1. The document discusses number systems and number conversions between binary, decimal, octal and hexadecimal numbering systems. It explains how to convert numbers between these bases using direct and indirect methods. 2. Binary operations like addition, subtraction, multiplication and division are covered. Methods for 1's complement and 2's complement subtraction are explained. 3. Combinational logic and digital logic gates are briefly introduced.

1. Digital Electronics
Prof. Manjunath V Gudur
Department of Electronics and Communication
Department Of Electronics & Communication
Engineering

2. Boolean Algebra and Logic Circuits:Binary numbers, Number Base
Conversion, octal & Hexa Decimal Numbers, Complements, Basic
definitions, Axiomatic Definition of Boolean Algebra, Basic Theorems
and Properties of Boolean Algebra, Boolean Functions, Canonical and
Standard Forms, Other Logic Operations, Digital Logic Gates (Text 3:
1.2, 1.3, 1.4, 1.5,2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7)
Combinational logic: Introduction, Design procedure, Adders- Half
adder, Full adder (Text 3:4.1, 4.2, 4.3)
Basic Electronics - ELN 2
Module-4

3. Each number system is associated with a base or radix
Ø The decimal number system is said to be of base or radix 10.
Ø A number in base r contains r digits 0,1,2,...,r-1.
Basic Electronics - ELN 3

4. Counting/Quantities
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5. Basic Electronics - ELN
• Number Conversions
• Arithmetic Operations
• Complements
5

6. Basic Electronics - ELN 6
Following are the possible conversions among bases

7. Decimal to Binary
Absolute
value
conversion
Fraction
value
conversion
MSD ( Most Significant Digit ) is same as MSB ( Most Significant Bit )
LSD ( Least Significant Digit ) is same as LSB ( Least Significant Bit )

8. Decimal to Octal
Absolute
value
conversion
Fraction
value
conversion
Convert (25)10 = (? )8
25
3 - 1
8
8
Convert (0.23)10 = ( ? )8
So (25)10 = (31)8
MSD
LSD
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9. Decimal to Hexadecimal
Absolute
value
conversion
Fraction
value
conversion
Convert (31)10 = (? )16
31
1 - F
16
16
Convert (0.65)10 = ( ? )16
So (31)10 = (1�)16
MSD
LSD
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10. Basic Electronics - ELN 10

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Number: An-1.....A1A0 . A-1A-2....A-m

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13. Binary to Decimal
(1101.101)2 = (?)10
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(1101.101)2 = (13.625)10

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Octal to Decimal
(5632.471)8 = (?)10

15. Hexa-decimal to Decimal
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16. • Octal and hexadecimal number system provides
convenient way of converting large binary numbers into
more compact and smaller groups.
• There are various ways to convert a binary number into
octal and hexadecimal number. You can convert using
direct methods or indirect methods.
• Binary to decimal and decimal octal/hexadecimal number
(indirect method)
• Direct method uses grouping.
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17. Basic Electronics - ELN 17
OCT Binary
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111
Hex Binary
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
Hex Binary
8 1000
9 1001
A 1010
B 1011
C 1100
D 1101
E 1110
F 1111

19. Ø Take binary number
Ø Separate the binary digits into groups of three bits from
binary point to the left for integer part and to the right for
fraction part.
Ø Then replace each group of binary number from its
equivalent octal digits. That will be octal equivalent of
given binary number.
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Note that you can append any number of 0’s in leftmost bit (or before the
most significant bit) for integer part and append any number of 0’s after
the rightmost bit for fraction part for completing the group of 3 bits, this
does not change value of input binary number.

20. Basic Electronics - ELN 20
Binary to Octal
Octal to Binary
Convert (125.62)8 to Binary
Convert (10101101.0111)2 to Octal

21. Ø Take binary number
Ø Separate the binary digits into groups of four bits from
binary point to the left for integer part and to the right for
fraction part.
Ø Then replace each group of binary number from its
equivalent hexadecimal digits. That will be hexadecimal
equivalent of given binary number.
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Note that you can append any number of 0’s in leftmost bit (or before the
most significant bit) for integer part and append any number of 0’s after
the right most bit for fraction part for completing the group of 4 bits, this
does not change value of input binary number.

22. Binary to Hexadecimal Hexadecimal to Binary
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23. Octal to Hexadecimal
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24. Hexadecimal to Octal
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25. Basic Electronics - ELN 25
Binary arithmetic is used in digital systems mainly because the
numbers are stored in binary format in most computer systems.
All arithmetic operations such as addition, subtraction, multiplication,
and division are done in binary representation of numbers.
•Binary Addition
• Binary Subtraction
• Binary Multiplication
• Binary Division

26. Basic Electronics - ELN 26
A B A+B
0 0 0
0 1 1
1 0 1
1 1 10
A + B Sum Carry Out
0 + 0 0 0
0 + 1 1 0
1 + 0 1 0
1 + 1 0 1
A + B + Carry in Sum Carry Out
0 + 0 + 1 1 0
0 + 1 + 1 0 1
1 + 0 + 1 0 1
1 + 1 + 1 1 1

27. Basic Electronics - ELN 27
Carry: 1 1 1
(6 0) 10 Augend 1 1 1 1 0 0
+ (4 2) 10 Addend 1 0 1 0 1 0
(1 0 2)10 Sum: 1 1 0 0 1 1 0
Carry: 1 1 1
(2 8) 10 Augend 1 1 1 0 0
+ (1 4) 10 Addend 0 1 1 1 0
( 4 2)10 Sum: 1 0 1 0 1 0
1. Add (60)10 with (42)10 using binary addition
2. Add (11100)2 and (1110)2 using binary addition

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Carry: 1 1 1 1
(5 . 7 5) 10
equivalent decimal is 1 0 1 . 1 1
+ (7 . 5 0) 10
equivalent decimal is 1 1 1 . 1 0
(1 3 . 2 5)10 Sum: 1 1 0 1 . 0 1
3. Add (101.11)2 and (111.10)2 using binary addition

29. Basic Electronics - ELN 29
Complements are used in the digital computers in order to simplify
the subtraction operation and for the logical manipulations.
For each radix-r system (base-r) there are two types of complements
they are,
1. r’s complement and
2. (r-1)’s complement.
Therefore for binary number system with r = 2, it has
1. 1’s complement and
2. 2’s complement.

30. Basic Electronics - ELN 30
Subtractions by 1’s Complement method:
The algorithm to subtract two binary numbers (i.e,A - B) using 1’s complement is explained as
follows:
Ø Take 1’s complement of ‘B’ subtrahend
Ø Result = ‘A’ (minuend) + 1’s complement of ‘B’ subtrahend.
Ø In step 2 if there is carry (1) out of MSB bit, then the result is positive and is in true form.
Add that carry to the least significant bit (LSB) of the above result to get final result.
Ø In step 2 if there is no carry out of MSB bit , then the result is negative and is in the 1’s
complement form. (To find the actual difference between ‘A’ and ‘B’ take 1’s complement
of this result and put minus sign infront of the number.)
(Note:
1. The 1's complement of a number is found by changing all 1's to 0's and all 0's to 1's. This is called as taking
complement or 1's complement.
2. Subtrahend is a number that to be subtracted from the another number, i.e., minuend.)

31. Basic Electronics - ELN 31
A = 05 = 1 0 1 (Minuend) ----------------------------------------------------> 1 0 1
- - - +
B = 04 = 1 0 0 (subtrahend) ----> 1’s complement of subtrahend ----> 0 1 1
01 EAC 1 0 0 0
+ 1
0 0 1
The result is positive
There is a carry out of MSB, the result is positive
and hence add this carry to LSB to get final result.
1. Perform (05)10 - (04)10 using binary 1’s complement subtraction.
we can write it as A + (-B) = 5 + (-4).
Now we have to represent +4 in binary and take 1’s complement of it to represent -4.

