1. Digital Logic & Design
Lecture 02

2. Recap Last lecture discussion Decimal Number Systems
Caveman Base 5 Number System
Binary Number System
Number System Conversion
Today’s lecture discussion
Decimal-Binary Conversion
In the second part of the last lecture we discussed three Number systems.
The decimal number system, a hypothetical base 5 caveman number system and the binary number system used by digital systems.
All three number systems are positional number systems. Roman number system is not a positional number system.
Number System Conversion.
I hope you had the chance to practice with caveman and binary numbers and calculating their decimal equivalents.
Today we would be starting our discussion on number systems by looking at binary to decimal and decimal to binary conversion.
Learning to convert numbers from binary to decimal and decimal to binary is important as all real world values are represented in terms of decimal numbers and all digital systems represent values in terms of binary.
Thus you would like to know how to represent in binary and how the binary output of a digital system can be represented in decimal.

3. Binary to Decimal Conversion
Sum-of-Weights
Expression base number & weights
Sum terms
Paper and pencil method
Sum of non-zero terms
Mental Arithmetic, quick method
Sum of weights of non-zero terms
Lets first look at the conversion from binary number system to decimal number system.
In the last lecture I had mentioned that there are two methods to convert a binary number to decimal.
The Sum-of-Weights method and the sum of non-zero terms.
Lets consider both methods.
We have already been converting binary numbers to decimal using the Sum-of-Weights.
Converting the binary number using the Sum-of-Weights method requires writing an expression in terms of the base number and weights.
(1 x 24) + (0 x 23) + (0 x 22) + (1 x 21) + (1 x 20)
Solving the expression results in the terms 16, 0, 0, 2 and 1 which add up to 19.

4. Binary to Decimal Conversion
Sum-of-Weights
100112
(1 x 24) + (0 x 23) + (0 x 22) + (1 x 21)
+ (1 x 20)
Terms 16, 0, 0, 2 and 1 19
Converting the binary number using the Sum-of-Weights method requires writing an expression in terms of the base number and weights.
(1 x 24) + (0 x 23) + (0 x 22) + (1 x 21) + (1 x 20)
Solving the expression results in the terms 16, 0, 0, 2 and 1 which add up to 19.

5. Binary to Decimal Conversion
Add weights of non-zero terms
Weights increase/decrease by power of 2
= = 19
= /2 + 1/8
= /8 =
A quicker method is to add the weights of non-zero terms.
What are the weights of the non-zero terms for the binary number ?
If you remember the weight of the binary bits increase by a factor of two starting from the least significant bit.
Thus the weights of non-zero terms are 16, 2 and 1. Adding the weights of non-zero terms results in 19.
Adding the weights of non-zero terms involves mental arithmetic and is quick way of converting binary numbers into decimal.
Consider another example of converting to decimal using sum of non-zero terms.
The weights of the non-zero terms are 8, 2, 1, ½ and 1/8 which add up to
You should practice the adding of non-zero terms to find decimal equivalents of binary numbers as it is a convenient and quick way to convert binary into decimal.

6. Decimal to Binary Conversion
Sum-of-Weights method used in reverse
Highest binary weight less than the decimal number
Subsequent smaller weights that add up to decimal number
Repeated division by 2
Paper and pencil method
Number repeatedly divided by 2
Conversion from decimal to binary can be done using two different methods.
Sum of weights method is used in reverse involves mental arithmetic and allow for quick conversion of small decimal numbers.
The repeated division by 2 is a paper and pencil method used for converting large decimal numbers to binary.
In the Sum-of-weights method the conversion starts by noting the highest weight that is smaller than the decimal number to be converted.

7. Decimal to Binary Conversion using Sum-of-Weights
number
Weight
Result after subtraction
Binary Bit
392
256
= 136 b8
136
128
= 8 b7 8 64 b6 32 b5 16 b4
8-8 =0 b3 4 b2 2 b1 1 b0
Consider the decimal number 392 which is to be converted into binary. The highest weight that is smaller than 392 is 256. The next highest weight is 512 which is not smaller than 392.
As the next step in the conversion process 1 is written at bit position 8 to represent the weight 256.
Why bit position 8? Because bit position 8 represents the weight 256.
The weight 256 is subtracted from 392 to result in 136.
The process of determining the next highest weight which is smaller than 136 is determined, which in this particular case is 128.
Thus another 1 is written at bit position 7 to represent the weight 128.
The weight 128 is subtracted from the previous result of subtraction 136, which results in 8.
The process of comparing the result with the next highest weight is repeated.
The weight 64 is greater than the result 8, thus a zero is added to the binary bits ‘11’
Successive lower weights 32 and 16 both are larger than 8. Thus two more binary bits are set to ’00’.
The next lower weight is exactly equal to 8. The corresponding binary bit is set to ‘1’.
The result 8 is subtracted from the weight 8 resulting in 0.
The next three weights are all greater than 0, therefore three 0 bits are added to signify 0 weights.
The binary number is

