5-bit LFSR
1.Decade counter2.Four-bit binary counter3.Decade counter again4.Slow decade counter5.Counter 1-126.Counter 10007.4-digit decimal counter8.12-hour clock9.Hdlbits博文分布10.4-bit shift register11.Left/right rotator12.Left/right arithmetic shift by 1 or 8
13.5-bit LFSR
14.3-bit LFSR15.32-bit LFSR16.Shift register17.Shift register(2)18.3-input LUT19.Rule 9020.Rule 11021.Conway's Game of Life 16x1622.Simple FSM1(asynchronous reset)23.Simple FSM1(synchronous reset)24.Simple FSM2(asynchronous reset)25.Simple FSM2(synchronous reset)26.Simple state transition 327.Simple one-hot state transition 328.Simple FSM 3(asynchronous reset)29.Simple FSM 3(synchronous reset)30.Design a Moore FSM31.Lemmings 132.Lemmings 233.Lemmings 334.Lemmings 435.One-hot FSM36.PS/2 packet parser37.PS/2 packet parser and datapath38.Serial receiver39.Serial receiver and datapath40.Serial receiver with parity checking41.Sequence recognition42.Q8:Design a Mealy FSM43.Q5a:Serial two's complementer(Moore FSM)44.Q5b:Serial two's complementer(Moore FSM)45.Q3a:FSM46.Q3b:FSM47.Q3c:FSM logic48.Q6b:FSM next-state logic49.Q6c:FSM next-state logic50.Q6:FSM51.Q2a:FSM52.Q2:One-hot FSM equations53.Q2a: FSM54.Q2b:Another FSM55.Counter with period 100056.4-bit shift register and down counter57.FSM:Sequence 1101 recognizer58.FSM:Enable shift register59.FSM:The complete FSM60.The complete timer61.FSM:One-hot logic equations62.UARTA 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.
module top_module(
input clk,
input reset, // Active-high synchronous reset to 5'h1
output [4:0] q
);
always@(posedge clk)begin
if(reset)begin
q<=5'h1;
end
else begin
q[4]<=1'b0^q[0];
q[3]<=q[4];
q[2]<=q[3]^q[0];
q[1]<=q[2];
q[0]<=q[1];
end
end
endmodule
这道题过多的背景知识并不想去了解了,看图说话即可有答案
合集:
Verilog学习
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