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.
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. 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 carry=0, so result = 002
1+0 = 1, with carry=0, so result = 012
0+1 = 1, with carry=0, so result = 012
1+1 = 0, with carry=1, so result = 102
Note that the result is two bits, the rightmost bit is called the sum, and the left bit is called the carry
To add the numbers 0610=01102 and 0710=01112 (answer=1310=11012) 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 1101 |
The only difficulty adding unsigned binary 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 10010 |
The 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). The same problem can occur with decimal numbers: if you add the two digit decimal numbers 65 and 45, the result is 110 which is too large to be represented in 2 digits.
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 |
0011 (carry) 1011 +0011 1110 |
| Decimal | Signed Binary |
-4 -3 -7 |
1100 (carry) 1100 +1101 1001 |
In 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 0110 |
Obviously 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.
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
|
0110 (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
0111100
|
In 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
10110110
|
As 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
11110000
|
This 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.
This section is not complete
| ← | Comments or Questions? |
Erik Cheever Professor Emeritus Engineering Department Swarthmore College |