Binary Arithmetic
Before going through this section, make sure you understand about the representation of numbers in binary. You can read the page on numeric representation to review.
Table of contents
This document will introduce you to the methods for adding and multiplying binary numbers. In each section, the topic is developed by first considering the binary representation of unsigned numbers (which are the easiest to understand), followed by signed numbers and finishing with fractions (the hardest to understand). For the most part we will deal with
Adding unsigned numbers in binary is quite easy. Recall that with 4 bit numbers we can represent numbers from 0 to 15. Addition is done exactly like adding decimal numbers, except that you have only two digits (0 and 1). The only number facts to remember are that
0+0 = 0, with no carry,
1+0 = 1, with no carry,
0+1 = 1, with no carry,
1+1 = 0, and you carry a 1.so to add the numbers 06_{10}=0110_{2} and 07_{10}=0111_{2} (answer=13_{10}=1101_{2}) we can write out the calculation (the results of any carry is shown along the top row, in italics).
Decimal Unsigned Binary 1 (carry) 06 +07 13 110 (carry) 0110 +0111 1101The only difficulty adding unsigned numbers occurs when you add numbers that are too large. Consider 13+5.
Decimal Unsigned Binary 0 (carry) 13 +05 18 1101 (carry) 1101 +0101 10010The result is a 5 bit number. So the carry bit from adding the two most significant bits represents a results that overflows (because the sum is too big to be represented with the same number of bits as the two addends).
Adding signed numbers is not significantly different from adding unsigned numbers. Recall that signed 4 bit numbers (2's complement) can represent numbers between -8 and 7. To see how this addition works, consider three examples.
Decimal Signed Binary -2 +3 1 1110 (carry) 1110 +0011 0001
Decimal Signed Binary -5 +3 -2 011 (carry) 1011 +0011 1110
Decimal Signed Binary -4 -3 -7 1100 (carry) 1100 +1101 1001In this case the extra carry from the most significant bit has no meaning. With signed numbers there are two ways to get an overflow -- if the result is greater than 7, or less than -8. Let's consider these occurrences now.
Decimal Signed Binary 6 +3 9 110 (carry) 0110 +0011 1001
Decimal Signed Binary -7 -3 -10 1001 (carry) 1001 +1101 0110Obviously both of these results are incorrect, but in this case overflow is harder to detect. But you can see that if two numbers with the same sign (either positive or negative) are added and the result has the opposite sign, an overflow has occurred.
Typically DSP's, including the 320C5x, can deal somewhat with this problem by using something called saturation arithmetic, in which results that result in overflow are replaced by either the most positive number (in this case 7) if the overflow is in the positive direction, or by the most negative number (-8) for overflows in the negative direction.
There is no further difficult in adding two signed fractions, only the interpretation of the results differs. For instance consider addition of two Q3 numbers shown (compare to the example with two 4 bit signed numbers, above).
Decimal | Fractional Binary |
-0.25 +0.375 0.125 |
1110 (carry) 1110 +0011 0001 |
Decimal | Fractional Binary |
-0.625 +0.375 -0.25 |
011 (carry) 1011 +0011 1110 |
Decimal | Fractional Binary |
-0.5 -0.375 -0.875 |
1100 (carry) 1100 +1101 1001 |
If you look carefully at these examples, you'll see that the binary representation and calculations are the same as before, only the decimal representation has changed. This is very useful because it means we can use the same circuitry for addition, regardless of the interpretation of the results.
Even the generation of overflows resulting in error conditions remains unchanged (again compare with above)
Decimal Fractional Binary 0.75 +0.375 1.125 110 (carry) 0110 +0011 1001
Decimal Fractional Binary -0.875 -0.375 -1.25 1001 (carry) 1001 +1101 0110
Multiplying unsigned numbers in binary is quite easy. Recall that with 4 bit numbers we can represent numbers from 0 to 15. Multiplication can be performed done exactly as with decimal numbers, except that you have only two digits (0 and 1). The only number facts to remember are that 0*1=0, and 1*1=1 (this is the same as a logical "and").
Multiplication is different than addition in that multiplication of an n bit number by an m bit number results in an n+m bit number. Let's take a look at an example where n=m=4 and the result is 8 bits
Decimal Binary 10 x6 60 1010 x0110 0000 1010 1010 +0000 0111100In this case the result was 7 bit, which can be extended to 8 bits by adding a 0 at the left. When multiplying larger numbers, the result will be 8 bits, with the leftmost set to 1, as shown.
Decimal Binary 13 x14 182 1101 x1110 0000 1101 1101 +1101 10110110As long as there are n+m bits for the result, there is no chance of overflow. For 2 four bit multiplicands, the largest possible product is 15*15=225, which can be represented in 8 bits.
There are many methods to multiply 2's complement numbers. The easiest is to simply find the magnitude of the two multiplicands, multiply these together, and then use the original sign bits to determine the sign of the result. If the multiplicands had the same sign, the result must be positive, if the they had different signs, the result is negative. Multiplication by zero is a special case (the result is always zero, with no sign bit).
As you might expect, the multiplication of fractions can be done in the same way as the multiplication of signed numbers. The magnitudes of the two multiplicands are multiplied, and the sign of the result is determined by the signs of the two multiplicands.
There are a couple of complications involved in using fractions. Although it is almost impossible to get an overflow (since the multiplicands and results usually have magnitude less than one), it is possible to get an overflow by multiplying -1x-1 since the result of this is +1, which cannot be represented by fixed point numbers.
The other difficulty is that multiplying two Q3 numbers, obviously results in a Q6 number, but we have 8 bits in our result (since we are multiplying two 4 bit numbers). This means that we end up with two bits to the left of the decimal point. These are sign extended, so that for positive numbers they are both zero, and for negative numbers they are both one. Consider the case of multiplying -1/2 by -1/2 (using the method from the textbook):
Decimal Fractional Binary -0.5 x0.5 -0.25 1100 x0100 0000 0000 +111100 11110000This obviously presents a difficulty if we wanted to store the number in a Q3 result, because if we took just the 4 leftmost bits, we would end up with two sign bits. So what we'd like to do is shift the number to the left by one and then take the 4 leftmost bit. This leaves us with 1110 which is equal to -1/4, as expected.
On a 16 bit DSP two Q15 numbers are multiplied to get a Q30 number with two sign bits. On the 320C50 there are two ways to accomplish this. The first is to use the p-scaler immediately after the multiplier, or the postscaler after the accumulator. to shift the result to the left by one
This section is not complete
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