3/12/2024 0 Comments System verilog decimal to binary![]() ![]() So as you'd expect your integer part is 0101 = 5. So like when decoding a normal binary integer, you start at the decimal point and work left for the integer part, the first bit is your 1, then your 2, then 4, then 8. Now fixed point is different, you essentially store N bits of integer, and M bits of fraction. Multiplication isn't too bad, but addition is very expensive, and that's one of your most common operations, so that's not great. So your decimal point moves, you have more accuracy (more fractional part) with smaller values than with larger values, hence floating point.įloating point maths is expensive in digital design, it takes a lot of resources to do this maths. Meaning you can represent a large amount of very small numbers, but you loose accuracy with large numbers. Because the mantissa is a constant number of bits, the precision varies with the size of the number. ![]() Meaning you can represent very small numbers (. The exponent is stored biased, so you can go from say -127 to +127. Your floating point number contains a sign bit, several bits of mantissa and several bits of exponent. Where M is your "mantissa" (in binary), E is your exponent, and A is 1 or 0, typically 1 for a normal number, but can be 0 for "denormals". In software you often use floating point, where your number is representad as: +/- A.MMMMMM * 2 E. There are several ways to do this, each of which has advantages and disadvantages. Basically you need a way to store a representation of a non integer number. The resulting circuit is a purely combinational circuit (the code could have been written linearly, without using a for loop, but it would have been much longer)."decimal point" is not a thing. Note the for loop does not directly describe a circuit – rather, it describes how the circuit components that will do the required shifting and adding are to be assembled. Here, a for loop is used to make the code more compact. ![]() Verilog code to implement the double-dabble is shown below. An illustration of the double-dabble algorithm The figure below illustrates the process (the blue boxes around BCD digits show BCD digits that are >=5, and therefore need 3 to be added). Adding three to any BCD digit greater than five does two things: first, at the next shift, the 3 that was added becomes 6, and that accounts for the difference in binary and BCD codes (BCD uses 10 binary codes, and binary uses 16) and second, adding 3 forces the MSB of the BCD digit to a 1, where it is “carried out” and into the next digit. Since BCD digits cannot exceed nine, a pre-shift number of five or more would result in a post-shift number of ten or more, which cannot be represented. This works because every left shift multiplies all BCD digits by two. After every shift, all BCD digits are examined, and 3 is added to any BCD digit that is currently 5 or greater. The binary number is left-shifted once for each of its bits, with bits shifted out of the MSB of the binary number and into the LSB of the accumulating BCD number. The “double dabble” algorithm is commonly used to convert a binary number to BCD. The Verilog code below illustrates converting a 4-digit BCD number to it’s binary equivalent. To find the binary equivalent, each BCD digit is multiplied by its weighted magnitude: 9 x 10^2 + 8 * 10^1 + 7 * 10^0, or 9 * 100 + 8 * 10+ 7 * 1. Consider the BCD number 987, stored as three 4-bit BCD codes: 1001 for 9 (digit 2), 1000 for 8 (digit 1), and 0111 for 7 (digit 0). ![]() Each BCD digit in a given number contributes a magnitude equal to the digit multiplied by its weight, and each digit’s weight is equal to 10 raised to the power of the digit’s position in the number. BCD numbers are representations of decimal (base 10) numbers, and like all modern number systems, BCD numbers use positional weighting. ![]()
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