# EE 110 Practice Problems for Final Exam: Solutions

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1 EE 1 Practice Problems for Final Exam: Solutions 1. Finite State Machines: Sequence Recognizer You want to build a finite state machine that will recognize the sequence x = and output the sequence z = as this sequence occurs. In other words, output z = 0 when first receiving x = 0. Then output z = 0 if the next bit of x = 1; output z = 0 again if the following bit of x = 1. Finally, if the last (fourth) bit of x = 0, output z = 1. More simply, output z = 0 until the sequence x = is received, at which time output z = 1. 1(a). First, draw and label the transitions in the state bubble diagram below. The states are already labeled (but state bit values have not been assigned). Allow overlap of sequences. Build as a Moore machine. Include the input bits of x and output bits of z. START z=0 GOT0 0 GOT 0 GOT 1 GOT1 0 Note: This sequence recognizer allows for overlap of sequences, and thus outputs z = 1 whenever the sequence x = is recognized with overlap of sequences. If we do not allow for overlap, then the transitions from state GOT will go to state START for both x = 0 and x = 1. Without overlap, whenever the sequence x = is input, the output will be z =. 1(b). Write the state table from the state bubble diagram. Fill in the table given below. Use the following state assignments: START=0, GOT0=1, GOT=0, GOT1=1 and GOT=1. All unused states should go to the START state and output z = 0. Assume you will build the circuit with J-K flip-flops. (next page) 1

2 EE 1 Practice Problems for Final Exam: Solutions, Fall 28 2 Present State Input Output Next State Flip-Flop Inputs z Q 2 Q 1 Q 0 J 2 K 2 J 1 K 1 J 0 K X 0 X 1 X X 0 X 0 X X 0 X X X 1 X X X X 1 1 X X X 0 1 X X X 1 X X X 1 X X 1 0 X 1 X X 1 1 X 0 X X 1 0 X X X 1 0 X X X 1 X 1 0 X X 1 X 1 0 X X 1 X 1 X X 1 X 1 X 1 1(c). Find the minimized logic equations for output z and flip-flop inputs J 2, K 2, J 1, K 1 and J 0, K 0 ; use K-maps if needed. Output z: Output z can be determined directly from the truth table, as z = 1 only when Q 2 = 1, = 0 and Q 0 = 0. Therefore, z = Q 0. You may check with a K-map if you prefer. z = Q 0 K-map for J 2 :

3 EE 1 Practice Problems for Final Exam: Solutions, Fall 28 3 From the K-map above, we obtain the following expression for J 2 : J 2 = K 2 : Note that K 2 is either X or 1 for every entry of the truth table. Therefore, we may set K 2 = 1. K 2 = 1 K-map for J 1 : From the K-map above with two groupings, we obtain the following expression for J 1 : J 1 = Q 2 + Q 2 This expression may be further simplified by recognizing that Q 2 Q 0 + Q 2 Q 0 = Q 2 Q 0 ; J 1 = x(q 2 ) K-map for K 1 : From the K-map above with three groupings, we obtain the following expression for K 1 : K 1 = Q 0 + Q 2 + x

4 EE 1 Practice Problems for Final Exam: Solutions, Fall 28 4 K-map for J 0 : 1 0 X X 1 1 X X 0 0 X X 1 0 X X From the K-map above with two groupings, we obtain the following expression for J 0 : J 0 = x + K-map for K 0 : X X 1 0 X X 1 1 X X 1 1 X X 1 1 From the K-map above with three groupings, we obtain the following expression for K 0 : K 0 = + Q 2 + x 1(d). outputs. Draw the corresponding circuit diagram with J-K flip-flops. Label all inputs and (next page) Note: The black dots in the figure indicate a connection between wires. If there is no black dot, there is no connection between wires that cross.

5 EE 1 Practice Problems for Final Exam: Solutions, Fall 28 5 x NOT X +5V CLK J 2 K 2 Q 2 Q 2 J 1 K 1 J 0 Q 0 K 0 Q 0 z 2. State Bubble Diagram of Mealy Machine Redraw the state bubble diagram using a Mealy machine design. Be sure to label the transitions and bubbles. You may name your states whatever you like. Again allow overlap of sequences. How many states and flip-flops do you need for the Mealy design? x/z=0/0 START BIT1 1/0 0/0 x/z=1/0 0/1, 1/0 BIT3 0/0 1/0 BIT2 Only four states are required for the Mealy machine design. Two state bits are thus needed to determine all the states. This means that only two flip-flops are required to implement the Mealy machine design of the sequence recognizer. 3. Analysis of FSM Circuit Design Below is a simplified block diagram of a heart pacemaker. The output p from the pacemaker pulses high if the heart does not contract within a certain time. The input s indicates whether the heart has contracted (s = 1) or not (s = 0). The input t comes from a timer,

6 EE 1 Practice Problems for Final Exam: Solutions, Fall 28 6 which counts for the expected time between contractions (approximately 1 second). When t = 1, the timer has counted up to 1 second, and the heart should have contracted. If the heart has not contracted within that time, the pacemaker sends a pulse p = 1. (There is also a timer reset control coming from the pacemaker, which we ignore for this problem.) The inputs to the pacemaker circuit are s (from the heart) and t (from a timer - not shown). The output of the pacemaker circuit is p, which goes to the heart. Arrows indicate the inputs and output. Heart p s NOT s Pacemaker D 1 D 0 Q 0 t CLK Q 0 Note: The black dots in the figure indicate a connection between wires. If there is no black dot, there is no connection between wires that cross. 3(a). Is this a Mealy or a Moore machine? Why? This is a Mealy machine. The output p depends on the input s, as well as the sequential logic output states and Q 0 and the input t from a timer clocked synchronously with FF1 and FF0. This means that the output p can change at times other than on a rising clock edge, if the input s changes. The input s is not clocked, and is not constrained to change only on a rising clock edge. 3(b). Write the equations for output p and the flip-flop inputs D 1 and D 0 from the block diagram. Output p: p = Q 0 s t D input for Flip-flop 1, D 1 : D 1 = p = Q 0 s t

7 EE 1 Practice Problems for Final Exam: Solutions, Fall 28 7 D input for Flip-flop 0, D 0 : D 0 = Q 0 + s 3(c). Fill in the state table below, using the equations obtained in part 3(b). Present State Inputs Output Next State Q 0 s t p Q 1 Q

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