K-map -- Boolean Simplification (pretty long)

Discussion in 'Homework Help' started by Hawkeye87, Nov 19, 2008.

  1. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
    52
    0
    Ok, i have a lab to do an i need to see if i can get my answers checked on here by you smart smart grateful people. Here is the lab requirements......


    PROBLEM SPECIFICATION:
    Design, build, implement, test, and document a digital logic circuit, which will implement an intersection traffic light controller. Refer to figure 4-74 and problem 4-30 in the text (10th edition) for specifications relating to the controller.

    Design a simplified logic circuit to control the traffic light. There should be two outputs, NS and EW, which go LOW (0) when the corresponding light is to be green (note that this may be different than the book). Note: like all traffice lights one light is green and the other direction is red, that is the EW logic is the inverse of the NS logic.

    REQUIRED:

    1)Define the truth-tables for the problem.
    2)Simplify the SOP of the truth table using K-maps
    3)Draw the circuit diagram for the problems using standard circuit symbols. Label all pin numbers, inputs, outputs, and intermediate values on all your drawings.
    4)Implement your circuit on the Cadet designer board and verify it operations with complete testing - Verify the operations by observing the output logic level for the input conditions from the truth tables.
    5)Write-up the results in the standard laboratory format to be turned in on Tuesday.

    A green is represented as a low which is a 0
    A red is represented as a high and as a 1
    (that's how its set up on our cadet board)



    The question from the book gave the specifications for setting up the lab as follows......
    A 4 lane intersection has vehicle sensors on each of the North West South and East roads. (see diagram) each labeled A, B, C, D.

    1) The East West light will be green whenever both lanes C and D are occupied.
    2) The East West light will be green whenever either C or D is occupied but lanes A and B are not both occupied.
    3) The North South light will be green whenever both lanes A and B are occupied but C and D and not both occupied.
    4) The North South light will be green when either A or B is occupied while C and D are both vacant.
    5) The East West light will be green when no vehicles are present.


    I'll attach the diagram and truth table and what i did so far for the output equations.


    Truth Table
    A B C D -EW- NS
    0 0 0 0 --0 --1
    0 0 0 1 --0 --1
    0 0 1 0 --0 --1
    0 0 1 1 --0 --1
    0 1 0 0 --1 --0
    0 1 0 1 --0 --1
    0 1 1 0 --0 --1
    0 1 1 1 --0 --1
    1 0 0 0 --1 --0
    1 0 0 1 --0 --1
    1 0 1 0 --0 --1
    1 0 1 1 --0 --1
    1 1 0 0 --1 --0
    1 1 0 1 --1 --0
    1 1 1 0 --1 --0
    1 1 1 1 --0 --1

    Then from the K-maps that i set up i got the equations.....
    The (') means (A bar) or invert.
    EW = BC'D' + AC'D' + ABC' + ABD'
    NS = A'B' + A'C'D + B'C'D + A'C + B'C + CD
    Can you simplify the equations more? Or is the truth table set up right?
    THANK YOU KINDLY!!
    Ok i just need to know if i got the boolean equations simplified correctly.
     
    Last edited: Nov 19, 2008
  2. blazedaces

    Active Member

    Jul 24, 2008
    130
    0
    Your EW equation is correct, but your NS is not.

    Re-do your K-map. Remember that if you CAN group by fours, you should, and that you can group 1's that have been grouped already. And in case you forgot for some reason, the sides all touch, and the corners too, though that won't help you here...

    Even if you didn't do it that way, you can look at your equation and see it can be simplified further by boolean algebra...

    Good luck,

    -blazed
     
  3. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
    52
    0
    hahaha boolean algebra sucks so much. k maps is easier. i'll redo the k map
     
  4. Hawkeye87

    Thread Starter Active Member

    Oct 7, 2008
    52
    0
    ok how about
    A'B' + A'D + CD + A'C + B'D +B'C
    does that sound better?
     
  5. blazedaces

    Active Member

    Jul 24, 2008
    130
    0
    *Nods head*

    Yup, that's it. Good job.

    -blazed
     
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