CSE370 Final Exam Solution (7 June 1999)
 



Please read through the entire examination first! This exam was designed to be completed in 100 minutes (one hour and 40 minutes) and, hopefully, this estimate will be reasonable. There are 5 problems for a total of 100 points. The point value of each problem is indicated in the table below.

Each problem is on a separate sheet of paper. Write your answer neatly in the space provided. If you need more space (you shouldn’t), there are a couple of extra sheets at the end of the exam, but please make sure that you indicate clearly the problem to which the comments apply. Do NOT use any other paper to hand in your answers.

The exam is oepn book and notes.  Please do not ask or provide anything to anyone else in the class during the exam.  Make sure to ask clarification questions early so that both you and the others may benefit as much as possible form the answers.

Good luck.



 
Name:
 

ID #:


 



 
 
Problem
Max Score
Score
1
10
 
2
25
 
3
15
 
4
25
 
5
25
 
TOTAL
100
 

 

1. Combinational Logic (10 points)

You are to design a circuit that takes a 4-bit number as input (I8, I4, I2, I1) and generates an output which is 1 if the input number is one of the prime numbers between 0 and 15 inclusive.  You may assume that the input is never 0, 1, or 2. The most significant bit of the input is I8. Please use the Karnaugh map below for this problem.

(a) Write this function as a list of minterms (S notation).
 

S m(3, 5, 7, 11, 13) + d (0, 1, 2)


(b) List the prime implicants for this function.
 

I8'I4', I8'I1, I4I2'I1, I4'I2I1


(c) List the essential prime implicants for this function.
 

I8'I1, I4I2'I1, I4'I2I1


(d) Find the minimal sum of products expression for this function.
 

I8'I1 + I4I2'I1 + I4'I2I1


(e) Draw a circuit diagram for the mimized function using only AND, OR, and NOT gates.
 
 
 
 
 
 
 

2. PAL/PLA/ROM Implementations of Combinational Logic (25 points)

You are to implement three functions (F2, F3, F5) of the same input (I8, I4, I2, I1 as in the problem above) that indicate when the binary is divisible by 2, 3, and 5, respectively.  Use the three K-maps below to specify your functions (you may continue to assume that the input is never 0, 1, or 2).
 
 

(a) Implement the three functions above using the read-only memory (ROM) shown below. Place Xs where connections need to be made in the OR plane.  (5 points)
 


 
 
 

(b) Implement the three functions above using the programmable array logic (PAL) shown below. Place Xs where connections need to be made in the AND plane. Note that all outputs must be a sum of no more than three product terms but that you have the ability to form a fourth output that can be fed back into the array as an input. NOTE: please write the product term that each AND gate implements inside the AND gate symbol.  (10 points)

     


 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

(c) Implement the three functions above using the programmable logic array (PLA) shown below. Place Xs where connections need to be made in the AND and OR planes. Note that there can be no more than eight total product terms for all three functions. NOTE: please write the product term that each AND gate implements inside the AND gate symbol.  (10 points)
 
 

















3. Sequential Logic (15 points)

Determine the fastest possible clock period (the time between consecutive clock edges) for the circuit shown below.  You can assume that the gate delay of the NAND and NOR gates is 2, the XOR gate's is 3, the inverter's is 1, the setup and hold times for the flip-flops are both 1, and the propagation time of the flip-flop is 2.  Explain how you derived your value.
 
 


min clock period = FF prop time + comb. logic delay + setup time
min clock period = 2 + max(XOR + NOR, NAND + NOT + NOR) + 1
min clock period = 2 + max(3 + 2, 2 + 1 + 2) + 1
min clock period = 2 + max(5, 5) + 1
min clock period = 2 + 5 + 1
min clock period = 8

Recall that the hold time of a flip-flop overlaps the propagation time.


