# JK flip flop

Discussion in 'Homework Help' started by titanjaw, Apr 13, 2009.

1. ### titanjaw Thread Starter New Member

Apr 13, 2009
2
0
Hey guys,

Need some help with this question. You help is very much appreciated

Thanks,
T

• ###### JK flip flop.doc
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36.2 KB
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64

Apr 5, 2008
19,799
4,107
Hello,

I am not able to read the file.
Can you post the drawing in PNG format to the forum?

Greetings,
Bertus

3. ### Peace Frog Member

Apr 13, 2009
18
1
Alright, what you've posted is a .docx file, not a .doc file (docx files have very different headers when viewed in plaintext) You'll probably want to repost it in PDF if you want many more people to reply.

I'm assuming open office broke the truth table when I tried to view it and assume it wants all four cases instead of just three. I did a quick search on wikipedia and came up with a characteristic equation for JK flip flops that should help you. (Would have been a great help to me if someone would have shown it to me when I learned this stuff)

http://en.wikipedia.org/wiki/JK_flip_flop#JK_flip-flop
Qnext = J * Q' + K' + Q

This is with respect to the input and output of one flip-flop. A brief analysis of the zero zero state will prove that it holds true for each flip flop. All you have to do is use the initial values given by your truth table to derive the results of the outputs Q0 and Q1.

Regards,
Ben

4. ### titanjaw Thread Starter New Member

Apr 13, 2009
2
0
Hey guys ,

Attached is a pdf version of the same file. Thanks for the replies though ... Sorry for the grammatical mistake in my first post ...

File size:
13.7 KB
Views:
40
5. ### vvkannan Active Member

Aug 9, 2008
138
11
Hello titanjaw,

I think you know that a J-K flip toggles when both its input are high
so the first flip flop is always going to toggle when you give a clock pulse.

But for the second flip flop we may have both high or both low depending on the output of flipflop 1.
assume they are both 'low' initally .
so when both inputs are low the ouput Q(t+1) = Qt (i.e) it will not change after you give the clock pulse so Q(t+1) = 0 if Qt was 0.
In this way you can analyze the circuit and what you have is a 2 bit counter
(i.e) the output is going to be
0 1
1 0
1 1
0 0
you see ?It counts from 1 and back to 0.If we 3 flip flops it will count upto
111 .
You have line 1 already filled.Now give that as input to the seond line and find Q(t+1) and proceed similarly