Unit 1(stld)

 Unit-1 : Boolean Algebra
 Unit-2 : Minimization of Switching Functions
 Unit-3 : Combinational Logic Design
 Unit-4 : Programmable Logic Devices, Threshold Logic
 Unit-5 : Sequential Circuits
 Unit-6 : Algorithmic State Machines

 Digital Design: Morris Mano, PHI,2nd Edition.
 Switching & Finite Automata Theory-Zvi Kohavi,
TMH, 2nd Edition.

Octal :
16 = 8¹ x 2 + 8º x 0 => (16)10 = (20)8
32 = 8¹ x 4 + 8º x 0 => (32)10 = (40)8
20, 21, 22, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 37, 40
Hexadecimal :
16 = 16¹ x 1 + 16º x 0 => (16)10 = (10)16
32 = 16¹ x 2 + 16º x 0 => (32)10 = (20)8
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B , 1C, 1D, 1E, 1F, 20

1.2-) What is the exact number of bytes in a system
that contains (a) 32K byte, (b)64M bytes, and
(c)6.4G byte ?
(a) 32K byte:
1K = 2¹º = 1,024
32K = 32 x 2¹º = 32 x 1,024 = 32,768
32K byte = 32,768 byte

(b) 64M byte:
1M = 2²º = 1,048,576
64M = 64 x 2²º = 64 x 1,048,576 = 67,108,864
64M byte = 67,108,864 byte
(c) 6.4G byte:
1G = 2³º = 1,073,741,824
6.4G = 6.4 x 2³º = 6.4 x 1,073,741,824 = 6,871,747,674
6.4G byte = 6,871,747,674 byte

1.3-) What is the largest binary number that can be
expressed with 12 bits? What is the equivalent
decimal and hexadecimal ?
Binary:
(111111111111)2
Decimal:
(111111111111)2 = 1x 2º+ 1 x 2¹ + 1 x 2² +…..+ 1 x 2¹¹ + 1 x 2¹²
(111111111111)2 = 4,095
Hexadecimal:
(1111 1111 1111)2
F F F
= (FFF)16

1.4-) Convert the following numbers with the
indicated bases to decimal : (4310)5 , and (198)12 .
(4310)5 = 0 x 5º + 1 x 5¹ + 3 x 5² + 4 x 5³ = 0 + 5 + 75 + 500
(4310)5 = 580
(198)12 = 8 x 12º + 9 x 12¹ + 1 x 12² = 8 + 108 + 144
(198)12 = 260

1.7-) Express the following numbers in decimal :
(10110.0101)2 , (16.5)16 .
( 1 0 1 1 0 . 0 1 0 1 )2
4 3 2 1 0 -1 -2 -3 -4
(10110.0101)2 = 2 + 4 + 16 + (1/4) + (1/16)
(10110.0101)2 = 22.3125
( 1 6 . 5 )16
1 0 -1
(16.5)16 = 6 + 16 + (5/16)
(16.5)16 = 22.3125
= 2¹ + 2² + (2^4) +( 2^-2) + (2^-4)

1.8-) Convert the following binary numbers to
hexadecimal and to decimal : (a) 1.11010
(a) ( 1 . 1101 0 )2 = ( 1 . D )16 = 1 x 16º + D x (16^-1)
1 D 0 0 -1

1.9-) Convert the hexadecimal number 68BE to
binary and then from binary convert it to octal .
(68BE)16
Binary form:
(0110 1000 1011
1110)2=(0110100010111110)2
6 8 B E
Octal form:
(0 110 100 010 111 110)2
0 6 4 2 7 6 =(064276)8

(a) 1.10-) Convert the decimal number 345 to
binary in two ways :
Convert directly to binary;
Convert first to hexadecimal, then from
hexadecimal to binary. Which method is faster
?

Method 1:
(345)10
Number Divided by 2 Remainder
345 345/2=172 1
172 172/2=86 0
86 86/2=43 0
43 43/2=21 1
21 21/2=10 1
10 10/2=5 0
5 5/2=2 1
2 2/2=1 1

Method 2:
Number
Divided by
16
Remainder
345 345/16=21 9
21 21/16=1 5
(345)10=(159)16 (1 101 1001)2

1.11-) Do the following conversion problems :
(a) Convert decimal 34.4375 to binary .
(b) Calculate the binary equivalent of 1/3 out
to 8 places.
Then convert from binary to decimal. How
close is the result to 1/3 ?
(c) Convert the binary result in (b) into
hexadecimal. Then convert the result to
decimal . Is the answer the same ?

