Answer / sahil gupta

j k qpr qout s r
0 0 0 0 0 x
0 0 1 1 0 x
0 1 0 0 1 0 conversion table
0 1 1 0 1 0 by solve k -map qpr qnew s r j k t d
1 0 0 1 0 1 s=j(not q) 0 0 0 x 0 x 0 0
1 0 1 1 0 1 r=kq 0 1 1 0 1 x 1 1
1 1 0 1 x 0 1 0 0 1 x 1 1 0
1 1 1 0 x 0 1 1 x 0 x 0 0 1
x represents----->>"don't care"
then make k-map for S and R

Answer / raghu

by using 2 and gates and clk signal...

Answer / divya nair

first divide into two parts : the DESIRED part n the
PROCESSED part
the DESIRED part consist of J0,K0,Q0 n Q1
the PROCESSED part consist of S0 n R0
J0 K0 Q0 Q1 S0 R0
0 0 0 -------> 0 0 0
0 0 1 -------->1 0 0
0 1 0 -------->0 0 0
0 1 1 -------->0 0 1
1 0 0 -------->1 1 0
1 0 1 -------->1 0 0
1 1 0 -------->1 1 0
1 1 1 -------->0 0 1

THIS IS THE TRUTH TABLE FOR THE CONVERSION.
EXPLANATION AS FOLLOWS:
Q0=Q1=0,1 : NO TRANSITION STATE
Q0=0&Q1=1 : THE SET CONDITION,S0=1,R0=0
Q0=1&Q1=0 : THE RESET CONDITION,S0=0,R0=1

in NO TRANSITION STATE,the values of S0 n R0 equals 0

Answer / abhay srivastava

j k Qn Qn+1 s r
0 0 0 0 0 x
0 0 1 1 x 0
0 1 0 1 1 0
0 1 1 1 x 0
1 0 0 0 0 x
1 0 1 0 0 1
1 1 0 1 1 0
1 1 1 0 0 1



here x is don't care solve j k Qn

s=j.Qn
r=k.Qn(bar)

Answer / remila saraboth lousy

S=J.QBAR
R=K.Q


DATA INPUTS OUTPUTS INPUTS OUTPUTS
Q QBAR S R QN+1
J K
0 0 0 1 0 x 0
0 0 1 0 0 x 1
0 1 0 1 0 1 0
0 1 1 0 0 1 0
1 0 0 1 1 0 1
1 0 1 0 1 0 1
1 1 0 1 x 0 1
1 1 1 0 x 0 0

Answer / mike

s=Qn'k
r=QnJ

Answer / sathish

SIMPLE METHOD ,ONLY YOU KNOW FOR K-MAP

j k qpr qout s r
0 0 0 0 0 x
0 0 1 1 0 x
0 1 0 0 1 0 conversion table
0 1 1 0 1 0 qpr qnew s r j k t d
1 0 0 1 0 1 0 0 0 x 0 x 0 0
1 0 1 1 0 1 0 1 1 0 1 x 1 1
1 1 0 1 x 0 1 0 0 1 x 1 1 0
1 1 1 0 x 0 1 1 x 0 x 0 0 1
x represents don't care(0 OR 1)

Answer / saravanan

To convert SR ff to JK ff, the equation is S=(J*(~Q)) and
R=(K*(Q)). So by using these equation u can do the
connection now. Give the output of J&(Q bar) and give to S
input and the output of K&Q to R input.

Answer / bibha singh

j k q1 q2 s r
0 0 0 0 0 x
0 0 1 1 x 0
0 1 0 0 0 x
0 1 1 0 0 1
1 0 0 1 1 0
1 0 1 1 x 0
1 1 0 1 1 0
1 1 1 0 0 1

r=product of k and q1
s=product of j and compliment of q1

Answer / taposh chandra

q k y q(t+1)
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0

Q=jy'+k'y

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