Difference between revisions of "CSC270 Homework 5 Solution 2012"

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<onlysmith>
 
<onlysmith>
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=PDF=
 
[[Media:CSC270Homework5Solutions2012.pdf|Pdf]]
 
[[Media:CSC270Homework5Solutions2012.pdf|Pdf]]
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 +
=Source=
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<source lang="python">
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# Homework 5
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# Problems 1 and 2
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# CSC 270
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# 2012
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# D.T.
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# for problem #2, since we associated Red ON with the state where
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# Q0, Q1, and Q2 are 0, then all we need to do is clear D2, D1, and
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# D0 when the command is 1.  So we AND the functions of Problem 1 that
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# generate D2, D1, and D0 with NOT( cmd ).  No analysis necessary!
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#
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def NOT( a ):
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    return 1 - a
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 +
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def Problem1():
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    Q2 = 1
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    Q1 = 0
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    Q0 = 0
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 +
 +
    for step in range( 20 ):
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 +
        # the Q1 and Q2 outputs go through combinational logic to generate the new values
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        # of D1, D2, and the outputs G, Y, R...
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        D2 = NOT( Q2 ) & Q1 & NOT( Q0 )
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        D1 = ( Q0 & Q1 ) | ( Q2 ^ Q1 ) ^ Q0
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        D0 = NOT( Q2 ) & NOT( Q1 ) | ( Q2 & Q0 )
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        R = NOT( Q1 ) & NOT( Q0 )
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        G = NOT( R )
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        # show the stable circuit signals
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        print( "Q2Q1Q0 = %d %d %d | GR = %d %d" % ( Q2, Q1, Q0, G, R ) )
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 +
        # wait for the next clock tick (the user presses Enter)
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        #input( "> " )
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        # as soon as the clock has ticked, D1 and D2 get latched in the flipflops
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        # and Q1 and Q2 reflect the values captured.
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        Q0 = D0
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        Q1 = D1
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        Q2 = D2
 +
 +
def Problem2():
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    Q2 = 0
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    Q1 = 0
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    Q0 = 0
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 +
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    for cmd in [0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,0,1,1,0]:
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        # the Q1 and Q2 outputs go through combinational logic to generate the new values
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        # of D1, D2, and the outputs G, Y, R...
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        D2 = NOT(cmd) & ( NOT( Q2 ) & Q1 & NOT( Q0 ) )
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        D1 = NOT(cmd) & (( Q0 & Q1 ) | ( Q2 ^ Q1 ) ^ Q0 )
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        D0 = NOT(cmd) & (NOT( Q2 ) & NOT( Q1 ) | ( Q2 & Q0 ))
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        R = NOT( Q1 ) & NOT( Q0 )
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        G = NOT( R )
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 +
        # show the stable circuit signals
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        print( "cmd Q2Q1Q0 = %d %d %d %d | GR = %d %d" % ( cmd, Q2, Q1, Q0, G, R ) )
 +
 +
        # wait for the next clock tick (the user presses Enter)
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        #input( "> " )
 +
 +
        # as soon as the clock has ticked, D1 and D2 get latched in the flipflops
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        # and Q1 and Q2 reflect the values captured.
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        Q0 = D0
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        Q1 = D1
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        Q2 = D2
 +
 +
def main():
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    print( "Problem #1\n-------------" )
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    Problem1()
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    print( "\n\nProblem #2\n-------------" )   
 +
    Problem2()
 +
 +
main()
 +
 +
"""
 +
Copy of the output
 +
Problem #1
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-------------
 +
Q2Q1Q0 = 1 0 0 | GR = 0 1
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Q2Q1Q0 = 0 1 0 | GR = 1 0
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Q2Q1Q0 = 1 1 0 | GR = 1 0
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Q2Q1Q0 = 0 0 0 | GR = 0 1
 +
Q2Q1Q0 = 0 0 1 | GR = 1 0
 +
Q2Q1Q0 = 0 1 1 | GR = 1 0
 +
Q2Q1Q0 = 0 1 0 | GR = 1 0
 +
Q2Q1Q0 = 1 1 0 | GR = 1 0
 +
Q2Q1Q0 = 0 0 0 | GR = 0 1
 +
Q2Q1Q0 = 0 0 1 | GR = 1 0
 +
Q2Q1Q0 = 0 1 1 | GR = 1 0
 +
Q2Q1Q0 = 0 1 0 | GR = 1 0
 +
Q2Q1Q0 = 1 1 0 | GR = 1 0
 +
Q2Q1Q0 = 0 0 0 | GR = 0 1
 +
Q2Q1Q0 = 0 0 1 | GR = 1 0
 +
Q2Q1Q0 = 0 1 1 | GR = 1 0
 +
Q2Q1Q0 = 0 1 0 | GR = 1 0
 +
Q2Q1Q0 = 1 1 0 | GR = 1 0
 +
Q2Q1Q0 = 0 0 0 | GR = 0 1
 +
Q2Q1Q0 = 0 0 1 | GR = 1 0
 +
 +
 +
Problem #2
 +
-------------
 +
cmd Q2Q1Q0 = 0 0 0 0 | GR = 0 1
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cmd Q2Q1Q0 = 0 0 0 1 | GR = 1 0
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cmd Q2Q1Q0 = 0 0 1 1 | GR = 1 0
 +
cmd Q2Q1Q0 = 0 0 1 0 | GR = 1 0
 +
cmd Q2Q1Q0 = 0 1 1 0 | GR = 1 0
 +
cmd Q2Q1Q0 = 0 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 0 0 0 1 | GR = 1 0
 +
cmd Q2Q1Q0 = 0 0 1 1 | GR = 1 0
 +
cmd Q2Q1Q0 = 1 0 1 0 | GR = 1 0
 +
cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
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cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
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cmd Q2Q1Q0 = 0 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 0 0 0 1 | GR = 1 0
 +
cmd Q2Q1Q0 = 0 0 1 1 | GR = 1 0
 +
cmd Q2Q1Q0 = 1 0 1 0 | GR = 1 0
 +
cmd Q2Q1Q0 = 1 0 0 0 | GR = 0 1
 +
cmd Q2Q1Q0 = 0 0 0 0 | GR = 0 1
 +
"""</sourcr>
 +
 
</onlysmith>
 
</onlysmith>
  

Revision as of 07:29, 13 March 2012

--D. Thiebaut 18:24, 12 March 2012 (EDT)


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