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I Prepare for Entrance Exam on PhD, CS Filed. I ran into a previous question with following options (A to D from one to forth lines):

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The question is:

if the input is always has valid values (no invalid inputs), the the second line is True, How Can help me how this will be reached ?!

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  • \$\begingroup\$ ensure a valid input to the final inverter. \$\endgroup\$ – Jasen May 15 '16 at 7:08
  • \$\begingroup\$ @Jasen Yes it's ensures. \$\endgroup\$ – Michle Niaye May 15 '16 at 7:10
  • \$\begingroup\$ out=0 is the correct answer \$\endgroup\$ – Jasen May 15 '16 at 7:11
  • \$\begingroup\$ Confusing, The answer sheet says (2) \$\endgroup\$ – Michle Niaye May 15 '16 at 7:12
  • \$\begingroup\$ ah, yeah, it see it now, the best way to show the solution to this is to draw one of them grids (I forget the name) and remove the illegal inputs, you're then left with something that simplifies to a.b \$\endgroup\$ – Jasen May 15 '16 at 7:44
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If \$c=0\$, then the top tri-state is hi-Z. The lower tri-state must be on (\$a=1\$) for the output to be well-defined, so \$out=c=0\$.

If \$c=1\$, then the top tri-state is enabled and:

a. \$a = 0\$ so that the bottom tri-state is off and \$out=a\cdot b = 0\$; or
b. \$a = 1\$ and the top and bottom tri-state must have the same output for the state to be valid, that is to say, \$\overline{c} = \overline{a\cdot b}\$. Given the known values, \$b = 1\$ is required and \$out = 1\$

This analyses all possible valid combinations (based on the valid possible configurations of the tri-state buffers). We can see that the output is 1 when \$a\cdot c\$, and 0 otherwise. Thus,

\$out = a \cdot c\$

This is perhaps not the most efficient way of analysing it, or the most generalisable (truth tables and then a Karnaugh map reduction would be more general), but it seems a better approach to understanding what's going on intuitively.

Point b above is potentially dangerous in actual implementation, due to shoot-through in the case that \$a = c = 1\$ and \$b = 0\$.

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There are eight possible input states, of which three are invalid: (a,b,c) = (0,0,0); (0,1,0); and (1,0,1). The first two of these give a high impedance state at the input to the final inverter, and the third gives a 0/1 conflict at the final inverter input.

Construct the truth table and it's clear that \$out=a.c\$ is the correct answer.

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  • \$\begingroup\$ is there another quick way to determine this? \$\endgroup\$ – Michle Niaye May 15 '16 at 9:08
  • \$\begingroup\$ A truth table with only three inputs gives a quick enough solution in a multiple choice quiz. You could, however, discard options 1 and 3 by inspection. \$\endgroup\$ – Chu May 15 '16 at 13:45

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