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I have the following function
x’y’zw’v’ + x’yz’w’v’ + x’yzw’v’ + xy’z’w’v’ + xy’zw’v
I should implement it using only 4x1 MUX (more than one allowed, of course). I'm clueless on how could I combine 4x1 MUXes to handle 5 variables.
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1\$\begingroup\$ check out my answer to electronics.stackexchange.com/questions/468736/… \$\endgroup\$vicatcu– vicatcu2021年05月09日 03:11:03 +00:00Commented May 9, 2021 at 3:11
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\$\begingroup\$ vicatcu answers your question. If xyzw represent the address of the MUX then each input is either 0,1,v or v' depending on the truth table. \$\endgroup\$Paul Ghobril– Paul Ghobril2021年05月09日 05:00:20 +00:00Commented May 9, 2021 at 5:00
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\$\begingroup\$ If you know the concept of how to combine 2:1 to make 4:1 , you should know this too. \$\endgroup\$Mitu Raj– Mitu Raj2021年05月09日 06:12:37 +00:00Commented May 9, 2021 at 6:12
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\$\begingroup\$ I've read the answer by vicatcu but I'm still struggling to conceptualize the problem with 5 variables. I mean, would you write the truth table of the expression complete or somehow would you break it into smaller pieces in order to be able to represent them with 4:1 MUXes that can afterwards be coupled together? \$\endgroup\$neil_huygens– neil_huygens2021年05月09日 13:46:03 +00:00Commented May 9, 2021 at 13:46
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\$\begingroup\$ group the rows of the truth table into pairs... you will see a pattern in the input space where the last input is related to the output in one of those four ways, and the other inputs are constant (0..15) iin each group. Right? \$\endgroup\$vicatcu– vicatcu2021年05月09日 14:01:27 +00:00Commented May 9, 2021 at 14:01