Count15
module top_module ( input clk, input reset, // Synchronous active-high reset output [3:0] q); always@(posedge clk) begin if(reset) q <= 4'd0; else q <= q + 1'b1; end endmodule
Count10
题目给的答案把q清零和reset放在了一个if里if (reset || q == 9)。
module top_module ( input clk, input reset, // Synchronous active-high reset output [3:0] q); always@(posedge clk) begin if(reset) q <= 4'd0; else q <= (q<9)?(q+1):0; end endmodule
Count1to10
module top_module ( input clk, input reset, output [3:0] q); always@(posedge clk) begin if(reset) q <= 4'd1; else q <= (q<10)?(q+1):1; end endmodule
Countslow
module top_module ( input clk, input slowena, input reset, output [3:0] q); always@(posedge clk) begin if(reset) q <= 4'd0; else if(slowena) q <= (q<9)?(q+1):0; end endmodule
Exams/ece241 2014 q7a
这道题感觉怪怪的，一开始我写的c_load = (Q==12)|reset;由于题目说了count4模块中的load优先级比enable高，所以enable为0时仍可以置数。
而题目貌似是希望load和enable不能冲突，enable为0时就不应该置数了，所以此时的load应该为0，所以我又把信号与上了一个enable。
module top_module ( input clk, input reset, input enable, output [3:0] Q, output c_enable, output c_load, output [3:0] c_d ); // count4 the_counter (clk, c_enable, c_load, c_d , Q ); assign c_d = 1; assign c_enable = enable; assign c_load = enable&(Q==12)|reset; endmodule
Exams/ece241 2014 q7b
把三个BCD计数器级联在一起即可。一开始我写的c_enable[2]=(Q2==9);OneHertz=(Q3==9)；而Q2、Q3都会持续多个周期，只有Q1是一个一直变的信号。
module top_module ( input clk, input reset, output OneHertz, output [2:0] c_enable ); // wire [3:0]Q1; wire [3:0]Q2; wire [3:0]Q3; bcdcount counter0 (clk, reset, c_enable[0],Q1); bcdcount counter1 (clk, reset, c_enable[1],Q2); bcdcount counter2 (clk, reset, c_enable[2],Q3); assign c_enable[0]=1'b1; assign c_enable[1]=(Q1==9); assign c_enable[2]=(Q2==9)&&(Q1==9); assign OneHertz=(Q3==9)&&(Q2==9)&&(Q1==9); endmodule
Countbcd
感觉写的稍微有点复杂，利用了verilog里写多个if优先下面的语句的特点，实际不会对一个数重复操作。
实际例化多个BCD计数器会简单很多。
module top_module ( input clk, input reset, // Synchronous active-high reset output [3:1] ena, output [15:0] q); always@(posedge clk) begin if(reset) q <= 'd0; else begin q[3:0] <= q[3:0] + 1; if(ena[1])begin q[3:0] <= 0; q[7:4] <= q[7:4]+1; end if(ena[2])begin q[3:0] <= 0; q[7:4] <= 0; q[11:8] <= q[11:8]+1; end if(ena[3])begin q[3:0] <= 0; q[7:4] <= 0; q[11:8] <= 0; q[15:12] <= (q[15:12]<9)?q[15:12] +1:0; end end end assign ena[1] = (q[3:0]==9); assign ena[2] = (q[7:4]==9)&&ena[1]; assign ena[3] = (q[11:8]==9)&&ena[2]; endmodule
Count clock
这道题稍微有点复杂，难度也不算特别大，但是需要耐心debug，把情况全部考虑清楚。
我提交了五次才通过，可恶。
module top_module( input clk, input reset, input ena, output reg pm, output reg[7:0] hh, output reg[7:0] mm, output reg[7:0] ss); always@(posedge clk) begin if(reset)begin pm <= 1'b0; hh <= {4'd1,4'd2}; mm <= 0; ss <= 0; end else if(ena)begin ss[3:0] <= ss[3:0] +1'b1; if(ss[3:0] == 9)begin ss[3:0] <= 'd0; ss[7:4] <= (ss[7:4]==5)?0:(ss[7:4]+1); end if(ss[3:0] == 9&&ss[7:4] == 5)begin if(mm[3:0]<9) mm[3:0] <= mm[3:0] + 1; else begin mm[3:0] <= 'd0; mm[7:4] <= (mm[7:4]==5)?0:(mm[7:4]+1); end end if(ss[3:0] == 9&&ss[7:4] == 5&&mm[3:0] == 9&&mm[7:4] == 5)begin if(hh[3:0]<9&&hh[7:4] == 0) hh[3:0] <= hh[3:0] + 1; else if(hh[3:0]<2&&hh[7:4] == 1) hh[3:0] <= hh[3:0] + 1; else begin if(hh[7:4] == 1) hh[3:0] <= 'd1;//注意这里是1 else hh[3:0] <= 'd0; hh[7:4] <= (hh[7:4]==1)?0:1; end end if(ss[3:0] == 9&&ss[7:4] == 5&&mm[3:0] == 9&&mm[7:4] == 5&&hh[3:0]==1&&hh[7:4]==1) pm<=~pm; end end endmodule