(6pts) using one 74x169 and three inverters, design a counter with the counting sequence 4, 3, 2, 1, 0, 11, 12, 13, 14, 15, 4, 3 ...

Answers

Answer 1

The frequency of the clock signal will determine the rate at which the counter counts.

To design a counter with the counting sequence 4, 3, 2, 1, 0, 11, 12, 13, 14, 15, 4, 3 ..., we need a modulo-16 counter that counts from 4 to 15 and then wraps around to 4 again. We can use a 74x169 counter chip for this purpose. The 74x169 is a 4-bit synchronous, reversible, up/down counter that can count up or down depending on the state of its up/down input (U/D). We need to modify the counter to count down from 4 to 0 and then count up from 11 to 15.

To implement this, we can use three inverters to generate the complement of the U/D input. We can then connect the complemented U/D input to the carry input (CI) of the counter, which will cause the counter to count down when the complemented U/D input is high and count up when it is low. To make the counter count from 4 to 15 instead of 0 to 15, we can preset the counter to 4 using the preset input (P) of the counter.

The following is the schematic for the counter:

+-|P      CP    |------+

   | |             |      |

   | +------+------|------|-+

   |        |      |      | |

   |        |      |      | |

   |        |      |      | |

   | +------+      |      | |

   | |             |      | |

   +-|U/D    QD    |------+ |

     |             |        |

     +-------------+        |

   +-|U/D'   Qa    |--------+

   | +-------------+

   |

   |

   |       +--------+

   +-------|  INV1  |

           +--------+

           |

           |

           |       +--------+

           +-------|  INV2  |

                   +--------+

                   |

                   |

                   |       +--------+

                   +-------|  INV3  |

                           +--------+

where CP is the clock input, P is the preset input, QD is the output of the counter, Qa is the complemented output of the counter, U/D is the up/down input, and U/D' is the complemented up/down input.

The counting sequence will be as follows:

When the counter is preset to 4 and the complemented U/D input is low, the counter will count up from 4 to 15.

When the counter reaches 15, it will wrap around to 4 and continue counting up.

When the counter reaches 4 again, the complemented U/D input will be high and the counter will count down from 4 to 0.

When the counter reaches 0, it will wrap around to 15 and continue counting down.

When the counter reaches 11, it will wrap around to 4 and start counting up again.

Therefore, the counting sequence will be: 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, ...

Note that this counter will require a clock signal to operate. The frequency of the clock signal will determine the rate at which the counter counts.

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Step-by-step explanation:

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The time, when the two vans leave at the same time = Every 1 hour

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7. The number of times the Nustafa Van leaves Polomolok so that the two vans leave together is given by the number of multiples 20 minutes in 60 minutes = 60 minutes/(20 minutes) = 3 times.

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The given parameters of the rate of departure of the two vans are;

The rate of departure of Smasher Van bound for Koronadal City = Every 15 minutes

The rate of departure of Nustafa Van= Every 20 minutes

5. The time, 'x', when the two vans leave at the same time is given by the Least Common Multiple (LCM) of 15 and 20 as follows;

Multiples of 15 = 15, 30, 45, 60

Multiples of 20 = 20, 40, 60

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The time, when the two vans leave at the same time = Every 1 hour

6. The number of times the Smasher Van leaves Polomolok so that the two vans leave together is given by the number of multiples 15 minutes in 60 minutes = [tex]\frac{60}{15}=4\text{times}[/tex]

7. The number of times the Nustafa Van leaves Polomolok so that the two vans leave together is given by the number of multiples 20 minutes in 60 minutes = [tex]\frac{60}{30}=3\text{times}[/tex]

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