- 2070 Magh · 5 marks
Design ROM for X,Y,Z to F1,F2 table
Short answer
8x2 ROM with 3-to-8 decoder. F1 = w0+w1+w5+w7, F2 = w2+w3+w5+w6.
Design a ROM that will store the following words in the corresponding addresses.
| X | Y | Z | F1 | F2 |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 |
The table has 3 address inputs, so there are 2^3 = 8 words, each 2 bits wide. The ROM is therefore 8 x 2.
Stored contents
| Address (X,Y,Z) | Word (F1,F2) |
|---|---|
| 000 (word line 0) | 10 |
| 001 (word line 1) | 10 |
| 010 (word line 2) | 01 |
| 011 (word line 3) | 01 |
| 100 (word line 4) | 00 |
| 101 (word line 5) | 11 |
| 110 (word line 6) | 01 |
| 111 (word line 7) | 10 |
Design
Use a 3-to-8 decoder (the address lines select exactly one word line) and one OR gate per output bit. Connect a word line to an OR gate only where that word stores a 1 in that column (a programmable connection at the crosspoint).
| Output | Word lines ORed (addresses with a 1) | Output expression |
|---|---|---|
| F1 | 0, 1, 5, 7 | F1 = w0 + w1 + w5 + w7 |
| F2 | 2, 3, 5, 6 | F2 = w2 + w3 + w5 + w6 |
X,Y,Z (address)
| +-----------+ w0..w7
+--->| 3-to-8 |===> word lines ===> [OR] --> F1
| decoder | [OR] --> F2
+-----------+
Each OR input is connected (by a transistor/fuse) to a word line only when that bit of the word is 1; the connection is left open for 0. When an address is applied, the decoder raises one word line and the OR gates output the stored word.
✓ Answer checked by Claude Opus 5.5 (high effort)
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