Digital Electronics And Microprocessor Question And Answers |

Digital Electronics And Microprocessor Question and Answers

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Q1) There are four propositions under which Anshul goes to office a) If it rains and Anshul gets his father’s car. b) If it does not rain and there is no cricket match c) If it rains and there is important meeting d) If it does not rain and there is a cricket match All the four propositions are represented in the order as ABCD with A as the MSB and D as the LSB representing Car’s availability as A, important meeting as B, match taking place as C, and Raining as D.
i. Find an algebraic expression in SOP form that assumes the value 1 whenever Anshul meets the condition of going to office.
ii. Simplify the above expression and suggest a simpler set of requirements.
iii. Show the realization if function developed in part I using 4 X1 Mux where AB is the input to the selection line.

Answer :

To represent the given conditions in algebraic expressions, we’ll use the following notation:

– A represents Car’s availability.
– B represents important meeting.
– C represents a cricket match.
– D represents raining.

Now, let’s express the given conditions in algebraic expressions in Sum of Products (SOP) form:

i. Algebraic expression in SOP form for Anshul going to the office:

We want to find an expression that is true whenever Anshul meets the condition of going to the office. This condition is represented by “It does not rain and there is no cricket match.”

So, the expression for Anshul going to the office in SOP form is:
(¬D) ⋅ (¬C)

ii. Simplifying the above expression and suggesting a simpler set of requirements:

To simplify the expression, we can use Boolean algebra rules:

(¬D) ⋅ (¬C)

Using De Morgan’s Law:
¬(D + C)

So, the simplified expression for Anshul going to the office is ¬(D + C), which means Anshul goes to the office if it is neither raining nor there is a cricket match.

iii. Show the realization of the function developed in part i using a 4×1 Mux where AB is the input to the selection line:

A 4×1 Multiplexer (Mux) has 4 data inputs (D0, D1, D2, D3), 2 selection lines (A and B), and 1 output (Y). We’ll use AB as the input to the selection lines.

The function we want to realize is ¬(D + C). Let’s represent this using the 4×1 Mux:

– D3: ¬(00) – This represents the condition when neither D nor C is true.
– D2: 0 – We don’t need this condition for our expression.
– D1: 0 – We don’t need this condition for our expression.
– D0: 1 – This represents the condition when neither D nor C is true.

Now, we connect AB to the selection lines as follows:
– A is connected to the least significant bit (LSB) D0.
– B is connected to the most significant bit (MSB) D3.

So, the 4×1 Mux realization is:

D0 = ¬(00)
D1 = 0
D2 = 0
D3 = ¬(00)

And AB selects between D0 and D3 based on the values of A and B. The output Y will be ¬(D + C) as desired.

This Mux implementation ensures that Anshul goes to the office when the conditions ¬(D + C) are met, which means neither it is raining (D is false) nor there is a cricket match (C is false).

Q2) a. Using 2’s complement method, find the result of subtraction of numbers given in base 10 to show the result in binary. (5)10-(12)10=(?)2

Answer :

To find the result of subtracting (5)10 from (12)10 using the 2’s complement method and express the result in binary, follow these steps:

Step 1: Convert both numbers to binary: (5)10 in binary is 0101. (12)10 in binary is 1100.

Step 2: Find the 2’s complement of the number to be subtracted, which is (5)10. Inverting all the bits (changing 0s to 1s and 1s to 0s) gives us 1010.

Step 3: Add the 2’s complement of (5)10 to (12)10: 1100

1010

(10110)2

Step 4: Since the result has an extra bit, discard the carry bit on the left, leaving us with (0110)2.

So, (12)10 – (5)10 = (0110)2 in 2’s complement binary notation.

Q2 b. What is the range of decimal numbers in 2’s complement 5 bit signed representation?

Answer :

In a 5-bit 2’s complement signed representation, the range of decimal numbers that can be represented is from -16 to 15.

Here’s how the range is determined:

The leftmost bit (the most significant bit, MSB) is the sign bit. When it’s 0, the number is positive, and when it’s 1, the number is negative.
The remaining 4 bits represent the magnitude of the number in binary form.

