358 in binary single floating point using excess 4

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Discussion Overview

The discussion revolves around expressing the number 358 in binary single floating point format using excess 4 notation. Participants explore the representation of the number in normalized format, addressing the challenges of using excess 4 for the exponent and the structure of the floating point representation.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that 358 can be represented as 1.01100110 with an exponent of 8, but expresses confusion about using excess 4 notation since 8+4=12 cannot be represented with 4 bits.
  • Another participant corrects that 12 can indeed be represented in 4 bits as 1100.
  • A different participant suggests that the representation should include 1 bit for the sign, 3 bits for the exponent (to represent values from -4 to +3 in excess-4), and 28 bits for the mantissa, concluding that the exponent would be 0 (which is 100 in excess-4) for the value 328.
  • Some participants emphasize the need for the answer to be in a specific format (SEEEMMMMMMMMMMMM) and in normalized format, indicating that they find the excess-4 notation challenging.
  • One participant reiterates the format and provides a potential representation of 358 as 0100000101100110, detailing the sign, exponent, and mantissa components.

Areas of Agreement / Disagreement

Participants express differing views on how to correctly apply the excess 4 notation and the structure of the floating point representation. There is no consensus on the final representation of 358 in this format, and confusion remains regarding the application of excess 4.

Contextual Notes

Participants have not fully resolved the implications of using excess 4 notation, and there are varying interpretations of how to structure the floating point representation. The discussion includes assumptions about the bit allocation for the sign, exponent, and mantissa.

francisg3
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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.
 
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12 can be represented in 4 bits:

1100 = 0*20 + 0*21 + 1*22 + 1*23 = 12
 
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
 
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.
 
francisg3 said:
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.

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
 

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