Obtain the simple logical expression of the following. Draw the corresponding switching circuit.
(∼ p ∧ q) ∨ (∼ p ∧ ∼ q) ∨ (p ∧ ∼ q)
Solution
Using the laws of logic, we have,
(∼ p ∧ q) ∨ (∼ p ∧ ∼ q) ∨ (p ∧ ∼ q)
≡ [∼ P ∧ (q ∨ ∼ q)] ∨ (p ∧ ∼ q) .......(By Distributive Law)
≡ (∼ p ∧ T) ∨ (p ∧ ∼ q) .......(By Complement Law)
≡ ∼ p ∨ (p ∧ ∼ q) ..........(By Identity Law)
≡ (∼ p ∨ p) ∧ (∼ p ∧ ∼ q) ........(By Distributive Law)
≡ T ∧ (∼ p ∧ ∼ q) ..........(By Complement Law)
≡ ∼ p ∨ ∼ q ............(By Identity Law)
Hence, the simple logical expression of the given expression is ∼ p ∨ ∼ q.
Let p: the switch S1 is closed
q: the switch S2 is closed
∼ p: the switch S1′ is closed or the switch S1 is open
∼ q: the switch S2′ is closed or the switch S2 is open.
Then the corresponding switching circuit is:
