# Logic Gates

 Question 1
The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is
 A 4 B 5 C 6 D 7
GATE EC 2016-SET-3   Digital Circuits
Question 1 Explanation: Question 2
The output of the combinational circuit given below is A A+B+C B A(B+C) C B(C+A) D C(A+B)
GATE EC 2016-SET-1   Digital Circuits
Question 2 Explanation:
\begin{aligned} y &=A B C \oplus A B \oplus B C \\ &=[\overline{A B C} \cdot A B+A B C \cdot \overline{A B}] \oplus B C \\ &=[(\bar{A}+\bar{B}+\bar{C}) \cdot A B+A B C \cdot(\bar{A}+\bar{B})] \oplus B C \\ &=(A B \bar{C}) \oplus(B C) \\ &=\overline{A B \bar{C}} \cdot B C+A B \bar{C} \cdot \overline{B C} \\ &=(\bar{A}+\bar{B}+C) \cdot B C+A B \bar{C} \cdot(\vec{B}+\bar{C}) \\ &=\bar{A} B C+B C+A B \bar{C} \\ &=B C(\bar{A}+1)+A B \bar{C}=B C+A B \bar{C} \\ &=B(C+A \bar{C})=B(C+A) \end{aligned}
 Question 3
A universal logic gate can implement any Boolean function by connecting sufficient number of them appropriately. Three gates are shown. Which one of the following statements is TRUE? A Gate 1 is a universal gate. B Gate 2 is a universal gate. C Gate 3 is a universal gate. D None of the gates shown is a universal gate
GATE EC 2015-SET-3   Digital Circuits
Question 3 Explanation:
Universal gate is a gate by which every other gate can be realized.
Gate 1 and Gate 2 are basic gates and can not be used as universal gates.
 Question 4
In the figure shown, the output Y is required to be $Y=AB+\bar{C}\bar{D}$.The gates G1 and G2 must be, respectively, A NOR, OR B OR, NAND C NAND, OR D AND, NAND
GATE EC 2015-SET-2   Digital Circuits
Question 4 Explanation: \begin{aligned} Y &=A B+\bar{C} \bar{D} \\ G_1 &=\text { NOR Gate } \quad \bar{A} \text{ NOR } \bar{B}=A B \\ G 2 &=\text { OR GATE } \quad A B+\bar{C} \bar{D} \end{aligned}
 Question 5
A 3-input majority gate is defined by the logic function M(a,b,c) = ab + bc + ca . Which one of the following gates is represented by the function $M(M\overline{(a,b,c)},M(a,b,\bar{c}),c)$ ?
 A 3-input NAND gate B 3-input XOR gate C 3-input NOR gate D 3-input XNOR gate
GATE EC 2015-SET-1   Digital Circuits
Question 5 Explanation:
\begin{aligned} M(a, b, c) &=a b+b c+c a \\ \overline{M(a, b, c)} &=\frac{a b+b c+c a}{a b \cdot b c} \cdot \overline{c a} \\ &=(\bar{a}+\bar{b})(\bar{b}+\bar{c})(\bar{c}+\bar{a}) \\ M(a, b, \bar{c}) &=a b+b \bar{c}+\bar{c} a \end{aligned}
\begin{array}{l} \begin{aligned} M(\overline{M(a, b, c)}, M(a, b, \bar{c}) c) &=\\ \left[\begin{array}{l}(\overline{a b} \cdot \overline{b c} \cdot \overline{c a})(a b+b \bar{c}+\overline{c a}) \\ +(a b+b \bar{c}+\bar{c} a)(c)+(\overline{a b} \cdot \overline{b c} \cdot \bar{a} a \mid)\end{array}\right]\end{aligned} \\ =\left[\begin{array}{l} (\bar{a}+\bar{b})(\bar{b}+\bar{c})(\bar{c}+\bar{a})(a b+b \bar{c}+\bar{c} a) \\ +a b c+(\bar{a}+\bar{b})(\bar{b}+\bar{c})(\bar{c}+\bar{a}) c \end{array}\right] \\ =(\bar{a}+\bar{b})(\bar{b}+\bar{c})(\bar{c}+\bar{a})[a b+b \bar{c}+\bar{c} a+c]+a b c \\ =(\bar{a}+\bar{b})(\bar{b}+\bar{c})(\bar{c}+\bar{a})[a+b+c]+a b c \\ =(\bar{a} \bar{b}+\bar{b} \bar{c}+\bar{c} \bar{a})[a+b+c]+a b c \\ F=a \bar{b} \bar{c}+b \bar{c} \bar{a}+c \bar{a} \bar{b}+a b c \\ F=A \oplus B \oplus C \end{array}
 Question 6
All the logic gates shown in the figure have a propagation delay of 20 ns. Let A = C = 0 and B = 1 until time t = 0. At t = 0, all the inputs flip (i.e., A = C = 1 and B = 0) and remain in that state. For t > 0, output Z = 1 for a duration (in ns) of ______________. A 20 B 30 C 40 D 50
GATE EC 2015-SET-1   Digital Circuits
Question 6 Explanation: Z is '1' for 40 n-sec.
