parameter參數(shù)傳遞的16位乘法器

1、parameter參數(shù)傳遞的16位乘法器 選用課后習(xí)題設(shè)計(jì)題目：設(shè)計(jì)一個(gè)4位乘法器，為此首先設(shè)計(jì)一個(gè)4位加法器，用例化語(yǔ)句調(diào)用這個(gè)加法器，用移位相加的方式完成乘法。并以此為基礎(chǔ)，使用parameter參數(shù)傳遞的功能，設(shè)計(jì)一個(gè)16位乘法器。加法器采用行波進(jìn)位的方法。4位加法器設(shè)計(jì)如下：module ripple_carry_adder(x, y, cin, sum, cout); parameter N = 4; input cin; input N-1:0 x, y; output N-1:0 sum; output cout; reg cout; reg N-1:0 sum; reg qN:

2、0; always (x or y or cin)begin:ADDER integer i; q0 = cin; for(i=0; i<=N-1; i=i+1)begin qi+1 = (xi&yi) | (xi&qi) | (yi&qi); sumi = xi yi qi; end cout = qN; endendmodule以此為基礎(chǔ)的4位乘法器如下：module top(load, rst, clk, dataa, datab, sum); parameter n = 4; input load; input rst; input clk; input

3、n-1:0 dataa; input n-1:0 datab; output 2*n-1:0 sum; parameter 1:0 state_s1 = 0, state_s2 = 1; reg 1:0 next_state; reg 2*n-1:0 dataa_register; reg n-1:0 datab_register; reg 2*n-1:0 sum_tmp,sum_tmp2; wire cout; assign sum = sum_tmp; always (negedge rst or posedge clk) begin: integer i; if (rst = 1'

4、;b0) begin for (i = 0; i <= 2 * n - 1; i = i + 1) begin dataa_registeri <= 1'b0; sum_tmpi <= 1'b0; end for (i = 0; i <= n - 1; i = i + 1) datab_registeri <= 1'b0; next_state <= state_s1; end else case (next_state) state_s1 : if (load = 1'b1) begin dataa_registern -

5、1:0 <= dataa; datab_register <= datab; for (i = 0; i <= 2 * n - 1; i = i + 1) sum_tmpi <= 1'b0; next_state <= state_s2; end state_s2 : begin if (datab_register0 = 1'b1) / sum_tmp <= sum_tmp + dataa_register;sum_tmp<=(n-1)'b0,cout,sum_tmp2n-1:0; if (datab_register = 4

6、'h0) next_state <= state_s1; else begin datab_registern - 2:0 <= datab_registern - 1:1; datab_registern - 1 <= 1'b0; dataa_register2 * n - 1:1 <= dataa_register2 * n - 2:0; dataa_register0 <= 1'b0; end end endcase end ripple_carry_adder #(N = 4) u0(.x(sum_tmp), .y(dataa_re

7、gister), .cin(1'b0), .sum(sum_tmp2), .cout(cout);endmodule以此為基礎(chǔ)的16位乘法器為：module top(load, rst, clk, dataa, datab, sum,ready); parameter n = 16; input load; input rst; input clk; input n-1:0 dataa; input n-1:0 datab; output 2*n-1:0 sum; output ready; parameter 1:0 state_s1 = 0, state_s2 = 1; reg 1

8、:0 next_state; reg 2*n-1:0 dataa_register; reg n-1:0 datab_register; reg 2*n-1:0 sum_tmp; reg ready; wire 2*n-1:0 sum_tmp2; wire cout; integer i; assign sum = sum_tmp; always (negedge rst or posedge clk) begin if (rst = 1'b0) begin ready<=1'b0; for (i = 0; i <= 2 * n - 1; i = i + 1) be

9、gin dataa_registeri <= 1'b0; sum_tmpi <= 1'b0; end for (i = 0; i <= n - 1; i = i + 1) datab_registeri <= 1'b0; next_state <= state_s1; end else case (next_state) state_s1 : if (load = 1'b1) begin dataa_register2*n - 1:n<=0; dataa_registern - 1:0 <= dataa; datab_r

10、egister <= datab; ready<=1'b0; for (i = 0; i <= 2 * n - 1; i = i + 1) sum_tmpi <= 1'b0; next_state <= state_s2; end state_s2 : begin if (datab_register0 = 1'b1) / sum_tmp <= sum_tmp + dataa_register; sum_tmp<=sum_tmp2; if (datab_register = 4'h0) begin next_state

11、<= state_s1; ready<=1'b1; end else begin datab_registern - 2:0 <= datab_registern - 1:1; datab_registern - 1 <= 1'b0; dataa_register2 * n - 1:1 <= dataa_register2 * n - 2:0; dataa_register0 <= 1'b0; end end endcase end ripple_carry_adder#(2*n) u0(.x(sum_tmp), .y(dataa_register