View source for CSC270 Exercises: Karnaugh Maps, Nands, Nors

--[[User:Thiebaut|D. Thiebaut]] 10:16, 6 February 2011 (EST)
----

=NANDs and NORs=

* The current homework asks you to implement the circuit below with NAND gates only.  For this exercise, implement  it with NOR gates only.<br /><br /><center>[[File:CSC270LogicCircuitHomework1.jpg|250px]]</center><br /><br />
** First approach: find the boolean expression for the output, and rewrite it using only expressions of the form  '''not(a + b)''' or '''not a and not b'''.<br />
** Second approach: solve it graphically (more information will be provided in class)<br />
** Third approach: use the minterm/maxterm form and go from there.

=Karnaugh Maps=

* After a quick introduction in class, reduce the following boolean functions to their simplest forms

** f( a, b, c ) = &Sigma; ( 0, 1, 2, 3, 7 )

** g( a, b, c, d ) = &Sigma; ( 0, 1, 2, 3, 7, 8, 10, 12, 13, 14, 15 )

** h( a, b, c, d ) = &Sigma; ( 0, 1, 5, 6, 7, 11, 12,  14, 15 )

** i( a, b, c, d ) =&Pi; ( 2, 3, 14 )

** j( a, b, c, d, e ) = &Sigma; ( 0, 1, 2, 3, 7, 8, 12, 13, 14, 15, 16, 17, 18, 24, 28, 29, 30, 31 )

=Majority Voter=

* A majority voter is a circuit receiving 3 inputs and comparing them, and reporting as output the majority of the three inputs.   When 2 or 3 inputs are 1, the circuit outputs 1.  When 2 or 3 inputs are 0, it outputs 0.   In cases when one of the inputs is different from the other 2, the majority voter will output 2 bits representing the index of the ''wrong'' input.


:For example, assume the 3 inputs are labeled '''b0''', '''b1''', and '''b2'''.   If '''b0'''=0, '''b1'''=1, and '''b2'''=0, the voter will output 0, because 2 inputs are outputting this value, and the index of the wrong input is 01, corresponding to the index of '''b1'''.


* Implement the majority voter with ANDs, ORs, and inverters, and XORs if necessary.


* Implement it with NANDs only.

<br />
<br />

<br />

<br />

<br />

<br />

<br />
[[Category:CSC270]][[Category:Exercises]]