[MsSykes]

PSA - Has it been misunderstood that X = 1 actually means the output needs to be 1?

Or is the question wrong in turning T off and sending this into both AND gates? (Or turning R off?)



Thanks in advance for any helpful comments.
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[macosxnerd101]

Moved to Computer Science.

The question (a), for example, states that the output should be 1 if (A = 1 or B = 1) or (A = 0 or B = 1).

[NeoTifa]

Are you supposed to be simplifying it or just drawing it as is? I can help you with the first one a little bit, but I'd like to see work on your end before helping you some more.

A|B|X
------
0|0|0
0|1|1
1|0|0
1|1|1

What do you see here? Is there a pattern? How can you simplify the equation any more?

[MsSykes]

Here's my working.

NeoTifa, I don't understand your truth table - it doesn't belong to any of the 7 gates I've been working with. Have i missed something?

Attached image(s)

• 

[MsSykes]

Here's a combined image of logic circuit and truth table.

My initial confusion was that how can X=1 if all three conditions output 0. Have I misunderstood that X=1 means it outputs 1?

Attached image(s)

• 

[Skydiver]

Are you trying to solve a different problem than the question that you posted?

Let's start off with you initial expression:

Quote

(R = 0) OR (C = 1 AND T = 0) OR (R = 1 AND T = 0)

I see R, C, and T. The problem with the turbine has S, T, and W. Can you tell us how your parameters map to the parameters in the problem?

Next issue is that the conditions in the turbine problem look something like this:

Quote

(turbine speed <= 1000 AND bearing temperature > 80C)
OR (turbine speed > 1000 AND wind velocity > 120KPH)
OR (bearing temperature <= 80C AND wind velocity > 120KPH)

but your expression looks like this:

Quote

(R = 0)
OR (C = 1 AND T = 0)
OR (R = 1 AND T = 0)

Never mind the details. Just the general structures don't match up.

[MsSykes]

I'm so sorry! I posted the wrong pic - here's the right question to the solution posted. So, this is the question I think has a mistake of some kind..

Attached image(s)

• 

[sepp2k]

Your solution looks correct to me (now that we have the right question).

MsSykes, on 13 February 2015 - 04:19 PM, said:

My initial confusion was that how can X=1 if all three conditions output 0.

I'm not sure what you mean by "how" here. If all inputs are 0, the output will be 1. That is correctly reflected in your truth table and your diagram. So what's problematic about that?

Quote

Have I misunderstood that X=1 means it outputs 1?

No, that's what it means.

[MsSykes]

Thank you sepp2k and Skydiver for your recent replies.

Is this right?
All inputs, in any circuit diagram, start off as '1'. They are then inverted (to '0') depending on the question. (So, T was '1' originally, but we inverted it to '0' before it was input to the two AND gates, and R was inverted to '0' before it was input to the final OR gate, as per scenario.)

And is this right (which is what is the main confusion point for me)?
X = 1 simply means 'the alarm will go off' rather than 'this is was the binary digit that comes out of the final OR gate which has 0 and 0 as the inputs.'

(I hope they are )

[sepp2k]

MsSykes, on 15 February 2015 - 02:37 PM, said:

All inputs, in any circuit diagram, start off as '1'.

No, that's not right at all. They're inputs. They can start off as anything (well, 0 or 1). And depending on what they do start off as, you get a different output (well usually, it is of course also possible to have a circuit whose output doesn't depend on its inputs at all, but that'd be boring).

Quote

X = 1 simply means 'the alarm will go off' rather than 'this is was the binary digit that comes out of the final OR gate which has 0 and 0 as the inputs.'

X = 1 means that the alarm will go off, yes. X is also the value of the final OR gate, so if the inputs of the final OR gate are 0 and 0, X will also be 0 and therefore the alarm will not go off.

[MsSykes]

sepp2k, on 15 February 2015 - 11:14 AM, said:

MsSykes, on 15 February 2015 - 02:37 PM, said:

All inputs, in any circuit diagram, start off as '1'.

No, that's not right at all. They're inputs. They can start off as anything (well, 0 or 1). And depending on what they do start off as, you get a different output (well usually, it is of course also possible to have a circuit whose output doesn't depend on its inputs at all, but that'd be boring).

If they don't necessarily start off as 1, why is it correct to invert T and R? By inverting them, to make them equal 0, that means they were 1 to start with no?

Quote

X = 1 simply means 'the alarm will go off' rather than 'this is was the binary digit that comes out of the final OR gate which has 0 and 0 as the inputs.'

X = 1 means that the alarm will go off, yes. X is also the value of the final OR gate, so if the inputs of the final OR gate are 0 and 0, X will also be 0 and therefore the alarm will not go off.

But the inputs to the final OR gate are 0 and 0 - the scenario says the alarm will go off. I've interpreted that your last sentence is contradicting the scenario.




*sorry, should have multi-quoted

[sepp2k]

MsSykes, on 15 February 2015 - 07:43 PM, said:

But the inputs to the final OR gate are 0 and 0

The inputs to the final OR gate are D and B. Whether or not they're 0 or 1 depends on the values of C, R and T.

Quote

the scenario says the alarm [I]will[/i] go off.

No, it's saying that it will go off [i]if[/il] one of the given conditions is met.

This post has been edited by sepp2k: 15 February 2015 - 12:27 PM

[MsSykes]

sepp2k, on 15 February 2015 - 11:14 AM, said:

MsSykes, on 15 February 2015 - 02:37 PM, said:

All inputs, in any circuit diagram, start off as '1'.

No, that's not right at all. They're inputs. They can start off as anything (well, 0 or 1). And depending on what they do start off as, you get a different output (well usually, it is of course also possible to have a circuit whose output doesn't depend on its inputs at all, but that'd be boring).

If they don't necessarily start off as 1, why is it correct to invert T and R? By inverting them, to make them equal 0, that means they were 1 to start with no?

When you're given a question, with a scenario, for which you have to draw the logic circuit diagram, then you will be assuming the default state for each input to be '1' which is why you 'leave them as they are' (i.e. simply draw a connecting line from input to the gate) for the cases of 'input = 1', as per scenario condition, and why you 'invert them from their original state' (i.e. add a NOT gate in between input and the next gate) for the cases of 'input = 0', as per scenario condition. This is for when you are drawing the original diagram.

This makes sense to me. And is working for every question I'm doing. I think what I'm trying to communicate is being misunderstood.



Thank you for your time everyone. />

[NeoTifa]

Oh, man, shoulda checked this thread sooner .

The truth table I posted was for the question about the X=1. It was a truth table for the whole equation (which it turns out I messed up anyways . That's what I get for posting half asleep). You could expand it also.

A|B|AvB|A'^B|(AvB)v(A'^B )=X
---------------------------
0|0| 0 | 0 | 0
0|1| 1 | 1 | 1
1|0| 1 | 0 | 1
1|1| 1 | 0 | 1

This post has been edited by NeoTifa: 16 February 2015 - 10:10 AM

[sepp2k]

MsSykes, on 16 February 2015 - 06:02 PM, said:

If they don't necessarily start off as 1, why is it correct to invert T and R?

You invert T and R because you need T=0 and R=0 as the inputs of your gates. That is if you feed the T signal into one of your gates, then the operand of that operator in your formula would be T=1. If you invert it first, the operand becomes T=0. That doesn't mean that T is 1 or that T is 0 after the inversion - all it means is that you now check whether T is 0 instead of checking whether it is one.

Quote

By inverting them, to make them equal 0, that means they were 1 to start with no?

You don't make them equal to anything - you just check what they are equal to. Nothing you do can change what the values of T, R and C actually are.