The output of a digital circuit (y) is given by this expression: y = (C+ B'A) (A + B + D)' Where A, B, C and D represent inputs. Draw a circuit of above equation using OR, AND and NOT gate and hence…

Digital Circuit and Truth Table for $y=(A^{\prime }{B}^{\prime }+B^{\prime }A)((A+B{)}^{\prime }+C)$

Step 1: Simplify the Boolean Expression

The given expression is: $$y=(A^{\prime }{B}^{\prime }+B^{\prime }A)((A+B{)}^{\prime }+C)$$

First, simplify each part:

1. First Parentheses: $(A^{\prime }{B}^{\prime }+B^{\prime }A)$

   • This is the Exclusive-NOR (XNOR) of $A$ and $B$, which can be written as: $$A^{\prime }{B}^{\prime }+B^{\prime }A=(A\odot B{)}^{\prime }$$
   • Alternatively, it can be implemented using NAND gates or directly as: $$(A\oplus B{)}^{\prime }=(A^{\prime }{B}^{\prime }+A{B}^{\prime })$$

2. Second Parentheses: $((A+B{)}^{\prime }+C)$

   • $(A+B{)}^{\prime }$ is the NAND of $A$ and $B$.
   • Adding $C$ gives: $$(A+B{)}^{\prime }+C=(A\cdot B{)}^{\prime }+C$$

Step 2: Circuit Implementation

The expression can be implemented using AND, OR, and NOT gates as follows:

1. First Term: $A^{\prime }{B}^{\prime }+B^{\prime }A$

   • Use NOT gates to get $A^{\prime }$ and $B^{\prime }$.
   • Compute $A^{\prime }{B}^{\prime }$ using an AND gate.
   • Compute $B^{\prime }A$ using another AND gate.
   • Combine them using an OR gate.

2. Second Term: $(A+B{)}^{\prime }+C$

   • Compute $A+B$ using an OR gate.
   • Apply a NOT gate to get $(A+B{)}^{\prime }$.
   • Combine with $C$ using an OR gate.

3. Final AND Operation

   • The outputs of the two terms are fed into a final AND gate to get $y$.

Step 3: Truth Table

The truth table for $y=(A^{\prime }{B}^{\prime }+B^{\prime }A)((A+B{)}^{\prime }+C)$ is constructed as follows:

Step 4: Circuit Diagram (Mermaid)

graph TD
    A["Input A"] --> NOT1["NOT"]
    B["Input B"] --> NOT2["NOT"]
    C["Input C"]

    NOT1 --> AND1["AND"]
    NOT2 --> AND1
    AND1 --> OR1["OR"]

    A --> OR2["OR"]
    B --> OR2
    OR2 --> NOT3["NOT"]

    NOT3 --> OR3["OR"]
    OR3 --> AND2["AND"]

    B --> AND3["AND"]
    NOT1 --> AND3
    AND3 --> OR1

    OR1 --> AND2
    C --> OR3

    AND2 --> Y["Output y"]

Final Truth Table Output

The output $y$ is 1 only in the following cases:

• $A=0,B=0,C=0$
• $A=0,B=0,C=1$
• $A=0,B=1,C=1$
• $A=1,B=0,C=1$