How can a Verilog counter be designed to convert a 50MHz clock signal to a 1Hz clock signal?

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One way to design a Verilog counter to convert a 50MHz clock signal to a 1Hz clock signal is to use a divide-by-50,000,000 counter. This counter will count up to a certain value and then reset back to zero, thus generating a clock signal with a frequency equal to the input clock signal divided by the count value.

The Verilog code for a divide-by-50,000,000 counter can be written as follows:

module clk_divider(
  input clk_50mHz,
  output reg clk_1Hz
);

reg [24:0] count;

always @(posedge clk_50mHz) begin
  if (count == 50000000) begin
    count <= 0;  // Reset the counter
    clk_1Hz <= ~clk_1Hz;  // Invert the output clock
  end else begin
    count <= count + 1;  // Increment the counter
  end
end

endmodule

Explanation: - The clk_divider module has an input clk_50mHz clock signal and an output clk_1Hz clock signal. - Inside the module, we declare a register count with a width of 25 bits (24:0) to store the current count value. - We use an always block that triggers on the positive edge of the clk_50mHz input clock signal. Inside the block, we check if the counter has reached the maximum count value of 50,000,000. If yes, we reset the counter to zero and invert the clk_1Hz output clock signal. If not, we simply increment the count register. - The inverted output clock signal is achieved by using the bitwise NOT (~) operator on clk_1Hz.

Note that this implementation assumes that the clock signals are regular and have a 50% duty cycle. If this is not the case, additional logic may be necessary to ensure the output clock signal has the desired characteristics.