Laws of Boolean Algebra

Boolean Algebra

The most obvious way to simplify Boolean expressions is to manipulate them in the same way as normal algebraic expressions are manipulated. With regards to logic relations in digital forms, a set of rules for symbolic manipulation is needed in order to solve for the unknowns.

A set of rules formulated by the English mathematician George Boole describe certain propositions whose outcome would be either true or false. With regard to digital logic, these rules are used to describe circuits whose state can be either, 1 (true) or 0 (false). In order to fully understand this, the relation between the AND gate, OR gate and NOT gate operations should be appreciated.

	AND Form	OR Form
Identity Law	A ● 1 = A	A + 1 = A
Zero and one Law	A ● 0 = 0	A + 0 = 1
Inverse Law	A ● A’ = 0	A + A’ = 1
Idempotent Law	A ● A = A	A + A = A
Commutative Law	A ● B = B ● A	A + B = B + A
Associate Law	A ● (B ● C) = (A ● B) ● C	A + (B + C) = (A + B) + C
Distributive Law	A + (B ● C) = (A + B) ● (A + C)	A ● (B + C) = (A ● B) + (A ● C)
Absorption Law	A (A + B) = A	A + A ● B = A A + A’B = A + B
DeMorgan’s Law	(A’ ● B’) = A’ + B’	(A’ + B’) = A’ ● B’
Double Complement Law	X’’ = X

De Morgan’s Law

Tips: Break the line, change the sign

De Morgan’s Law Logic Gates

by: Muhammad Nasruddin Rosli

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Muted Professor

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Number System Conversion

Number System Conversion

Here we only focus on the conversion of decimal, binary and hexadecimal.

Figure below show the Decimal/Hexadecimal/Binary Conversion

Decimal to binary

Convert 156₁₀ to binary

2|156    0

2|78      0

2|39      1                      

2|19      1

2|9         1

2|4         0

2|2         0

2|1         1

And we read the answer from the bottom to top, therefore 156₁₀ in binary is 10011100_2.

Decimal to hexadecimal

Convert 1432₁₀ to hexadecimal

Weight	163	162	161	160
Value represented	4096	256	16	1
Hexadecimal		5	9	8

1432 – (256x5) = 152

152 – (16x9) = 8

8 – (1x8) = 0

Therefore 1432₁₀ = 598₁₆

Binary to hexadecimal

Convert 11110000.0011₂ to hexadecimal

1111	0000	.	0011
8421	8421	.	8421
1(8) +1(4) +1(2) +1(1) = 15(F)	0(8) +0(4) +0(2) +0(1) = 0	.	0(8) +0(4) +1(2) +1(1) = 3

11110000.0011₂ = F0.3₁₆

Hexadecimal to binary

E186₁₆ = ?

E	1	8	6
1110	0001	1000	0110

Therefore E186₁₆ = 1110000110000110₂

Binary to hexadecimal

11010₂ = ?

Binary	1	1	0	1	0
Weight	24	23	22	21	20
Value	16	8	4	2	1

11010₂ = 1(16) + 1(8) + 0(4) + 1(2) + 0(1)

                = 26

by: Muhammad Nasruddin Rosli

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Muted Professor

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KARNAUGH MAP

What is Karnaugh map?

Karnaugh map or K-Map provides a simple a straight-forward method of minimizing Boolean expressions. The only limitation is that will be ineffective for more than four input.

The K-Map also can also be described as a grid-like representation of truth table. The rows and columns correspond to the possible values of the function's inputs. Each cell represents the outputs the function for those possible inputs.

karnaugh map two variable

Example :

The truth table contain two 1s. The k-map must have both of them. Locate the first 1 in the 2nd row of the truth table above.

• The truth table AB address
• Locate the cell in the K-map having the same address
• Place a 1 in that cell

Repeat the process for the 1 in the last line in the truth table.

K-map for three variable

Example for three variable.

Mapping the four p-terms yields a group of four, which is cover by one variable C.