8. Decimal to Binary Conversion
number
Quotient after division
Remainder after division
392
196
(b0) 98
(b1) 49
(b2) 24
(b3) 12
(b4) 6
(b5) 3
(b6) 1
(b7)
(b8)
Repeated Division-by-2 method allows decimal numbers of any magnitude to be converted into binary.
The original number is divided by 2, its quotient and the remainder is noted.
The quotient obtained in the first step is again divided and the new quotient and remainder value noted.
The process of repeatedly dividing the previous quotient value and noting the remainder continues until the quotient value becomes 0.
The remainders noted in the repeated division are written out in reverse to represent the binary equivalent of the decimal number.

9. Binary-Decimal fraction conversion
Binary to Decimal Conversion
Sum-of-Weights method
Weights decrease by a factor of 2
weights ½, ¼, 1/16
Sum up to
Decimal to Binary Conversion
Repeated Multiplication by 2
example
Converting fractions in decimal number system to Binary and vice versa is also important.
Binary to Decimal conversion uses the Sum-of Weight method used earlier.
The weight of binary bits constituting the fraction part decreases by a factor of 2 starting from the bit to the immediate left of the decimal point.
Thus has the weights ½, ¼ and 1/16 for the non-zero terms.
Adding the non-zero term weights results in 13/16 or

10. Decimal-Binary fraction conversion
Decimal to Binary Conversion
Repeated multiplication by 2
Number
Mult. By 2
Integer
0.8125
1.625
(b-1)
0.625
1.250
(b-2)
0.250
0.500
(b-3)
1.000
(b-4)
Converting fractions from decimal to binary requires repeated multiplication by 2.
The decimal number to be converted into binary is multiplied by 2 which results in
The integer part ‘1’ is noted and the fraction part is multiplied by 2. The result is
The integer part ‘1’ is noted and the fraction part is multiplied by 2. The result is 0.5.
The integer part ‘0’ is noted and the fraction part is multiplied by 2. The result is 1.
Since the fraction part is 0, therefore there is no further need for multiplying with 2.
The integer part is noted and the noted integer values are written out preceded by a decimal point to give

11. Binary Arithmetic Binary Addition Binary Subtraction
Binary Multiplication
Binary Division
We have looked at binary to decimal conversions and decimal to binary conversions.
Now we know how to convert integers as well as fractions from one number system to the other.
Now lets look at simple arithmetic operations such as add, subtract, multiply and divide.
We should be able to perform simple arithmetic operations such as add, subtract, multiply and divide on binary numbers, as we normally do with decimal numbers.
Arithmetic operations performed on binary numbers are in fact exactly like the operations performed on decimal numbers.
Let us start by looking at the binary addition operation

12. Binary Addition 1st digit 2nd digit Sum Carry 1
Four Basic rules for binary addition
1st digit
2nd digit
Sum
Carry 1
Looking at the results the first three additions give a result 0, 1 and 1 respectively which can be represented by a single binary digit (bit).
The fourth addition results in the number 2, which is represented in binary as 102.
Thus two digits (bits) are required. The sum bit is 0 and the carry bit is 1.
This is similar to the addition of in decimal. The answer is 12 which can not be represented by a single digit, thus two digits are required. The number 2 is the sum part and 1 is the carry part.
Addition of multiple binary numbers

13. Binary Addition Carry 1 10 1st Number 2nd Number 3rd Number 4th Number
2nd Number
3rd Number
4th Number
Result
Similar to Decimal addition any number of binary numbers having any number of digits can be added together.
Thus the number 1011, 110, 1000 and 11 can be added together. Most significant digits (bits) of second and fourth numbers are assumed to be zero.
At Bit 2 the carry is 10, is it possible to have a 2-bit carry?
Consider adding the number 19, fifteen times.
Adding the first digit 9, fifteen times results in is the Sum and 13 is the carry.
The carry 13 added with fifteen 1s of the second digit result in 28.
The answer is 285.

