# 358 in binary single floating point using excess 4

1. Jan 15, 2010

### francisg3

i need to express 358 in binary single floating point format using excess 4 notation...i can do it easily using the 32-bit standard representation with the excess 127 exponent. i know that 358 is 101100110 therefore i assume floating point would be 1.01100110 with an exponent of 8. i know excess 4 is the base exponent added to 4 but 8+4=12 which cannot be represented with 4 bits. any help would be greatly appreciated! thanks.

2. Jan 17, 2010

12 can be represented in 4 bits:

1100 = 0*20 + 0*21 + 1*22 + 1*23 = 12

3. Jan 18, 2010

### zgozvrm

I would guess this would mean that you have 1 bit for the sign, 3 bits for the exponent (3 bits to represent the values from -4 to +3 in excess-4), and the remaining 28 bits for the mantissa.

The value 328 can be represented within the given 28 bits, so there is no need to move the decimal point, therefore the exponent is 0 (which equates to 100 in excess-4).

So, 328 should be represented as:
0 100 0000 0000 0000 0000 0001 0110 0110
or 40000166 hex

4. Jan 18, 2010

### francisg3

well the answer is supposed to be in this format:
SEEEMMMMMMMMMMMM
the answer is supposed to be in normalized format. The sign is easy (i know it's 0) it's the rest I'm having trouble with. I can do excess-127 notation with no problem, it's the excess-4 business I'm not catching.

5. Jan 18, 2010

### zgozvrm

Easy enough then (you didn't state that it was to be 16-bit, so I assumed 32-bit)

The value would then be:
0100000101100110

the sign is 0
the exponent is 100 (or 0 in excess-4)
the mantissa is 000101100110