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In a small railway station, there are three platforms, #1, #2, #3. Up and down trains can enter in platform number #2 and #3, but platform #1 is only devoted to up trains. Design a logic circuit using basic gates for train entry into the station with proper truth table.

I'm not able to completely understand what the exactly the question is asking me to do? What shall be the output operation in such a case? Be it up trains or down trains, there ARE trains entering the platforms all the time. So, if the output shows 0, what would that signify? But I need to show both up and down trains, so I took 1 to indicate the up ones and 0 to indicate the ones which were down. But, what is my operation here? What shall I find out? How do I exclude all down trains from platform 1? Shall I remove the four of the combinations which involve 0( if 0 is taken as down)?

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  • \$\begingroup\$ I’m voting to close this question because homework questions require effort to be shown and a specific point of difficulty to be raised. \$\endgroup\$
    – Neil_UK
    Commented Jun 11, 2020 at 6:25
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    \$\begingroup\$ If you have been struggling for a few days surely you have some work with reasoning behind it, correct? You can edit the question with the work you have so far, after that people are more likely to want to help. \$\endgroup\$
    – Yet Another Michael
    Commented Jun 11, 2020 at 7:05
  • \$\begingroup\$ @Michael Thanks a lot. I'm completely new here and didn't know how things get done here. Thanks for helping me out! \$\endgroup\$
    – user255006
    Commented Jun 11, 2020 at 7:17
  • \$\begingroup\$ @Neil_UK Sir, I've edited the question. Would you care to give an answer? \$\endgroup\$
    – user255006
    Commented Jun 11, 2020 at 7:46
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    \$\begingroup\$ doing 'logic' is easy. Mapping the world onto logic is much more difficult. Your task is for 'train entry'. A train may not enter a platform if one is already there, so 'train on platform x' would be useful boolean variables, that get set when a train enters, and reset when a train leaves. 'Train may enter platform x from direction y' would be useful outputs, as logic functions of whether platform x is empty, and direction y is permissible for x. I notice Michael has an answer, which involves more parts than minimalist interpretation. See how you get on with this and his as a start. \$\endgroup\$
    – Neil_UK
    Commented Jun 11, 2020 at 9:00

1 Answer 1

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Digital circuits can represent yes or no. Here I am defining an output of '1' can means the given station can accept a given train. The question is vague, but I'm assuming that each station can only fit one train in it at a time.

So let's breakdown your problem into inputs and outputs. With those inputs and outputs we can get a truth table, and by extension a logic circuit. One input can be if a train is 'up' or 'down'. One output can be if station 1 can accept a train or not. If a train is down, we know that station 1 can not accept it. So on your input output table, if the input train is 'down' (let's call that 0), the output for station 1 is always no entry ('0').

So now we can add some more inputs. If the train is 'up' (let's call that 1), depending on the if the station is empty or full it may be either 'no entry' (0) or 'entry' (1). So we can make a 'station 1 empty' signal. If station 1 is empty, and the train is 'up', station 1 can accept the train (output a 1). If the station is full, you can fit another train, so the station can't accept the train (output a 0).

Repeat this process for the other stations and organize it into a table. Once you do that it's just a matter of truth table -> digital circuit, which I am positive that has been answered thoroughly on this site and others.

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  • \$\begingroup\$ Thanks a lot sir! Your explanation really helped me. I think there was a lot of confusion regarding the question but you put it so nicely I'd no difficulty further in solving it. \$\endgroup\$
    – user255006
    Commented Jun 11, 2020 at 10:25
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    \$\begingroup\$ @user255006 If the answer helped, then thank the answer by accepting the answer! \$\endgroup\$
    – StainlessSteelRat
    Commented Jun 11, 2020 at 22:38

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