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GATE 2017 EE – Question 46

Analog and Digital Electronics · VCOs and timers, combinatorial and sequential logic circuits, multiplexers, demultiplexers · 2 marks · Multiple choice

The output expression for the Karnaugh map shown below is

AB \ CD00011110
000000
011001
111011
100000
  1. $B\bar{D} + BC\bar{D}$
  2. $B\bar{D} + AB$
  3. $\bar{B}D + ABC$
  4. $B\bar{D} + ABC$

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Correct answer: (D) $B\bar{D} + ABC$

Explanation

The 1s at $AB = 01$ and $11$ with $D = 0$ form a group of four, which is $B\bar{D}$. The remaining 1 at $ABCD = 1111$ combines with the neighbouring 1 at $1110$ to give $ABC$. So the output is $B\bar{D} + ABC$.