For the positive range:

The smallest positive number is 00000 (0 in decimal).
The largest positive number is 01111 (15 in decimal).

For the negative range:

The smallest negative number is 10000 (-16 in decimal).
The largest negative number is 11111 (-1 in decimal).

So, the range of decimal numbers in a 5-bit 2’s complement signed representation is from -16 to 15.

Q3) For the combinational circuit shown a). Find the Boolean expression b) Give the implementation with minimum number of logic gates A’ U1 D AND2 A U2 D’ AND2 A U3 c’ D’ AND3 A U4 C D AND 4 U5 y=f(A,B,C,D) OR4

Answer :

let’s analyze the given combinational circuit and find the Boolean expression for the output Y and then provide its implementation using the minimum number of logic gates.

Given the circuit :

A’ U1 D AND2
A U2 D’ AND2
A U3 C’ D’ AND3
A U4 C D AND4
U5
OR4

a) Finding the Boolean Expression for Y:

To find the Boolean expression for Y, we need to analyze the logic gates in the circuit. Let’s break it down step by step:

U1: A’ (NOT A)
U2: A
U3: A
U4: A
U5: C’
AND2_1: D AND (A’ U1)
AND2_2: D’ AND (A U2)
AND3: (C’ D’) AND (A U3)
AND4: (C D) AND (A U4)
OR4: OR4 takes inputs from AND2_1, AND2_2, AND3, and AND4.

Now, let’s express each of these gates in Boolean terms:

U1: A’ = !A
U2: A
U3: A
U4: A
U5: C’
AND2_1: D AND (A’ U1) = D AND (!A)
AND2_2: D’ AND (A U2) = D’ AND A
AND3: (C’ D’) AND (A U3) = (C’ D’) AND A
AND4: (C D) AND (A U4) = (C D) AND A
OR4: OR4 takes inputs from AND2_1, AND2_2, AND3, and AND4.

Now, we can express the overall Boolean expression for Y:

Y = OR4 = (AND2_1) OR (AND2_2) OR (AND3) OR (AND4) = ((D AND (!A)) OR (D’ AND A) OR ((C’ D’) AND A) OR ((C D) AND A))

b) Implementation with the Minimum Number of Logic Gates:

To minimize the number of logic gates, we can simplify the Boolean expression further using Boolean algebra. Here’s a simplified implementation:

Y = ((D AND (!A)) OR (D’ AND A) OR ((C’ D’) AND A) OR ((C D) AND A))

Using Boolean algebra, we can simplify this expression:

Y = (D AND (!A)) OR (D’ AND A) OR ((C’ D’) AND A) OR ((C D) AND A)

We can distribute A in the last two terms:

Y = (D AND (!A)) OR (D’ AND A) OR (C’ D’ A) OR (C D A)

Now, we can see that the common factor in the last two terms is A. We can factor it out:

Y = (D AND (!A)) OR (D’ AND A) OR (A(C’ D’ + C D))

Next, we can simplify the last term using the distributive law:

Y = (D AND (!A)) OR (D’ AND A) OR (A(C’ + D’))

Finally, we can implement this simplified expression using the minimum number of logic gates. The exact implementation may vary depending on the specific gates available, but it should include AND gates and OR gates as required by the expression.

Q4) Design an RS flip-flop using D flipflop. Show the following in your answer a. Truth table b. State equation c. Circuit diagram

Answer :

design an RS flip-flop using D flip-flops. Here’s the requested information:

a. Truth table:

R	S	Q(t)	Q(t+1)
0	0	0	0
0	1	0	1
1	0	1	0
1	1	0	X

In the truth table, R stands for the Reset input, S stands for the Set input, Q(t) represents the current state, and Q(t+1) represents the next state. “X” denotes an invalid or don’t care state.

b. State equation: Q(t+1) = D

Where D is the input to the D flip-flop. This equation states that the next state Q(t+1) is equal to the D input.

c. Circuit diagram: To create an RS flip-flop using D flip-flops, you can use two D flip-flops and connect them as follows:

Connect the S input to the D input of the first D flip-flop.
Connect the R input to the complement (inverted) of the D input of the first D flip-flop.
Connect the Q output of the first D flip-flop to the D input of the second D flip-flop.
Connect the Q’ output of the first D flip-flop to the D input of the second D flip-flop.
The Q output of the second D flip-flop represents the Q(t+1) output of the RS flip-flop.

