Number Systems, Operations, and Codes

Question 1

Largest decimal number with n bits

Answer 1

With n bits, you can count up to a number equal to
(2raise to n) - 1.
Largest decimal number = (2 raise to n) - 1.

Question 2

Represent fractional bits using binary

Answer 2

Fractional numbers can also be represented in binary by placing bits to the right of the
the binary point, just as fractional decimal digits are placed to the right of the binary point.
The left-most bit(after decimal) is the MSB in a binary fractional number and has a weight of 2^(-1) = 1/2 = 0.5[equivalent decimal weight]
The fractional weights decrease from left to right by a negative power of two for each bit.

Question 3

Binary weight

Answer 3

All the bits to the left of the
binary point have weights that are positive powers of two, as previously discussed for whole
numbers. All bits to the right of the binary point have weights that are negative powers of
two, or fractional weights.

Question 4

Binary-to-Decimal Conversion

Answer 4

Add the weights of all 1s in a binary number to get the decimal value.

Question 5

Decimal whole number to binary:

Sum of weights:

Answer 5

determine the set of binary weights whose sum is equal to the decimal number

Question 6

Decimal whole number to binary:

Repeated division by 2:

Answer 6

To get the binary number for a given
decimal number, divide the decimal
number by 2 until the quotient is 0.
Remainders form the binary number.
 The first remainder to be produced is the LSB and last one is MSB.

Question 7

Converting Decimal Fractions to Binary:

Sum-of-Weights

Answer 7

The sum-of-weights method can be applied to fractional decimal numbers, as shown:
0.625 = 0.5 + 0.125 = 2^-1 + 2^-3 = 0.101
There is a 1 in the 2^-1
position, a 0 in the 2^-2
position, and a 1 in the 2^-3
position.

Question 8

Converting Decimal Fractions to Binary:

Repeated Multiplication by 2

Answer 8

to convert the decimal fraction 0.3125 to binary, begin by multiplying
0.3125 by 2 and then multiplying each resulting fractional part of the product by 2 until
the fractional product is 0 or until the desired number of decimal places is reached or stop when the fractional part is all zeros.
The carry digits, or carries, generated by the multiplications produce the binary number.
The first carry produced is the MSB, and the last carry is the LSB.

Question 9

4 Basic rules of binary addition:

Answer 9

0 + 0 = 0 Sum of 0 with a carry of 0
0 + 1 = 1 Sum of 1 with a carry of 0
1 + 0 = 1 Sum of 1 with a carry of 0
1 + 1 = 10 Sum of 0 with a carry of 1

Question 10

Binary addition, carry situation of adding 3 bits:

Answer 10

Carry bits
1 + 0 + 0 = 01 Sum of 1 with a carry of 0
1 + 1 + 0 = 10 Sum of 0 with a carry of 1
1 + 0 + 1 = 10 Sum of 0 with a carry of 1
1 + 1 + 1 = 11 Sum of 1 with a carry of 1

Question 11

Binary Subtraction basic rules:

Answer 11

0 - 0 = 0
1 - 1 = 0
1 - 0 = 1
10 - 1 = 1 [0 - 1 with a borrow of 1]

Question 12

Binary Multiplication, 4 basic rules:

Answer 12

0 * 0 = 0
 0 * 1 = 0
 1 * 0 = 0
 1 * 1 = 1
Binary multiplication of two bits is
the same as multiplication of the
decimal digits 0 and 1.

Question 13

Binary Division:

Answer 13

Division in binary follows the same procedure as division in decimal

Question 14

1’s complement:

Answer 14

The 1’s complement of a binary number is found by changing all 1s to 0s and all 0s to 1s. here zero has two representations, 0000 0000 and 1111 1111.. not appreciated.

Question 15

Logical circuit for 1’s complement:

Answer 15

Parallel inverters.

Question 16

2’s complement

Answer 16

2’s complement = (1’s complement) + 1
OR the alternative method is:
Change all bits to the left of the least significant 1 to get 2’s complement.
for zero an overflow would occur but that taken care then main 8 bits represent 0.

Question 17

Logical circuit for 2’s complement:

Answer 17

parallel inverters followed by an adder with a carry input [1].

Question 18

Get true form (uncomplemented) from 1’s or 2’s complement:

Answer 18

Simply repeat what u did to get the complement.

Question 19

Sign Bit

Answer 19

It’s the left-most bit in a signed binary number.

A 0 sign bit indicates a +ve number, and a 1 : -ve.

Question 20

Signed numbers:

Sign Magnitude form

Answer 20

In the sign-magnitude form, a negative number has the same magnitude bits as the corresponding positive number but the sign bit is a 1 rather than a zero.

Question 21

Signed numbers:

Answer 21

1s complement: 1s complement will be the negative number.

2s complement: 2s complement will be the -ve number.

Question 22

The decimal value of signed number in 1s complement form and 2s complement:

Answer 22

+ve numbers:
determined by summing the weights in all bit positions where there are 1s.
-ve numbers:
determined by assigning a negative value to the weight of the sign bit, summing all the weights where there are 1s, and
adding 1 to the result.
And in 2s complement same process but no need of adding one to the answer.

Question 23

For 2’s complement signed numbers, the range of values for n-bit numbers is?

Answer 23

Range = -( 2^[n-1] ) to +( 2^[n-1] - 1)

[1 sign bit and n-1 magnitude bits]

Question 24

Floating Point Unit

Answer 24

• the coprocessor to free up the CPU (to perform other tasks) and increase speed.
• Performs complicated math operations using floating point numbers.