(a) 34.4375
34 0.4375
34:2=17 r=0
17:2=8 r=1
8:2=4 r=0
4:2=2 r=0
2:2=1 r=0
34=(100010)2
0.4375*2=0.875 r=0
0.875*2=1.75 r=1
0.75*2=1.5 r=1
0.5*2=1.0 r=1
0*2=0 r=0
0.4375=(0.01110)2
34.4375=(100010.01110)2

(b) 1/3=0.3333…
0.33333*2=0.66666 r=0
0.66666*2=1.33332 r=1
0.33332*2=0.66664 r=0
0.66664*2=1.33328 r=1
.
.
.
0.3333…=(0.010101….)= 0+ ¼ + 0 + 1/8 +
0 + 1/32 +… =~0.33333…

(c)
0.010101010…=0.0101 0101 0101
(0.555..)16=5/16 +5/256 +5/4096 +…=~0.33203

1.12-) Add and multiply the following numbers
without
converting them to decimal.
(a) Binary numbers 1011 and 101 .
(a) 1011 (11) 1011(11)
101 (5) 101(5)
+__________ x_____
10000(16) 1011
0000
+ 1011
_________
110111 (55)

1.13-) Perform the following division in binary :
1011111 ÷ 101 .
(1011111)2=95
(101)2=5
95/5=19 (10011)2
1011111 101
101 10011
000111
101
0101
101
0000

1.14-) Find the 9’s- and the 10’s-complement of
the following decimal numbers :
(a) 98127634 (b) 72049900 (c) 10000000 (d)
00000000 .
9’s comlements :
(a) 99999999-98127634=01872365
(b) 99999999-72049900=27950099
(c) 99999999-10000000=89999999
(d) 99999999-0000000=99999999

10’s complements
(a)100000000- 98127634= 01872366
(b)100000000-72049900=27950100
(c)100000000-10000000=90000000

1.16-) Obtain the 1’s and 2’S complements of the
following binary numbers :
(a)11101010 (b)01111110 (c)00000001
(d)10000000
1’s complements:
(a) 00010101 (b)10000001 (c)11111110 (d)01111111
2’s complement :
(a) 00010110 (b)10000010 (c)11111111 (d)10000000

1. Axiomatic definition of Boolean algebra
2. Binary operators
3. Postulates and Theorems
4. Switching functions
5. Canonical forms and standard forms
6. Simplification of switching functions using
theorems

1. Axiomatic definition of Boolean algebra
2. Binary operators

Postulate 2 (a) x+0 = x (b) x.1 = x
Postulate 5 (a) x+x’ = 1 (b) x.x’ = 0
Theorem 1 (a) x+x = x (b) x.x = x
Theorem 2 (a) x+1 = 1 (b) x.0 = 0
Theorem3, involution (x’)’ = x
Postulate3, commutative (a) x+y = y+x (b) xy = yx
Theorem4, associative (a) x+(y+z)=(x+y)+z (b) x(yz) = (xy)z
Postulate4, distributive (a) x(y+z)=xy+xz (b) x+yz = (x+y)(x+z)
Theorem5, DeMorgan (a) (x+y)’ = x’y’ (b) (xy)’ = x’+y’
Theorem6, absorption (a) x+xy = x (b) x(x+y)=x

x y x.y x y x+y x x’
0 0 0 0 0 0 0 1
0 1 0 0 1 1 1 0
1 0 0 1 0 1
1 1 1 1 1 1
x.(y+z) = (x.y)+(x.z)
x y z Y+z x.(y+z) x.y x.z (x.y)+x.z
0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0
0 1 0 1 0 0 0 0
0 1 1 1 0 0 0 0
1 0 0 0 0 0 0 0
1 0 1 1 1 0 1 1
1 1 0 1 1 1 0 1
1 1 1 1 1 1 1 1

Operator Precedence
1.( )
2.NOT
3.AND
4.OR

x y z F1 F2 F3 F4
0 0 0 0 0 0 0
0 0 1 0 1 1 1
0 1 0 0 0 0 0
0 1 1 0 0 1 1
1 0 0 0 1 1 1
1 0 1 0 1 1 1
1 1 0 1 1 0 0
1 1 1 0 1 0 0

x
y
z
F1
z
y
F2
x
(a) F1 = xyz’
(b) F2 = x+y’z
(c) F3 = x’y’z+x’yz+xy’
F3
x
y
z

F4
(c) F4 = xy’+x’z
x
y
z
Implementation of Boolean Function with GATES

1. x+x’y = (x+x’)(x+y)
= 1.(x+y)=x+y
2. x(x’+y) = xx’+xy
= 0+xy=xy
3. x’y’z+x’yz+xy’
= x’z(y’+y)+xy’
= x’z+xy’
4. xy+x’z+yz (Consensus Theorem)
=xy+x’z+yz(x+x’)
=xy+x’z+xyz+x’yz
=xy(1+z)+x’z(1+y)
=xy+x’z
5. (x+y)(x’+z)(y+z)=(x+y)(x’+z)
by duality from function 4

