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The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?
Standard Format for B and F as given below :
B=
F=
Standard Format for B and F as given below :
B=
F=
- a)6.02(27 - 30) dB
- b)6.02(28 - 31) dB
- c)6.02(27 - 31) dB
- d)6.02(28 - 30) dB
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The difference in the dynamic range of 32-bit binary number(B) and flo...
Concept:
The dynamic range in
In fixed point, unsigned integer representation using N-bit, the range of Max to Min is 2N to 1
In fixed-point signed integer representation using N-bit, the range of Max to Min number is 2N-1 to 1
Calculation:
The dynamic range of a 32-bit binary number B is:
Exponent = e – 127 varies from –127 to 128,
e = 0 = all bits (b30 ---- b23) are zero.
Now, the dynamic range is determined by the size of the exponent, which is
1 × 2-127(min) to 2128 (max).
Dynamic range:
Difference in Dynamic Ranges = 6.03 (28 – 31)
So, Option (2) is correct.
The dynamic range in
In fixed point, unsigned integer representation using N-bit, the range of Max to Min is 2N to 1
In fixed-point signed integer representation using N-bit, the range of Max to Min number is 2N-1 to 1
Calculation:
The dynamic range of a 32-bit binary number B is:
Exponent = e – 127 varies from –127 to 128,
e = 0 = all bits (b30 ---- b23) are zero.
Now, the dynamic range is determined by the size of the exponent, which is
1 × 2-127(min) to 2128 (max).
Dynamic range:
Difference in Dynamic Ranges = 6.03 (28 – 31)
So, Option (2) is correct.
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The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?Standard Format for B and F as given below :B=F=a)6.02(27 - 30) dBb)6.02(28 - 31) dBc)6.02(27 - 31) dBd)6.02(28 - 30) dBCorrect answer is option 'B'. Can you explain this answer?
Question Description
The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?Standard Format for B and F as given below :B=F=a)6.02(27 - 30) dBb)6.02(28 - 31) dBc)6.02(27 - 31) dBd)6.02(28 - 30) dBCorrect answer is option 'B'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?Standard Format for B and F as given below :B=F=a)6.02(27 - 30) dBb)6.02(28 - 31) dBc)6.02(27 - 31) dBd)6.02(28 - 30) dBCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?Standard Format for B and F as given below :B=F=a)6.02(27 - 30) dBb)6.02(28 - 31) dBc)6.02(27 - 31) dBd)6.02(28 - 30) dBCorrect answer is option 'B'. Can you explain this answer?.
The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?Standard Format for B and F as given below :B=F=a)6.02(27 - 30) dBb)6.02(28 - 31) dBc)6.02(27 - 31) dBd)6.02(28 - 30) dBCorrect answer is option 'B'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?Standard Format for B and F as given below :B=F=a)6.02(27 - 30) dBb)6.02(28 - 31) dBc)6.02(27 - 31) dBd)6.02(28 - 30) dBCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The difference in the dynamic range of 32-bit binary number(B) and floating-point number(F) is?Standard Format for B and F as given below :B=F=a)6.02(27 - 30) dBb)6.02(28 - 31) dBc)6.02(27 - 31) dBd)6.02(28 - 30) dBCorrect answer is option 'B'. Can you explain this answer?.
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