__Signed Number System__

In signed number systems, a number is represented with two parts; one the sign bit, and the rest bits denote the magnitude of the number. Signed number are represented using the following representation:

- Sign and Magnitude number system:
- 1’s complement Number System
- 2’s Complement Number System

## Sign and Magnitude number system

In this system, the MSB is set to 0(zero) to indicate +ve number and is set to 1(one) to indicate the -ve number. Signed numbers are represented as:

## Complete the following Table

+7 | Sign | Magnitude | ||||

+7 | 0 | 1 | 1 | 1 | ||

-6 | 1 | 1 | 1 | 0 | ||

+25 | ||||||

-24 | ||||||

+14 | ||||||

-15 | ||||||

-10 | ||||||

+0 | ||||||

-0 |

## Complement method:

Complement of a number: Complement of a number is done to represent its -ve / or positive representation. Thus the complement of a +ve number is the -ve number, and complement of -ve number is +ve number. There are different types of -ve number representation using complement method:

a. 1’s Complement System

b. 2’s Complement System

c. 9’s Complement System

d. 10’s Complement System

e. r’s and (r-1)’s Complement System

### 1’s Complement System:

Obtained by complementing each bit including the sign bit(MSB). It can also be obtained by subtracting the binary equivalent of the given number from all 1’s.

Decimal |
1’s Complement |
|||

+0 | 0 | 0 | 0 | 0 |

+1 | 0 | 0 | 0 | 1 |

+2 | 0 | 0 | 1 | 0 |

+3 | 0 | 0 | 1 | 1 |

-0 | 1 | 1 | 1 | 1 |

-1 | 1 | 1 | 1 | 0 |

-2 | 1 | 1 | 0 | 1 |

-3 | 1 | 1 | 0 | 0 |

**Example: ****
** Represent -56 in 1’s complement system

+56=0111000

-56= 1000111 (by complementing each bit of +56)

or subtract the binary number from binary all 1’s as given below

1111111

-0111000

————-

1000111

————–

**2’s Complement System:**

2’s complement is obtained by taking 1’s complement and adding 1 to it.

examples

-2 = 1’s complent of +2 and adding 1 to it

= (0010)’

1101

+ 1

——————–

1101 =-2

——————–

2’s Complement Code

Deimal | Sign | Magnitude | ||||

+0 | 0 | 0 | 0 | 0 | ||

+1 | 0 | 0 | 0 | 1 | ||

+2 | 0 | 0 | 1 | 0 | ||

+3 | 0 | 0 | 1 | 1 | ||

-0 | 0 | 0 | 0 | 0 | ||

-1 | 1 | 1 | 1 | 1 | ||

-2 | 1 | 1 | 1 | 0 | ||

-3 | 1 | 1 | 0 | 1 |

Advantage of 2’s complement:

There only one code for +0 and -0

Addition and subtraction becomes easy

Represent -56 in 2’s complement system

+56= 0111000

1’s complement of +56 is obtained by complementing each bit of +56. So

-56=Â 1000111 (in 1’s complement)

2’s Complement= 1’s complement + 1

Therefore, -56 = 1000111+ 1= 1001000

or

1111111

-0111000 (subtract binary of +56) from all 1s

————-

1000111

+1 ( Add 1 to get the 2’s complement )

————–

1001000 ( this is the 2’s complement representation of -56)

————

**9’s Complement System:**

9’s complement is obtained by taking subtracting the given decimal/or BCD) from a number of all 9’s.

Ex. : Represent -7656 in 9’s complement system

99999

– 07656 (subtract binary of +56 from all 9s)

————-

92343 ( This is the 9’c complement representation of -7656).

**10’s Complement System:**

10’s complement is obtained by taking 9’s complement and adding 1 to it.

Ex. : Represent -7656 in 10’s complement system

99999

– 07656 (subtract binary of +56 from all 9s

————-

92343

+ 1 ( Add 1 to get the 10’s complement )

————–

92344 ( this is the 10’s complement representation of -7656)

————

**(r-1)’s Complement System:**

In general the (r-1)’s complement of a number is obtained as follows:

subtract the given number from r^{n+1} – 1; where n is the number of digits in the given number and r is the base of the given number system.

**r’s Complement System:**

In general the r’s complement of a number is obtained as follows:

subtract the given number from r^{n+1} – 1 and add 1 to the result. Here n is the number of digits in the given number and r is the base of the given number system.

**Range of numbers**

Number system-> | Sign & Magnitude | 1’s Complement | 2’s Complemet |

Range | +/- 0 to +/- 2^{n-1} – 1 |
+/- 0 to +/-Â 2n-1 – 1 | -1 to – 2n-1 0 toÂ + 2 ^{n-1} – 1 |

**Numericals:**

Add +65 and -26; +12 and -15 using 1’s and 2’s complement

- code the two number with equal number of bits

+26 = 00011010

+65 =01000001

1’s complement of +26 = -26 = 11100101

-26 =+ 11100101

+65 = 01000001

Add **1 -> Carry**

00100110

Add the carry+1

**Therefore (65-26) =00100111 which is + 39**

**Explain the concept of underflow and Overflow in different number systems**

**Overflow:**If the range of number exceeds the size of the number, we say that their is an overflow:

If the size of the register is 4 bit, and if the addition of two numbers results in a number greater than 16 for an unsigned number or greater than 8 for signed number. The exact value is given in following table:

Sign & Magnitude | 1’s Complement | 2’s Complemet | |

Maximum allowable range | +/- 0 to +/- 2^{n-1} – 1 |
+/- 0 to +/- 2^{n-1} – 1 |
-1 to – 2n-1Â 0 to + 2 ^{n-1} – 1 |

Overflow | Result > + 2^{n-1} – 1 |
Result > + 2^{n-1} – 1 |
Result > + 2^{n-1 }– 1 |

Underflow | Result < – 2^{n-1} – 1 |
Result < – 2^{n-1} – 1 |
Result <Â – 2^{n-1} |

**Examples:**

Consider two 4 bit numbers to be added are + 3 and + 5, the result after addition as we know will be = +8 = 1000, but .

Sign & Magnitude | 1’s Complement | 2’s Complemet | |

+3 = 0011 +5 = 0101 = 1000 |
OverflowResult > + 2 ^{4-1}– 1 Also we note that the addition of +ve numbers results in a –ve number (MSB=1) |
OverflowResult > + 2 ^{4-1} – 1 Also we note that the addition of +ve results in a –ve number (MSB=1) |
OverflowResult > + 2 ^{4-1} -1 Also we note that the addition of +ve numbers results in a –ve number (MSB=1) |

**Underflow Example:**

If the result of additional / subtraction results in a number that is smaller than the one that can be represented in a given size of register

Consider two 4 bit numbers as -3 and 5, on subtracting 5 from -3 we get the answer = -8 , which is (11000) in sign and magnitude and requires 5 bits, and 10111 in 1’s complement system and 1000 in 2’s complement system, so looking at the range of various representation we conclude as stated in table below:

Sign & Magnitude | 1’s Complement | 2’s Complemet | |

Underflow | Result < – 2^{n-1} – 1 |
Result < – 2^{n-1} – 1 |
Result < – 2n-1 |

For 4 bit number the result of add/sub should be : | < -7 | < – 7 | Â < -8 |

-3-5=-8 | Underflow | Underflow | Within RanUnderflow |

** **

**How to find an Overflow: **

Check the MSB and the carry bit, if there is a carry in to the MSB bit but no carry out of MSB then there is an overflow. i.e. The XOR of MSB and Carry will give the status of overflow:

MSB (xor) Carry = 0 No overflow

MSB (xor) Carry = 1 , then there is an overflow.