In this question, you will design a circuit for an 8x1 memory, where writing to the memory is accomplished by shifting-in bits, and reading is "random access", as in a typical RAM. You will then use the circuit to realize a 3-input logic function.文章来源:https://www.toymoban.com/news/detail-618552.html
First, create an 8-bit shift register with 8 D-type flip-flops. Label the flip-flop outputs from Q[0]...Q[7]. The shift register input should be called S, which feeds the input of Q[0] (MSB is shifted in first). The enable input controls whether to shift. Then, extend the circuit to have 3 additional inputs A,B,C and an output Z. The circuit's behaviour should be as follows: when ABC is 000, Z=Q[0], when ABC is 001, Z=Q[1], and so on. Your circuit should contain ONLY the 8-bit shift register, and multiplexers. (Aside: this circuit is called a 3-input look-up-table (LUT)).文章来源地址https://www.toymoban.com/news/detail-618552.html
module top_module (
input clk,
input enable,
input S,
input A, B, C,
output Z );
reg [7:0] Q;
always@ (posedge clk)
if(enable)
Q <= {Q[6:0],S};
always@ (*)
case({A, B, C})
3'b000: Z <= Q[0];
3'b001: Z <= Q[1];
3'b010:Z <= Q[2];
3'b011:Z <= Q[3];
3'b100:Z <= Q[4];
3'b101:Z <= Q[5];
3'b110:Z <= Q[6];
3'b111:Z <= Q[7];
default: Z <= Q[0];
endcase
endmodule
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