1: Divide by 5 clock
module clk_divn(clk,reset, clk_out);
input clk;
input reset;
output clk_out;
reg [2:0] pos_count, neg_count;
wire [2:0] r_nxt;
always @(posedge clk)
if (reset)
pos_count <=0;
else if (pos_count ==4) pos_count <= 0;
else pos_count<= pos_count +1;
always @(negedge clk)
if (reset)
neg_count <=0;
else if (neg_count ==4) neg_count <= 0;
else neg_count<= neg_count +1;
assign clk_out = ((pos_count > (5>>1)) | (neg_count > (5>>1)));
endmodule
2 : DIVIDE by 8 circuit verilog code
module clk_div (clk,reset, clk_out);
input clk;
input reset;
output clk_out;
reg [2:0] r_reg;
wire [2:0] r_nxt;
reg clk_track;
always @(posedge clk or posedge reset)
begin
if (reset)
begin
r_reg <= 0;
clk_track <= 1'b0;
end
else if (r_nxt == 3'b100)
begin
r_reg <= 0;
clk_track <= ~clk_track;
end
else
r_reg <= r_nxt;
end
assign r_nxt = r_reg+1;
assign clk_out = clk_track;
endmodule
*in SystemVerilog Problem1: 1) Study the clock divider lecture. 2) Implement the following designs in Systemverilog...
Clock Divider can i get some simple explanation ( what I'm suppose to understand from this) my lecturer explains it but I honestly don't understand what statements he's trying to make my understanding : there's a frequency input of 512 Mhz, since we know 8 bit counter can count up to 256, it will do it once before it rolls overload (???) can someone please clarify and point out the important facts that i should be undertanding please and...
Clock Divider can i get some simple explanation ( what I'm suppose to understand from this) my lecturer explains it but I honestly don't understand what statements he's trying to make my understanding : there's a frequency input of 512 Mhz, since we know 8 bit counter can count up to 256, it will do it once before it rolls overload (???) can someone please clarify and point out the important facts that i should be undertanding please and...
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