Electronic – Multiplication of two binary numbers in fixed point arithmetic

computer-arithmeticfixed-point

I'm performing some operations with fractional numbers in a 16-bit FIXED-POINT processor.

I have to multiply the numbers \$ x=-6.35 \$, represented in \$ Q_{11} \$, and \$ y=-0.1 \$, represented in \$ Q_{14}\$.

First I represent the numbers in the respective notation in binary. The MSB is the sign bit.

So \$ x=11001.10100110011 \$ and \$ y=11.11100110011001 \$. I know the binary point is just in our mind and the processor treats this numbers as integers.

Ok then we multiply the numbers and get \$ x*y=11001000000100010011111001111011 \$. We eliminate the repeated sign bit and save the 16 MSB and represent the result in the appropriate format \$ Q_{10}\$: \$ x*y=100100.0000100010\$. This number corresponds to \$ – 27.966796875\$. But this doesn't make any sense, the result should be \$ 0.635\$.

What is going on here? Why is the result different? Am I missing something?

EDIT:
So I realized that it was pointed out that I was performing unsigned multiplication. Taking what was said now I'm doing signed multiplication:

$$1100110100110011$$
$$1111100110011001$$
$$—————-$$
$$\color{blue}{1111111111111111}1100110100110011$$
$$\color{blue}{000000000000000}0000000000000000\text{_}$$
$$\color{blue}{00000000000000}0000000000000000\text{__}$$
$$\color{blue}{1111111111111}1100110100110011\text{___}$$
$$\color{blue}{111111111111}1100110100110011\text{____}$$
$$\color{blue}{00000000000}0000000000000000\text{_____}$$
$$\color{blue}{0000000000}0000000000000000\text{______}$$
$$\color{blue}{111111111}1100110100110011\text{_______}$$
$$\color{blue}{11111111}1100110100110011\text{________}$$
$$\color{blue}{0000000}0000000000000000\text{_________}$$
$$\color{blue}{000000}0000000000000000\text{__________}$$
$$\color{blue}{11111}1100110100110011\text{___________}$$
$$\color{blue}{1111}1100110100110011\text{____________}$$
$$\color{blue}{111}1100110100110011\text{_____________}$$
$$\color{blue}{11}1100110100110011\text{______________}$$
$$\color{blue}{0}0011001011001101\text{_______________}$$
$$—————-$$
$$\color{red}{1}\color{green}{0}{0010001000111111010101001111011}$$

The last carry (red) is eliminated. Then the repeated sign bit (green) is also shifted to the left, adding an extra 0 bit at the end. Finally we save the 16 most significant bit and represent the result in \$ Q_{10}\$:

$${0010001000111111}$$

But this is \$ 8.5615234375 \$, not \$ 0.635 \$… So we are getting closer but the result is still off. Any ideas?

Best Answer

I have to multiply the numbers \$ x=-6.35 \$, represented in \$ Q_{11} \$, and \$ y=-0.1 \$, represented in \$ Q_{14}\$. So \$ x=11001.10100110011 \$ and \$ y=11.11100110011001 \$

I have just calculated the corresponding values for \$x\$ and \$y\$ using scaling values \$2^{11} \$ and \$2^{14}\$ respectively, and found them to be correct in 2's complement signed notation. So there is nothing wrong in your starting point.

I know the binary point is just in our mind and the processor treats this numbers as integers. Ok then we multiply the numbers and get \$ x*y=11001000000100010011111001111011 \$

This is where things went wrong. You would have been absolutely correct if you were multiplying two unsigned numbers (unsigned arithmetic). It is straight forward. But if one of the numbers is signed, then you have to do signed arithmetic, which is a little different.

There are two points I would like to highlight:

  • When you multiply two signed numbers \$x\$ and \$y\$, each partial product has to be treated as signed and you have to properly sign-extend each partial product.
  • If the multiplier(\$y\$) is a negative number, the last partial product has to be computed differently. It should be computed as the 2's complement of the sign-extended multiplicant (\$x\$).

Based on the two above points I will illustrate a simple example \$x* y\$ for \$x=-3\$ and \$y = -2\$, in 3-bit 2's complement notation (pardon my naive formatting):

$$x=-3=101_2$$ $$y=-2=110_2$$ So \$x * y\$ in signed arithmetic will go like: $$101$$ $$110$$ $$---$$ $$\color{blue}{000}000$$ $$\color{blue}{11}101\text{_}$$ $$\color{blue}{0011}\text{__}$$ $$-------$$ $$\color{red}{1}{000110}$$

Notice how partial products are sign-extended (blue color).

Notice that since \$y\$ is negative, the last partial product was calculated as 2's complement of sign-extended \$x\$ which is \$ = 0011_2\$.

The 7th bit can be disregarded (red color), because the answer is only 6 bits when two 3-bit numbers are multiplied. The rest 6 bits hence form the signed result (product). Therefore, the answer is \$x*y = 000110_2=\color{green}{+6_{10}}\$.

For the same \$ x\$ and \$y\$, if you do unsigned arithmetic, you will get \$x*y= 011110_2=\color{red}{+30_{10}}\$ !!

Conclusion

In your example, while multiplying two signed binary numbers, you probably did (if hand-derived/calculator) unsigned arithmetic, not signed arithmetic, hence the wrong results.

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