pMOS

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pMOS nMOS

• pMOS is open when input is high.
• nMOS is closed when input is high.
• Always pMOS at top (high) and nMOS at botom
  (ground)
• implement F in pMOS and F in nMOS
• eg. FABCAB

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Sum-Product Product-of-Sum

• Sum of Product
• see entries with value 1.
• Ensure output 1 for those inputs So OR
• ABAB
• Product of Sum
• see entires with value 0
• Ensure output 0 for those input so AND
• (AB)(AB)

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DeMorgans Law

• X Y X . Y
• Thus, the negation doesnt distribute directly
  it also changes the or to an and
• Example
• X . Y X Y

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DeMorgans law on circuits

• You can do DeMorgans law directly on the circuit

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Proving completeness

• You are given a set of operators. You are asked
  to show that you can implement any function using
  these operators. The simplest way to do this is
  to show that the set of operators you are given
  reduces reduces to a set of operators that we
  already know is complete, for example NOT, AND,
  OR.
• This means we have to show how to implement NOT,
  AND and OR using only the operators in the set
  you are given.

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Example of completeness

• Show that NOR is complete.
• Solution
• Implement NOT
• A (AA) A nor A
• Implement AND
• AB (A B) (A nor A) nor (B nor B)
• Implement OR
• A B ((A B)) (A nor B) nor (A nor B)

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Simplification

• Some important rules for simplification (how do
  you prove these?)
• AB AB A
• A AB A
• Note that you can use the rules in either
  direction, to remove terms, or to add terms.
  Indeed, sometimes you need to add some terms in
  order to get to the simples solution.

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Examples

• Simplify abc abc abc
• abc abc abc
• abc abc abc abc ac bc
• Show that X XY X Y
• X XY
• X(1 Y) XY
• X XY XY
• X Y

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Examples (contd)

• SimplifyWX XY XZ WYZ
• WX XY XZ WYZ
• WX XY XZ WYZX WYZX
• WX(1 YZ) XY XZ(1 WY)
• WX XY XZ

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Examples (contd)

• Prove the consensus theorem, which says
• XY XZ YZ XY XZ
• Solution
• XY XZ YZ
• XY XZ (X X)YZ
• XY XZ XYZ XYZ
• XY XYZ XZ XZY
• XY(1 Z) XZ(1 Y)
• XY XZ

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Long example

• Simplify
• ABCD ABCD ABD ABCD ABCD ACD
  BCD
• ACD(B B) ABD(1 C) ABCD ACD
  BCD
• ACD ABD ABCD ACD BCD
• ACD BD(A AC) ACD BCD
• ACD BD(A C) ACD BCD (Since X
  XY X Y)
• ACD ABD (BCD ACD) BCD
• ACD ABD (BCD ACD ABC) BCD
• (Added ABC by consensus)
• ACD (ABD ABC BCD) (ABC BCD
  ACD)
• ACD ABD ABC BCD

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Proof of Duality

• Duality says If you have an equation that holds,
  and you change all the ANDs to ORs, the Ors to
  ANDs, the 0s to 1s, and the 1s to 0s, then
  you get another equation that holds
• Example
• A 1 1 Thats certainly true
• Dual is A . 0 0. Wow, This holds aswell!
• Alright, lets prove that this is true. Actually,
  youll prove it

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Proof of Duality (contd)

• Definitions
• If E is an expression, let ED be the dual of D.
• ie f( A, B, C, , 0, 1, , )D f( A, B, C, ,
  1, 0, , )
• If E is an expression, let EID be ED, but with
  all the variables inverted.
• ie f( A, B, C, , 0, 1, , )ID f( A, B,
  C, , 1, 0, , )
• Example
• (A 1)D A . 0, however (A 1)ID A . 0
• (A B)D A . B, however (A B)ID A . B
• What is EID?
• Lets take an example (A B)ID A . B (A
  B) (By DeMorgans)
• So, it would look like EID is just E Well,
  thats true, and this is called the generalized
  version of DeMorgans (proof is by induction, we
  wont do it)
• f( A, B, C, , 0, 1, , ) f( A, B, C, ,
  1, 0, , )

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Proof of Duality (contd)

• Now, we are equipped to prove the duality
  theorem.
• Lets say that the original equation is
• S T, where S and T are expressions.
• Lets now invert both sides of our equation
• S T
• But inverting is the same as taking the ID, so
• SID TID
• Now, let S f( A, B, C, , 0, 1, , ), and T
  g( A, B, C, , 0, 1, , )
• Thus
• f( A, B, C, , 0, 1, , )ID g( A, B, C, , 0,
  1, , )ID
• Or, by the definition of ID
• f( A, B, C, , 1, 0, , ) g( A, B, C,
  , 0, 1, , )

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Proof of Duality (contd)

• Now, do variable replacement a A, b B,
  etc. Thus
• f( a, b, c, , 1, 0, , ) g( a, b, c, , 0,
  1, , )
• Now, do another variable replacement A a, B
  b, etc. Thus
• f( A, B, C, , 1, 0, , ) g( A, B, C, , 0,
  1, , )
• By the definition of dual, this is
• f( A, B, C, , 0, 1, , )D g( A, B, C, , 0,
  1, , )D
• Or
• SD TD