The document explains the concepts of computer organization and architecture, highlighting their significance in designing and operating computer systems. It discusses the five functional components of a computer, including input devices, memory units, arithmetic and logic units, output devices, and control units. The document also covers the von Neumann architecture and the application of boolean algebra and logic gates in digital systems, including methods for simplifying boolean expressions using Karnaugh maps.

1. Unit-1
Boolean Algebra and Logic Gates

2. Computer Organization and Architecture
Computer Architecture-It describes how a computer system is designed. It is a
blueprint for design. So, while designing a computer system, architecture is
decided first.
Computer Organization-It describes how a computer system works. Computer
organization is decided after the architecture.
EXAMPLE: Say you are in a company that manufactures cars, design and all low-
level details of the car come under computer architecture (abstract,
programmers view), while making it’s parts piece by piece and connecting
together the different components of that car by keeping the basic design in
mind comes under computer organization (physical and visible).

3. Conti…

4. Functional blocks of a computer
 A computer consists of five functionally components:
 1. input device
 2. memory unit
 3. arithmetic and logic unit
 4. output device
 5. control unit

5. Conti…

6. Conti…
 Input Unit-Computers accept coded information through input units. The most
common input device is the keyboard. Whenever a key is pressed, the
corresponding letter or digit is automatically translated into its corresponding
binary code and transmitted to the processor. Microphones can be used to
capture audio input which is then sampled and converted into digital codes for
storage and processing. Similarly, cameras can be used to capture video input.
 Memory Unit: The function of the memory unit is to store programs and data.
There are two classes of storage, called primary and secondary.
 Primary Memory: Primary memory, also called main memory, is a fast memory
that operates at electronic speeds. Programs must be stored in this memory
while they are being executed.The memory consists of a large number of
semiconductor storage cells, each capable of storingone bit of information.

7. Conti…
 Secondary Storage: Although primary memory is essential, it tends to be
expensive and does not retain information when power is turned off. Thus
additional, less expensive, permanent secondary storage is used when large
amounts of data and many programs have to be stored, particularly for
information that is accessed infrequently. Access times for secondary storage
are longer than for primary memory. A wide selection of secondary storage
devices is available, including magnetic disks, optical disks (DVD and CD), and
flash memory devices.

8. Conti…
 Arithmetic and Logic Unit: Most computer operations are executed in the
arithmetic and logic unit (ALU) of the processor. Any arithmetic or logic
operation, such as addition, subtraction, multiplication, division, or
comparison of numbers, is initiated by bringing the required operands into
the processor, where the operation is performed by the ALU. For example, if
two numbers located in the memory are to be added, they are brought into
the processor, and the addition is carried out by the ALU. The sum may then
be stored in the memory or retained in the processor for immediate use.
When operands are brought into the processor, they are stored in high-speed
storage elements called registers. Each register can store one word of data.

9. Conti…
 Output Unit: The output unit is the counterpart of the input unit. Its function is to
send processed results to the outside world. A familiar example of such a device is
a printer. Some units, such as graphic displays, provide both an output function,
showing text and graphics, and an input function, through touchscreen capability.
 Control Unit: The memory, arithmetic and logic, and I/O units store and process
information and perform input and output operations. The operation of these units
must be coordinated in some way. This is the responsibility of the control unit. The
control unit is effectively the nerve center that sends control signals to other units
and senses their states. Control circuits are responsible for generating the timing
signals that govern the transfers and determine when a given action is to take
place. In practice, much of the control circuitry is physically distributed
throughout the computer. A large set of control lines (wires) carries the signals
used for timing and synchronization of events in all units.

10. Von Neumann Architecture
 The Von-Neumann Architecture or Von-Neumann model is also known as
“Princeton Architecture”. This architecture was published by the
Mathematician John Von Neumann in 1945. Von Neumann architecture is the
design upon which many general purpose computers are based. This
architecture implemented the stored program concept in which the data and
instructions are stored in the same memory. This architecture consists of a
CPU(ALU, Registers, Control Unit), Memory and I/O unit.

