CSC270 Homework 9 Solution

Problem #1

No problem on this one. Everybody got the schematics correctly.

The point of the second part of the question was that it is much easier to change software once the microprocessor system is built, whereas with a flip-flop based sequencer, it's much harder to redesign and change the wiring.

Problem #2

It would work. We know that when 8000 comes out on the address bus, and we're in the 2nd half of a cycle, E is 1, A15 is 1, A14 is 0, and A13 is 0.

We know the 7442 is wired as shown here:

• A3 of the 7442 is connected to A13 of the 6811
• A2 of the 7442 is connected to A14 of the 6811
• A1 of the 7442 is connected to A15 of the 6811
• A0 of the 7442 is connected directly to E of the 6811 (no inverter!)
• Ouput 3 of the 7442 (Pin 4) is connected to the clock input of the 74LS74 flipflop.

Therefore, we have A3=0, A2=0, A1=1, and A0=1, which is 3 in decimal. So Y3 is activated. Which is the signal sent to the clock input of the 74LS74.

The circuit works! And we don't need the 04 inverter on E!

Problem #3

The displacements were 0D, 5F, 8E, and 9B.