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in Digital Logic by Junior (911 points) | 214 views

3 Answers

+3 votes
Best answer

Consider a state 100 representing 001 as Q2 is the rightmost in figure. The FFs are triggered on rising edge (0-1). Ignoring the flip flop delays as it is asynchronous counter,

111 - next state 011, CLK1 changed from 1-0
011 - next state 101, CLK1 changed from 0-1, CLK2 changed from 1-0
101 - next state 001, CL1 changed from 1-0, 
001 - next state 110, CLK1 changed from 0-1, CLK2 changed from 0-1
110 - next state 010, CLK1 changed from 1-0
010 - next state 100, CLK1 changed from 0-1, CLK2 changed from 1-0
100 - next state 000, CLK1 changed from 1-0
000 - next state 111, CLK1 changed from 0-1, CLK2 changed from 0-1

So, cycle repeats 7-6-5-4-3-2-1-0-7 mod 3 down counter. 

So, A is the answer.

by Veteran (422k points)
edited by
+1
Q2 is MSB given in questn then how is it pssbl??
+2
ans should be (a) always down counter in fact it is mod 8 down counter..
+1
yes its a down counter .
0
Yes. I missed "negative" triggering. Corrected now.
+1
its positive edge triggered and count will be like this  

000 > 111 > 110 > 101 > 100 > 011 > 010 > 001
+2 votes

sir , plzz check it in above explanation i hav done this question for negative edge triggering so it will give up counter but if i will do it by positive edge triggering then we will get down counter 

so answer will be always down couter from any initial state 

by Active (5k points)
+1
so option will be (a)
0

two another way to draw a down counter

0
Yes, you are correct. I had missed the "rising edge" part.
0 votes
always down bcz Q of previous is connected to clk of next which is +edge triggrd..
by (313 points)
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