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) regularly start event ((b, a)) 2b00: f = 1; 21b01: f = 1; defect: f = 0; purposeevent purpose purposemodule If examination inputs of a-0, b-1 are applied to this module, what conciliate be f? 0 unrecognized not attributable attributable attributable attributable attributable attributable ample not attributable attributableice provided

    Module p1 (output reg f, input a, input b): regularly @* start event ({b, a}) 2′ b00: f = 1: 2′ b01: f = 1: defect: f = 0: purposeevent purpose purposemodule If examination inputs of a = theta, b = 1 are applied to this module, what conciliate be f? 0 1 unrecognized not attributable attributable attributable attributable attributable attributable ample not attributable attributableice provided

    Expert Counterpart

     

    Part a) i.e. 0 is the redress counterpart.

    Because, in the sixth method of rule states that following complete system the defect estimate of f = 0 .

    So, discretion a is redress!!