{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

d07hwk6sol - ECE2022 Homework 6 D07 Name €00“ dieing...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE2022 Homework 6 D07 Name: €00“ dieing ECE Box Number: Each problem is worth the number of points shown. Clearly show the final answer to each problem by drawing a heavy square around it. This assignment is due Monday, April 30, bring to class — no later! O>< CLOCK 0 Problem 1 - 30 points total For the sequential machine shown: (3) Determine Boolean expressions for next state logic and output logic from the circuit. That is7 express D1, D2 and F as logic expressions in terms of Q1, Q2 and X . (b) Fill in the state table, given below. ('01) DI: (6f’X}@L Present Next State Input State Output Q1 Q2 X D1 D2 F 0 0 0 C) i 1 0 0 1 O I ( 0 1 O ‘ 0 1 O 1 1 l ' | 1 0 0 0 t O 1 0 1 c) i O 1 1 0 0 ( i 1 1 1 L l I (c) Draw a state transition diagram for this sequential machine. X/F Problem 2 A 20 points total This problem investigates the state transition diagram shown. The flip flop output values will be used as the finite state machine outputs7 so, no other output value is shown in the state diagram, nor do you have to generate any entries in the table for it. The values shown inside the “bubbles” are QA and QB in that order. (a) Fill in the state table below. Present Next State Input State QA QB X DA DB 0 0 0 0 \ O 0 1 0 O 0 1 0 t 0 0 1 1 I \ 1 0 0 <6 l 1 0 1 (9 O 1 1 0 O O 1 1 1 O l Problem 3 A 10 points total A Read-Only Memory (ROM) will be used to compute a full adder operation by adding two 2—bit binary numbers and a carry value from a preceding operation to form a 3 bit result. That is, the inputs are X1, X0 the most and least significant bits of the first operand, X, Y1, Y6 the most and least significant bits of the second operand, Y, and a carry bit Co. The results are Z2, Z1, Zo. For example, if X1 —1,X0 — 0, Y1 — 0,Y0 — 1 and Co = 1, then Z2 2 1, 21 = 0,Z0 = 0. For the assignment of address lines and output lines shown below on the right, fill out this table with the values to be stored in the ROM at each address to produce the desired results. OOOOOOAUO 00000000 4-0 PM Ha ...
View Full Document

{[ snackBarMessage ]}