EE 244 - Digital Logic Design. Quiz 2 Solutions

1.

Solution:

The months of the year are coded in four variables, vjyo, such that January is 0000, February is 0001,...., and December is 1011. The remaining 4 combinations are never used. (Remember: 30 days has September, April, June, and November. All the rest have 31, except February....) Show a truth table for a function, c, that is 1 if the month has 31 days and 0 if it does not. Use a capital "X" for don't care conditions.

v j y o c0 0 0 0 ____ 0 0 0 1 ____ 0 0 1 0 ____ 0 0 1 1 ____ 0 1 0 0 ____ 0 1 0 1 ____ 0 1 1 0 ____ 0 1 1 1 ____ 1 0 0 0 ____ 1 0 0 1 ____ 1 0 1 0 ____ 1 0 1 1 ____ 1 1 0 0 ____ 1 1 0 1 ____ 1 1 1 0 ____ 1 1 1 1 ____

First, write in the Don't Care conditions

v j y o c0 0 0 0 0 0 0 1 0 0 1 0

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0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 X1 1 0 1 X1 1 1 0 X1 1 1 1 X

Next, fill in which months have 30 days, reading directly from the problem statement

v j y o c0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 00 1 0 0 0 1 0 1 00 1 1 0 0 1 1 1 1 0 0 0 01 0 0 1 1 0 1 0 01 0 1 1 1 1 0 0 X1 1 0 1 X1 1 1 0 X1 1 1 1 X

Finally, fill in the rest of the table, noting the exception given for February v j y o c0 0 0 0 10 0 0 1 00 0 1 0 10 0 1 1 00 1 0 0 10 1 0 1 00 1 1 0 1

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2.

0 1 1 1 11 0 0 0 01 0 0 1 11 0 1 0 01 0 1 1 11 1 0 0 X1 1 0 1 X1 1 1 0 X1 1 1 1 X

Correct answer is:

v j y o c0 0 0 0 10 0 0 1 00 0 1 0 10 0 1 1 00 1 0 0 10 1 0 1 00 1 1 0 10 1 1 1 11 0 0 0 01 0 0 1 11 0 1 0 01 0 1 1 11 1 0 0 X1 1 0 1 X1 1 1 0 X1 1 1 1 X

For the following circuit,

i. find an algebraic expression ii. put it in sum of products form.

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Solution:

3.

Solution:

r= ____

= ____

r=qb(u + g) + q(b + ug)=qub + qbg + qb + qug

Correct answer is: r= qb(u + g) + q(b + ug)

= qub + qbg + qb + qug

Reduce the expression to a minimum SOP form, using P1 through P12.

h = bd a' + bda m + dam + bd a + dm (3 terms, 6 literals)

A. am + md + mb

B. d' + m + a

C. d' + dm + a

D. bd + dm + da

h = bd a' + bda m + dam + bd a + dm= bd + dm + dam P9a, P12a

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4.

Solution:

= bd + d(m + am') P8a = bd + d(m + a) P10a = bd + dm + da P8a

Correct answer is: D

Reduce the following expressions to a minimum SOP form.

F = k dg + d g + kpg + kpdg + pd g + k d g

A. k g + d g + kpg

B. g + d g + kpg

C. k g + d g + pg

D. kg + d g + kpg

= k dg + d g + kpg + kpdg + pd g + k d g= k g + d g + kpg + kpdg P9a, P12a = k g + d g + kpg + kpd P10a = k g + d g + kpg P13a

Correct answer is: A

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