CSE 591 Hardware Synthesis with VHDL
Examination #1

I. (15 pts.) Sketch schematics for the following constructs. Unless specified otherwise, assume signals of type std_logic or std_logic_vector of the appropriate size.

      SIGNAL enable : std_ulogic;       
      SIGNAL a, b : boolean; 
      SIGNAL X, Y, Z : std_logic;

 
a.   IF enable = '0'   
      THEN Y <= X;                                                           
      END IF;

b.   IF enable = '1'   
      THEN Z <=  X;
      ELSE  Z <= Y;
      END IF;

c.   IF a OR b
      THEN Z <= X; 
      ELSIF  a AND b   
            THEN Z <= not X;  
            ELSE   Z <= Y;  
      END IF;

There is actually an optimization of this if you realize that a or b contains a and b and the middle elsif clause will never be used, but that was not required for this problem.

II. (15 pts.) Sketch schematics for the following constructs. Unless specified otherwise, assume signals of type std_logic or std_logic_vector of the appropriate size.

a.  IF  enable = '1'       
    THEN Z <= X;   
    ELSIF enable = '0' 
        THEN  Z <= not X;   
        ELSE Z <= 'X';
    END IF;

Z < = 'X' is not used by the synthesis tool. You may also recognize the circuit above can be replaced by XNOR.

EVERYONE GETS 5 FREE POINTS FOR PART B. The question shown below is the one intended, and the solution is shown on the right.

b. IF  sel = "00"
    THEN  Y <= X;   
    ELSE Y <= 'Z';                                                        
    END IF;     
    IF  sel = "01" and a   
    THEN  Z <= X;   
    ELSE Z <= Z';  
    END IF;

c.  SIGNAL p : integer range 0 to 3; 
    SIGNAL Y :  std_logic_vector  (0 to 1);
     
    CASE p IS 
    when 0 => Y <= "10" after Tprop;  
    when 1 => Y <=   X   after Tprop;  
    when 2 => Y <=   Y   after Tprop;  
    when 3 => Y <= "01" after  Tprop;  
    END CASE;

III. (15 pts.) Sketch schematics for the following constructs. Unless specified otherwise, assume signals of type std_logic or std_logic_vector of the appropriate size. Indicate clearly whenever you are showing buses what the width and connection is.

a.  WITH regsel SELECT  
    Z <= A after Tprop WHEN "00"   
        B after Tprop WHEN  "01"   
        C after Tprop WHEN "10"   
        D after Tprop WHEN "11"  
        "XX" after Tprop WHEN others;

where A, B, C, D, Z may be any width vectors as long as they are compatible.

b.   SIGNAL  pbus : std_logic  BUS;                            
    a1:  BLOCK  ( enable = '0' )   
        pbus <= guarded X after Tprop;   
        END BLOCK a1;                                                
    a2:  BLOCK  ( enable = '1' )   
        pbus <= guarded Y after Tprop;   
        END BLOCK a2;

c.  SIGNAL Z : std_logic_vector( 0 to 2) BUS;  
    SIGNAL X, Y : std_logic_vector( 0 to 2);  
    SIGNAL a, b : std_logic;  

    a0:  BLOCK  ( a = '1')  
        Z <= X after Tprop;                                                
        END BLOCK;   

    a1:  BLOCK ( b = '1')  
        Z <= Y after Tprop;  
        END BLOCK;

The signals are not guarded, so a and b do not contribute here.

IV. (15 pts.) How does a synthesis tool use each of the following pieces of information:

a. SIGNAL x : 0 to 63;

	Signal X is either a bus or a register of 6 bits.

b. CONSTANT xxxs : std_logic_vector (0 to 7) := (others => 'X');

	XXXs represents "XXXX_XXXX" which is useful for simulation, 
            but not for synthesis.

c. SIGNAL Y : std_logic_vector ( X'range ) BUS;


The signal Y will be synthesized as a bus whose width is the 
same as for signal X (either a bus or a register) and uses the 
same indexing scheme to referenceits bits as is used for signal X.

d. ASSERT clock'event and clock = '1' and J'stable(3 ns) and K'stable(3 ns)

ASSERTIONS are not synthesized.

e. Z <= X(N-1) & X( 0 to N-2) -- X(0 to N-1)

Z gets the value of signal X right rotated by one place.  
The high bit of Z gets the low bit of X, and all the rest of the 
bits Z(i) get X(i-1).