14. Binary Subtraction 1st digit 2nd digit Difference Borrow 1
Four Basic rules for binary subtraction
1st digit
2nd digit
Difference
Borrow 1
The 2nd digit is subtracted from the first.
The result of the first, third and fourth subtraction operations result in a difference value where no borrow is required.
The second subtraction subtracts 1 from 0 for which a borrow is required to make the first digit equal to 2. The answer is 1.
This is similar to what is done in decimal where 7 can not be subtracted from 1 therefore a 1 is borrowed from the next most significant digit to allow 7 to be subtracted from 11.
Thus if only the least significant digit of the decimal numbers is considered subtracting 7 from 1 results in a borrow of 1 and a difference of 4.

15. Binary Subtraction
Borrow 1
1st Number
2nd Number
Result

16. Binary Multiplication
Four Basic rules for binary multiplication
1st digit
2nd digit
Product 1
In binary multiplication anything multiplied by 0 results in 0 just like in decimal multiplication.
Example of Binary Multiplication

17. Binary Multiplication
1101 (13)
x (5)
1st product term 1101
2nd product term
3rd product term
Product (65)
Binary multiplication is identical to the decimal multiplication.
The first product term results in 1101
The second product term results in 0000 and is written shifted left by 1 bit
The third product term results in 1101 and is written shifted left by 2 bits
The product result is sum of all the partial product terms

18. Multiplication by shifting left
Decimal 29 shifted left by one digit
290
Shift left 1 digit is multiply by 10
Binary (29) shifted left by one bit
(58)
Shift left 1 bit is multiply by 2
Multiplication operations can be performed by shifting a number towards left.
For example decimal 29 shifted left by 1 digit is like multiplying by 10
Shifting 29 by two digits towards left is multiplying by 102 or 100
Each shift left operation in decimal is equivalent to multiplying a number by 10.
In a binary number system, each left shift is equivalent to multiplying by 2.
Shifting twice towards left is which is multiply by 4.

19. Binary Division 10
101 | 1101
101
011
000 11
Binary division follows the same rules as in decimal division.
The binary division divides (13) by 1012 (5) which results in a quotient value of 102 (2) and a remainder 112 (3) which is in confirmation with equivalent decimal division.

20. Division by shifting right
Decimal 29 shifted right by one digit
2.9
Shift left 1 digit is divide by 10
Binary (29) shifted left by one bit
(14.5)
Shift left 1 bit is divide by 2
Division operations can be performed by shifting a number towards right.
For example decimal 29 shifted right by 1 digit is like dividing by 10
Shifting 29 by two digits towards right is dividing by 102 or 100
Each shift right operation in decimal is equivalent to dividing a number by 10.
In a binary number system, each right shift is equivalent to dividing by 2.
Shifting twice towards right is which is divide by 4.

21. Signed and Unsigned Numbers
Unsigned Binary Numbers
Signed Binary Numbers
Most significant bit represents sign
0 represents a positive number
1 represents a negative number
Digital systems should be able to handle positive and negative numbers.
So far we have only considered positive binary numbers.
How does the digital system represent the + or – symbol?
To handle positive and negative binary numbers, the digital system sets aside the most significant digit (bit) to represent the sign
MSB set to 1 indicates a negative number
MSB set to 0 indicates a positive number
Thus +13 and -13 are represented as and respectively.
Digital systems handle two different types of binary numbers, signed and unsigned.
Thus represents -13 in signed binary and 29 in unsigned binary.
The digital system has to be told beforehand to deal with a binary number as signed or unsigned.

22. 2’s Complement form 1’s complement form 2’s complement form
Binary number (13)
1’s complement 10010
2’s complement (-13)
Informing the digital system beforehand to deal with a number as signed or unsigned is inconvenient
Signed binary numbers are represented in their 2’s complement form.
A 2’s complement of a binary number is achieved by first taking the 1’s complement of a number followed by its 2’s complement.
The 1’s complement of a binary number is obtained by simply inverting each bit.
The 2’s complement of a binary number is obtained by adding a 1 to the 1’s complement of the original number.
In a 2’s complement form all negative binary numbers are represented in their 2’s complement form
All such negative numbers have their most significant bit set to 1 signifying a negative number.
All positive numbers are represented in their original form.
Their most significant bit is a 0 specifying a positive number.