Here’s a simple text representation of the circuit:

S —–|>— D —|>— Q
R —–|>— ~D –|>— Q

Where “S” is the Set input, “R” is the Reset input, “D” is the input to the first D flip-flop, “~D” is the inverted input, “Q” is the output of the second D flip-flop, and “|>” represents a D flip-flop.

This circuit configuration ensures that the RS flip-flop operates according to the truth table and state equation provided.

Q5) Show the implementation of 4 to 16 decoder using any number of 2 to 4 decoder where all 2 to 4 decoders have high enable

Answer :

To implement a 4-to-16 decoder using 2-to-4 decoders with active-high enables, you can cascade four 2-to-4 decoders. Here’s a high-level description of the implementation:

Use a 2-to-4 decoder to decode the two most significant bits of the 4-bit input.
Use another 2-to-4 decoder to decode the two least significant bits of the 4-bit input.
Connect the enable inputs of both 2-to-4 decoders to the enable input of the 4-to-16 decoder.

Here’s the logical diagram for this implementation:

4-to-16 Decoder
_______
| |
A[3]–|2-to-4 | O0
A[2]–|2-to-4 | O1
A[1]–|2-to-4 | = O2
A[0]–|2-to-4 | …
EN— |2-to-4 | O15
|_______|

Here, A[3:0] represents the 4-bit input to the 4-to-16 decoder, and EN is the enable input for all the 2-to-4 decoders. The outputs O0 to O15 represent the 16 possible combinations of the 4-bit input.

Each of the 2-to-4 decoders takes the two bits of the 4-bit input and generates a 4-bit output, where only one of the output lines will be high (active) depending on the input combination. The enable input EN is used to control whether the 2-to-4 decoders are active or not. When EN is high, the 2-to-4 decoders are enabled and will generate their outputs based on the input bits.

This configuration allows you to create a 4-to-16 decoder using 2-to-4 decoders with active-high enables.

Q6) Realize the given function f(A,B,C)=(0,3,5,6) using only 2 NAND gates and one 3 to 8 decoder implemented using NAND gates

Answer :

To implement the given function f(A,B,C) = (0,3,5,6) using only 2 NAND gates and a 3-to-8 decoder implemented using NAND gates, you can follow these steps:

First, you need to create a truth table for the function f(A,B,C) based on the given outputs (0,3,5,6). The truth table will have 8 rows (for all possible combinations of A, B, and C).

A B C f(A,B,C)

0 0 0 0

0 0 1 –

0 1 0 –

0 1 1 3

1 0 0 –

1 0 1 5

1 1 0 –

1 1 1 6
Now, you can see that the function has four non-zero outputs (3, 5, 6) for different input combinations. You will need to use the 3-to-8 decoder to select these specific combinations and feed them into the NAND gates to generate the desired outputs.
Use the 3-to-8 decoder to generate the following select signals based on the truth table:
- Select line 3 (for output 3): A=0, B=1, C=1
- Select line 5 (for output 5): A=1, B=0, C=1
- Select line 6 (for output 6): A=1, B=1, C=0
Implement these select lines using NAND gates. For example, to create the select line for output 3 (A=0, B=1, C=1), you can connect NAND gates as follows:
- NAND gate 1: Inputs A and C
- NAND gate 2: Input B and the output of NAND gate 1
- The output of NAND gate 2 is the select line for output 3.
Repeat the same process to generate the select lines for outputs 5 and 6.
Now, you have three select lines based on the 3-to-8 decoder’s inputs. Use these select lines to select the desired outputs (3, 5, and 6) and connect them to the output NAND gates accordingly.
Implement the output NAND gates with the selected lines as inputs to generate the final function outputs (0, 3, 5, 6).

This way, you can implement the function f(A,B,C) using only 2 NAND gates and a 3-to-8 decoder implemented using NAND gates, achieving the specified outputs.

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