Answer / kalgu jinu

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 / suraj kumar

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
then solve k-map for s and r
in case of s three variable j,k and Qn like three bit k-map solve
so s=j.Qn
r=k.Qn'
then make a circuit of sr flip-flop and join Qn and Qn+1 with sr flip-flop and make jk flip-flop.
sr latch

Answer / gaurav kumar

An SR flip flop can be converted to D flip flop by shorting
S terminal through a Not gate or inverter and applying it to
R terminal and the S terminal is kept the same as in SR flip
flop.

To convert SR to JK flip flop an extra input to the gate is
applied to first gate in both cases. That is in extra to the
input and clock at S an input is also given from complement
of Q to get J at S terminal and at K terminal an input from
Q is given in extra to input and clock.

JK can be converted to T by shorting J and K terminal of JK
flip flop.

Answer / shabnam

By using clk sgnls 2 both the gates...

Answer / omar

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=k 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 / radhar

by using clock to the rs f/f we can convert it to jk f/f

Answer / amit

simply by feedback
and the s and q'
similarly and the r and q
and fed it to recpective s,r nodes

State assumptions made for analyzing ideal op-amp.

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