HBLbits_Verilog Basic_Exams/ece241 2014 q3
For the following Karnaugh map, give the circuit implementation using one 4-to-1 multiplexer and as many 2-to-1 multiplexers as required, but using as few as possible. You are not allowed to use any other logic gate and you must use a and b as the multiplexer selector inputs, as shown on the 4-to-1 multiplexer below.
module top_module (
input c,
input d,
output [3:0] mux_in
);
always@(*) begin
case({c,d})
2'b00: mux_in=4'b0100;
2'b01: mux_in=4'b0001;
2'b11: mux_in=4'b1001;
2'b10: mux_in=4'b0101;
endcase
end
// assign mux_in = (c&d)?4'b1001:(c)?4'b0101:(d)?4'b0001:4'b0100;
endmodule
input c,
input d,
output [3:0] mux_in
);
always@(*) begin
case({c,d})
2'b00: mux_in=4'b0100;
2'b01: mux_in=4'b0001;
2'b11: mux_in=4'b1001;
2'b10: mux_in=4'b0101;
endcase
end
// assign mux_in = (c&d)?4'b1001:(c)?4'b0101:(d)?4'b0001:4'b0100;
endmodule
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