The document discusses the arithmetic logic unit (ALU) and its operations. It describes: 1) The ALU performs basic arithmetic operations like addition, subtraction, incrementing and decrementing on numeric data stored in registers. 2) Binary adders and adder-subtractors are used to implement addition and subtraction operations. They are built by connecting full adders in series. 3) An arithmetic circuit can perform the basic arithmetic operations as well as logic operations by setting control signals appropriately. 4) A status register is used to store flags indicating the results of arithmetic and logic operations like carry, overflow, zero etc. This allows instructions to conditionally execute based on previous results.

1. Chapter 9
Arithmetic Logic unit

3. Arithmetic Microperations
• Arithmetic Microoperations perform arithmetic
operation on numeric data stored in registers.
The basic arithmetic micro operations are:-
• Addition
• Subtraction
• Increment
• Decrement
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4. 4
Binary Adder
FA
FA
FA
FA C0
A0
B0
S0
A1
B1
S1
A2
B2
S2
A3
B3
S3
C1
C2
C3
C4
4-bit binary adder (connection of
FAs)
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5. 5
Binary Adder-Subtractor
FA
FA
FA
FA
C0
A0
B0
S0
A1
B1
S1
A2
B2
S2
A3
B3
S3
C1
C2
C3
C4
4-bit adder-subtractor
M
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6. 6
Binary Incrementer
C S
x y
HA
C S
x y
HA
C S
x y
HA
C S
x y
HA
S0
S1
S2
S3
C4
1
A0
A1
A2
A3
4-bit Binary Incrementer
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7. Arithmetic Circuit
The basic arithmetic micro operations (addition,
subtraction, increment and decrement) can be
performed in one composite arithmetic circuit
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8. A
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9. Working
This arithmetic circuit can perform 8 operations among
them some are :-
Addition:-
• When S1 S0 = 0 0, the value of B is applied to the Y
inputs of the adder.
• If Cin = 0, the output D = A + B.
• If Cin =1, output D = A + B + 1.
• Both cases perform the add microoperation with or
without adding the input carry.
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10. Subtraction
• When S1 S0 = 0 1, the value of B is applied to the
Y inputs of the adder.
• If Cin = 1, then D = A + B + 1. this produces A plus
the 2’s complement of B, which is equivalent to a
subtraction of A – B.
• when Cin = 0, then D = A + B. this is equivalent to
a subtract with borrow, that is , A – B – 1.
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11. Increment
• When S1 S0 = 1 0, the inputs from B are neglected ,
and instead, all 0’s are inserted into the y inputs. The
output becomes D = A + 0 + Cin.
• This gives D = A when Cin = 0 and D = A + 1 when Cin
= 1.
• In the first case we have a direct transfer from the
input A to output D and in the second case, the value
of A is incremented by 1.
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12. Decrement
• When S1 S0 = 1 1, all 1’s are inserted into the Y
inputs of the adder to produce the decrement
operation D = A – 1 when Cin = 0. this is because a
number with all 1’s is equal to the 2’s complement
of 1 (the 2’s complement of binary 0001 is 1111).
• Adding number A to the 2’complement of 1
produces F = A + 2’s complement of 1 = A – 1 when
Cin = 1, then D = A – 1 + 1 = A, which causes a direct
transfer from input A to output D.
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13. Logic Microoperations
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• Logic microoperations are bit-wise operations, i.e.,
they work on the individual bits of data.
• It is useful for bit manipulations on binary data.
• Useful for making logical decisions based on the bit
value. There are, in principle, 16 different logic
functions that can be defined over two binary input
variables.
• However, most systems only implement four of
these – AND , OR , XOR and Complement/NOT

14. Hardware implementation
• Hardware implementation of logic
microoperations requires that logic gates be
inserted be each bit or pair of bits in the
resisters to perform the required logic
operation.
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16. Shift Micro-Operations
• There are three types of shifts
• Logical shift
• Circular shift
• Arithmetic shift
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17. Circular Shift operation
• In a circular shift the serial input is the bit that is
shifted out of the other end of the register.
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18. Arithmetic Shift Operation
• An arithmetic shift is meant for signed binary
numbers (integer).
• An arithmetic left shift multiplies a signed number by
two and an arithmetic right shift divides a signed
number by two.
• The main distinction of an arithmetic shift is that it
must keep the sign of the number the same as it
performs the multiplication or division
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An left arithmetic shift operation must be checked for
the overflow

