Fixed Point Multiplication
Fixed point addition and subtraction are straightforward. Additions and subtractions are performed using integer operations. For example, if two 16 bit Q15 format numbers are added, the result is a Q15 number. But what about fixed about multiplication? What happens if two Q15 numbers are multiplied?
Let’s try an example. Take 0.5 multiplied by 0.25. In Q15 the number 0.5 is represented (in hexadecimal) as 0x8000 times 0.5 or 0x4000. Similarly, 0.25 is 0x2000. When we multiply these together, the product is 0x08000000. Obviously the result is not a Q15 number since the number of bits required is more than 16. The expected product, 0.125, is 0x1000 in Q15.
To see what is going on, define the following two Q15 numbers a and b:
where and are the integer representations of our numbers (0x4000 and 0x2000 in our example). The product of a and b is:
From the above, it can be seen that the product is a Q30 number. Going back to our example, 0x4000 times 0x2000 is 0x08000000, which is 0.125 times .
A general rule when multiplying a Qm format number by a Qn format number, is that the product will be a Q(m+n) number. The number of bits required to represent the product is at least (n+m) for unsigned multiplication and (n+m+1) for signed (twos complement) multiplication.
For the more general case of a Qa.b number times a Qc.d number, the product is Q(a+c).(b+d). The number of bits needed for the result is (a + b + c + d + 1) for signed numbers (and one less for unsigned numbers).
Consider the example of a Q16 unsigned multiplication between the two largest unsigned numbers that can be represented. The largest Q16 number is 65535/65536 = 0.9999847412109375. The product is 0xffff times 0xffff or 0xfffe0001. The result is a Q32 number requiring at least 32 bits. If we divide by then we get 0.99996948265470564365386962890625, the expected result.
There are a number of things that are done with the product of a multiplication, depending on the application. Some of the commonly seen options are:
- Convert the product to a different Q format.
- Use the product in the resulting Q format.
- Add the product to a running sum in an accumulator register.
- Convert the product to a different Q format, then add to a running sum.
Let’s look at some of these options for the case of signed multiplication using Q15 format numbers. For case 1, assume we want to multiply two Q15 numbers and get a Q15 result. The required operation is to take the Q30 product, and shift it right by 15 bits. The result can then be stored in 16 bits. There is also the option of rounding the product before shifting out the lower 15 bits (I may discuss rounding in a future post). Some CPU architectures are better set up to shift the product left by 1, and then store the upper 16 bits. This is almost exactly the same as shifting right by 15 bits and keeping the lower 16 bits.
Multiply-accumulate (MAC) operations are used a lot in many DSP algorithms. Many processors have one or more dedicated accumulator registers for this purpose (often with 32 or 40 bits). For the case of Q15 multiplies, each Q30 product can be summed to the accumulator.
I have seen a lot of code that shifts each product left by 1 when performing the MAC operations. Some DSP chips can do the left shift in hardware using a special mode of the ALU. In this case, the value in the accumulator is in Q31 format. Although very common, this method has a greater chance of overflow problems since each product is effectively two times bigger. I think this method became popular because certain older DSP chip architectures required the storing of the high 16 bits of the accumulator or product register, rather than having a single cycle instruction allowing a shift by 15 bits.
In summary, because multiplication operations are often a chief component of signal processing implementations, it is important to understand how they work. This is especially true for fixed point operations, where one must know the effect of multiplication on the format of the numbers themselves.