6 Sequential Logic

„I Know What You Did Last Cycle“

6.1 First Terminology

The most important term for the upcoming topics is the word state. But what is a state?
It is a unique situation, where the possible next steps (= possible next states), the inner behavior, or the outputs are distinguishable from other situations. Here are some practical examples:

• Being happy or being sad, are two different states, since the inner behavior is different (this least often also to a different output).
• Similarly, an empty memory (or hard drive) is in a different state compared to a filled one.
• A traffic light showing green has an output distinguishable from red, or yellow.

Sequential logic is used to describe logic circuits that show internal states („stateful logic“), and therefore have at least one memory element (= flip-flop).

The following terminology is used in the upcoming explanations:

• The input vector $\vec{X}$ represents the $k$ inputs $X_0 ... X_{k-1}$.
• The output vector $\vec{Y}$ represents the $l$ outputs $Y_0 ... Y_{l-1}$.
• The state vector $\vec{Z}$ represents the $m$ inputs $Z_0 ... Z_{m-1}$.
• The subscript $(n)$ or $n$ marking the current point in time and therefore e.g. the current state $Z_0(n)$.
• The subscript $(n+1)$ or $n+1$ marking the next upcoming point in time and therefore e.g. the next state $Z_0(n+1)$.
• Sequential logic circuits are also called Finite State Machines (FSM) or sometimes also shortened to „state machines“.

The Abbildung 12 shows the different terms in an abstract diagram. The „current“ output values are here the values, which are shown after the delay time of the gates (about some nanoseconds).

The principle interior of the black box in Abbildung 12 was already shown in one practical application in the previous chapter: We saw, that we need the combination of combinatorial logic and some storage components. Additionally, both have to be connected by feeding back some of the outputs back to the combinatorial logic. The output bits $\vec{Y}$ can result either from the combinatorial logic or the flip-flops. This is shown in Abbildung 13.

One simple example of sequential logic is shown in Abbildung 4. There, the combinatorial logic is explicitly shown. Depending on the input $X$ the output $\vec{Y}$ shows an up-counting 2-bit value counting $0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow 0 \rightarrow ...$. This is a simple state machine that will be used in the net chapters

6.2 Classical State Machine Types

The up-counter in the previous sub-chapter was able to count from $0$ to $3$. But what can we do to count differently, like $ 7, 6, 1, 5$? To understand this, a simpler situation will be investigated. So, let's look at how one can create an up-counter counting $1, 2, 3, 4$.

6.2.1 Moore Machine

The first idea might be to use what we already have: an up-counter, which facilitates 2 flip-flops to result in 2-bit output. The wanted new state machine needs 3 bits for the output, since the binary representation of our outputs is $001_2$,$010_2$,$011_2$,$100_2$.

A simple idea is to take the 2-bit up-counter and add a combinatorial logic in behind it. This logic shall convert the 2-bit up-counter output $00_2$ into $001_2$, the $01_2$ into $010_2$, $10_2$ into $011_2$ and $11_2$ into $101_2$. This can be logic can be created by:

• writing down the truth table
• putting the values into a Karnaugh map
• extracting the formula with a view onto the implicants
• generating the circuit with gates

When this is done, the result looks like Abbildung 5

This resembles a so-called Moore Machine.

6.2.2 Mealy Machine

When looking at Abbildung 5 a bit more in detail, one can see, that the outputs $Y_0$ and $Y_1$ just equals the output of the first combinatorial logic circuit. This is not surprising: the input logic circuit shows the $\vec{Z}(n+1)$ and this is for the counter always the stored value plus one, except when the maximum is reached.

With this information, the state machine in Abbildung 5 can be simplified by using the outputs of the input circuit for $Y_0$ and $Y_1$. This is shown in Abbildung 6.

6.2.3 Medvedev Machine

The Abbildung 8 shows the principle differences in the architecture of the state machines.

6.3 State Diagram, State Transition Diagram

6.3.1 Motivation

The diagrams of different states are well known from physics for example the state diagram (or better: phase diagram) of water, where its three states are: solid ice, liquid water, and gaseous steam. The possible state transitions are due to temperature increase or decrease.

In Abbildung 9 image (1) the states of water are shown on the temperature axis. When only the state transitions are relevant, the states are simplified to a circle, showing the state name and behavior. The transitions are depicted as arrows, where the needed condition is written onto (See Abbildung 9 image (2) ). This diagram is called state transition diagram.

For matter not only the dimension „temperature“ is important, but also the „pressure“. The full phase diagram is shown in Abbildung 10 image (1). By this, another variable is available and more transitions. These can be drawn into the state transition diagram (Abbildung 10 image (2)).

6.3.2 Simple logic Example

In Germany, often one has to pay for entering the toilet. An example of such an entrance control system is shown in Abbildung 11. In this (artificial) example, one can pay either 50ct or 1€.
Once paid, the turnstile will release and one can enter. Once the turnstile was pushed the entrance is closed again.

