2
\$\begingroup\$

Have a few problems with my assignment.

enter image description here

Ive derived a truth table and come up with this :

TT

Can someone verify if the above is correct?

Ive then got these 2 logic expressions :

Y = A'D' + C'D' + B'D'

Z = D+ B

What I want to know is (if the above is correct) Is it possible to implement the above circuit using fewer logic gates?

\$\endgroup\$
1
\$\begingroup\$

You have a major mistake. Look carefully at U1A. It's a NAND gate. Go back and look up what a NAND gate does.

You seem to have the function of the other two gates right.

\$\endgroup\$
  • \$\begingroup\$ Sorry that was an error when I made up the schematic. That is supposed to be an AND gate \$\endgroup\$ – Chic2015 Oct 12 '17 at 11:04
  • \$\begingroup\$ Ive rectified this and posted the correct circuit below \$\endgroup\$ – Chic2015 Oct 12 '17 at 11:09
0
\$\begingroup\$

Is it possible to implement the above circuit using fewer logic gates?

You have three different outputs, N, Y, and Z. This means that you need at least three output gates to implement the functions. So no, you can't reduce the number of gates. However, a 3-input AND gate can be made out of 2-input logic NAND gates (or equivalently NOR gates) followed by a 2-input AND gate like below:

schematic

simulate this circuit – Schematic created using CircuitLab

The function Y is simply \$ Y=\overline{ABC + D}\$. Using De-Morgan's laws we can re-write Y as \$ Y=\overline{ABC}.\overline{D}\$. Another form of writing Y can be this.

$$ \overline{Y}=ABC+D= \overline{\overline{A}+ \overline{B}+ \overline{C}}+\overline{\overline{D}}$$ Take \$ \alpha=\overline{A}+ \overline{B}+ \overline{C}\$ and \$ \beta=\overline{D}\$. Again using De-Morgan's laws we get

$$ \overline{Y}=\overline{\alpha}+\overline{\beta}=\overline{\alpha. \beta}$$ Therefore,

$$Y=\alpha.\beta$$ or $$Y=(\overline{A}+ \overline{B}+ \overline{C}).\overline{D}$$

It can also be expanded out $$Y=\overline{A}.\overline{D}+\overline{B}.\overline{D}+\overline{C}.\overline{D}$$

\$\endgroup\$
  • \$\begingroup\$ Thank you. Are my boolean expressions ok? Could I simplify the equation for Y? Y = C' D' + B' D' + A' D' \$\endgroup\$ – Chic2015 Oct 12 '17 at 14:50
  • \$\begingroup\$ Would this be correct? Y= A.B.C.D' \$\endgroup\$ – Chic2015 Oct 12 '17 at 14:51
  • \$\begingroup\$ @Chic2015 see the update \$\endgroup\$ – dirac16 Oct 12 '17 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.