I drew the map on the right, but what I drew doesn't work for what the question is asking me. I think I did something very wrong, and I don't really understand what this question is asking me. Am i suppose to re arrange the binary inputs somehow?
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1$\begingroup$ Could you please label the inputs w,x,y,z in the diagram ? $\endgroup$– GauravCommented Feb 27, 2014 at 8:16
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2$\begingroup$ Don't use images as main content of your post. Not only is it lazy, it also makes your question impossible to search and inaccessible to the visually impaired; we don't like that. Please transcribe text and maths (note that you can use LaTeX) and don't forget to give proper attribution to your sources! $\endgroup$– D.W. ♦Commented Feb 27, 2014 at 9:46
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1 Answer
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I assume $w,x,y,z$ are inputs to the circuit given. What we need is $$\sum_{w,x,y,z}(2,3,8,9)$$
Put this in a Karnaugh Map and you will get the simplified equation as $w^\prime x^\prime y + wx^\prime y^\prime$. Now, let $a,b,c,d$ be the inputs to the circuit that you have given. Output of first OR gate will be $a+b$. Let it be called $f$. Output of second OR gate will be $c+d$. Let it be called $s$. Output of the XOR gate is $fs^\prime+sf^\prime$. Substitue the values of $f$ and $s$ in the equation and use the equality $(\alpha+\beta)^\prime=\alpha^\prime \beta^\prime$The output will be $$a^\prime b^\prime\cdot(c+d)+c^\prime d^\prime\cdot(a+b)$$.
Compare this with the equation that we require and we find that $$a=w$$$$b=x$$$$c=x$$$$d=y$$ These are the order in which inputs should be given.
Karnaugh Map that you have drawn is correct assuming that $w,x,y,z$ are inputs to the circuit.
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$\begingroup$ Hey thanks for your answer. The karnaugh map represents the OR-XOR circuit (fig.1) where the input from top to bottom is w,x,y,z. so when w=1,x=0,y=0,z=0 (the top right cell in the map) then F=1. isnt that how the Karnaugh map works? I am unsure on how you get that simplified equation, and I dont know how you get that output equation $\endgroup$ Commented Feb 28, 2014 at 2:32
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$\begingroup$ also, isnt the simplified equation wx'y'+yw'x' ? i.imgur.com/AZM3HWp.png $\endgroup$ Commented Feb 28, 2014 at 4:18
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$\begingroup$ Yeah, I was wrong. I edited the answer $\endgroup$– nitishchCommented Feb 28, 2014 at 8:33
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$\begingroup$ @user14864 Yeah, logically that's how Karnaugh Map is to be filled. See some standard logic design book to know how to simplify the Karnaugh Map $\endgroup$– nitishchCommented Feb 28, 2014 at 13:41
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$\begingroup$ Ok now, how did you find the equation a'b'⋅(c+d)+c'd'⋅(a+b) ? $\endgroup$ Commented Mar 1, 2014 at 0:53
