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# Written Hwk8 - Written Hwk#8 Solution Problem 8.1(a(i The...

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Written Hwk #8 Solution Problem 8.1 (a) (i) The following next-state table also includes the output z and the excitations T 1 and T 2 for use in part (ii), J 1 , K 1 and J 2 , K 2 for use in part (iii), and S 1 , R 1 and S 2 , R 2 for use in part (iv). Q1 Q2 x Q1+ Q2+ z T1 T2 J1 K1 J2 K2 S1 R1 S2 R2 0 0 0 0 1 0 0 1 0 x 1 x 0 x 1 0 0 0 1 0 0 0 0 0 0 x 0 x 0 x 0 x 0 1 0 1 1 0 1 0 1 x x 0 1 0 x 0 0 1 1 0 0 1 0 1 0 x x 1 0 x 0 1 1 0 0 0 1 1 1 1 x 1 1 x 0 1 1 0 1 0 1 0 0 1 1 0 x 1 0 x 0 1 0 x 1 1 0 1 1 0 0 0 x 0 x 0 x 0 x 0 1 1 1 1 0 1 0 1 x 0 x 1 x 0 0 1 (ii) From the table in part (i) we get the following minimal SOP expressions: z = Q1Q2' + Q2x T1 = Q1Q2' + Q1'Q2x' T2 = Q2'x' + Q2x = Q2 XNOR x (iii) From the table in part (i) we get the following minimal SOP expressions: z = Q1Q2' + Q2x J1 = Q2x' K1 = Q2' J2 = x' K2 = x (iv) To draw the ROM implementation, we need only know the ROM inputs (x, Q1, Q2) and the ROM outputs (z, S1, R1, S2, R2): the ROM size will be 8 x 5. We don't even need to know the SR values! However, in order to specify the ROM programming table, we do need the z, S1, R1, S2, R2 values, already determined back in part (i).

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ROM Programming Table address contents Q1 Q2 x z S1 R1 S2 R2 0 0 0 0 0 x 1 0 0 0 1 0 0 x 0 x 0 1 0 0 1 0 x 0 0 1 1 1 0 x 0 1 1 0 0 1 0 1 1 0 1 0 1 1 0 1 0 x 1 1 0 0 x 0 x 0 1 1 1 1 x 0 0 1 (b) This circuit recognizes the sequences 01, 001, 0010, and 0011. [001 is redundant, since it also fits 01.] Problem 8.2.
Stat e Meanin g A - (nothing has been read) B 0 has just been read C 1 has just been read D 01 has just been read E 011 has just been read F 0111 has just been read G 01110 has just been read (b) Using the "window on the input sequence" means we shift the input values into a shift register. We want to be able to look at the most recent 6 bits, so we need 5 flip-flops.

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Written Hwk8 - Written Hwk#8 Solution Problem 8.1(a(i The...

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