Question 3 of 6: 3-input / 4-output logic: K-map minimization and PAL vs PLA

Question 3 (12 marks) — 3-input / 4-output logic: K-map minimization and PAL vs PLA

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four output columns as functions of $(X,Y,Z)$. Find. minimal SoP for each output and the better programmable-logic architecture. Approach. Map each column separately, then compare the number of distinct product terms against the number of term instances to decide PAL vs PLA.

Minimized outputs (part a). On-sets: $A=\sum m(1,3,4,7)$, $B=\sum m(0,3,5,6,7)$, $C=\sum m(0,2,3,4,6)$, $D=\sum m(2,3,4,5,7)$ (minterm index $=4X+2Y+Z$). The four maps give:

$$A=\overline{X}Z+YZ+X\overline{Y}\,\overline{Z}\qquad B=\overline{X}\,\overline{Y}\,\overline{Z}+XY+XZ+YZ$$$$C=\overline{Z}+\overline{X}Y\qquad D=\overline{X}Y+X\overline{Y}+XZ$$

PAL vs PLA (part b). List the product terms used by all four outputs and count sharing:

There are 12 term instances but only 9 distinct product terms: $YZ$ is shared by $A$ and $B$, $XZ$ by $B$ and $D$, and $\overline{X}Y$ by $C$ and $D$ — three sharing events. A PLA, whose AND array is fully programmable and shared by every OR output, exploits this reuse: 9 AND terms feed all four sums. A PAL dedicates a fixed block of AND terms to each output and cannot share, so it would need all 12 terms (and any output limited to fewer product terms per OR could overflow). Choose the PLA — the shared product terms $YZ,XZ,\overline{X}Y$ make the shared AND plane the economical choice.

PLA implementation. Program the AND plane with the 9 distinct terms (1 = true input, 0 = complemented input, – = not connected) and connect each OR output to the terms it uses (✓):

This is a 3-input, 9-product-term, 4-output PLA; the OR plane carries 12 connections, one per term instance counted above.