digital logic design

( i have a lot questios so please help me)

Proof that:
A¡¦.B + B¡¦.C¡¦ + A.B + B¡¦.C = 1

Q3 [2 points], According to DeMorgan theorem, the complement of
W . X + Y . Z is W¡¦ + X¡¦ . Y¡¦ + Z¡¦

Yet both functions are 1 for WXYZ = 1110. How can both function and its complement be 1 for the same input combination? What is wrong here?

Q4 [2 points], Write a truth table for the following logic function:
F = A¡¦ . B . (C . B . A¡¦ + B . C¡¦)

Q5 [3 points], List the octal and hexadecimal numbers from 16 to 32. Using the last A, B, C for the last three digits, list the numbers from 8 to 28 in base 13

Q6 [6 points „» each 1.5 point], Do the following SHOW YOUR WORK (IF YOU DO NOT SHOW YOUR WORK YOU WILL EARN ZERO MARK):

b)(27.315)10 =( )2= ( )8

c)What would be the base of the number system if (BEE)r = (2699)10

d)(1.10010)2 = ( )10 = ( )16


Q7 [5 points], Do the followings:
a) Perform the subtraction on the given unsigned numbers using 10¡¦s complement subtraction of the subtrahend (SHOW YOUR WORK)
(125-1800) , (6428 - 3409)

b) Perform the following operations in 10¡¦s complement
(-9286) + (+801) , (-9286) + (-801)

c) [2 points] Convert the decimals +49 and +29 to binary, using the signed-2¡¦s complement representation and enough binary digits to accommodate the numbers. Then perform the binary equivalent of (+29) + (-49) , (-29) + (-49)

Q1: Proof that A'.B + B'.C' + A.B + B'.C = 1

To prove this expression, we can use Boolean algebra and simplification techniques.

First, let's expand the expression using the distributive property:
A'.B + B'.C' + A.B + B'.C = A'.B + A.B + B'.C + B'.C'

Next, let's simplify the expression:
A'.B + A.B = (A' + A).B = 1.B = B

B'.C + B'.C' = B'(C + C') = B'(1) = B'

Therefore, the simplified expression becomes:
B + B' = 1

This proves that A'.B + B'.C' + A.B + B'.C equals 1.

Q2: Both functions being 1 for the same input combination

In this case, it appears that both the function and its complement are equal to 1 for the input combination WXYZ = 1110. This seems contradictory since the complement of a function should give the opposite result.

The likely explanation for this is that there is an error in either the original function or its complement. Double-check the expressions for any mistakes in logic or typos. Additionally, verify if the function and its complement are defined correctly for all possible input combinations. It is possible that there might be an oversight or an error in understanding the problem.

If you are certain that both the function and its complement are correctly represented and still yield 1 for the same input combination, then it could be a rare coincidence. However, it is always recommended to review the expressions and calculations to ensure accuracy.

Q3: Truth table for F = A'.B . (C . B . A' + B . C')

To create a truth table for the given logic function, we need to consider all possible input combinations for the variables A, B, and C.

First, list all possible combinations of A, B, and C:

A B C
---
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

Next, substitute these combinations into the given function and calculate the corresponding output F:

A B C F
------
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1

The resulting truth table for the logic function F = A'.B . (C . B . A' + B . C') is as follows:

A B C F
------
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1

Q4: Octal and hexadecimal numbers

To list the octal and hexadecimal numbers from 16 to 32, we can convert the decimal numbers to the desired base.

Octal:
Decimal - Octal
16 - 20
17 - 21
18 - 22
19 - 23
20 - 24
21 - 25
22 - 26
23 - 27
24 - 30
25 - 31
26 - 32
27 - 33
28 - 34
29 - 35
30 - 36
31 - 37
32 - 40

Hexadecimal:
Decimal - Hexadecimal
16 - 10
17 - 11
18 - 12
19 - 13
20 - 14
21 - 15
22 - 16
23 - 17
24 - 18
25 - 19
26 - 1A
27 - 1B
28 - 1C
29 - 1D
30 - 1E
31 - 1F
32 - 20