2021年6月25日 星期五

HDLBits 5-bit LFSR(Lfsr5)

 HDLBits 5-bit LFSR(Lfsr5)

linear feedback shift register is a shift register usually with a few XOR gates to produce the next state of the shift register. A Galois LFSR is one particular arrangement where bit positions with a "tap" are XORed with the output bit to produce its next value, while bit positions without a tap shift. If the taps positions are carefully chosen, the LFSR can be made to be "maximum-length". A maximum-length LFSR of n bits cycles through 2n-1 states before repeating (the all-zero state is never reached).

The following diagram shows a 5-bit maximal-length Galois LFSR with taps at bit positions 5 and 3. (Tap positions are usually numbered starting from 1). Note that I drew the XOR gate at position 5 for consistency, but one of the XOR gate inputs is 0.
Lfsr5.png

Build this LFSR. The reset should reset the LFSR to 1.


module top_module(
    input clk,
    input reset,    // Active-high synchronous reset to 5'h1
    output [4:0] q
); 
    always @(posedge clk) begin
        if (reset)
            q<=5'h1;
        else
            begin
            //q <= {q[0],q[4],q[3]^q[0],q[2:1]};
            q[0] <= q[1];
            q[1] <= q[2];
            q[2] <= q[3] ^ q[0];
            q[3] <= q[4];
            q[4] <= q[0];
        end
    end
    
endmodule

//=============原網址解法======================

module top_module(
input clk,
input reset,
output reg [4:0] q);

reg [4:0] q_next; // q_next is not a register
// Convenience: Create a combinational block of logic that computes
// what the next value should be. For shorter code, I first shift
// all of the values and then override the two bit positions that have taps.
// A logic synthesizer creates a circuit that behaves as if the code were
// executed sequentially, so later assignments override earlier ones.
// Combinational always block: Use blocking assignments.
always @(*) begin
q_next = q[4:1]; // Shift all the bits. This is incorrect for q_next[4] and q_next[2]
q_next[4] = q[0]; // Give q_next[4] and q_next[2] their correct assignments
q_next[2] = q[3] ^ q[0];
end


// This is just a set of DFFs. I chose to compute the connections between the
// DFFs above in its own combinational always block, but you can combine them if you wish.
// You'll get the same circuit either way.
// Edge-triggered always block: Use non-blocking assignments.
always @(posedge clk) begin
if (reset)
q <= 5'h1;
else
q <= q_next;
end

endmodule

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