SUMITA ARORA Class 11 CS ASSIGNMENT CHAPTER-3 TYPE-B PART-3

SUMITA ARORA Class 11 CS ASSIGNMENT CHAPTER-3 TYPE-B PART-3

26. Derive the boolean expression for the output F at the network shown below:

Sol: The derived expression is: \( \overline{ (\overline{A}\overline{B}+CD) } \). The expression can be further reduced as follow:
= \( \overline{ (\overline{A}\overline{B}+CD) } \) – Apply Demorgan’s Law
= \(  (\overline{\overline{A}}+\overline{\overline{B}}) (\overline{C}+\overline{D})  \) – Apply the Involution law
= \( (A+B) (\overline{C}+\overline{D}) \) – Apply Distribution law
= \( (\overline{C}+\overline{D})A+(\overline{C}+\overline{D})B \) – Apply Distribution
= \( A\overline{C}+A\overline{D}+(\overline{C}+\overline{D})B \) – Apply Distribution
= \( A\overline{C}+A\overline{D}+B\overline{C}+B\overline{D} \)

27. What function is implemented by the circuit show:

(a) \( \overline{x}\overline{y}+z \)     (b) \( (\overline{x}+\overline{y})z \)
(c) \( \overline{x}\overline{y}z \)         (d)  \( \overline{x}+\overline{y}+z \)
(e) None of these
Sol: The derived function is  \(\overline{(xy)}+z\). Applying Demorgan’s law we get the answer (d) \( \overline{x}+\overline{y}+z \)

28. What function is implemented by the circuit shown below:

(a) \( x+y+z \)   (b) \( x+y+\overline{z} \)   (c) \( \overline{x}\overline{y}z \)
(d) \( \overline{x}+\overline{y}+\overline{z} \)   (e) none of these

Sol: The derived function is \( \overline{ \overline{x+y} z} \). The derived function can be further simplified as below:
= \( \overline{ \overline{x+y} z} \)  – Apply Demorgan’s law
= \( \overline{\overline{x+y}}+\overline{z} \) – Apply Involution law
= \( x+y+\overline{z} \)
So the answer is (b)

29. What function is implemented by the circuit shown below:

(a) \( x\overline{z}+y \)  (b) \( xz+y \)   (c) \( \overline{x}z+\overline{y} \)  (d) \(\overline{x}\overline{y}+\overline{y}\overline{z} \)  (e) \( \overline{x}\overline{y}+\overline{y}z \)

Sol: The derived function is \( (\overline{x}\overline{y}\overline{y})+(\overline{y}z) \). By applying Idempotent law, we get the answer (e) \( \overline{x}\overline{y}+\overline{y}z \)

30. Draw the logic circuit diagram for expressions (a) \((\overline{A}+BC) (\overline{B}+\overline{C}A) \)  (b) \(A\overline{B}+\overline{B}\overline{C}+ABC \)

Sol: (a)

The expression can be reduced as below:
= \((\overline{A}+BC) (\overline{B}+\overline{C}A) \) – Apply Distribution
= \( (\overline{B}+\overline{C}A)\overline{A}+(\overline{B}+\overline{C}A)BC \) – Apply Distribution
= \( \overline{A}\overline{B}+\overline{A}\overline{C}A+(\overline{B}+\overline{C}A)BC \) – Apply the Complement Law​
= \( \overline{A}\overline{B}+0+(\overline{B}+\overline{C}A)BC \) – Apply Identity law
= \( \overline{A}\overline{B}+(\overline{B}+\overline{C}A)BC \) – Apply Distribution
= \(\overline{A}\overline{B}+BC\overline{B}+BC\overline{C}A \) – Apply Identity law
= \(\overline{A}\overline{B}+BC\overline{C}A \) – Apply Complement law
= \(\overline{A}\overline{B}+0 \) – Apply Identity law
= \(\overline{A}\overline{B} \)

Converting the reduced expression into Circuit Diagram:

(b) \(A\overline{B}+\overline{B}\overline{C}+ABC \)

Sol:

The function can be further reduced as below:
= \(A\overline{B}+\overline{B}\overline{C}+ABC \) – Apply the distributive law (AB+AC=AB+C)
= \(\overline{B}\overline{C}+A(BC+\overline{B} \) – Apply the absorption law \( (AB+\overline{A} = B+\overline{A}) \)
= \(\overline{B}\overline{C}+A(C+\overline{B}) \)  – Apply Distribution
= \(\overline{B}\overline{C}+AC+A\overline{B} \) – Apply Consensus
= \(\overline{B}\overline{C}+AC \)

The new circuit diagram will be: