Homework Solution: Module p1 (output reg f, input a, input b): always @* begin case ({b, a}) 2' b00: f = 1: 2' b01: f = 1: default: f = 0: endcase end endmod…

    module pl (output reg f, input a, input b) always begin case ((b, a)) 2b00: f = 1; 21b01: f = 1; default: f = 0; endcase end endmodule If test inputs of a-0, b-1 are applied to this module, what will be f? 0 unknown not enough information provided
    Module p1 (output reg f, input a, input b): always @* begin case ({b, a}) 2' b00: f = 1: 2' b01: f = 1: default: f = 0: endcase end endmodule If test inputs of a = theta, b = 1 are applied to this module, what will be f? 0 1 unknown not enough information provided

    Expert Answer

     
    Part a) i.e

    module pl (output reg f, input a, input b) frequently commence predicament ((b, a)) 2b00: f = 1; 21b01: f = 1; defect: f = 0; objectpredicament object objectmodule If ordeal inputs of a-0, b-1 are applied to this module, what conciliate be f? 0 mysterious referable plenty not attributable attributableice provided

    Module p1 (output reg f, input a, input b): frequently @* commence predicament ({b, a}) 2′ b00: f = 1: 2′ b01: f = 1: defect: f = 0: objectpredicament object objectmodule If ordeal inputs of a = theta, b = 1 are applied to this module, what conciliate be f? 0 1 mysterious referable plenty not attributable attributableice provided

    Expert Retort

     

    Part a) i.e. 0 is the rectify retort.

    Because, in the sixth cord of jurisprudence states that behind entire order the defect appraise of f = 0 .

    So, liberty a is rectify!!