Other useful videos related to Floating Point Representation: 1) IEEE 754: Single Precision and Double Precision Format: th-cam.com/video/e_J9lXnU_vs/w-d-xo.htmlsi=aBlp8v7sD2UY6tiQ 2) Fixed Point vs Floating Point Numbers: th-cam.com/video/zVM8NKXsboA/w-d-xo.htmlsi=8yIaPa14s4jwg9Cw Link for the Digital Electronics (Playlist): bit.ly/31gBwMa
Hello from Germany. Very nice explenation! I only have one question: What would the numer at minute 5:29 be in Decimal (its the -0.00011 * 2 ^-126). Could you give me a Calculation method for that pls
Yes, sure. First, you need to separate the given HEX number in 32 bit binary number. Then write it in a different segments. (like sign, exponent and mantissa). Here, since exponent is zero and mantissa is non-zero, it means the given number represents de-normalized number. So, it will be in the form ± 0.000 x 2 ^-126. That means here, exponent is 2^-126. Now, here since the mantissa is 00011. So, overall number will be 0. 00011 x 2^-126. More over since sign bit is 1, so number is negative number. Therefore, the equivalent decimal number is - 0.00011 x 2^-126. I hope, it will clear your doubt.
I think there is a mistake, in case of normalized numbers, the biased exponent can be between 0x01 to 0xFE, but in case of denormal number the exponent is 0, which makes the actual exponent equal to 2^-127 not 2^126. To differentiate whether the number has preceding one of not is identified by exponent value which is 1 for all normalized numbers but 0 for denormal numbers. Correct me if I am wrong.
The de normalized numbers are used to represent numbers which are smaller than smallest possible normalized numbers. In single precision format, the smallest positive normalized number is 1. 0 x 2^-126. So, to represent numbers between 0 and this smallest number this denorms are used. The thing of the biased exponent holds true only for normalized numbers. For denorms, the exponent is always -126. And the numbers are represented as 0. BBB x 2^-126. So, with this representation, the numbers will be less than the smallest possible normalized numbers. For example, 0.11 x 2^-126. I hope, it will clear your doubt.
Other useful videos related to Floating Point Representation:
1) IEEE 754: Single Precision and Double Precision Format:
th-cam.com/video/e_J9lXnU_vs/w-d-xo.htmlsi=aBlp8v7sD2UY6tiQ
2) Fixed Point vs Floating Point Numbers:
th-cam.com/video/zVM8NKXsboA/w-d-xo.htmlsi=8yIaPa14s4jwg9Cw
Link for the Digital Electronics (Playlist):
bit.ly/31gBwMa
Very well explained 👌🏻👌🏻👍🏻
Thank you 🙏👍
You are awesome 🎉🎉
Many thanks for such an interesting explanation.
Hello from Germany. Very nice explenation! I only have one question: What would the numer at minute 5:29 be in Decimal (its the -0.00011 * 2 ^-126). Could you give me a Calculation method for that pls
Yes, sure. First, you need to separate the given HEX number in 32 bit binary number. Then write it in a different segments. (like sign, exponent and mantissa). Here, since exponent is zero and mantissa is non-zero, it means the given number represents de-normalized number. So, it will be in the form ± 0.000 x 2 ^-126. That means here, exponent is 2^-126. Now, here since the mantissa is 00011. So, overall number will be 0. 00011 x 2^-126. More over since sign bit is 1, so number is negative number. Therefore, the equivalent decimal number is - 0.00011 x 2^-126. I hope, it will clear your doubt.
very good explanation. Thanks for helping!!! :)
I think there is a mistake, in case of normalized numbers, the biased exponent can be between 0x01 to 0xFE, but in case of denormal number the exponent is 0, which makes the actual exponent equal to 2^-127 not 2^126.
To differentiate whether the number has preceding one of not is identified by exponent value which is 1 for all normalized numbers but 0 for denormal numbers.
Correct me if I am wrong.
The de normalized numbers are used to represent numbers which are smaller than smallest possible normalized numbers. In single precision format, the smallest positive normalized number is 1. 0 x 2^-126.
So, to represent numbers between 0 and this smallest number this denorms are used.
The thing of the biased exponent holds true only for normalized numbers. For denorms, the exponent is always -126. And the numbers are represented as 0. BBB x 2^-126.
So, with this representation, the numbers will be less than the smallest possible normalized numbers. For example, 0.11 x 2^-126. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS if For denorms, the exponent is always -126, then how it represents the exponent being all 0?
Great video!
awesome