Normalized Floating Point & Representation of +46.25 in IEEE 754 32-bit (7 Marks)

1. When is Floating Point Number Normalized?
A floating point number $N = \pm M \times B^{E}$ (mantissa $M$, base $B=2$, exponent $E$) is said to be normalized when the mantissa is adjusted so that the most significant digit is non-zero, i.e. $1 \le M < 2$ for binary. In binary this means form $1.xxxxx... \times 2^{E}$ (leading 1 before binary point).

Purpose: (i) unique representation, (ii) maximum precision (no leading zeros wasted), (iii) efficient exponent use.
Example: $0.00101 \times 2^{5}$ is unnormalized, normalized is $1.01 \times 2^{2}$.

Denormalized/subnormal numbers have exponent zero and leading 0 to represent very small numbers near zero.

2. Floating Point Representation:
General format: Sign (S) + Exponent (E with bias) + Mantissa/Fraction (M).
IEEE 754 single precision (32 bits): | S (1) | Exponent (8) | Mantissa (23) |

• $S=0$ for +ve, $1$ for -ve
• Biased exponent: $E_{biased}=E_{true}+127$ (bias 127)
• Mantissa stores fractional part after hidden 1 (normalized). Value $= (-1)^S \times 1.M \times 2^{(E-127)}$

Double precision 64-bit uses 1-11-52 with bias 1023.

3. Represent +46.25 in 32-bit:

Step 1: Binary conversion
$46_{10}=101110_2$ ($32+8+4+2$)
$0.25=0.01_2$
So $46.25=101110.01_2$

Step 2: Normalize
$101110.01 = 1.0111001 \times 2^{5}$ (shift binary point 5 left)
True exponent $E=5$, Mantissa $M=0111001$

Step 3: Encode
$S=0$ (+ve)
$E_{biased}=5+127=132_{10}=10000100_2$
Mantissa 23 bits: $01110010000000000000000$ (pad zeros to 23 bits, after the hidden 1, we store $0111001$)

Result:
0 | 10000100 | 01110010000000000000000

Grouping into Hex for verification:
$0100\ 0010\ 0011\ 1001\ 0000\ 0000\ 0000\ 0000 = 0x42390000$

Value $= +1.0111001 \times 2^5 = +46.25$. Normalized because leading 1 is implicit.