Number Systems Operations
 

Addition

Adding unsigned numbers

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 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 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).

Adding signed numbers

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
  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.

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.

 

Adding fractions

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

 

Multiplication

Multiplying unsigned numbers

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.

Multiplying signed numbers

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).

Multiplying fractions

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.

 

 

The 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
 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).

Adding signed numbers

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
  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.

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.

 

Adding fractions

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

 

Multiplication

Multiplying unsigned numbers

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.

Multiplying signed numbers

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).

Multiplying fractions

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.

 

 

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).

Adding signed numbers

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
  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.

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.

 

Adding fractions

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

 

Multiplication

Multiplying unsigned numbers

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.

Multiplying signed numbers

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).

Multiplying fractions

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.

 

 

Decimal Signed Binary 
 
   -5
   +3
   -2
 
 011  (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.

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.

 

Adding fractions

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

 

Multiplication

Multiplying unsigned numbers

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.

Multiplying signed numbers

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).

Multiplying fractions

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.

 

 

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.

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.

 

Adding fractions

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

 

Multiplication

Multiplying unsigned numbers

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.

Multiplying signed numbers

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).

Multiplying fractions

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.

 

 

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.

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.

 

Adding fractions

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

 

Multiplication

Multiplying unsigned numbers

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.

Multiplying signed numbers

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).

Multiplying fractions

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.

 

 

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.

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.

 

Adding fractions

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 lo

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