32. Basic Electronics - ELN 32
A = 05 = 1 0 1 (Minuend) ----------------------------------------------------> 1 0 1
- - - +
B = 04 = 1 0 0 (subtrahend) ----> 1’s complement of subtrahend ----> 0 1 1
01 EAC 1 0 0 0
+ 1
0 0 1
The result is positive
There is a carry out of MSB, the result is positive
and hence add this carry to LSB to get final result.
2. Perform (05)10 - (04)10 using binary 1’s complement subtraction.
we can write it as A + (-B) = 5 + (-4).
Now we have to represent +4 in binary and take 1’s complement of it to represent -4.

33. Basic Electronics - ELN 33
3. Subtract 1000.012 from 1011.102 using binary 1’s complement subtraction.
A = 1011.10 = 11.50 --------------------------------------------------> 1 0 1 1 . 1 0
- - +
B = 1000.01 = 08.25 ---> 1’s complement of subtrahend ---> 0 1 1 1 . 1 0
03.25 1 0 0 1 1 . 0 0
+ 1
0 0 1 1 . 0 1
There is a carry out of MSB, the result is positive
and hence add this carry to LSB to get final result.

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35. Basic Electronics - ELN 35
Subtractions by 2’s Complement method:
The algorithm to subtract two binary numbers (i.e,A - B) using 2’s complement is explained as
follows:
Ø Take 2’s complement of ‘B’ subtrahend.
Ø Result = ‘A’ (minuend) + 2’s complement of ‘B’ subtrahend.
Ø In step 2 if there is carry (1) out of MSB bit then the result is positive and is in true form. in
this case carry is discarded..
Ø In step 2 if there is no carry out of MSB bit , then the result is negative and is in the 2’s
complement form. (To find the actual difference between ‘A’ and ‘B’ take 2’s complement
of this result)
Note:
Ø 2’s complement = 1’s complement+1
Ø Adding end-around carry-bit occurs only in 1’s complement arithmetic operations but not in 2’s complement
arithmetic operations

36. A = 05 = 1 0 1 (Minuend) To get 2’s complement of subtrahend:
- - -
B = 04 = 1 0 0 (subtrahend) --------------------> 1’s complement------> 0 1 1
01 + 1
2’s complement of subtrahend: 1 0 0
Now add Minuend with 2’s complement of subtrahend to get the result:
1 0 1
+
1 0 0
1 0 0 1
Basic Electronics - ELN 36 36
There is a carry out of MSB, discard
the carry and the result is positive.
Final Result = 001
1. Perform (5)10 - (4)10 using binary 2’s complement subtraction.

37. 2. Perform (04)10 - (05)10 using binary 2’s complement subtraction.
A = 04 = 1 0 0 (Minuend) To get 2’s complement of subtrahend:
B = 05 = 1 0 1 (subtrahend) --------------------> 1’s complement---> 0 1 0
- 01 + 1
2’s complement of subtrahend: 0 1 1
Now add Minuend with 2’s complement of subtrahend to get the result:
1 0 0
+
0 1 1
1 1 1
Now take 2’ complement of 111 = 000+1= -001.
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There is no carry out of MSB, result is
negative. To get the actual difference
take the 2’s complement of the result.

38. Basic Electronics - ELN 38
3. Perform 2810 - 1910 using 2’s complement method.

39. 4. Subtract 1000.012 from 1011.102 using binary 2’s complement subtraction.
A = 1011.10 (Minuend) To get 2’s complement of subtrahend:
- -
B = 1000.01 (subtrahend) ----------------> 1’s complement----------> 0 1 1 1 . 1 0
+ 1
2’s complement ----------> 0 1 1 1 . 1 1
Now add Minuend with 2’s complement of subtrahend to get the result:
1 0 1 1 . 1 0
+
0 1 1 1 . 1 1
1 0 0 1 1 . 0 1
Basic Electronics - ELN 39
There is a carry out of MSB, discard
the carry. Result obtained is positive.