Verilog grammar basics HDL bits training 06

Combinational Logic:Basic Gates

One 、Wire

• RTL Code

module top_module (
    input in,
    output out);
    
    assign out = in;

endmodule

Two 、GND

• RTL Code

module top_module (
    output out);
    
    assign out = 1'b0;

endmodule

3、 ... and 、NOR

module top_module (
    input in1,
    input in2,
    output out);
    
    assign out = ~(in1 | in2);

endmodule

Four 、Another gate

• RTL Code

module top_module (
    input in1,
    input in2,
    output out);
    
    assign out = in1 & (~in2);

endmodule

5、 ... and 、Two gates

• RTL Code

module top_module (
    input in1,
    input in2,
    input in3,
    output out);
    
    assign out = ~(in1 ^ in2) ^ in3;

endmodule

6、 ... and 、More logic gates

• RTL Code

module top_module( 
    input a, b,
    output out_and,
    output out_or,
    output out_xor,
    output out_nand,
    output out_nor,
    output out_xnor,
    output out_anotb
);
    assign out_and = a & b;
    assign out_or = a | b;
    assign out_xor = a ^ b;
    assign out_nand = ~(a & b);
    assign out_nor = ~(a | b);
    assign out_xnor = ~(a ^ b);
    assign out_anotb = a & (~b);

endmodule

• Simulation oscillogram
  

7、 ... and 、7420 chip

• RTL Code

module top_module ( 
    input p1a, p1b, p1c, p1d,
    output p1y,
    input p2a, p2b, p2c, p2d,
    output p2y );
    
    assign p1y = ~(p1a & p1b & p1c & p1d);
    assign p2y = ~(p2a & p2b & p2c & p2d);

endmodule

• Simulation oscillogram
  

8、 ... and 、Truh tables

This problem can be hard solved , The box case Statement lists all Probably , Of course , What this topic wants to examine is the knowledge points of listing logical expressions according to the truth table , It is recommended to use Karnaugh map method to simplify

• case sentence RTL Code

module top_module( 
    input x3,
    input x2,
    input x1,  // three inputs
    output f   // one output
);
    always @(*)
        case({
    x3,x2,x1})
            3'd2,3'd3,3'd5,3'd7:
                f = 1'b1;
            3'd0,3'd1,3'd4,3'd6:
                f=1'b0;
            default: f = 1'b0;
        endcase

endmodule

• Logical expression RTL Code

module top_module( 
    input x3,
    input x2,
    input x1,  // three inputs
    output f   // one output
);
    assign f = ((~x3)&x2&(~x1))|((~x3)&x2&x1)|(x3&(~x2)&x1)|(x3&x2&x1);
    
endmodule

Nine 、Two-bit equality

• RTL Code

module top_module ( input [1:0] A, input [1:0] B, output z ); 
    
    assign z = (A == B)? 1'b1:1'b0;

endmodule

Ten 、Simple circuit A

• RTL Code

module top_module (input x, input y, output z);
    
    assign z = (x ^ y) & x;

endmodule

11、 ... and 、Simple circuit B

• RTL Code

module top_module ( input x, input y, output z );
    
    assign z = (x == y)? 1'b1:1'b0;

endmodule

• Simulation oscillogram
  

Twelve 、Combine circuits A and B

This question needs to use the sentences of the first two questions to pave the way

• RTL Code

module top_module (input x, input y, output z);
    
    wire z_A;
    wire z_B;
    assign z_A = (x ^ y) & x;
    assign z_B = (x == y)? 1'b1:1'b0;
    assign z = (z_A | z_B) ^ (z_A & z_B);

endmodule

13、 ... and 、Ring or vibrate

You can see the code by looking at the simulation waveform

• RTL Code

module top_module (
    input ring,
    input vibrate_mode,
    output ringer,       // Make sound
    output motor         // Vibrate
);
    always @(*)
    if(ring & !vibrate_mode)begin
       	motor = 1'b0; 
        ringer = 1'b1;
    end
    else if(vibrate_mode & ring)begin
       	ringer = 1'b0;
        motor = 1'b1;
    end
    else begin
    	ringer = 1'b0;
        motor = 1'b0;
    end
        
endmodule

• Simulation oscillogram
  

fourteen 、Thermostat

Make a thermostat control unit

• RTL Code

module top_module (
    input too_cold,
    input too_hot,
    input mode,
    input fan_on,
    output heater,
    output aircon,
    output fan
); 
    assign heater = mode & too_cold & (!aircon);
    assign aircon = (!mode) & too_hot & (!heater);
    assign fan = heater | aircon | fan_on;

endmodule

• Simulation oscillogram
  
  

15、 ... and 、3-bit population count

• RTL Code

module top_module( 
    input [2:0] in,
    output [1:0] out );
    
    integer i;
    always @(*)begin
       out = 1'b0;
        for(i=0;i<3;i=i+1)
            if(in[i])
                out = out + 1'b1;
        	else
                out = out;
    end
        

endmodule

• Simulation oscillogram
  

sixteen 、Gates and vectors

• RTL Code

module top_module( 
    input [3:0] in,
    output [2:0] out_both,
    output [3:1] out_any,
    output [3:0] out_different );
    
    integer i;
    always @(*)begin
        for(i=0;i<3;i=i+1)
            begin
                out_both[i] = in[i] & in[i+1];
                out_any[3-i] = in[3-i] | in[2-i];
                out_different[i] = in[i] ^ in[i+1];
            end
    end
    assign out_different[3] = in[3] ^ in[0];

endmodule

• Simulation oscillogram
  

seventeen 、Even longer vectors

• RTL Code

module top_module( 
    input [99:0] in,
    output [98:0] out_both,
    output [99:1] out_any,
    output [99:0] out_different );

    integer i;
    always @(*)begin
        for(i=0;i<99;i=i+1)
            begin
                out_both[i] = in[i] & in[i+1];
                out_any[99-i] = in[99-i] | in[98-i];
                out_different[i] = in[i] ^ in[i+1];
            end
    end
    assign out_different[99] = in[99] ^ in[0];
    
endmodule