 Question 7
In the circuit shown in the figure, if C = 0, the expression for Y is A $Y=A\bar{B}+\bar{A}B$ B Y=A+B C $Y=\bar{A}+\bar{B}$ D Y=AB
GATE EC 2014-SET-4   Digital Circuits
Question 7 Explanation: Output of gate $1: \bar{C}$
Output of gate $2: \rightarrow(\overline{A+B})$
Output of gate $3: \rightarrow(\overline{A+B}+C)$
Output of gate $4: \rightarrow AB$
Output of gate $5: \rightarrow(\overline{A+B}+C)+A B$
Output of gate 6 is output Y i.e.
\begin{aligned} Y &=\overline{\bar{C} \cdot(\overline{A+B}+C)+A B} \\ &=C+\overline{(\overline{A+B}+C+A B)} \end{aligned}
Using Demorgan's theorem
\begin{aligned} &=C+(\overline{\overline{A+B}}) \cdot \bar{C} \cdot(\overline{A B})\\ &=C+(A+B) \cdot \bar{C} \cdot(\bar{A}+\bar{B}) \end{aligned}
Given in question C=0
\begin{aligned} \text{So, }Y&=0+(A+B) \cdot \overline{0} \cdot(\bar{A}+\bar{B}) \\ &=\bar{A} B+A \bar{B} \end{aligned}
 Question 8
The output F in the digital logic circuit shown in the figure is A $F=\bar{X}YZ+X\bar{Y}Z$ B $F=\bar{X}Y\bar{Z}+X\bar{Y}\bar{Z}$ C $F=\bar{X}\bar{Y}Z+XYZ$ D $F=\bar{X}\bar{Y}\bar{Z}+XYZ$
GATE EC 2014-SET-1   Digital Circuits
Question 8 Explanation: \begin{aligned} A &=X \oplus Y \\ B &=A \odot Z=A \cdot Z+\bar{A} \bar{Z} \\ &=Z(X \oplus Y)+(X \odot Y) \bar{Z} \\ F &=A B \\ &=(X \oplus Y)[Z(X \oplus Y)+\bar{Z}(\overline{X \oplus Y})] \\ &=Z[(X \oplus Y)(X \oplus Y)+\bar{A}(\overline{X \oplus Y})(X \oplus Y)] \\ &\begin{aligned} \text { as we know } A \cdot A &=A \\ A \cdot \bar{A} &=0 \\ &=Z[(X \oplus Y)+0] \\ &=\bar{X} Y Z+X \bar{Y} Z \end{aligned} \end{aligned}
 Question 9
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles
 A an AND gate B an OR gate C an XOR gate D a NAND gate
GATE EC 2013   Digital Circuits
Question 9 Explanation:
Truth table of XOR gate
$\begin{array}{cc|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}$
So, from the XOR gate truth table it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch.
 Question 10
The output Y in the circuit below is always '1' when A two or more of the inputs P, Q, R are '0' B two or more of the inputs P, Q, R are '1' C any odd number of the inputs P, Q, R is '0' D any odd number of the inputs P, Q, R is '1'
GATE EC 2011   Digital Circuits
Question 10 Explanation: $Y= PO+ PR+ RO$
There are 10 questions to complete.