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enter image description here

For the truth table above, abc is the input and xyz is the output.

I do know how to draw the diagram using a decoder in case of when I have only one output. How can I draw the diagram using a decoder when I have 3 outputs for this case?

This is what I am guessing enter image description here

After I apply the advice, here is what I have

enter image description here

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  • \$\begingroup\$ Start by showing what you tried to solve it. It helps us to see where you're running into trouble, or just to feel like we're not doing your homework for you. \$\endgroup\$ – Andrew Macrae May 15 at 15:59
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    \$\begingroup\$ @andrewMacrae Yes, I just updated it with my guess! \$\endgroup\$ – Sangmin Choi May 15 at 16:11
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    \$\begingroup\$ For z, you do not need a decoder, as in z = c. \$\endgroup\$ – StainlessSteelRat May 15 at 16:42
  • \$\begingroup\$ @andrewMacrae I appreciate your help! I tried based on what you said! \$\endgroup\$ – Sangmin Choi May 15 at 16:49
  • \$\begingroup\$ @stainlesssteelrat Got it. I could see z = c. What does it mean by the way? Do I still need to keep the or gate for z like above? \$\endgroup\$ – Sangmin Choi May 15 at 17:22
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it looks like you nearly have it.

X is high if the input decodes to the '100 value, Y is high if it decodes to '010, '011, '101 or '110 and Z is high if it decodes to '001 '011 '101 or '111.

There is one error in your drawing: X, Y and Z are already your outputs. You don't need to "or" them together at the end.

Label your outputs X,Y and Z and lose the last or gate

Then take a look at the drawing and see if you can understand how the ABC decoders are redundant and how you can reuse the outputs from just one decoder to compute X, Y and Z so you only need one decoder.

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