23. Addition and Subtraction with 2’s Complement
There are four cases of addition
Both numbers are positive
Both numbers are negative
the carry generated from the msb is discarded
One number is positive and its magnitude is larger than the negative number
the carry generated from the msb is discarded
One number is positive and its magnitude is smaller than the negative number
By using signed number based on 2’s complement the addition operation serves to add and subtract numbers.

24. Range of Numbers Maximum Range Number of digits Decimal number example
Binary number example
Overflow
The maximum range of values that can be represented in any number system depends upon the number of digits assigned to represent the value.
A 6-digit car odometer can only count up to and then it rolls back to
Similarly an 8-digit calculator can only handle integer numbers of the magnitude
A calculator that reserves the most significant digit to write + or – can only handle a maximum range of integer numbers from to
A 3-bit unsigned binary number can have values ranging between 000 and 111.
The total number of values are 8.
A simple formula can used to determine the total number of unsigned value represented by n binary bits. 2n
The magnitude of the largest binary number that can be represented is one less than the total number or range of values, 2n -1
If 100 and 111 unsigned numbers are added which give a result 1011, this result is considered to be out of range as 4 bits are required.

25. Range of Binary Numbers
Decimal
Number
Sign Magnitude
2n-1, 2n-1 -1
2’s Complement
Unsigned
2n, 2n - 1
0000
000 1
0001
001 2
0010
010 3
0011
011 4
0100
100 5
0101
101 6
0110
110 7
0111
111
Binary numbers can be represented in three different formats or methods
Unsigned
Signed
2’s complement
In the unsigned method only positive numbers can be represented.
To represent positive numbers from 0 to 7, a 3-bit binary combination is required.
In the signed method an extra bit, the most significant bit is used to represent the sign of the number.
0 as the sign bit represents a positive number
1 as the sign bit represents a negative number
Thus the signed representation of numbers 0 to 7 requires 4-bit binary combination.
The 2’s complement representation only converts the negative values into their respective 2’s complement form, the positive values remain in their original form.
Four bits are required to represent positive numbers 0 to 7 in their 2’s complement form.

26. Range of Binary Numbers
Decimal
Number
Sign Magnitude
2n-1-1, -(2n-1 -1)
2’s Complement
2n-1, -2n-1
Unsigned -8
1000 -7
1111
1001 -6
1110
1010 -5
1101
1011 -4
1100 -3 -2 -1
In the unsigned form, negative numbers can not be represented thus numbers -1 to -8 have no unsigned binary representation
In the signed representation the most significant bit is used to represent the sign of the number.
Thus 4-bit signed binary representation is used to represent numbers -1 to -7.
The most significant sign bit is 1 for all the negative numbers
-8 can not be represented using 4-bits as the number 8 in its unsigned form requires 4-bits Representing -8 in signed format requires 5 bits
In the 2’s complement form the numbers -1 to -8 can be represented in their 2’s complement form using 4 bits.
A 4-bit sign magnitude number can handle a number range between -7 and can not be represented as 5-bits are required
A 4-bit 2’s complement based signed number range is between -8 to +7.
Let us find out the best way to store numbers. Since both negative and positive numbers are required, thus the unsigned binary representation is not appropriate as negative numbers can not be represented.
The signed magnitude method although capable of representing both negative and positive numbers however the total range of numbers is one less than the 2’s complement method.
Secondly, direct addition and subtraction of numbers using the signed magnitude representation does not give the actual result. It has to be converted back into signed magnitude format.
2,s complement based representation is the most efficient in terms of range of numbers and the fact that add and subtract operations directly result in a value which is represent able in 2’s complement number system without any translation or change.

27. Summary Binary to Decimal Conversion Sum-of-Weights
Sum of non-zero terms
Decimal to Binary Conversion
Sum-of-Weights (in reverse)
Repeated Division by 2

28. Summary Binary to Decimal fraction conversion Sum-of-Weights
Repeated Multiplication by 2

29. Summary Binary Addition Binary Subtraction Binary Multiplication
Multiplication by shift left operation
Binary Division
Division by shift right operation

30. Summary Unsigned Binary Signed Binary Sign Bit 2’s Complement
Range of Binary Numbers

31. Lecture No. 2 Number Systems We end today’s lecture with this summary.
As I mentioned at the end of my last lecture, you have to practice with binary numbers and conversion from one number system to the other.
You are well aware of the decimal number system. You should start thinking in terms of the Binary Number system.
In the next lecture we would be starting our discussion be looking at an alternate method of representing very large binary numbers having an integer part and a fraction part.
The alternate method uses the floating point notation to represent large numbers.
Good luck.