# Question 6.1: Analyze the implementing functionality of a SSC, as shown in......

Analyze the implementing functionality of a SSC, as shown in Figure 6.2.1.

Step-by-Step
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Step 1 Derive the exciting equations and output equations directly from logic diagram in Figure 6.2.1.
There are two J-K flip-flops in the SSC. According to logic diagram, the exciting equations of the two J-K flip-flops are

$J_{1}={\overline{{X}}}\cdot{\overline{{Q}}}_{2n},\quad K=1$                      (6.2.1)

$J_{2}={\overline{{X}}}Q_{\mathrm{1n}},\quad K=X$                    (6.2.2)

The output equation can be expressed as

$Y=\bar{X}Q_{2n}\bar{Q}_{1n}$                            (6.2.3)

Step 2 Derive the state equation of each flip-flop by substituting the exciting equations into the characteristic equation of each flip-flop.
The characteristic equation of J-K flip-flops is

$Q_{n+1}=J{\bar{Q}}_{n}+{\bar{K}}Q_{n}$                           (6.2.4)

By substituting the excitation eqs. (6.2.1) and (6.2.2) into the characteristic eq. (6.2.4), respectively, two resulting state equations can be derived as follows:

$Q_{1n+1}=J_{1}{Q}_{1n}+{K}_{1}Q_{1n}=\;\bar{X}\cdot {Q}_{2n}{Q}_{1n}$                    (6.2.5)

$Q_{2n+1}=J_{2}{\bar{Q}}_{2n}+{\bar{K}}_{2}Q_{2n}= \bar{X}Q_{1n}{\bar{Q}}_{2n}+\bar{X}Q_{2n}=\bar{X}(Q_{1n}+\ Q_{2n})$                    (6.2.6)

Note that these eqs. (6.2.5) and (6.2.6), which express the next state as a function of present states and inputs, are also called transition equations.

Step 3 Construct the state table from the state equations and output equation.
A state table is essentially a truth table in which some of the inputs are current states, and the outputs include next state, along with other outputs. For each combination of the level of input X and present state $Q_{2n}$ , $Q_{1n}$ , the corresponding next state $Q_{2n+1}$ , $Q_{1n+1}$ can be calculated by using eqs. (6.2.5) and (6.2.6), and the corresponding output Y can be also obtained from output eq. (6.2.3). The resulting value of next state and the output corresponding each input combination are filled in the corresponding row in the state table as shown in Table 6.2.1.

Alternatively, we can simply use Q and $Q^{+}$ for representing present state and next state, respectively, so the state table can be simplified as Table 6.2.2.

Step 4 Draw a state diagram from the state table and deduce the implementing functionality of the given sequential logic circuits.
A state diagramis another way to illustrate the behavior of a SSC. Each state is denoted by a circle in the state diagram. The number of states in a state table or a state diagram will equal $2^{m}$ , where m is the number of flip-flops. Each combination of the flip-flop values in a circle represents a state.

Each arrow stands for a state transition of the sequential circuit, corresponding to a row in the state table. The number of arrows will equal to  $2^{m}\times2^{k}$ , where k is the number of binary input signals. A label of the form “X/Y” is attached to each arrow corresponding to this transition. Each arrow is labeled with the “X/Y,” where X denotes the input while Y denotes the output of combinational circuit, which cause the transition from present state (the source of the arrow) to next state (the destination of the arrow). In case of Example 6.1, there are two J-K flip-flops in the SSC, so the number of possible states is four. That is, $Q_{2}Q_{1}$ can be equal to 00, 01, 10, or 11. Each state is denoted by a circle with the value of $Q_{2}Q_{1}$ inside, as shown in Figure 6.2.2. There are total eight arrows since the given sequential circuit only has one input, X. Each arrow between two circles denotes a transition of the sequential circuit, corresponding to a row in the state table. For example, the first row in the state table represents one transition from present state 00 ( $Q_{2}\,Q_{1}$ ) to next state 01 ( $Q_{2}\,Q_{1}$ ) and the corresponding output Y is a 0 when X is a 0. In the state diagram, one arrow attached to 0/0 starts from state 00 and end at state 01. Similarly, you can draw the other transition in the state diagram.

 Table 6.2.1: A state table. Input Present state Next state Output X $Q_{2n}$ $Q_{1n}$ $Q_{2n+1}$ $Q_{1n+1}$ Y 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 1 1 1 0 0 0
 Table 6.2.2: A simplified state table. Next state Output Present state X = 0 X=1 X=0 X=1 $Q_{2}$ $Q_{1}$ ${Q_{2}}^{+}$ ${Q_{1}}^{+}$ ${Q_{2}}^{+}$ ${Q_{1}}^{+}$ Y Y 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 0 0

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