20. Cont…
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An overflow flip-flop V can be used to detect an
arithmetic shift-left overflow. V = Rn-1 ⊕ Rn-2
If V = 0, there is no overflow but if V = 1, overflow is
detected

21. Hardware Implementation of Shift
Microoperations
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22. Status Register
• A status register, flag register, or condition code register
(CCR) is a collection of status flag bits for a processor.
• An example is the FLAGS register of the x86 architecture or
flags in a program status word (PSW) register.
• The status register is a hardware register that contains
information about the state of the processor.
• Individual bits are implicitly or explicitly read and/or
written by the machine code instructions executing on the
processor.
• The status register lets an instruction take action on the
outcome of a previous instruction.
• Typically, flags in the status register are modified as effects
of arithmetic and bit manipulation operations.

23. Status Register

24. Status register

25. Error Correction codes
• Hamming code is a liner code that is useful for error detection up to
two immediate bit errors.
• In Hamming code, the source encodes the message by adding
redundant bits in the message. These redundant bits are mostly
inserted and generated at certain positions in the message to
accomplish error detection and correction process.
• Hamming code method is effective on networks where the data
streams are given for the single-bit errors.
• Hamming code not only provides the detection of a bit error but
also helps you to indent bit containing error so that it can be
corrected.
• The ease of use of hamming codes makes it best them suitable for
use in computer memory and single-error correction.

26. How to Encode a message in
Hamming Code
The process used by the sender to encode the message includes the
following three steps:
1. Calculation of total numbers of redundant bits.
2. Checking the position of the redundant bits.
3. Lastly, calculating the values of these redundant bits.
When the above redundant bits are embedded within the message, it is
sent to the user.
Step 1) Calculation of the total number of redundant bits.
Let assume that the message contains:
n– number of data bits
p – number of redundant bits which are added to it so that np
can indicate at least (n + p + 1) different states.
Here, (n + p) depicts the location of an error in each of (n + p)
bit positions and one extra state indicates no error. As p bits can indicate
2p states, 2p has to at least equal to (n + p + 1).

27. Cont…
Step 2) Placing the redundant bits in their correct position.
The p redundant bits should be placed at bit positions of
powers of 2. For example, 1, 2, 4, 8, 16, etc. They are referred
to as p1 (at position 1), p2 (at position 2), p3 (at position 4), etc.
Step 3) Calculation of the values of the redundant bit.
• The redundant bits should be parity bits makes the number
of 1s either even or odd.
• The two types of parity are ?
• Total numbers of bits in the message is made even is called
even parity.
• The total number of bits in the message is made odd is
called odd parity.

28. Cont….
• Here, all the redundant bit, p1, is must calculated as the
parity. It should cover all the bit positions whose binary
representation should include a 1 in the 1st position
excluding the position of p1.
• P1 is the parity bit for every data bits in positions whose
binary representation includes a 1 in the less important
position not including 1 Like (3, 5, 7, 9, …. )
• P2 is the parity bit for every data bits in positions whose
binary representation include 1 in the position 2 from right,
not including 2 Like (3, 6, 7, 10, 11,…)
• P3 is the parity bit for every bit in positions whose binary
representation includes a 1 in the position 3 from right not
include 4 Like (5-7, 12-15,… )

29. Cont….
Decrypting a Message in Hamming code
• Receiver gets incoming messages which require to
performs recalculations to find and correct errors.
The recalculation process done in the following steps:
• Counting the number of redundant bits.
• Correctly positioning of all the redundant bits.
• Parity check
Step 1) Counting the number of redundant bits
• You can use the same formula for encoding, the number of
redundant bits
2p ? n + p + 1
• Here, the number of data bits and p is the number of
redundant bits.

30. Cont….
Step 2) Correctly positing all the redundant bits
Here, p is a redundant bit which is located at bit
positions of powers of 2, For example, 1, 2, 4, 8, etc.
Step 3) Parity check
Parity bits need to calculated based on data bits and
the redundant bits.
p1 = parity(1, 3, 5, 7, 9, 11…)
p2 = parity(2, 3, 6, 7, 10, 11… )
p3 = parity(4-7, 12-15, 20-23… )

31. Conversion from RS FF into JK FF
• Step 1 : For conversion of SR Flip flop to JK Flip flop at first we
have to make combine truth table for SR flip flop and JK Flip
Flop. In bellow see the combine truth table of SR flip flop and
JK Flip Flop.

32. Accumulator register