Question

Design a BCD counter with four T flip-flops.

- The state table should have the present state, next state, output, minterm, and flip-flop inputs. The output signal Y = 0 only during the counter transition from 1001 to 0000, otherwise, Y = 1 (for each valid input).

- The input equation for T_Q4, T_Q2 and T_Q1 in SOP.     

- The equation of the output signal Y in SOP.     

Answer 1

(a.) The state table of the given BCD counter using T flip flop is as follows:-

PRESENT STATE NEXT STATE O/P MIN TERM T FLIP FLOP

Q8	Q4	Q2	Q1	Q8	Q4	Q2	Q1	Y	min term	TQ8	TQ4	TQ2	TQ1
0	0	0	0	0	0	0	1	1	0	0	0	0	1
0	0	0	1	0	0	1	0	1	1	0	0	1	1
0	0	1	0	0	0	1	1	1	2	0	0	0	1
0	0	1	1	0	1	0	0	1	3	0	1	1	1
0	1	0	0	0	1	0	1	1	4	0	0	0	1
0	1	0	1	0	1	1	0	1	5	0	0	1	1
0	1	1	0	0	1	1	1	1	6	0	0	0	1
0	1	1	1	1	0	0	0	1	7	1	1	1	1
1	0	0	0	1	0	0	1	1	8	0	0	0	1
1	0	0	1	0	0	0	0	0	9	1	0	0	1

In above state table, there are 4 inputs, which makes total 16 combinations (0-15), out of these 16 combinations BCD system only supports starting  10 combinations (0-9) , therefore the next state of the next 6 combinations (min terms 10-16) and output Y can not be predicted and hence we assume there values X (don't care).

(b.) The state equation of TQ4,TQ2,TQ1 are as follows:-

(c.) The equation Y in SOP form is as follows:-

Q2Q1 08 Q4 OU Il 00 t 3 2. 4 5 7 6 X 12 Х 13 is 14 10 xil 8 9 X 10 Toy = Q2 Qi Q2Q1 Q8Q4 il 0 3 2 01 4 5 7 6 х 12 Х 151 Х X 8 9 x 10 TO2 = Q8 Oi 022, 00 68 Qy 10 1 o - 3 2 01 - 5 1 Х 12 13 Х Х is 10 T X'io 9 10 โ91 = 1

Q2Q1 3804 3 ol ५ S 7 6 x ll 12 15 4 XX 8 lo Y = Oo + By Qi