V. (15 pts.) Sketch the unoptimized circuit (4 pts.) and then explaining your steps transform the original circuit into an optimized result. (4 pts).

a.  Y := X(0);    
    FOR i in 1 to 3 LOOP    
        IF X(i) = '1' THEN Y <= not Y;    
    END LOOP;

The inverted X with the multiplexor at each stage is equivalent to XOR. We then balance the XOR gates as shown below to minimize the worst case input-output delay for the circuit.

Vb. SIGNAL Y : std_logic_vector (0 to 3);    
    SIGNAL opcode : std_logic_vector (0 to 1);    

    WITH  opcode SELECT     
    Y <= "0001" WHEN "11",    
        "0010" WHEN "10",    
        "0100" WHEN "01",    
        "1000" WHEN "00";

The leftmost multiplexor appears below. Constant propagation allows us to replace any NAND gate with a '0' input with a '1' on the output signal line and to delete any '1' input to an NAND gate. This eliminates gates b, c, d below, and the top input to gate a. This then allows us to drop the bottom three inputs to gate e, leaving an inverter. The NAND followed by the inverter can be replaced by an AND gate.

Repeating this for the other three stages and collecting redundant inverters from opcode(0) and opcode(1) gives us this final optimized circuit.

VI. (15 pts.) Given the following state transition diagram, write a VHDL description that separates the clock, the next state logic, and the output function into separate processes. Use an enumerated type for the state object.

ENTITY FSM port ( X : in bit;  
                  clock, reset : in bit;
                  Z : out bit ) is
end FSM;

ARCHITECTURE enum OF FSM is
    type State is ( s0, s1, s2 );
    signal current_state, next_state : State;
begin

Ck_reg:  process ( clock )
    begin
        if (reset = '1') then current_state <= s0;
        elsif (clock'event and clock='1' and clock'last_value = '0') then
                current_state <= next_state;
        end if;
    end process;

Next_logic: process (X, current_state)
    begin
        case current_state is
        when s0 =>  if X='1' then next_state <= s1 
                             else next_state <= s0;
        when s1 =>  if X='1' then next_state <= s2 
                             else next_state <= s0;
        when s2 =>  if X='1' then next_state <= s2 
                             else next_state <= s1;
        end case;
    end process;

Output_logic:  process ( current_state )
-- Note that the Z output is function only of the current_state.
begin
    case current_state is
    when s0 | s1 =>  Z <= '0';
    when s2     => Z <= '1';
    end case;
end process;

end enum;

VI.b. Draw a schematic (using D flipflops) similar to the circuit that either Synopsys or Autologic should produce from such a description. Assume a "one hot" encoding of the states.

VII. (10 pts.) Asynchronous design

Even though Synopsys and Autologic won't synthesize this description, using the form using a "one hot" encoding of the implicit states developed in class this semester to support the representation of algorithmic state machines (flowchart approaches to describing state machines), what should a synthesis tool produce?

    ARCHITECTURE a OF b IS  
    SIGNAL  request, send,  ack : boolean;  
    SIGNAL  Data_in, Cvec : std_logic_vector (0 to 3); 
    BEGIN

    PROCESS BEGIN  
        send <= false;  
        WAIT UNTIL request;  
        send <= true;     
        Cvec <= Data_in;  
        WAIT UNTIL ack; 
        END PROCESS;    
    END a;

Copyright 1995, Ben M. Huey
Copying this document without the permission of the author is prohibited and a violation of international copyright laws.

Revised 12/11/95

CSE 591 Hardware Synthesis with VHDL Examination #1

CSE 591 Hardware Synthesis with VHDL
Examination #1