11. Conti…
Von Neumann Architecture

12. Conti…

13. Boolean Algebra and Logic Gates
 It is used to perform the logical operations in digital computer.
 In digital computer True represent by ‘1’ and False represent by ‘0’
 Logic gates are the basic building blocks of any digital system. It is an
electronic circuit having one or more than one input and only one output. The
relationship between the input and the output is based on a certain logic. Based
on this, logic gates are named as AND gate, OR gate, NOT gate, NAND gate,
XOR gate, XNOR gate etc.

14. AND Gate
Logic diagram
Truth Table
A circuit which performs an AND operation is shown in figure. It has n
input (n >= 2) and one output.

15. OR Gate-
Logic diagram
Truth Table
A circuit which performs an OR operation is shown in figure. It has n input (n >=2)
and one output.

16. NOT Gate-
NOT gate is also known as Inverter. It has one input A and one output Y.
Logic diagram
Truth Table

17. NAND Gate-
A NOT-AND operation is known as NAND operation. It has n input (n >=2)
and one output.
Logic diagram
Truth Table

18. XOR Gate-
XOR or Ex-OR gate is a special type of gate. It can be used in the half adder, full adder
and subtractor. The exclusive-OR gate is abbreviated as EX-OR gate or sometime as
X-OR gate. It has n input (n >= 2) and one output.
Logic diagram
Truth Table

19. XNOR Gate-
Logic diagram
Truth Table
XNOR gate is a special type of gate. It can be used in the half adder, full adder and
subtractor. The exclusive-NOR gate is abbreviated as EX-NOR gate or sometime as X-
NOR gate. It has n input (n >= 2) and one output.

20. Boolean laws-

21. Examples-
 1) Minimize the following Boolean expression using Boolean identities
F(A,B,C) = (A+B)(A+C)

22. Conti…
1) Minimize the following Boolean expression using Boolean
identities
F(A,B,C) = (A+B)

23. There are two methods to minimize Boolean
expressions
The Karnaugh Map also called as K Map is a graphical representation that
provides a systematic method for simplifying the boolean expressions.
For a boolean expression consisting of n-variables, number of cells
required in K Map = 2n
cells.

24. Two Variable K Map-
 Two variable K Map is drawn for a boolean expression consisting of two variables.
 The number of cells present in two variable K Map = 22
= 4 cells.
 So, for a boolean function consisting of two variables, we draw a 2 x 2 K Map.
Two variable K Map may be represented as-
Here, A and B are the two variables of the given boolean function.

25. Three Variable K Map-
 Three variable K Map is drawn for a boolean expression consisting of three variables.
 The number of cells present in three variable K Map = 23
= 8 cells.
 So, for a boolean function consisting of three variables, we draw a 2 x 4 K Map.
Three variable K Map may be represented as-

26. Four Variable K Map-
 Four variable K Map is drawn for a boolean expression consisting of four variables.
 The number of cells present in four variable K Map = 24
= 16 cells.
 So, for a boolean function consisting of four variables, we draw a 4 x 4 K Map.
Here, A, B, C and D are the four variables of the given boolean function.
Four variable K Map may be represented as-

27. Conti…

28. Karnaugh Map Simplification Rules-
 To minimize the given boolean function,
 We draw a K Map according to the number of variables it contains.
 We fill the K Map with 0’s and 1’s according to its function.
Then, we minimize the function in accordance with the following rules.
Rule-01:
We can either group 0’s with 0’s or 1’s with 1’s but we can not group 0’s and 1’s together.
X representing don’t care can be grouped with 0’s as well as 1’s.
NOTE
 There is no need of separately grouping X’s i.e. they can be ignored if all 0’s and 1’s are
already grouped.

29. Conti…
Rule-02:
Groups may overlap each other.
Rule-03:
We can only create a group whose number of cells can be represented in the power of 2.
In other words, a group can only contain 2n
i.e. 1, 2, 4, 8, 16 and so on number of cells.
Example-

30. Conti...
Rule-04:
Groups can be only either horizontal or vertical.
We can not create groups of diagonal or any other shape.