(A+B+C)’ = (A+X)’
= A’X’
= A’.(B+C)’
= A’.(B’C’)
= A’B’C’
(A+B+C+D+…..Z)’ = A’B’C’D’…..Z’
(ABCD….Z)’ = A’+B’+C’+D’+….+Z’
Example using De Morgan’s Theorem (Method-1)
F1 = x’yz’+x’y’z
F1’ = (x’yz’+x’y’z)’
= (x+y’+z)(x+y+z’)
F2 = x(y’z’+yz)
F2’= [x(y’z’+yz)]’
= x’+(y+z)(y’+z’)

F1 = x’yz’ + x’y’z
Dual of F1 = (x’+y+z’)(x’+y’+z)
Complement  F1’ = (x+y’+z)(x+y+z’)
F2 = x(y’z’+yz)
Dual of F2=x+[(y’+z’)(y+z)]
Complement =F2’= x’+ (y+z)(y’+z’)

Minterm or a Standard Product
n variables forming an AND term provide 2n possible
combinations, called minterms or standard products (denoted as
m1, m2 etc.).
Variable primed if a bit is 0
Variable unprimed if a bit is 1
Maxterm or a Standard Sum
n variables forming an OR term provide 2n possible
combinations, called maxterms or standard sums (denoted as
M1,M2 etc.).
Variable primed if a bit is 1
Variable unprimed if a bit is 0

MINTERMS MAXTERMS
x y z Term Designation Term Designation
0 0 0 x’y’z’ m0 x+y+z M0
0 0 1 x’y’z m1 x+y+z’ M1
0 1 0 x’yz’ m2 x+y’+z M2
0 1 1 x’yz m3 x+y’+z’ M3
1 0 0 xy’z’ m4 x’+y+z M4
1 0 1 xy’z m5 x’+y+z’ M5
1 1 0 xyz’ m6 x’+y’+z M6
1 1 1 xyz m7 x’+y’+z’ M7

x y z Function f1 Function f2
0 0 0 0 0
0 0 1 1 0
0 1 0 0 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
f1 = x’y’z+xy’z’+xyz =m1 + m4 + m7
f2 = x’yz+xy’z+xyz’+xyz = m3 + m5 + m6 +
m7

f1 = x’y’z+xy’z’+xyz
f1’ = x’y’z’+x’yz’+x’yz+xy’z+xyz’
f1 =(x+y+z)(x+y’+z)(x+y’+z’)(x’+y+z’) (x’+y’+z)
= M0.M2.M3.M5.M6
= M0M2M3M5M6
f2 = x’yz+xy’z+xyz’+xyz
f2’ = x’y’z’+x’y’z+x’yz’+xy’z’
f2 = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)
= M0 M1 M2 M4

Boolean functions expressed as a sum of
minterms or product of maxterms are said to
be in canonical form.
m3+m5+m6+m7 or M0 M1 M2 M4

Example: F = A+B’C
F = A(B+B’)+B’C(A+A’)
= AB+AB’+AB’C+A’B’C
= AB(C+C’)+AB’(C+C’)+AB’C+A’B’C
= ABC+ABC’+AB’C+AB’C’+AB’C+A’B’C
= A’B’C+AB’C’+AB’C+ABC’+ABC
= m1+m4+m5+m6+m7
F(A,B,C)=(1,4,5,6,7)
ORing of term AND terms of variables A,B &C
They are minterms of the function

Example: F = xy+x’z
F = xy+x’z
F = (xy+x’)(xy+z) distr.law (x+yz)=(x+y)(x+z)
= (x+x’)(y+x’)(x+z)(y+z)
= (x’+y)(x+z)(y+z)
= (x’+y+zz’)(x+z+yy’)(y+z+xx’)
=
(x’+y+z)(x’+y+z’)(x+z+y)(x+z+y’)(y+z+x)(y+z+x’
)
= (x+y+z)(x+y’+z)(x’+y+z)(x’+y+z’)
= M0 M2 M4 M5
F(x,y,z) = (0,2,4,5)
ANDing of terms Maxterms of the function (4 OR
terms of variables
x,y&z)

