Q21GATE 2011MCQ

The logic function implemented by the circuit below is (ground implies logic 0)

📖 Explanation

\begin{array}{lll} P & Q & F \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array}

$F= P \oplus Q$

Q22GATE 2010MCQ

The Boolean function realized by the logic circuit shown is

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📖 Explanation

\begin{aligned} F(A, B, C, D) &=\bar{A} \bar{B} C+\bar{A} B D+A \bar{B} \bar{C}+A B(\bar{C} \bar{L}\\ &=\bar{A} \bar{B} C(D+\bar{D})+\bar{A} B(C+\bar{C}) D \\ &+A \bar{B} \bar{C}(D+\bar{D})+A B \bar{C} \bar{D} \end{aligned}

Placing above minterms in Karnaugh map, So, $F=\Sigma m(2, 3, 5, 7, 8, 9, 12)$

Q23GATE 2009MCQ

Two products are sold from a vending machine, which has two push buttons P1 and P2. When a buttons is pressed, the price of the corresponding product is displayed in a 7-segment display. If no buttons are pressed, '0' is displayed signifying 'Rs 0'. If only P1 is pressed, '2' is displayed, signifying 'Rs. 2' If only P2 is pressed '5' is displayed, signifying 'Rs. 5' If both P1 and P2 are pressed, ' ' E is displayed, signifying 'Error' The names of the segments in the 7-segment display, and the glow of the display for '0', '2', '5' and 'E' are shown below. Consider (1) push buttons pressed/not pressed in equivalent to logic 1/0 respectively. (2) a segment glowing/not glowing in the display is equivalent to logic 1/0 respectively. What are the minimum numbers of NOT gates and 2-input OR gates required to design the logic of the driver for this 7-Segment display

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Q24GATE 2009MCQ

What are the minimum number of 2-to-1 multiplexers required to generate a 2-input AND gate and a 2-input Ex-OR gate

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📖 Explanation

\begin{aligned} Y &=\bar{S} \cdot I_{0}+S_{1} I_{1} \\ &=\bar{A} 0+A B \\ &=A B \end{aligned}

AND GATE Similarly EX OR gate required 2 MUX of 2 x 1

Q25GATE 2008MCQ

For the circuit shown in the following, $I_{0}-I_{3}$ are inputs to the 4:1 multiplexers, R(MSB) and S are control bits. The output Z can be represented by

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📖 Explanation

$Z=P R S+P Q R \bar{S}+P \bar{R} S+(P+\bar{Q}) \bar{R} \bar{S}$ Mapping above terms in Karnaugh map- $Z=P Q+P \bar{Q} S+\bar{Q} \bar{R} \bar{S}$

Q26GATE 2007MCQ

In the following circuit, X is given by

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📖 Explanation

Let the output of first MUX is Y

\begin{aligned} \text { So } \quad Y&=\bar{A} B+A \bar{B}=A \oplus B \\ X&=\bar{Y} C+Y \bar{C}=Y \oplus C \\ \text { So } \quad X&=A \oplus B \oplus C \\ &=A\bar{B}\bar{C}+\bar{A}B\bar{C}+\bar{A}\bar{B}C+ABC \end{aligned}

Q27GATE 2005MCQ

The Boolean function f implemented in the figure using two input multiplexes is

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📖 Explanation

\begin{aligned} f & =E \cdot A \\ E & =\bar{B} C+B \bar{C} \\ \therefore \quad f & =A \bar{B} C+A B \bar{C} \\ &\begin{array}{ccc|c} A&B&C&f\\ 1&0&1&1\\ 1&1&0&1 \end{array} \end{aligned}

Q28GATE 2004MCQ

The minimum number of 2 to 1 multiplexers required to realize a 4 to 1 multiplexers is

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📖 Explanation

\begin{aligned} \text { No. of } M U X &=\frac{4}{2}=2 \\ \frac{2}{2} &=1\\ \therefore \quad \text { Total }&=2+1=3 \end{aligned}

Q29GATE 2003MCQ

The circuit in the figure has 4 boxes each described by inputs P,Q,R and outputs Y,Z with Y = $P\oplus Q\oplus R$ and $Z=RQ+\overline{P}R+Q\overline{P}$ The circuit acts as a

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📖 Explanation

\begin{array}{l} Y=D=P \oplus Q \oplus R \\ Z=\text { Borrow }=R Q+\bar{P} R+Q \bar{P} \end{array}

Q30GATE 2003MCQ

The circuit shown in figure converts

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📖 Explanation

\begin{array}{l} W=a \\ x=a \oplus b \\ y=c \oplus x(a+b) \\ z=d \oplus y(a+b+c) \end{array}

By substituting given option in the Boolean equations of two circuit,: it shows Gray to binary code converter. The input =1010 and output =1100 The circuit is converting Gray code number to Binary code number.

Q31GATE 2003MCQ

Without any additional circuitry, an 8:1 MUX can be used to obtain

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