Digital circuit and logic design short answer questions, experts to help

Red light = f2 = one exception.

Yellow light = f1 = two abnormalities.

Green light = f0 = all normal.

Red + Yellow = F1 + F2 = All Exceptions.

Truth table. a b c f0 f1 f2

Logical. f0 = abc

f1 = a'b'c' + a'b'c + a'bc' + ab'c'

f2 = a'b'c' + a'bc + ab'c + abc'

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Off work.

Digital circuit and logic design short-answer questions, experts to help

The input logic variables a, b, and c represent the working conditions of the three devices, with 1" indicating normal and "0" indicating abnormality.

The red, yellow, and green indicators are used to indicate the working conditions of the three devices.

Use the output logical variables f2, f1, and f0 to represent the red, yellow, and green indicators, 1" means the light is on, and "0" means the light is off.

A green light indicates that everything is normal; A red light indicates that one is not normal; The yellow light indicates that the two units are abnormal; The red and yellow lights indicate that all three are abnormal.

List the truth table of the control circuit, write out the standard and/or expression of the output f2, f1, and f0, and select the appropriate integrated circuit to implement it.

It's so complicated, it's obviously a design topic!

How is it still a "short-answer question"?

The truth table and circuit diagram are as follows:

First, output a number for each gate, and then write the expression according to the gate logic;

y = f1*f2)'

f1 = f4+f5+f3)'

f2 = c*f3)'

f3 = f4+f5+c)'

f4 = a*f6)'

f5 = b*f6)'

f6 = a*b)'

Then substitute and sort it out, just simplify it as much as possible;

The y of this circuit is always equal to 1.

I've been learning this for too long, and I've forgotten it, my brother used to be a master.

Solution: baij1=q2' k1=1

j2=q1 k2=1

Z=Q2 Order: Q1=J1Q1'=q2'q1'

Secondary state q2 = j2q2'=q2'q1

State transition. q2q1/z

You can do self-starting, zhi output 3-way clock signal, if you have any opinions, welcome to discuss, learn together.

The check bit is the XOR of the others

p = 1, pb6 pb0 is an odd number of 1s, plus p = 1 is an even number of 1s;

p = 0, pb6 pb0 is an even number of 1s, plus p = 0 is also an even number of 1s.

So it's an even check.

1-7, regardless of the carry truth table, a, b are the input, s is the output a b s

Considering carrying, a, b, c1 are inputs, s, c2 are outputs, and truth tables;

a b c1 s c2

1-8 does not consider the debit truth table, a, b are the input (a-b), d is the output; a b d

0 1 1 (You need to borrow a seat from a high position, but the question does not require consideration.) ）1 0 1

Considering the borrowing, a, b, c1 are the input (a-b-c1), s, c2 are the output, c1 is the low borrowing, c2 is the borrowing to the high, d is the result, the truth table;

a b c1 d c2

This topic is the easiest, as long as you have learned to count electricity, you should be able to do it.

f=a'bc+ab'c+abc'+abc

abc'+abc)+(ab'c+abc)+(a'bc+abc): The term of the logical expression is repeated several times, and the result does not change.

ab(c+c')+ac(b+b')+bc(a+a') //：a+a'=b+b'=c+c'=1

ab+bc+ca

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