31. Conti…
Rule-05:
Each group should be as large as possible.
 Example-

32. Conti…
Rule-06:
Opposite grouping and corner grouping are allowed.
The example of opposite grouping is shown illustrated in Rule-05.
The example of corner grouping is shown below.
Example-

33. Conti…
Rule-07:
There should be as few groups as possible.
PROBLEMS BASED ON KARNAUGH MAP-
Problem-01: Minimize the following boolean function-
F(A, B, C, D) = Σm(0, 1, 2, 5, 7, 8, 9, 10, 13, 15)
Solution-
Since the given boolean expression has 4 variables, so we draw a 4 x 4 K Map. We fill the cells
of K Map in accordance with the given boolean function. Then, we form the groups in
accordance with the above rules.

34. Conti…
Then, we have-
Now,
F(A, B, C, D)
= (A’B + AB)(C’D + CD) + (A’B’ + A’B + AB + AB’)C’D + (A’B’ +
AB’)(C’D’ + CD’)
= BD + C’D + B’D’

35. Conti…
 Thus, minimized boolean expression is-
F(A, B, C, D) = BD + C’D + B’D’
Solution-
Since the given boolean expression has 4 variables, so we draw a 4 x 4 K Map.
We fill the cells of K Map in accordance with the given boolean function.
Then, we form the groups in accordance with the above rules.

36. Conti…
Now,
F(A, B, C, D) = (A’B’ + A’B + AB + AB’)(C’D + CD) + (A’B’ + AB’)(C’D’ + C’D)
= D + B’C’
Thus, minimized boolean expression is-
F(A, B, C, D) = B’C’ + D
Problem-03:
Minimize the following boolean function-
F(A, B, C, D) = Σm(1, 3, 4, 6, 8, 9, 11, 13, 15) + Σd(0, 2, 14)
Solution-
Since the given boolean expression has 4 variables, so we draw a 4 x 4 K Map.
We fill the cells of K Map in accordance with the given boolean function.
Then, we form the groups in accordance with the above rules.

37. Conti…
Now,
F(A, B, C, D)
= (AB + AB’)(C’D + CD) + (A’B’ + AB’)(C’D + CD) + (A’B’ +
AB’)(C’D’ + C’D) + (A’B’ + A’B)(C’D’ + CD’)
= AD + B’D + B’C’ + A’D’
Thus, minimized boolean expression is-
F(A, B, C, D) = AD + B’D + B’C’ + A’D’

38. Conti…
Problem-04:
Minimize the following boolean function-
F(A, B, C) = Σm(0, 1, 6, 7) + Σd(3, 5)
Solution-
Since the given boolean expression has 3 variables, so we draw a 2 x 4 K Map.
We fill the cells of K Map in accordance with the given boolean function.
Then, we form the groups in accordance with the above rules.

39. Conti…
Now,
F(A, B, C) = A'(B’C’ + B’C) + A(BC + BC’)
= A’B’ + AB
Thus, minimized boolean expression is-
F(A, B, C) = AB + A’B’
NOTE-
It may be noted that there is no need of considering the quad group.
This is because even if we consider that group, we will have to consider the other two
duets.
So, there is no use of considering that quad group.

40. Conti…
Problem-05:
Minimize the following boolean function-
F(A, B, C) = Σm(1, 2, 5, 7) + Σd(0, 4, 6)
Solution-
Since the given boolean expression has 3 variables, so we draw a 2 x 4 K Map.
We fill the cells of K Map in accordance with the given boolean function.
Then, we form the groups in accordance with the above rules.

41. Conti…
Now,
F(A, B, C) = (A + A’)(B’C’ + B’C) + A(B’C’ + B’C + BC + BC’) + (A + A’)(B’C’ + BC’)
= B’ + A + C’
Thus, minimized boolean expression is-
F(A, B, C) = A + B’ + C’
Problem-06:
Minimize the following boolean function-
F(A, B, C) = Σm(0, 1, 6, 7) + Σd(3, 4, 5)

42. Conti…
Solution-
Since the given boolean expression has 3 variables, so we draw a 2 x 4 K Map.
We fill the cells of K Map in accordance with the given boolean function.
Then, we form the groups in accordance with the above rules.

43. Conti…
Now,
F(A, B, C) = (A + A’)(B’C’ + B’C) + A(B’C’ + B’C + BC + BC’)
= B’ + A
Thus, minimized boolean expression is-
F(A, B, C) = A + B’