Question

a) A counter is designed to go through the sequence : 1,3,5,7,0,2,5,6, repeat, Using JK flip- flops:

(i) Construct the state table.

(ii) Draw the circuit.

Answer 1

And! that counts counter a have to We design by using the sequence 1,3,5,7,0, 2,5,6 the JK flip flops. to 3 to use J K flip flops We need because, the this counter implement here is gequence count que maximum it is equal to 7. 8 and less than We can represent in binary by using (7) (!!!).. 3 bit 102 the state table, for to write > We need the excitation table of we that require flip flop. JK 1,3,5,7,0,2,5,6 * Count sequence :- binary ir Count sequence in 001, OQI, 101, 111, 000,010, 101, 110

Encitation Table of J K flip flop Present state Pyeri Input J K 'Nent state (Qnt) J Qi K ☺ X o o Х 0 x - X o (i) State Table FIE inputs Nent state Present state QA ар, Qc Q(AH) QB+1) Ql+h) JAKA JB Ka JC Ke X 0 0 x 1 ох O. o D 0 0 Х 1 X Xo 0 0 1 o I X - 0 X X 1 xo - 0 0 1 х х X Х x XX X o 0 X 1 O IX xo O x 1 0 1 X 1 1 0 0 X 1 x 1 XI G

JA = Em (2,3) teal 4,5,6,7) KA Em (6,7) + Ed(0, 1, 2, 3, 4) 1 JB = Em (0,1,5.) + Ed (2,3,4,6,7) KB: Em (2,3,6,7) +E. (0,1,4,5) Je=Em (216) teal 1,3,4,5,7) Ke= Em (7)& Ea(0, 2, 416) to these equations: minimize K-maps Use For JA :- JA = Em (2,3)+ E2(4,5,6,7) QB Qc. 10 01 QA X Х x x QB .. Jau QB For for KA!- Kas Em (617) +Ed(0,1,2,314) A QB Qc X QA x Х KA=QB 1 QB

For JB - JB: Em (0,1,5) + Ed (2,3,416,7) QB Qc QA 00 ol 10 0 li х X Х X Х + JB=1 for KB!- KB = Em (2,316,7) + Ed (0,1,4,5) QB Qc QA 10 00 01 x X + 1 Х x KB-1 For Icio Em (2,6) + Ed ( 1,3,4,5,7) QB Qc 10 11 O! QA 00 Х X 0 х X QB x I. Jc = QB

For Kci- Ke = Em (7) + Ed (0,?,4,6) QB Qc QA 01 I 10 x 0 Х X X 11 Kc = QA QB counter using Jk flip flops (11) Implementing the JA QA QA JA clock. Q -QA *lk FFA KA KA QB JB @ QB logic1 J Elk FFB Dj K ба KB Qe Q Qc J FFc Ack Q K Qc QA