The document discusses various types of microoperations including register transfer, arithmetic, logic, and shift microoperations. It provides details on how basic arithmetic operations like addition, subtraction, and increment/decrement are implemented. Logic operations like AND, OR, XOR, and NOT are also covered. Shift operations include logical, circular, and arithmetic shifts. The hardware implementation of these microoperations uses basic logic gates and circuits like half adders, full adders, and multiplexers.

1. REGISTER TRANSFER
AND
MICROOPERATIONS
PART 2
Dr. Prasenjit Dey

2. Types of Microoperations
There are 4 types of microoperations
1. Register transfer microoperations
2. Arithmetic microoperations
3. Logic microoperations (bit manipulations on non-numeric data)
4. Shift microoperations(bit manipulations on non-numeric data)
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3. Arithmetic Microoperations
The basic arithmetic microoperations are: addition, subtraction,
increment, decrement.
Addition Microoperation:
R3 ←R1+R2
Subtraction Microoperation:
R3 ←R1-R2 or R3 ←R1+R2+1
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1’s complement
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4. Arithmetic Microoperations cont.
One’s Complement Microoperation:
R2 ←R2
Two’s Complement Microoperation:
R2 ←R2+1
Increment Microoperation:
R2 ←R2+1
Decrement Microoperation:
R2 ←R2-1
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5. Used for 1-bit addition
Let, x and y are two 1-bit variables. Then, the truth table of the half
adder will be
If we design this truth table in a k-map, then we will get the following
equations corresponding to the carry and sum
Which can be designed by this digital circuit
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carry = xy sum = xy’ + x’y
= x  y
x
y carry
sum
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0
x y carry sum
Hardware implementation: Half Adder

6. Used for n-bit addition
The truth table of the full adder is
If we design this truth table in a k-
map, then we will get the following
equations corresponding to the
carry and sum
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0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1
cn = xy + xcn-1+ ycn-1
= xy + (x  y)cn-1
s = x’y’cn-1+x’yc’n-1+xy’c’n-1+xycn-1
= x  y  cn-1 = (x  y)  cn-1
x
y
cn-1
x
y
cn-1
cn s
x
y
cn-1
S
cn
x y cn-1 cn s
0
0
1
0
0
1
1
1
0
1
0
1
1
0
1
0
Hardware implementation: Full Adder

7. Binary Adder using Full Adder
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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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FA
FA
FA
FA
A3A2A1A0 => 1 0 1 0
+B3B2B1B0=> 0 1 0 0
1 1 1 0

8. Binary Adder-Subtractor
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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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M=0: Addition
M=1: Subtraction
B3 B2 B1 B0
B3 B2 B1 B0
=0 1
M=1: A+B+1
M=1: A+B+0

9. Binary Adder-Subtractor
For unsigned numbers, this gives A – B if A≥B or the 2’s complement of
(B – A) if A < B
(example: 3 – 5 = -2= 1110)
For signed numbers, the result is A – B provided that there is no
overflow. (example : -3 – 5= -8) 1101
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C3
C4
V =
1, if overflow
0, if no overflow
Overflow detector for signed numbers
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10. Binary Incrementer/Decrementer
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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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 Binary Incrementer can also be implemented using a counter
 A binary Decrementer can be implemented by adding 1111 to the
desired register each time!
A+(1111)
=A+2’s of 1
=A-1

11. Arithmetic Circuit
This circuit performs 4 distinct arithmetic operations and the
basic component of it is the parallel adder
The output of the binary adder is calculated from the following
arithmetic sum:
D = A + Y + Cin
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12. Arithmetic Circuit cont.
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B0
3 2 1 0 S1 S0
4×1 MUX
FA
FA
FA
FA Cin
D0
D1
D2
D3
C1
C2
C3
Cout
B0
1 0 S1 S0
B1
3 2 1 0 S1 S0
4×1 MUX
B1
1 0 S1 S0
B2
3 2 1 0 S1 S0
4×1 MUX
B2
1 0 S1 S0
B3
3 2 1 0 S1 S0
4×1 MUX
B3
1 0 S1 S0
A0
A1
A2
A3
4-bit Arithmetic Circuit
X0
Y0
X1
Y1
X2
Y2
X3
Y3
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S0S1(00), cin(0): A+B
S0S1(01), cin(1): A+B+1
S0S1(10), cin(1): A+0+1
S0S1(11), cin(0): A + (-1)
A+(1111)
=A+2’s of 1
=A-1

13. Logic Microoperations
There are in general 4 types of logical microoperations
OR Microoperation
Symbol: , +
Gate:
Example: 1001102  10101102 = 11101102
P+Q: R1←R2+R3, R4←R5 R6
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OR OR

14. Logic Microoperations cont.
AND Microoperation
Symbol: 
Gate:
Example: 1001102  10101102 = 00001102
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PQ: R1←R2R3

15. Logic Microoperations cont.
Complement (NOT) Microoperation
Symbol:
Gate:
Example: 01010012= 10101102
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16. Logic Microoperations cont.
XOR (Exclusive-OR) Microoperation
Symbol: 
Gate:
Example: 1001102  0101102 = 1100002
1 0 0 1 1 0
0 1 0 1 1 0
1 1 0 0 0 0
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17. Complex Logic Microoperations
Selective-set Operation
Used to force selected bits of a register into logic-1 by using the OR
operation
Example: 01002  10002 = 11002
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0 1 0 0
1 1 0 0
1 1 0 0

18. Complex Logic Microoperations cont.
Selective-complement (toggling) Operation
Used to force selected bits of a register to be complemented by using
the XOR operation
Example: 00012  10012 = 10002
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0 0 0 1
1 0 0 1
1 0 0 0

