Question: Part A (10 + 5= 15 points). A BCD to 7 segment decoder has 4 inputs (X, Y, Z, W) and 7 outputs (a, b,

 Part A (10 + 5= 15 points). A BCD to 7

Part A (10 + 5= 15 points). A BCD to 7 segment decoder has 4 inputs (X, Y, Z, W) and 7 outputs (a, b, c, d, e f, g) that select the correspondent segments of the LED display represented in the figure (a). The numeric representation of the decimal numbers is given in the figure (b) shown below. 12345b1090 (a) Segment designation display (b) Numeric representation of decimal numbers on the LED (a) Considering the decoder's outputs to being in "don't care" states (marked with"d or X") for any of the 6 unused input combinations (i) Find the minimized expression for the output e only (i) Draw the logic diagram of the two-level NOR circuit that implements the above expression (c) (b) Redo part (a) considering that your decoder has to display the letter E (Error) on the 7-segment display if any of the unused combinations is presented to the decoder's inputs Note: You can use any type of gates with any number of inputs you may need. Assume, as well, that the input variables are available in both true and complemented form

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