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8.8: Exercises

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    27668
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    1. Implement the 1 bit 4-to-1 MUX in Figure 8.5.1.
    2. Implement a 1 bit 4-to-1 MUX using the 74139 decoder chip introduced in section 7.4. This will require both the 74139 decoder and 7404 inverter chip.
    3. Implement the 1 bit 4-to-1 MUX using the 74139 decoder chip introduced in section 7.4, but do not use an inverter on the 74139 output. Instead use the enable low outputs from the 74139 chip directly. This allows the circuit to be implemented using only 3 chips, a 74139, a 7402, and a 7432 chip.

      HINT: Remember deMorgan's Law, AB = (A'+B')'.

    4. Implement a 1 bit 4-to-1 MUX using the 74153 chip, as in section 8.3.
    5. Explain how a 1 bit 4-to-1 MUX can calculate any binary Boolean function. Because the MUX can calculate the result of any Boolean function, we call the MUX a univeral operation.
    6. In Logism implement an 8 bit 4-to-1 MUX using 8 4-to-1 MUXes.
    7. In Logisim implement an 8-to-1 MUX using 2 4-to-1 MUXes and a 2-to-1 MUX.
    8. In Logisim implement an 8-to-1 MUX using 4 2-to-1 MUXes and a 4-to-1 MUX.
    9. In Logisim implement a circuit similar to the one in Figure 8.6.1, but which produces output which is the opposite of the input switch (e.g. the LEDs are 0 when the switch is 1, and 1 when the switch is 0). Change the program in the breadboard to match this new logic.

    This page titled 8.8: Exercises is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Charles W. Kann III via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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