Logic Gates and Logic Circuits — practice questions

(i) To draw the circuit, we identify the gates needed: two AND gates, one NOT gate, and one OR gate.

• The first AND gate takes inputs A and B.
• The NOT gate takes input B to produce $\neg{B}$.
• The second AND gate takes inputs $\neg{B}$ and C.
• The OR gate takes the outputs of the two AND gates to produce the final output X.

(A diagram would be drawn here showing A and B into an AND gate; B into a NOT gate, then its output and C into a second AND gate; the outputs of both AND gates into an OR gate, which gives X.)

(ii) The truth table requires columns for all inputs (A, B, C), any intermediate steps ($$\neg{B}$,$A \cdot B$,$\neg{B} \cdot C$), and the final output (X). There are$2^3 = 8$possible input combinations.$

Marking: [1] for correct AND/OR structure. [1] for correct NOT gate placement. [1] for correct labelling. [1] for correct column headings in truth table. [1] for correct $\neg{B}$ column. [1] for correct intermediate columns. [1] for correct final X column.

We can simplify the expression using Boolean algebra, specifically the distributive law.

1. Original Expression: $X = (A + B) \cdot (A + \neg{B})$
2. Apply Distributive Law ($P \cdot Q$ where $P=(A+B)$ and $Q=(A+\neg{B})$ is not the easiest way. Instead, let's expand the brackets like in normal algebra): $X = A \cdot A + A \cdot \neg{B} + B \cdot A + B \cdot \neg{B}$ [Distributive Law]
3. Apply Idempotent Law ($A \cdot A = A$) and Complement Law ($B \cdot \neg{B} = 0$): $X = A + A \cdot \neg{B} + A \cdot B + 0$
4. Apply Annulment Law ($A+0 = A$): $X = A + A \cdot \neg{B} + A \cdot B$
5. Factor out A (Distributive Law in reverse): $X = A \cdot (1 + \neg{B} + B)$
6. Apply Complement Law ($B + \neg{B} = 1$) and then Annulment Law ($1+1=1$): $X = A \cdot (1 + 1) = A \cdot 1$
7. Apply Identity Law ($A \cdot 1 = A$): $X = A$

Simplified Expression: $X = A$

Circuit Diagrams:

• Original Circuit: Input A and B go into an OR gate. Input A and a NOT of B go into a second OR gate. The outputs of these two OR gates go into an AND gate to produce X. This requires two OR gates, one NOT gate, and one AND gate.
• Simplified Circuit: The output X is simply connected directly to the input A. This requires no gates at all.

This demonstrates how simplification can dramatically reduce the complexity and cost of a circuit.