Question 25

Single precision Double precision Extended precision

Answer 25

32bit floating point numbers - Single precision 64 - Double precision 80 - Extended precision

Question 26

Floating point Number.

Answer 26

has 1. Mantissa [ shows magnitude, between 0-1, fractional number] 2. Exponent [ number of places the point has to move, it's the power of 10]

Question 27

Overflow

Answer 27

When the addition of same signed bits and 7 magnitude bits are not enough to represent the resulting the magnitude it changes the sign bit to adjust and creates an overflow situation.

Question 28

Addition vocab:

Answer 28

Addend + augend = sum

Question 29

Subtraction vocab: and function and how to:

Answer 29

minuend - subtrahend = difference Basically, the subtraction operation changes the sign of the subtrahend and adds it to the minuend. To subtract two signed numbers, take the 2’s complement of the subtrahend and add. Discard any final carry bit.

Question 30

Multiplication: | sirect addition method only

Answer 30

``` Multiplicand * multiplier = product Multiplication is equivalent to adding a number to itself a number of times equal to the multiplier. In the direct addition method, you add the multiplicand a number of times equal to the multiplier ```

Question 31

Partial product method multiplication : | just method, not rules.

Answer 31

The partial products method is perhaps the more common one because it reflects the way you multiply longhand. The multiplicand is multiplied by each multiplier digit beginning with the least significant digit. The result of the multiplication of the multiplicand by a multiplier digit is called a partial product. Each successive partial product is moved (shifted) one place to the left and when all the partial products have been produced, they are added to get the final product

Question 32

Rules or steps of partial product:

Answer 32

1. if both same sign, product same sign otherwise product sign -. 2. Make sure numbers in the true or uncomplemented form, negative numbers tend to be in 2s complement form. 3.Starting with the least significant multiplier bit, generate the partial products. When the multiplier bit is 1, the partial product is the same as the multiplicand. When the multiplier bit is 0, the partial product is zero. Shift each successive partial product one bit to the left. 4.Add each successive partial product to the sum of the previous partial products to get the final product. 5.If the sign bit that was determined in step 1 is negative, take the 2’s complement of the product. If positive, leave the product in true form. Attach the sign bit to the product.

Question 33

Division, whats result and how to get it:

Answer 33

Dividend / Divisor = Quotient the quotient is the number of times that the divisor will go into the dividend. Step 1: determine the sign of quotient for the final result, set it to 0, make sure both numbers are in true form. 3. Subtract the divisor from the dividend using 2’s complement addition to get the first partial remainder and add 1 to the quotient. If this partial remainder is positive, go to step 3. If the partial remainder is zero or negative, the division is complete.

Question 34

Hexadecimal

Answer 34

Base 16, 0 to F. Most digital systems process binary data in groups that are multiples of four bits, making the hexadecimal number very convenient because each hexadecimal digit represents a 4-bit binary number

Question 35

Binary to Hexadecimal conversion

Answer 35

Break the binary number into 4-bit groups, starting at the right-most bit and replace each 4-bit group with the equivalent hexadecimal symbol.

Question 36

Hexadecimal-to-Binary Conversion

Answer 36

To convert from a hexadecimal number to a binary number, reverse the process and replace each hexadecimal symbol with the appropriate four bits.

Question 37

Hexadecimal to decimal:

Answer 37

multiply the decimal value of each hexadecimal digit by its weight and then take the sum of these products. The weights of a hexadecimal number are increasing powers of 16 (from right to left) .

Question 38

Decimal-to-Hexadecimal Conversion

Answer 38

Repeated division of a decimal number by 16 will produce the equivalent hexadecimal number, formed by the remainders of the divisions. The first remainder produced is the least significant digit (LSD). Each successive division by 16 yields a remainder that becomes a digit in the equivalent hexadecimal number. Note that when a quotient has a fractional part, the fractional part is multiplied by the divisor to get the remainder.

Question 39

Octal

Answer 39

Base 8. 0-7. Each octal digit can be represented by 3 binary bit.

Question 40

Octal to decimal conversion:

Answer 40

multiplying each | digit by its weight and summing the products.

Question 41

Decimal to Octal

Answer 41

Repeated division by 8. Each successive division by 8 yields a remainder that becomes a digit in the equivalent octal number. The first remainder generated is the least significant digit (LSD).

Question 42

Octal to Binary:

Answer 42

To convert an octal number to a binary number, simply replace each octal digit with the appropriate three bits.

Question 43

Binary to Octal

Answer 43

Start with the right-most group of three bits and, moving from right to left, convert each 3-bit group to the equivalent octal digit. If there are not three bits available for the left-most group, add either one or two zeros to make a complete group. These leading zeros do not affect the value of the binary number.

Question 44

Binary coded decimal

Answer 44

A way to represent each of binary digits with binary code. 8421 BCD has a 4 bit value for each decimal. 0: 0000 1: 0001 : 9: 1001 4 bit can form 16 values. Invalid codes are the 6 values we don't use: 1010, 1011, 1100, 1101, 1110, 1111.

Question 45

BCD to decimal and | decimal to BCD

Answer 45

Start at the right-most bit and break the code into groups of four bits. Then write the decimal digit represented by each 4-bit group. replace each decimal digit with the appropriate 4-bit code, to get the BCD.