 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

4. Implementation of Sequential Logic (25 points)

(a) Derive a state diagram for an asynchronous Mealy finite state machine that outputs a 1 whenever it sees the pattern 1110, 0111, 0110, or 1111 on its single input.  Do NOT consider overlapping patterns, instead look at each 4 bits as distinct.  Use the skeletal state diagram provided below and fill in the correct output value on each arc.  (2 points)
 


 

(b) Minimize the number of states in your state machine using the set-partition method discussed in lecture.  List all equivalent states and draw a new state diagram (keep it in the same style as much as you can, 0 transition on the left, 1 transition on the right).  (5 points)

(ABCDEFGHJKLMNPQ)

(ABCDEFGHJKMNP)  (LQ)

(ABCDFHJKMNP) (EG) (LQ)

(ADFHJKMNP) (BC) (EG) (LQ)

(A) (DFHJKMNP) (BC) (EG) (LQ)

(A) (DF) (HJKMNP) (BC) (EG) (LQ)


 
 
 
 
 
 
 
 
 

(c) Derive a state table for your minimized state machine.  Keep it small by using don't cares.  (3 points)
 
 

Input
Present State
Next State
Output
-
A
B
0
0
B
D
0
1
B
E
0
-
D
H
0
0
E
H
0
1
E
L
0
-
H
A
0
-
L
A
1

 

(d) Determine a minimum-width (smallest possible number of bits) state assignment and explain how you came up with your encoding.  (4 points)
 
 

state code  reason
A 000  most common state, keeps number of 1s in truth table down
B 001  first bit has been seen
D 010  keep number of 1s down (like E, both from B)
E 011  second bit has been seen (like D, both from B)
H 100 keep number of 1s down
L 111  third bit has been seen

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

(e) Write a pseudo-Verilog decription (as close as possible to Verilog syntax but don't worry about being precise) of your minimized state machine.  Make sure to make your encoding clear.  (5 points)
 

module FSM (Input, Reset, Output)
input Input, Reset;
output Output;
reg Output;
reg present_state[2:0];
reg next_state[2:0];
'define A 3b'000;
'define B 3b'001;
'define D 3b'010;
'define E 3b'011;
'define H 3b'100;
'define L 3b'111;

always @(posedge clk) begin
   if (Reset) then present_state = 'A;
              else present_state = next_state;
end

always @(Input or present_state) begin
   Output = 0;
   case (present_state)
      'A: begin
             next_state = 'B;
          end
      'B: begin
             if (Input) next_state = 'E
                else next_state = 'D;
          end
      'D: begin
             next_state = 'H;
          end
      'E: begin
             if (Input) next_state = 'L;
                else next_stae = 'H;
          end
      'H: begin
             next_state = 'A;
          end
      'L: begin
             next_state = 'A;
             Output = 1;
          end
      default: display("error");
   endcase
end
endmodule


(f) Derive the logic to implement the state machine.  Do not bother drawing a circuit, simply write down the Boolean equations for the output and each of the state bits.  The input is named In and the output is named Out.  The state bits should have names such as P1, P2, ... and N1, N2, ... for present and next state, respectively.  (6 points)
 

Encoded state table:
Input
Next State
Present State
Output
-
000
001
0
0
001
010 
0
1
001
011
0
-
010 
100
0
0
011
100
0
1
011
111
0
-
100
000
0
-
111
000
1
OUT = P1P2P3
N1 = P1’P2
N2 = P1’P2’P3 + P1’P2P3IN
N3 = P1’P2’P3’ + P1’P3IN

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

5. Finite State Machine Design (25 points)

The Ethernet physical layer protocol encodes data using a method called Manchester encoding.  In Manchester encoding, a "zero" is transmitted by first sending a 1 and then a 0, while a "one" is transmitted by sending a 0 and then a 1.  When no packet is being sent, the line is kept at 1.  Every packet starts with a 0 bit.  The figure below provides an example of a packet containing the data 010101011000111001101.  The tick marks below the waveform represent clock periods, those above represent "data periods" which correspond to two clock periods.


 

Derive a Moore state machine that decodes an Ethernet packet.  The input should be called E (for Ethernet) and there should be two outputs, one for the data decoded (D for data) and one for whether it is should be ignored or not (V for valid).  Note that when the line is quiet, consecutive bits will all be 1.  Therefore, you should make the decoder "reset" whenever there are a number of consecutive 1s inconsistent with normal data transmission.  Be sure to handle the error condition of too many consecutive bits that are 0, as well.  Your machine should keep V low and wait for two consecutive 1s to "reset".
 
 


 
 
 
 
 


This state diagram can then be reduced via state minimization to: