the DE board utilizes common-anode seven-segment displays, thus; a 0 is placed in the truth table anywhere a segment needs to be on, and a 1 anywhere a segment needs to be off. Note that the HEX1 display will only be blank or -, so its g segment is the only one that needs to be in the truth table. WXYZ Signed decimal 0 001 1 011 1 010 1 1 1 0 1 1 1 O 101 1 1 O 1 0 1001 1 0 0 1 10 0 1 1 0 0 1 1 1 0 1 0 -1 HEX Displays go fo e0 do co bo a0 g1

This is the next part of the worksheet but I totally get if only the truth table may be solved. Thank you!

For each of the 8 outputs in the truth-table, construct a K-Map to determine the
simplest Boolean expression. Be sure to look for XOR and XNOR simplifications to
reduce you final gate count. For example: XY’Z + XYZ’ = X(Y xor Z). Since segments a-f
of HEX1 will always be off, k-maps are necessary.