BC+C'D)'
((W'X)'+(Y'+Z))'

Answer 1

1)
a)
0 10011001 11001101111100000000000
sign bit is 0(+ve)
exp bits are 10011001
   => 10011001
   => 1x2^7+0x2^6+0x2^5+1x2^4+1x2^3+0x2^2+0x2^1+1x2^0
   => 1x128+0x64+0x32+1x16+1x8+0x4+0x2+1x1
   => 128+0+0+16+8+0+0+1
   => 153
in decimal it is 153
so, exponent/bias is 153-127 = 26
frac bits are 110011011111

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.110011011111 * 2^26
1.110011011111 in decimal is 1.804443359375
   => 1.110011011111
   => 1x2^0+1x2^-1+1x2^-2+0x2^-3+0x2^-4+1x2^-5+1x2^-6+0x2^-7+1x2^-8+1x2^-9+1x2^-10+1x2^-11+1x2^-12
   => 1x1+1x0.5+1x0.25+0x0.125+0x0.0625+1x0.03125+1x0.015625+0x0.0078125+1x0.00390625+1x0.001953125+1x0.0009765625+1x0.00048828125+1x0.000244140625
   => 1+0.5+0.25+0.0+0.0+0.03125+0.015625+0.0+0.00390625+0.001953125+0.0009765625+0.00048828125+0.000244140625
   => 1.804443359375
so, 1.804443359375 * 2^26 in decimal is 121094144.0
so, 01001100111001101111100000000000 in IEEE-754 single precision format is 121094144.0
Answer: 121094144.0

b)
1 01101011 11001101010011000000000
sign bit is 1(-ve)
exp bits are 01101011
   => 01101011
   => 0x2^7+1x2^6+1x2^5+0x2^4+1x2^3+0x2^2+1x2^1+1x2^0
   => 0x128+1x64+1x32+0x16+1x8+0x4+1x2+1x1
   => 0+64+32+0+8+0+2+1
   => 107
in decimal it is 107
so, exponent/bias is 107-127 = -20
frac bits are 11001101010011

IEEE-754 Decimal value is 1.frac * 2^exponent
IEEE-754 Decimal value is 1.11001101010011 * 2^-20
1.11001101010011 in decimal is 1.80194091796875
   => 1.11001101010011
   => 1x2^0+1x2^-1+1x2^-2+0x2^-3+0x2^-4+1x2^-5+1x2^-6+0x2^-7+1x2^-8+0x2^-9+1x2^-10+0x2^-11+0x2^-12+1x2^-13+1x2^-14
   => 1x1+1x0.5+1x0.25+0x0.125+0x0.0625+1x0.03125+1x0.015625+0x0.0078125+1x0.00390625+0x0.001953125+1x0.0009765625+0x0.00048828125+0x0.000244140625+1x0.0001220703125+1x6.103515625e-05
   => 1+0.5+0.25+0.0+0.0+0.03125+0.015625+0.0+0.00390625+0.0+0.0009765625+0.0+0.0+0.0001220703125+6.103515625e-05
   => 1.80194091796875
so, 1.80194091796875 * 2^-20 in decimal is 1.7184647731482983e-06
so, 10110101111001101010011000000000 in IEEE-754 single precision format is -1.7184647731482983e-06
Answer: -1.7184647731482983e-06