Electronic – How to express a Boolean function with don’t cares (X) inputs

It's not too complex, I think, assuming you worked out the equation you wanted correctly (I'll assume you did okay there.) Start by looking at the equation for a 2-in MUX:

$$ \begin{align*} M_2(A,B,S) &= A\cdot \bar{S} + B\cdot S \end{align*} $$

From this, you can derive some useful results:

$$ \begin{align*} M_2(0,x,y) &= x\cdot y \\ M_2(x,0,y) &= x\cdot\bar{y} \\ M_2(x,y,0) &= x \\ M_2(1,x,y) &= x + \bar{y} \\ M_2(x,1,y) &= x + y \\ M_2(x,y,1) &= y \\ M_2(x,y,x) &= x\cdot y \\ M_2(x,y,y) &= x + y \\ M_2(0,0,x) &= 0 \\ M_2(0,1,x) &= x \\ M_2(1,0,x) &= \bar{x} \\ M_2(1,1,x) &= 1 \end{align*} $$

So it follows that:

$$ \begin{align*} F &= A B + B C + C D \\ x &= A B = M_2(A,B,A) \\ y &= B C = M_2(B,C,B) \\ z &= C D = M_2(C,D,C) \\ \therefore F &= x + y + z \\ F &= M_2(M_2(x,y,y),z,z) \end{align*} $$

In short, you will need (5) 2-in muxes:

^{simulate this circuit – Schematic created using CircuitLab}

There's also a nice symmetry there. Notice it?

ADDED: You asked about only being able to use 0, 1, or D as a mux data input source. I assume by this that you mean that all of A, B, C, and D can be used as mux selectors. (Otherwise, I don't think the result can be achieved.) So, this just means you need to use some of the other useful results I'd earlier mentioned. The simplest idea would be to just add three more 2-in muxes:

^{simulate this circuit}

I'm not sure if there is a way to optimize it further. I haven't examined all of the possibilities.

EDIT AGAIN: Yes! Using the OP's newly added solution, the following two just flow out. The left one answers his first part of the question, the right one answers his second part.

^{simulate this circuit}

EDIT AGAIN AGAIN: The ordering isn't complicated. It's just assigning the letters where they belong. The author took (A) to be the high order bit of a three-bit binary value, so it represents either \$0\cdot 2^2=0\$ or \$1\cdot 2^2=4\$; took (B) to be the middle bit of a three-bit binary value, so it represents either \$0\cdot 2^1=0\$ or \$1\cdot 2^1=2\$; and took (C) to be the low order bit of a three-bit binary value, so it represents either \$0\cdot 2^0=0\$ or \$1\cdot 2^0=1\$. A variety of different perspectives would work equally well. But that's the one they appear to have chosen.

So they now started with the first (left) tier, laid out (4) muxes controlled by (A), and stayed mentally convenient by numbering those muxes as ABC="x00", ABC="x01", ABC="x10", and for the bottom one ABC="x11".

Now, since for the top one, ABC="x00", this means that it accepts either "000"=0 or "100"=4. So for the "0" input of that mux (mux1), they looked into the table for ABC="000"=0 and placed the table entry into its "0" side input. For the "1" input of that mux, they looked into the table for ABC="100"=4 and placed that table entry into its "1" side input. (That table looks wrong here, as their should be a 0 in that box, confirmed by looking at the earlier expanded columns.)

The next mux down (mux2) is for ABC="x10" and therefore used ABC="010"=2 and ABC="110"=6; the next mux down (mux3) is for ABC="x01" and therefore used ABC="001"=1 and ABC="101"=5; and finally the last mux down (mux4) is for ABC="x11" and therefore used ABC="011"=3 and ABC="111"=7.

Both mux1 (ABC="x00") and mux2 (ABC="x10") are jointly fed to mux5. You can see here that B is the variation between these, 0 or 1, so that's how they hooked them up here. The output of mux5 will be ABC="xy0", where the first two bits have already been decoded and all that remains is to decode the C=0 situation. So the output of mux5 goes to the "0" input of mux7. Similarly, mux3 (ABC="x01") and mux4 (ABC="x11") are jointly fed to mux6. B again being the variation that mux6 selects between. The output of mux6 is always related to the C=1 case, and that is fed into the "1" input of mux7.

All that remains is for mux7 to pick between C=0 and C=1.

First of all, these are the steps you should follow in order to solve for SOP:

1. Write AND terms for each input combination which produce HIGH output.
2. Write the input variable if it is 1, and write the complement if the variable value is 0.
3. OR the AND terms to obtain the output function. In other words, add the AND terms together to produce your wanted output function.
4. SOP will have this form from the truth table given in the question: $$F = \overline{A}BC + A\overline{B}C + AB\overline{C} + ABC$$

The first term: $$\overline{A}BC$$ A is equal to 0 in the truth table when output F is equal to 1. The second term: $$A\overline{B}C$$ B is equal to 0 in the truth table when output F is equal to 1. The third term: $$AB\overline{C}$$ C is equal to 0 in the truth table when output F is equal to 1. The fourth term: $$ABC$$ A, B, C are all equal to 1 in the truth table when output F is equal to 1.

Secondly, these are the steps you should follow in order to solve for POS:

1. Write OR terms when the output F is equal to 0.
2. Write the input variable (A, B, C) if the value is zero, and write the complement if the input is 1.
3. AND the OR terms to obtain the output function. In other words, multiple the OR terms together to get the final output logic equation.
4. POS will have this form from the truth table given in the question: $$F=(A+B+C)(A+B+\overline{C})(A+\overline{B}+C)(\overline{A} + B + C)$$

The first term: $$(A+B+C)$$ A, B, and C are equal to zero and the output F is equal to zero as well. The second term: $$(A+B+\overline{C})$$ Output F is equal to zero but C is equal to 1. Hence why we take complement. The third term: $$(A+\overline{B}+C)$$ Output F is equal to zero but B is equal to 1. Hence why we take complement. The fourth term: $$(\overline{A}+B+C)$$ Output F is equal to zero but A is equal to 1. Hence why we take complement.

Essentially, you have to follow the three first steps in order to successfully for SOP and POS. You could have a lot more terms depending on your output results or the number of inputs you have.

Note that the results shown above are not in the reduced format. You could potentially reduce those Boolean-logic equations much more.