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Software Developers View of Hardware - Last Question (1 Viewer)

Bij

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Oct 23, 2005
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I did something like this:


1--NOT---AND
2---------/........\
..................... NAND---- Motor
3----------\....../
4----------AND


I think that works
 
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Anthrax

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i just swapped the AND and OR gate from the original around.

i think its right.

can someone double check ?
 

chaotichampster

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I did exactly the same as Bij and it works for every combination.

I wastn't entirely sure how they were going to distribute 4 marks for it though...

I mean you cant give out 4 marks for just drawing up a logic circuit, unless they were being overly generous.
 

breesy

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InFeRn0 said:
did u all read the question properly ... it said converting all other four-bit signals to 1 ...
It means any combination of other inputs will result in 1, rather than all other input's are 1. Admittedly a bit porrly worded. Seems like a lot of that this test. View of Hardware especially.

I think the client empowerment approach is a curve ball, the whole process indetifying designing implementing etc are part of the 'software development', thus a development approach is perfectly valid... - C
 
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Vex

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Here we go, just threw this together in paint:



You'll find that the last result for 0111 is 0. Highly unlikely there is another mixture that will go through and also output a 0, actually 99% sure there isnt.

Edit: There's yer damn truth table.. :p
 
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haboozin

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Vex said:
You'll find that the last result for 0111 is 0. Highly unlikely there is another mixture that will go through and also output a 0, actually 99% sure there isnt.

you should have done a deskcheck
it was only 16 different possibilities haha..

there will be quite a few different ways of doing this question
i did something similar to the first post (althoguth i didnt change the last gate to like an AND and a NOT instead of NAND for no reason :s
 

haboozin

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Bij said:
I did something like this:


1--NOT---AND
2---------/........\
..................... NAND---- Motor
3----------\....../
4----------AND


I think that works
bit.................Z.................X
1..2..3..4..Not1 AND 2|3 And 4|NOT(Z and X)
1..1..1..1..0.................1...........1
1..1..1..0..0.................0...........1
1..1..0..1..0.................0...........1
1..1..0..0..0.................0...........1
1..0..1..1..0.................1...........1
1..0..1..0..0.................0...........1
1..0..0..1..0.................0...........1
1..0..0..0..0.................0...........1
0..1..1..1..1.................1...........0
0..1..1..0..1.................0...........1
0..1..0..1..1.................0...........1
0..1..0..0..1.................0...........1
0..0..1..1..0.................1...........1
0..0..1..0..0.................0...........1
0..0..0..1..0.................0...........1
0..0..0..0..0.................0...........1


so thats def right
 

Vex

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haboozin said:
you should have done a deskcheck
it was only 16 different possibilities haha..
Bah fine, go up to the pic and have a look now, heh. I'm way too motivated to see how I went in this exam..(its my band 6 chance, my one and only..)
 

Vex

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jarro_2783 said:
I did that as well. It looks like there are three answers that work, and that must be an easy four marks.
That was a tough 4 marks though, I reckon alot of people would of had a mental block there.
 

Netterhead1

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1------not---------- nand---
2--------------and/
3-------and/
4------/

i think there are more than 3 ways of doing it. heaps more...
 

Tim Crockford

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snowboarder

there is also another way where bit 2,3 and 4 go into a NAND gate then meet bit 1 with an OR gate
 

Mumma

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im in year 11 but you could have easily worked it out using boolean algebra
 
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jarro_2783

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I think you'll find that boolean algebra isn't something anyone actually does unless you decide to look it up and learn it yourself.
 

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