HDLBits 系列（38）值得一看的状态机设计题目
目录
背景
这是这个系列中的一个状态机的题目，但是相比于给了你完整状态转移图之类的题目，这个题目还是稍微有点难的，我实在不知道该怎么给这个博客起个什么名字？
我在线等一个简单的方式去解决今天的问题，而如题所说，我用最无能的方式来解决这个问题，但简单的方式一定存在。
2019/12/16更新
今天一个帅兄弟给了我一个答案，很巧妙，这里十分感谢。我把它放到后面来给出。
原题复现
Consider a finite state machine with inputs s and w. Assume that the FSM begins in a reset state called A, as depicted below. The FSM remains in state A as long as s = 0, and it moves to state B when s = 1. Once in state B the FSM examines the value of the input w in the next three clock cycles. If w = 1 in exactly two of these clock cycles, then the FSM has to set an output z to 1 in the following clock cycle. Otherwise z has to be 0. The FSM continues checking w for the next three clock cycles, and so on. The timing diagram below illustrates the required values of z for different values of w.
Use as few states as possible. Note that the s input is used only in state A, so you need to consider just the w input.
我的方案
状态转移图
我的解决方案，有点无奈，多加了一些状态：
我的设计
根据此状态转移图给出我的设计：
-
module top_module (
-
input clk,
-
input reset, // Synchronous reset
-
input s,
-
input w,
-
output z
-
);
-
-
localparam A = 0, B = 1, S0 = 3, S1 = 4, S2 = 5, S3 = 6, S4 = 7, S5 = 8, S6 = 9, S7 = 10, S8 = 11, S9 = 12, S10 = 13, S11 = 14;
-
reg [3:0] state, next_state;
-
always@(*) begin
-
case(state)
-
A: begin
-
if(s) next_state = B;
-
else next_state = A;
-
end
-
B: begin
-
if(w) next_state = S1;
-
else next_state = S0;
-
end
-
S0: begin
-
if(w) next_state = S2;
-
else next_state = S4;
-
end
-
S1: begin
-
if(w) next_state = S9;
-
else next_state = S6;
-
end
-
S2: begin
-
if(w) next_state = S3;
-
else next_state = S5;
-
end
-
S3: begin
-
if(w) next_state = S1;
-
else next_state = S0;
-
end
-
S4: begin
-
next_state = S5;
-
end
-
S5: begin
-
if(w) next_state = S1;
-
else next_state = S0;
-
end
-
S6: begin
-
if(w) next_state = S7;
-
else next_state = S8;
-
end
-
S7: begin
-
if(w) next_state = S1;
-
else next_state = S0;
-
end
-
S8: begin
-
if(w) next_state = S1;
-
else next_state = S0;
-
end
-
S9: begin
-
if(w) next_state = S11;
-
else next_state = S10;
-
end
-
S10: begin
-
if(w) next_state = S1;
-
else next_state = S0;
-
end
-
S11: begin
-
if(w) next_state = S1;
-
else next_state = S0;
-
end
-
default: begin
-
next_state = A;
-
end
-
-
endcase
-
-
end
-
always@(posedge clk) begin
-
if(reset) state <= 0;
-
else state <= next_state;
-
end
-
assign z = (state == S10 || state == S7 || state == S3) ? 1 : 0;
-
-
-
-
-
-
endmodule
测试成功。
更新方案
今天群里的大佬给了一种简单的方法，状态转移图确实简单了，但是理解起来呢？
我之前用的方案是在B之后的三个周期内，列举w的值，取值情况有8种，然后添加更多的状态去解决这个问题，不得不说状态转移图看起来复杂很多，也绝对不是推荐的方案。
我等待的确实是今天的这个方案，通过计数，不需要添加更多的状态，这也是我一开始就想用的方法，只是计数的时序当时没有搞定，今天很感谢，群里的一位兄弟。
直接给出设计，通过代码就应该能看出来设计的思路：
-
module top_module (
-
input clk,
-
input reset, // Synchronous reset
-
input s,
-
input w,
-
output z
-
);
-
parameter A = 1'b0, B = 1'b1;
-
reg current_state;
-
reg next_state;
-
-
always@(posedge clk)begin
-
if(reset)begin
-
current_state <= A;
-
end
-
else begin
-
current_state <= next_state;
-
end
-
end
-
-
always@(*)begin
-
case(current_state)
-
A:begin
-
next_state = s ? B : A;
-
end
-
B:begin
-
next_state = B;
-
end
-
endcase
-
end
-
-
reg w_reg1;
-
reg w_reg2;
-
always@(posedge clk)begin
-
if(reset)begin
-
w_reg1 <= 1'b0;
-
w_reg2 <= 1'b0;
-
end
-
else if(next_state == B)begin
-
w_reg1 <= w;
-
w_reg2 <= w_reg1;
-
end
-
else begin
-
w_reg1 <= 1'b0;
-
w_reg2 <= 1'b0;
-
end
-
end
-
-
always@(posedge clk)begin
-
if(reset)begin
-
z <= 1'b0;
-
end
-
else if(next_state == B && counter == 2'd0)begin
-
if(~w & w_reg1 & w_reg2 | w & ~w_reg1 & w_reg2 | w & w_reg1 & ~w_reg2)begin
-
z <= 1'b1;
-
end
-
else begin
-
z <= 1'b0;
-
end
-
end
-
else begin
-
z <= 1'b0;
-
end
-
end
-
-
reg [1:0] counter;
-
always@(posedge clk)begin
-
if(reset)begin
-
counter <= 2'd0;
-
end
-
else if(counter == 2'd2)begin
-
counter <= 2'd0;
-
end
-
else if(next_state == B)begin
-
counter <= counter + 1'b1;
-
end
-
end
-
-
endmodule
巧妙之处在于将输入w延迟两拍之后进行判断，如果有两个w为1，则在下一个周期将输出z置位.
有的朋友，也许在状态机的设计中，习惯将第三段使用组合逻辑来实现，这个题目的第三段也可以使用组合逻辑，但是呢？确实也没有必要，因为状态机的第三段本身既可以使用组合逻辑，也可以使用时序逻辑，如果使用时序逻辑。
下面给出组合逻辑的方案：
-
module top_module (
-
input clk,
-
input reset, // Synchronous reset
-
input s,
-
input w,
-
output reg z
-
);
-
parameter A = 1'b0, B = 1'b1;
-
reg current_state;
-
reg next_state;
-
-
always@(posedge clk)begin
-
if(reset)begin
-
current_state <= A;
-
end
-
else begin
-
current_state <= next_state;
-
end
-
end
-
-
always@(*)begin
-
case(current_state)
-
A:begin
-
next_state = s ? B : A;
-
end
-
B:begin
-
next_state = B;
-
end
-
endcase
-
end
-
-
reg w_reg1;
-
reg w_reg2;
-
always@(posedge clk)begin
-
if(reset)begin
-
w_reg1 <= 1'b0;
-
w_reg2 <= 1'b0;
-
end
-
else if(next_state == B)begin
-
w_reg1 <= w;
-
w_reg2 <= w_reg1;
-
end
-
else begin
-
w_reg1 <= 1'b0;
-
w_reg2 <= 1'b0;
-
end
-
end
-
-
reg z_mid;
-
always@(*)begin
-
if(reset)begin
-
z_mid <= 1'b0;
-
end
-
else if(current_state == B && counter == 2'd0)begin
-
if(~w & w_reg1 & w_reg2 | w & ~w_reg1 & w_reg2 | w & w_reg1 & ~w_reg2)begin
-
z_mid <= 1'b1;
-
end
-
else begin
-
z_mid <= 1'b0;
-
end
-
end
-
else begin
-
z_mid <= 1'b0;
-
end
-
end
-
-
always@(posedge clk)begin
-
if(reset)begin
-
z <= 1'b0;
-
end
-
else begin
-
z <= z_mid;
-
end
-
end
-
-
-
reg [1:0] counter;
-
always@(posedge clk)begin
-
if(reset)begin
-
counter <= 2'd0;
-
end
-
else if(counter == 2'd2)begin
-
counter <= 2'd0;
-
end
-
else if(next_state == B)begin
-
counter <= counter + 1'b1;
-
end
-
end
-
-
endmodule
只需要将第三段改为组合逻辑，但是输出需要延迟一拍，为什么？看时序图。
FPGA/IC群推荐
有幸在这个群里认识了诸多俊杰，让我欣慰。
文章来源: reborn.blog.csdn.net，作者：李锐博恩，版权归原作者所有，如需转载，请联系作者。
原文链接：reborn.blog.csdn.net/article/details/103484037
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