Timeline for How to generate 2's complement of a 3-bit number using 3:8 active low decoder?
Current License: CC BY-SA 4.0
10 events
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| Nov 5, 2019 at 0:44 | history | edited | jonk | CC BY-SA 4.0 |
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| Nov 5, 2019 at 0:06 | comment | added | jonk | @ThePhoton It's active-low. I think. Double check me by reading the OP. But I think it is there. | |
| Nov 4, 2019 at 23:33 | comment | added | The Photon | Won't ANDing any two outputs of a 3-to-8 decoder always give 0? Or is this a decoder with active low outputs? | |
| Nov 4, 2019 at 23:13 | history | edited | jonk | CC BY-SA 4.0 |
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| Nov 4, 2019 at 22:31 | history | edited | jonk | CC BY-SA 4.0 |
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| Nov 4, 2019 at 21:39 | comment | added | jonk | @Crist I had no intention on optimizing. I merely wanted to pass on a way to visualize well enough to solve the problem without even paper. The above can be worked in the head, entirely. | |
| Nov 4, 2019 at 21:32 | comment | added | Cristobol Polychronopolis | That successfully inverts each input bit. However, a0 can go straight to the output y0 unmodified. You do need to carry if a0 is 0, so you can use the above product D0D2D4D6 (in conjunction with its complement, a0) to select between a1 and ~a1 for the y1 output. Replicate that trick to select between a2 and ~a2 for the y2 output. | |
| Nov 4, 2019 at 17:59 | history | edited | jonk | CC BY-SA 4.0 |
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| Nov 4, 2019 at 17:52 | vote | accept | CommunityBot | ||
| Nov 4, 2019 at 17:48 | history | answered | jonk | CC BY-SA 4.0 |