Digital Design_ Principles and Practices, Chegg Solution Manual_parte_388

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Step of 3 7.097E Conventional positive edge trigger D flip-flop responds to the input at the time of clock signal is changing from 0 to 1. Similarly, negative edge trigger D flip-flop responds to the input at the clock signals is changing from 1 to 0. Whereas, dual edge triggered D flip-flop responds to both edges implies for 0 to 1 as well as 1 to 0. Draw the general implementation of Dual edge trigger D flip-flop. D D Q 0 2 to 1 c c MUX Q D Q 1 c Figure 1 From the Figure 1, write the excitation equations for the double edge triggered D-flip flop. Step of 3 Construct a flow table for dual edge-triggered D-flip-flop using excitation equations. State CD 00 01 11 10 S1 S2,0 S4,0/S5,- S3,0 S2 S5,- S3,0S6,- S3 S4,0 S1,0 S4 S1,0 S2,0 S5 S1,0 S2,0/S7,- S3,0 S6 S1,0/S8,- S7,1 S3,0 S4,0 S7 S1,-/S8,1 S7,1 S6,1 S8 S1,0 S2,-/S7,1 S8,1 S5,1 S5,1 S6,1 S5,1 S3,-/S6,1 S7,1 S7,1 S4,-/S5,1 S3,- S8,1 Table 1 Here, represents clock, D represents input. Step of 3 The Table 1 is a fundamental flow table for a DET-D-FF. Note that simultaneous changes of D and C are treated as though one of these variables changed first, but that it does not matter which changed first. This is a realistic assumption as we can never rely on exact simultaneity for any pair of events. The option of treating a simultaneous change in either of two ways is left open for exploitation later in the design process. Identify the essential hazard in the Table 1. In the stable state S6, for CD = 00 if changes 1, the stable state reached is S6; but if changes from 0 to 1 then back to 0 and back to 1 again then the stable state reached is S3 a different state as shown in the Figure 2 State CD 00 01 11 10 S1 S1,0 S2,0 S4,0/S5,- S2 S1,0 S2,0 S5,- S3,0S6,- S3 S1,0 S2,0/S7,- S4,0 S3,0 S4 S1,0/S8,- S7,1 S4,0 S3,0 S5 S7,1 S5,1 S6,1 S6 S1,0 S2,-/S7,1 S5,1 S6,1 S7 S8,1 S7,1 S5,1 83,-/S6,1 S8 S8,1 S7,1 S3,- Figure 2 Hence there is an essential hazard from S6 at clock change.