2017-11-22 16:15:15
zhengxiaowei
125

Consider the circuit in Fig. 5.4 which is simply an XOR gate with one of the inputs being the output which is fed back, so making it an asynchronous sequential circuit.5 (Note that in this case, Z, the external output is the same as Y, the internal output.)

If A=0 and y=1 then the output 1, so the output produced, and fed back to the inputs, matches the input (i.e. y= Y) and the circuit is stable. Similarly, if and A=0, then Y=0, and so Y and the circuit is again stable. The fact that circuit is stable means that all of the variables will remain unchanged until the input A is changed (as this is the only variable that can be accessed, i.e. the only external input).

Now, if-4 is changed to 1 then if then y= 1; and if y= 1 then Y=0. Hence, y≠Y. This will clearly lead to an unstable circuit that will continually oscillate. For example, if y=0 the output will be 1, which will be fed back to y causing the output to go to 0 which will be fed back and so cause the output to go to 1 and so on. The speed of oscillation will be determined chiefly by the time it takes the signals to propagate through the XOR gate and back along the feedback path.

The Karnaugh map for yin terms of A and y is also shown in Fig. 5.4 and illustrates the operation of the circuit. For the circuit to be stable Y must equal y, therefore for the top row, y (and hence ^=0), the circuit will only be stable when the corresponding cell of the Karnaugh map has a 0 in it (i.e. the internal output Y =0). For the bottom row the circuit will only be stable when Y=\. We therefore see that only two stable conditions for this circuit exist. Firstly, A =0 and y=0 and secondly A=0 and y= 1, that is when A = 0 (the left hand column of the Karnaugh map) as we deduced above.

The Karnaugh map confirms the instability of the circuit when A=1, since nowhere in the right-hand column (A= 1) does y=Y. All of the remaining circuit analyses are based upon the use of such Karnaugh maps.