1 (341)

Prévia do material em texto

Step 1 of 4 7.050E Refer state table in Figure 7-49(a). Using the state table and the counting order from 000-110 the new transition table is obtained as shown in table 1. Q2Q1Q0 AB Z 00 01 11 10 000 001 001 010 010 0 001 011 011 010 010 0 010 001 001 100 100 0 011 011 011 110 010 1 100 001 101 100 100 1 101 011 011 110 010 1 110 001 101 100 100 1 Table 1 Step 2 of 4 Assuming a "minimal cost" treatment of unused states, the table 1 leads to the following K-maps for the excitation logic as shown in Figure 1. A A AB AB 00 01 11 10 00 01 11 10 00 1100 00 1100 K 01 1100 01100 02 02 11 1100 10 1100 10 1100 8 A A AB 00 01 11 10 00 01 11 10 00 0011 00 1 1 01 0 0 0 0 01 1111 02 11 11 dddd 10 0000 10 1111 8 A A AB AB 00 01 11 10 Q102 00 01 11 10 00 0 0 0 0 00 0 0 0 0 01 0 1 01 0 0 1 0 11 0 1 L 1 11 d d d Q1 10 0 1 1 10 0 0 0 B Figure 1 Step 3 of 4 The resulting excitation equations are, Step 4 of 4 From the excitation equations, the circuit realisation with the new equations requires one 2-input gate, one 5-input gate and six 3-input gates. So, this circuit is expensive.