After the mapping six p-terms above, identify the upper group four, pick up the lower two cell of group four by sharing the two with two more from the other group. Covering these two with a group of four give a simpler result. Since there are two group, there will be two p-terms in the sum-of-product result A'+B.

K-map four variable

Example for four variable :

The four cells above are a group of because they all have the Boolean variable B' and D' in common. In other word, B = 0 for the four cells, and D = 0 for the four cells. The other variables (A, B) are 0 in some cases, 1 in other cases with respect to the four corner cells. These variables (A, B) are not involved with this group of four. This single group comes out of the map as one product term for the simplified result : Out = B'C'

                                                                                                                                                                        ADAM HARRIS ~.~

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Muted Professor

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NEGATIVE NUMBER CONVERSION

this is
NEGATIVE NUMBER CONVERSION

territory !!!

negative number? conversion? sounds complicated huh? ;-)
don't worry. i'll make it easy to you guys.just do this steps.

STEP 1:     Find the binary number for the decimal number given

                  eg:    9  =  1001

STEP 2:     Change

                  eg:     1001  =  0110

STEP 3:     Add one (1)

        

                  eg:          0110

                           +  ____1

                                 0111

final answer:  0111

artikel ini dikarang oleh si dia. KLIK SINI ;-)

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Muted Professor

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Digital Logic : Logic Gates

Digital Logic

Logic Gates

There are many types of of logic gates . Each of it has its own Boolean Expression and Truth Table .

Each of the gates has its own name . Below are the gates :

AND gate 

     

A    B    F = AB 

0     0         0               Boolean Expression

0     1         0                     

1     0         0                      F = A.B           

1     1         1   

OR gate  

A   B     F = A+B

   0     0          0             Boolean Expression            

  0     1          1

  1     0          1                      F =A+B

  1     1          1

NOT gate

A    F = A’

0        1                       Boolean Expression

1        0                               F = A'                      

NAND gate

A   B     F = (AB)’

0     0          1                 Boolean Expression

0     1          1                 

1     0          1                        F = (A.B)'

1     1          0

NOR gate

A   B     F = (A+B)’

0     0           1               Boolean Expression

0     1           0                     

1     0           0                     F = (A+B)'

1     1           0

XOR gate

A    B         F

0     0         0               Boolean Expression

0     1         1                  

1     0         1                   F = A.B' + A'.B                

1     1         0

                                                                     Post by :

                                                                                                                                                   

                                                          IkhmalYazid
                               B031210153 > BITC S1G2
                                                                      

                 

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Muted Professor

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Number System : 2's Complement

NUMBER SYSTEM

2's Complement

2's complement representation allows the use of binary arithmetic operation on signed integers , yielding the correct 2's complement results .

As with 1's complement, the most significant bit is used to indicate the sign (0 = positive, 1 = negative), and positive numbers are represented in the same way. To negate a positive number, its 2's complement is used .


Example :

                       3810  = 1001102

There are two steps to negate the binary number . First step , assume the bits to 8-bits . Second step , 1's complement . Third step is ,  2's complement .


               1001102  => 001001102   ( the leftmost bit is called sign bits )


For 1's complement , invert the binary number (flip it) . Invert 0 to 1 and 1 to 0 .

                 001001102 => 110110012   ( The sign bits shows 1 = negative )

For 2's complement , add +1 to the 1's complement number .

                       110110012

              +    000000012

                      ___________________

           110110102    (2's complement Number)

= -3810






Example :


4310  -  2210  =  ?



                     4310  =  1010112 => 001010112                                      
                     
                    -2210   =  101102    => 000101102
                                              => 111010012     (flip the bits)
                                              => 111010102 (add +1 / 2's complement)





                                      001010112
                                +       111010102 
                                           _____________________

                                       1000101012                          

                      = 2110








                                                                                                Post by :
                                                                         
                                                            IkhmalYazid

                                                                                        B031210153 > BITC S1G2

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Muted Professor

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