Author's Note: 1a. By using a truth, the expressions AB C B B’ C’ AB’C AC (AC) AB’C+AC+BC’ (AB’C+AC+BC’)

Running head: ELECTRONICS
ELECTRONICS
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Author’s Note:

ELECTRONICS
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1a. By using a truth, table show that the expressions –
A B C B’ C’ AB’C AC (AC)’ BC’ AB’C+AC+BC’ (AB’C+AC+BC’)’
0 0 0 1 1 0 0 1 0 1 0
0 0 1 1 0 0 0 1 0 1 0
0 1 0 0 1 0 0 1 1 1 0
0 1 1 0 0 0 0 1 0 1 0
1 0 0 1 1 0 0 1 0 1 0
1 0 1 1 0 1 1 0 0 1 0
1 1 0 0 1 0 0 1 1 1 0
1 1 1 0 0 0 1 0 0 0 1
From the above truth table we show that (AB’C+AC+BC’)’ is not equivalent to AB’C is
equivalent to ABC
It can also be proved by utilizing the boolean algebra laws and minimizing the equation
as follows:
[(AB’C)’+(AC)’+BC’]’
=[A’+B+C’+A’+C’+BC’]’
=[A’+B+C’+(BC’)]’
=(A’+B+C’)’
=(A’BC’)’
=ABC

ELECTRONICS
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1b. Show by plotting truth tables that
In case of Nand with single input , only one value is taken as a input, as a result the input can
either 0 or 1 hence the truth table for NAND gate -
A A Output
0 0 1
1 1 0
Similarly ,
For the case Nor Gate , only one value is taken as a input, as a result the input can either 0 or 1
hence the truth table for NOR gate
A A Output
0 0 1
1 1 0
Now in case of Inverter /(NOT) gate the truth as follows
A Output
0 1
1 0
For above three tables output are same so we easily proved that NAND=NOR=Invereter

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2. Draw the logic circuit to represent the following Boolean expression using only NAND
gates.
F=[(AB)’+(BC)’+(AC’)’]’
F= (A+B’)’+(A’+B)’
3. In the ladder diagram shown in FIGURE 1, Y006 is a lamp, X and R references are
relay contacts. Write down the Boolean expression for the lamp to light.