The Abbildung 12 the state transition diagram is drawn.

• The two states are that (1) the turnstile is opened and one can go through and (2) the turnstile is closed and one cannot enter anymore.
• The transitions are given by the done actions: one can either insert a coin or push on the turnstile.

Important here are some additional considerations:

• For the state transition diagram one has to look for all possible transitions. So, also pushing a closed turnstile or inserting more coins have to take into account.
• A state transition diagram is not complete without a legend and without an beginning/reset point. The reset point is given by an arrow with „reset“ written on it

Out of this state transition diagram, one can create a table-like representation, see Abbildung 13.

the inputs, outputs, and states have to be encoded into binary, to investigate this table a bit more. How the binary value is connected to the outputs does not matter. We will choose the following coding:

• Encoding of the states: turnstile closed ≙ $Z=0$, turnstile opened ≙ $Z=1$,
• Encoding of the inputs: no coin inserted ≙ $Xc=0$, coin inserted ≙ $Xc=1$, turnstile not pushed ≙ $Xp=0$, turnstile pushed ≙ $Xp=1$,
• Encoding of the outputs: disallow entrance ≙ $Y=0$, allow entrance ≙ $Y=1$,

This table is shown in Abbildung 14 and is called state transition table.

Interestingly, the logic circuit for this state transition table was already part of the course: it is the RS flip-flop! When looking deeper into the table in Abbildung 14 one can substitute $Xc$ with $S$ (as in Set) and $Xp$ with $R$ (as in Reset) to directly get the truth table of the RS flip flop.

6.3.3 First Adaption for Up-Counter

In the previous sub-chapter (6.2) we had a look at different implementations of an up counter and in this chapter the way to represent a state machine via a state transition diagram. So one question is: what do these up-counters look like in the state transition diagram?

The answer is quite simple: there are 4 states, so 4 „circles“ are needed. These states need to have circular connections to get the wanted increase from $00_2=0_{10}$, to $01_2=1_{10}$, to $10_2=2_{10}$, to $11_2=3_{10}$ and then back to $00_2=0_{10}$. The first state shall be $00_2=0_{10}$, so after the reset, the state machine will get entered there. Finally, the output vector is similar to the state vector. With a deeper look at it: This is therefore in particular a Medvedev machine!

In Abbildung 15 both types of state machines are shown.

• In the Moore machine the already seen „halving of the circle“ is used to show the distinct state vector in the upper half and the output vector in the lower half.
• In the Medvedev machine only one value is written in the circle, since here the state vector is also the output vector.

The next step is to transform this into an up counter from $1...4$. For this simply the output has to be changed. This means the numbers in the „lower half of the circles“ have to be adapted (see in <imfref pic16>).

Looking at what we got, there is one part missing: the state machine does not have any input… So the input vectors have to be added to the transition (arrows). For an activatable counter, there only has to be one input $X$, which acts as an enable input.

Exercises

Exercise 6.1.1. State Transition Diagram I - fill the gaps

The following state transition diagram shall be given:

• There are two transitions marked with $A$ and $B$.
  What values do the inputs need to have to show all transitions explicitly?

Strategy

• Find the transitions wanted
• Look at which state these transitions start.
• Which other transitions start there?
• Which transition conditions are missing?

Solution

• transition A
  • Starts at state $000$
  • Also transition with $11$, $00$ starts here
  • $01$, $10$ are missing
• transition B
  • Starts at state $010$
  • Also transition with $0-$ starts here
  • $1-$ are missing

Result

• A: $01$, $10$
• B: $1-$

• How many flip-flops are necessary for such a Moore Machine?

Strategy

Each flipflop can store one Bit. Each stored bit can be used to address states. So, check the number of bits $i$ of states $Z_i$ („size of the state vector“).
Be aware, that one bit can address a maximum of 2 states, two bits a maximum of 4 states, three bits a maximum of 8 states, and so on.

Solution

• Number of bits $i$ of states $Z_i$ is given in the legend. The number of bits $i$ has also to fit the number of states.
• Here the legend shows $Z_2, Z_1, Z_0$, so there are 3 bits.
• Also the number of states in the diagram is 5. This can only be numbered with at least 3 bits.

Result

3

• Fill in the missing cells in the following state transition table:

Strategy

Check for each line:

• What is the start/present state (given by the bits $\color{red}{Z_2(n), Z_1(n), Z_0(n)}$ in the line)?
• Which transition is shown as start/present state (given by $\color{brown}{X_1, X_0}$)?

Out of this orientation, the next columns can be derived:

• At the end of the transition, the next state can be found (necessary for columns $\color{green}{Z_2(n+1), Z_1(n+1), Z_0(n+1)}$)
• Within the state symbol of the present state, the output values can be found (necessary for columns $\color{violet}{Y_2(n), Y_1(n), Y_0(n)}$)

Result