Represent $11.0011*2^{10}$ using the IEEE-$754$ standard for $32$-bit floating point representation.

$0$-Sign Bit

Is this answer is correct ? I am bit confused .. Please Help

  • $\begingroup$ This answer might help. $\endgroup$ – Regret May 29 '15 at 9:07
  • $\begingroup$ Your answer is incorrect. Use this to find the correct representation. $\endgroup$ – achille hui May 29 '15 at 9:09
  • $\begingroup$ @achillehui is the answer given by "gammatester" is correct ? Plz Help $\endgroup$ – Himanshu Chawla May 29 '15 at 10:44
  • $\begingroup$ @HimanshuChawla If $11.0011$ really mean a number in base $2$ (It is not clear in your question), then gammatester answer is the one you want. $\endgroup$ – achille hui May 29 '15 at 11:09
  • $\begingroup$ But if $11.0011$ is in binary, who is to say what the exponent $10$ means? $\endgroup$ – TonyK Dec 8 '16 at 13:29

Your answer is almost correct, you only missed a $0$ in the mantissa. Your number is $1.10011_2\times 2^{11}=3264_{10}$. The sign bit is $0$ the biased exponent is $11+127=138= 10001010_2$ and the mantissa without the implied leading bit is $10011000000000000000000_2$. Concatenated this gives $$11.0011_2\times 2^{10} \rightarrow01000101010011000000000000000000_2\\ = 0100\_0101\_0100\_1100\_0000\_0000\_0000\_0000_2 =454C0000_{16}$$ Check it e.g. with http://babbage.cs.qc.cuny.edu/IEEE-754.old/Decimal.html

  • $\begingroup$ Thank You @gammatester for giving the answer .... I m very thankful to you. $\endgroup$ – Himanshu Chawla May 29 '15 at 10:42

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