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OpenStudy (anonymous):
..
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OpenStudy (anonymous):
you can write the F1 as
F1=yx'(w+w')
am i right ?
OpenStudy (anonymous):
It looks right. .. Which rule did you apply and what about F2?
OpenStudy (anonymous):
i simply took common yx'
for F2 use De Morgan's Law
OpenStudy (anonymous):
F2 = (y' + xw)'
F2=(y')'.(xw)'
F2=y(x'+w')
OpenStudy (anonymous):
Now, yx'(w+w') + y(x'+w')?
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OpenStudy (anonymous):
you can simplify the F1 because
w+w'=1
F1=yx'(w+w')=yx'(1)=yx'
F1=yx'
OpenStudy (rsadhvika):
w+w' = 1
OpenStudy (anonymous):
F2?
OpenStudy (rsadhvika):
F2 cannot be simplified further. now we can take the sum expression
F1 + F2 = yx' + y(x'+w')
OpenStudy (anonymous):
Thank you VERY much guys. I appreciate your assistance! Is any one of you having insight on Karnaugh Maps?
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OpenStudy (rsadhvika):
yes
OpenStudy (rsadhvika):
btw, you see the sum expression can be simlified...
OpenStudy (anonymous):
Okay, I see.. .
OpenStudy (anonymous):
What's the final simplyfication please?
F1+F2= y.w'?
OpenStudy (anonymous):
Final simplification : F1 + F2 = yx' + yx' + yw
= yx' + yw
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OpenStudy (anonymous):
Thanks! I confused myself and figured it out later on. Just in time for submition.
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