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+10 votes

Design a logic circuit to convert a single digit BCD number to the number modulo six as follows (Do not detect illegal input):

  1. Write the truth table for all bits. Label the input bits $I_1, I_2, \ldots$ with $I_1$ as the least significant bit. Label the output bits $R_1, R_2\ldots$ with $R_1$ as the least significant bit. Use $1$ to signify truth.
  2. Draw one circuit for each output bit using, altogether, two two-input AND gates, one two-input  OR gate and two NOT gates.
asked in Digital Logic by Veteran (52k points)
edited by | 686 views

2 Answers

+8 votes
Best answer

\bf{I_4}&    \bf{I_3}&  \bf{I_2}&\bf{ I_1}& &\bf{R_3}& \bf{R_2} & \bf{R_1}\\\hline
0&0&0&0&\bf{0} &0&0&0\\\hline 0&0&0&1&\bf{1}& 0&0&1 \\ \hline    0&0&1&0&\bf{2}& 0&1&0 \\ \hline 0&0&1&1&\bf{3}& 0&1&1 \\ \hline 0&1&0&0&\bf{4}& 1&0&0 \\ \hline   0&1&0&1&\bf{5} &1&0&1  \\ \hline  0&1&1&0&\bf{6}& 0&0&0 \\ \hline0&1&1&1&\bf{7}& 0&0&1\\ \hline    1&0&0&0&\bf{8}& 0&1&0 \\ \hline 1&0&0&1&\bf{9}& 0&1&1 \\ \hline

  • $R_1 = I_1$
  • $R_2 = I_2\overline{ I_3} + I_4$
  • $R_3 = I_3\overline{I_2}$

This requires $2$ NOT gates, $2$ two-input AND gates and $1$ two-input OR gate. 

answered by (369 points)
edited by
Simply Y0=D
why other inputs not consider here that is after 9. i think input should go from 0 to 15.
No. of Logic Gates given in question would be insufficient, so how you are going to implement your expression in circuit diagram???

and for Expression why you did not use don't care for (10,11,12,13,14,15) ?
why you did not use don't care for (10,11,12,13,14,15) ?
wrong answer selected as best answer
R2=(i2)'(i4) + (i2)(i3')
+7 votes

After using Don't care (10,11,12,13,14,15) and after K-Map simplification you will get


R2=I2.I3' + I4

R3= I3.I2'

R4 = 0

Here, 2 input AND Gate used=2

          2 input OR Gate used=1

          NOT Gate used=2

answered by Active (1.2k points)

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