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Write out the truth table for the 4-line-to-2-line encoder that takes a four-line decimal signal and converts it to binary code. Design, and draw, the circuit to implement this encoder.

Computer Science

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Step-by-step solution

4 steps

Here's how to solve question 4b:

  1. Identify the original AND-OR-INVERT function and apply De Morgan's theorem.

    The standard AND-OR-INVERT (AOI) function with four inputs A, B, C, D is given by:

    Y=(A⋅B)+(C⋅D)‾Y = \overline{(A \cdot B) + (C \cdot D)}

    Applying De Morgan's theorem, X+Z‾=X‾⋅Z‾\overline{X+Z} = \overline{X} \cdot \overline{Z}:

    Y=(A⋅B)‾⋅(C⋅D)‾Y = \overline{(A \cdot B)} \cdot \overline{(C \cdot D)}

    Applying De Morgan's theorem again to each term, X⋅Z‾=X‾+Z‾\overline{X \cdot Z} = \overline{X} + \overline{Z}:

    Y=(A‾+B‾)⋅(C‾+D‾)Y = (\overline{A} + \overline{B}) \cdot (\overline{C} + \overline{D})

    This is the alternative form using the four input variables in their complemented form.
    The alternative form is: Y=(A‾+B‾)⋅(C‾+D‾)\boxed{\mathbf{Y = (\overline{A} + \overline{B}) \cdot (\overline{C} + \overline{D})}}

  2. Draw the logic circuit for the original AND-OR-INVERT function.

    The circuit for Y=(A⋅B)+(C⋅D)‾Y = \overline{(A \cdot B) + (C \cdot D)} consists of two AND gates, one OR gate, and one NOT gate (represented by the bubble on the OR gate output).

          A ---|      |
           | AND  |---
      B ---|      |   |
                       | OR   |--- O--- Y
      C ---|      |   |      |
           | AND  |---
      D ---|      |

  3. Draw the logic circuit for the alternative form.

    The circuit for Y=(A‾+B‾)⋅(C‾+D‾)Y = (\overline{A} + \overline{B}) \cdot (\overline{C} + \overline{D}) consists of four NOT gates, two OR gates, and one AND gate.

          A ---O---|      |
      B ---O---| OR   |---
                       | AND  |--- Y
      C ---O---|      |   |
      D ---O---| OR   |---

  4. Construct truth tables for both functions to prove their equivalence.

    The truth table for Y1=(A⋅B)+(C⋅D)‾Y_1 = \overline{(A \cdot B) + (C \cdot D)} and Y2=(A‾+B‾)⋅(C‾+D‾)Y_2 = (\overline{A} + \overline{B}) \cdot (\overline{C} + \overline{D}) is shown below. The columns for Y1Y_1 and Y2Y_2 are identical, proving their equivalence.

    ABCDA⋅BC⋅D(A⋅B)+(C⋅D)Y1A‾B‾C‾D‾A‾+B‾C‾+D‾Y2000000011111111000100011110111001000011101111001101101100100010000011011111010100011010111011000011001111011101101000100100000010111111100100010110111101000010101111101101100100100110010100011010110110100010010111010100001010111111100000000\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline A & B & C & D & A \cdot B & C \cdot D & (A \cdot B)+(C \cdot D) & Y_1 & \overline{A} & \overline{B} & \overline{C} & \overline{D} & \overline{A}+\overline{B} & \overline{C}+\overline{D} & Y_2 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}
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