F(A,B,C) = (1,4,5,6,7)
 sum of minterms
F’(A,B,C) = (0,2,3)
= m0+m2+m3
F(A,B,C) = (m0+m2+m3)’
= m0’.m2’.m3’
= M0 M2 M3
= (0,2,3)
 Product of maxterms
Similarly
F(x,y,z) = (0,2,4,5)
F(x,y,z) = (1,3,6,7)

Sum of Products (OR operations)
F1 = y’+xy+x’yz’
(AND term/product term)
Product of Sums (AND operations)
F2=x(y’+z)(x’+y+z’+w)
(OR term/sum term)
Non-standard form
F3=(AB+CD)(A’B’+C’D’)
Standard form of F3
F3=ABC’D’ + A’B’CD

x y F0 F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
Operator
symbols
+ ,  , 
F0 = 0 F1 = xy F2 = xy’ F3 = x
F4 = x’y F5 = y F6 = xy’ +x’y F7= x +y
F8 = (x+y)’ F9 = xy +x’y’ F10 = y’ F11 = x +y’
F12 = x’ F13 = x’ + y F14 = (xy)’ F15 = 1

 Equivalence is also known as equality, coincidence,
and exclusive NOR.
 16 logic operations are obtained from two variables
x & y
 Standard gates used in digital design are:
complement, transfer, AND, OR , NAND, NOR, XOR &
XNOR (equivalence).

NAME GRAPHIC
SYMBOL
ALGEBRIC
FUNCTION
TRUTH
TABLE
AND F=XY X Y F
0 0 0
0 1 0
1 0 0
1 1 1
OR F=X+Y X Y F
0 0 0
0 1 1
1 0 1
1 1 1
X
Y
F
X
Y
F

NAME GRAPHIC
SYMBOL
ALGEBRIC
FUNCTION
TRUTH
TABLE
Inverter
F=X’
X F
0 1
1 0
Buffer
F=X
X F
0 0
1 1
NAND F=(XY)’
X Y F
0 0 1
0 1 1
1 0 1
1 1 0
X F
X F
X F
Y

NAME GRAPHIC
SYMBOL
ALGEBRIC
FUNCTION
TRUTH
TABLE
X
NOR F=(X+Y)’
X Y F
0 0 1
0 1 0
1 0 0
1 1 0
Exclusive-OR
(XOR)
F=XY’+X’Y
= X  Y
X Y F
0 0 0
0 1 1
1 0 1
1 1 0
Exclusive-NOR
or
Equivalence
F=XY+X’Y’
=X Y
X Y F
0 0 1
0 1 0
1 0 0
1 1 1
Y F
X F
Y
X F
Y

Y (X Y) Z=(X+Y) Z’
Y
x
(X+Y)’
=XZ’+YZ
’
[Z+(X+Y)’]’
(Y+Z)’
(X ( Y Z)=X’(Y+ Z)
=X’Y+X’
Z
[X+(Y+Z)’]’
Z
X
Z
Demonstrating the nonassociativity of the NOR operator
(X  Y) Z  X (Y Z)

XY
Z
(X+Y+Z)
’
X
Y
Z
(XYZ)’
(a) There input NOR gate (b) There input NAND gate
A
B
C
D
E
F=[(ABC)’. (DE)’]’=ABC+DE
(c) Cascaded NAND gates
Multiple-input AND cascaded NOR and NAND gates

X
Y
Z F=X  Y  Z
(a) Using two input
gates
X
Y
Z
(b) Three input gates
(b) Three input exclusive OR gates
TRUTH TABLE
X Y Z F
0 0 0 0 1
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 0
XOR
XNOR
Odd
functio
n
Even
functio
n
F=X  Y  Z

Signal amplitude assignment and type of logic
1 H
0
L
0
1
H
L
LOGIC
VALUE
SIGNAL
VALUE
LOGIC
VALUE
SIGNAL
VALUE
Negative Logic
Positive Logic

DEMONSTRATION OF POSITIVE AND
X y z
1 1 0
1 0 1
0 1 1
0 0 1
Truth table for negative
logic
L=1 H=0
NEGATIVE LOGIC
x
z
y
Graphic symbol for negative
logic NOR gate
Same gate can function
+ive logic NAND or -ive logic NOR
+ive logic NOR or -ive logic
NAND

 F(A,B,C)=(1,4,5,6,7)
 F(A,B,C)=(1,2,3,6,7)
 F(x,y,z)=(1,4,5,6,7)
 F(x,y,z) = (0,2,4,5)

Unit 1(stld)

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Unit 1(stld)