19. Complex Logic Microoperations cont.
Insert Operation
Step1: mask the desired bits
Step2: OR them with the desired value
Example: Let, R1 = 0110 1010, and we desire to replace the leftmost 4
bits (0110) with 1110 then:
◦ Step1: 0110 1010  0000 1111
◦ Step2: 0000 1010  1110 0000
 R1 = 1110 1010
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20. Complex Logic Microoperations cont.
NAND Microoperation
Symbols:  and 
Gate:
Example: 1001102  10101102 = 11110012
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21. Complex Logic Microoperations cont.
NOR Microoperation
Symbols:  and 
Gate:
Example: 1001102  10101102 = 00010012
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22. Complex Logic Microoperations cont.
Set (Preset) Microoperation
Force all bits into 1’s by ORing them with a value in which
all its bits are being assigned to logic-1
Example: 1001102  1111112 = 1111112
Clear (Reset) Microoperation
Force all bits into 0’s by ANDing them with a value in which
all its bits are being assigned to logic-0
Example: 1001102  0000002 = 0000002
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23. Hardware Implementation
The hardware implementation of logic microoperations requires that
logic gates be inserted for each bit or pair of bits in the registers to
perform the required logic function
Most computers use only four (AND, OR, XOR, and NOT) from which all
others can be derived.
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24. Hardware Implementation cont.
S1 S0 Output Operation
0 0 O = Ai  Bi XOR
0 1 O = Ai  Bi OR
1 0 O = Ai  Bi AND
1 1 O = Ai Complement
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S1
S0
0
1
2
3
4×1
MUX
Oi
Ai
Bi
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25. Shift Microoperations
Used for serial transfer of data
Also used in conjunction with arithmetic, logic, and other data-
processing operations
The contents of the register can be shifted to the left or to the right
As being shifted, the first flip-flop receives its binary information from
the serial input
Three types of shift:
Logical Circular Arithmetic
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26. Shift Microoperations cont.
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r0
r1
r3
rn-1
r0
r1
r2
r3
rn-1
Shift Right
Shift Left
Serial Input Serial Output
Serial Output Serial Input
r2
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27. Shift Microoperations: Logical Shifts
Transfers 0 through the serial input
Logical Shift Right: R1←shr R1,1
Logical Shift Left: R2←shl R2,1
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Logical Shift Left
? 0
r0
r1
r2
r3
rn-1
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28. Shift Microoperations: Circular Shifts
Circulates the bits of the register around the two ends without loss of
information
Circular Shift Right: R1←cir R1,1
Circular Shift Left: R2←cil R2,1
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Circular Shift Left
r0
r1
r2
r3
rn-1
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29. Shift Microoperations: Arithmetic Shifts
Shifts a signed binary number to the left or right
An arithmetic shift-left multiplies a signed binary number by 2: ashl
(00100): 01000
An arithmetic shift-right divides the number by 2
 ashr (00100) : 00010
An overflow may occur in arithmetic shift-left, and occurs when the
sign bit is changed (sign reversal)
 ashl (10100) : 01001
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30. Shift Microoperations: Arithmetic Shifts cont.
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Sign
Bit
Arithmetic Shift Left
Sign
Bit
?
0
?
r0
r1
r2
r3
rn-1
r0
r1
r2
r3
rn-1
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ashr (00100) : 00010
ashl (10100) : 010010
Arithmetic Shift Right

31. An overflow flip-flop Vs can be used to detect an arithmetic shift-left
overflow
Vs = Rn-1  Rn-2
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Rn-2
Vs=
Rn-1 1  overflow
0  no overflow
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Shift Microoperations: Arithmetic Shifts cont.

32. Shift Microoperations cont.
Example: Assume R1=11001110, then:
◦ 1-bit Arithmetic shift right : R1 = 11100111
◦ 1-bit Arithmetic shift left : R1 = 10011100
◦ 1-bit Logical shift right : R1 = 01100111
◦ 1-bit Logical shift left : R1 = 10011100
◦ 1-bit Circular shift right : R1 = 01100111
◦ 1-bit Circular shift left : R1 = 10011101
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MSB
LSB

33. Hardware Implementation
A shift unit could be a bidirectional shift register with parallel
load.
Has drawbacks:
◦ Needs two pulses (the clock and the shift signal pulse)
◦ Not efficient in a processor unit where multiple number of registers
share a common bus
It is more efficient to implement the shift operation with a
combinational circuit
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34. Hardware Implementation cont.
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S 1 0 S 1 0 S 1 0 S 1 0
A3A2A1A0
Serial Input IR Serial Input IL
Select
0 for shift right
1 for shift left
H3 H2 H1 H0
MUX MUX MUX MUX
4-bit Combinational Circuit Shifter
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35. Arithmetic Logic Shift Unit
Individual registers performing the microoperations directly
Connect these set of registers with a common operational unit called
an Arithmetic Logic Unit (ALU)
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36. Arithmetic Logic Shift Unit cont.
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0
1
2
3
S3
S2
S1
S0
Bi
Ai
Ai+1
Ai-1
Enable
4×1
MUX
Ci
Ci+1
One stage of
arithmetic
circuit
One stage of
logic circuit
Di
Oi
Fi
shr
shl
One stage of
ALU
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s1s0: Arithmetic/logical
operations
s1s0 Arithmetic Logic
(00): Add XOR
(01): Subtract OR
(10): Increment AND
(11): Decrement NOT
s3s2 (00): Arithmetic operation
s3s2(01): Logic operation
s3s2(10): Right shift operation
s3s2 (11): Left shift operation