m(2) w1 w4 m(1) w3 PORT MAP signal mymux_package

m(2) w1 w4 m(1) w3 PORT MAP signal mymux_package.all NOR OR m(0) mycircuit Consider the 3-input function f = w1.w2.w3+w1′.w2′ + w2′.w3′. Give the Shannon expansion of f in terms of w1 and w3. Select one: a. f=w1′.w3′ [w2′] + w1′.w3 [w2′] + w1.w3′ [w2′] + w1.w3 [w2] b. f=w1′.w3′ [w2′] + w1′.w3 [w2′] + w1.w3′ [w2′] + w1.w3 [1] c. f=w1′.w3′ [w2] + w1′.w3 [w2] + w1.w3′ [w2] + w1.w3 [w2′] d. f = w1′.w3′ [w2] + w1′.w3 [0] + w1.w3′ [1] + w1.w3 [w2′]

The Correct Answer and Explanation is :

The Shannon expansion is a technique used in Boolean algebra to express a Boolean function ( f ) in terms of one variable and its complement. It allows us to write the function ( f ) as a sum of two terms based on whether a variable is 0 or 1.

Given the Boolean function:
[ f = w1 \cdot w2 \cdot w3 + w1′ \cdot w2′ + w2′ \cdot w3′ ]

We need to expand ( f ) in terms of ( w1 ) and ( w3 ).

Step 1: Write the Shannon expansion for ( w1 )

The Shannon expansion with respect to ( w1 ) is:
[ f = w1 \cdot f_{w1=1} + w1′ \cdot f_{w1=0} ]

Evaluate ( f_{w1=1} ) (when ( w1 = 1 )):

Substitute ( w1 = 1 ) into ( f ):
[ f_{w1=1} = (1 \cdot w2 \cdot w3) + (0 \cdot w2′) + (w2′ \cdot w3′) ]
[ f_{w1=1} = w2 \cdot w3 + w2′ \cdot w3′ ]

Evaluate ( f_{w1=0} ) (when ( w1 = 0 )):

Substitute ( w1 = 0 ) into ( f ):
[ f_{w1=0} = (0 \cdot w2 \cdot w3) + (1 \cdot w2′) + (w2′ \cdot w3′) ]
[ f_{w1=0} = w2′ + w2′ \cdot w3′ ]
Factorize ( w2′ ):
[ f_{w1=0} = w2′ \cdot (1 + w3′) = w2′ ]

Thus, ( f ) becomes:
[ f = w1 \cdot (w2 \cdot w3 + w2′ \cdot w3′) + w1′ \cdot (w2′) ]


Step 2: Expand in terms of ( w3 ):

Now, expand ( f ) with respect to ( w3 ):
[ f = w3 \cdot f_{w3=1} + w3′ \cdot f_{w3=0} ]

Evaluate ( f_{w3=1} ):

Substitute ( w3 = 1 ) into ( f ):
[ f_{w3=1} = w1 \cdot w2 + w1′ \cdot w2′ ]

Evaluate ( f_{w3=0} ):

Substitute ( w3 = 0 ) into ( f ):
[ f_{w3=0} = w1 \cdot w2′ + w1′ \cdot w2′ ]


Final Result:

After combining terms, the correct Shannon expansion is:
[ f = w1′ \cdot w3′ [w2′] + w1′ \cdot w3 [w2′] + w1 \cdot w3′ [w2′] + w1 \cdot w3 [w2] ]

Answer:

a. ( f = w1′ \cdot w3′ [w2′] + w1′ \cdot w3 [w2′] + w1 \cdot w3′ [w2′] + w1 \cdot w3 [w2] )


Explanation:

  • Shannon expansion decomposes a function into terms based on specific variable values.
  • Each term evaluates ( f ) for specific variable combinations.
  • Option a matches the result derived above.
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