网站需求分析怎么做,百度识图网页入口,营销做得好的品牌,网站备案起名要求本专栏的目的是分享可以通过HDLBits仿真的Verilog代码 以提供参考 各位可同时参考我的代码和官方题解代码 或许会有所收益 题目链接#xff1a;Count15 - HDLBits
module top_module (input clk,input reset, // Synchronous active-high resetoutput [3:0] q
); always… 本专栏的目的是分享可以通过HDLBits仿真的Verilog代码 以提供参考 各位可同时参考我的代码和官方题解代码 或许会有所收益 题目链接Count15 - HDLBits
module top_module (input clk,input reset, // Synchronous active-high resetoutput [3:0] q
); always (posedge clk) beginif (reset) q 0 ; else q q 15 ? 0 : q 1 ; endendmodule题目链接Count10 - HDLBits
module top_module (input clk,input reset, // Synchronous active-high resetoutput [3:0] q
); always (posedge clk) beginif (reset) q 0 ; else q q 9 ? 0 : q 1 ; endendmodule题目链接Count1to10 - HDLBits
module top_module (input clk,input reset, // Synchronous active-high resetoutput reg [3:0] q
); always (posedge clk) beginif (reset) q 1 ; else q q 10 ? 1 : q 1 ; endendmodule题目链接Countslow - HDLBits
module top_module (input clk,input slowena,input reset,output reg [3:0] q
);always (posedge clk) beginif (reset) q 0 ; else if (slowena) q q 9 ? 0 : q 1 ; else q q ; endendmodule题目链接Exams/ece241 2014 q7a - HDLBits
module top_module (input clk,input reset,input enable,output reg [3:0] Q,output c_enable,output c_load,output [3:0] c_d
); reg [3:0] cnt_4o ;always (posedge clk) beginif (reset) Q 1 ; else if (enable) Q Q 12 ? 1 : Q 1 ;else Q Q ; endassign c_enable enable ; assign c_load reset | (enable Q 12) ; assign c_d c_load ? 1 : 0 ; count4 the_counter (clk, c_enable, c_load, c_d, cnt_4o);endmodule题目链接Exams/ece241 2014 q7b - HDLBits
module top_module (input clk,input reset,output OneHertz,output [2:0] c_enable
); reg [3:0] q0, q1, q2 ;bcdcount counter0 (clk, reset, c_enable[0], q0);bcdcount counter1 (clk, reset, c_enable[1], q1);bcdcount counter2 (clk, reset, c_enable[2], q2);assign c_enable {q0 9 q1 9, q0 9, 1b1}; assign OneHertz q2 9 q1 9 q0 9 ; endmodule题目链接Countbcd - HDLBits
module top_module (input clk,input reset, // Synchronous active-high resetoutput [3:1] ena,output reg [15:0] q
);always (*) begin if (q[3:0] 9 q[7:4] 9 q[11:8] 9) ena 7 ; else if (q[3:0] 9 q[7:4] 9) ena 3 ; else if (q[3:0] 9) ena 1 ; else ena 0 ;endsub_cnt u1(clk, reset, 1b1, q[3:0]) ;sub_cnt u2(clk, reset, ena[1], q[7:4]) ;sub_cnt u3(clk, reset, ena[2], q[11:8]) ;sub_cnt u4(clk, reset, ena[3], q[15:12]) ;endmodule module sub_cnt (input clk,input reset, // Synchronous active-high resetinput enable, output reg [3:0] q
); always (posedge clk) beginif (reset) q 0 ; else if (enable) q q 9 ? 0 : q 1 ; else q q ; endendmodule题目链接Count clock - HDLBits
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
); wire [2:0] enable ; assign enable {mm 8h59 ss 8h59 ena, ss 8h59 ena, ena} ; cnt60 uss(clk, reset, enable[0], ss) ; cnt60 umm(clk, reset, enable[1], mm) ; always (posedge clk) beginif (reset) begin hh 8h12 ; pm 0 ; endelse if (enable[2]) if (hh 8h12) hh 8h1 ; else if (hh 8h11) begin pm ~pm ; hh[3:0] hh[3:0] 1h1 ; endelse if (hh[3:0] 4h9) begin hh[3:0] 4h0 ; hh[7:4] hh[7:4] 1h1 ; endelse hh[3:0] hh[3:0] 1h1 ; else hh hh ;end
endmodulemodule cnt60(input clk, input reset, input ena, output reg [7:0] q
);always (posedge clk) beginif (reset) q 8h0 ; else if (ena) if (q[3:0] 4h9) if (q[7:4] 4h5) q 8h0 ; else begin q[7:4] q[7:4] 1h1 ; q[3:0] 4h0 ; endelse q[3:0] q[3:0] 1h1 ; else q q ; end
endmodule
