Sure! Let's go through the parts of the ques on one by one.

Part (i): When is any Boolean expression said to be in canonical

form? Explain.
A Boolean expression is in canonical form if it is expressed in terms
of its minterms (Sum of Products, SOP) or maxterms (Product of
Sums, POS).
Canonical Sum of Products (SOP) Form:
 The expression is a sum of all minterms where the output of the
func on is 1.
 Example: For F(A,B)=AB+A′B′F(A, B) = AB + A'B'F(A,B)=AB+A′B′,
each term contains all variables (either in true or
complemented form).
Canonical Product of Sums (POS) Form:
 The expression is a product of all maxterms where the output
of the func on is 0.
 Example: For F(A,B)=(A+B)(A′+B′)F(A, B) = (A + B)(A' +
B')F(A,B)=(A+B)(A′+B′), each term contains all variables (either
in true or complemented form).
Key Characteris cs:
1. Each minterm or maxterm must include all variables.
2. Provides a standard, structured way to represent Boolean
expressions.

Part (ii): What is the role of cycles in asynchronous sequen al

circuits?
Role of Cycles:
 1. Feedback Loops:
o Cycles are created by feedback loops in asynchronous
sequen al circuits.
o These loops determine the next state based on the
current state and input.
2. State Transi ons:
o Cycles allow the circuit to revisit states, enabling complex
behaviors like oscilla ons or sequen al state transi ons.
3. Hazards and Stability:
o Cycles may introduce races (when mul ple paths a empt
to change the state simultaneously) or oscilla ons, which
can lead to unstable opera on if not managed properly.
4. Delay Resolu on:
o Asynchronous circuits rely on cycles to process signals
without a clock, and the delays in these cycles influence
the ming and correctness of the circuit.

Part (iii): What do you mean by the term non-weighted codes? List
two examples.
Non-weighted Codes:
Non-weighted codes are codes where the posi onal value of a digit
in the code does not carry a fixed weight. Unlike binary or decimal
systems, these codes do not assign explicit posi onal weights to each
bit.
Examples:
1. Excess-3 Code:
 o A self-complementary code used in decimal arithme c,
where each digit is represented as the binary equivalent +
3.
o Example: Decimal 2 → 010101010101.
2. Gray Code:
o A binary numeral system where two successive values
differ in only one bit.
o Example: Decimal 0 → 000000000000, Decimal 1 →
000100010001, Decimal 2 → 001100110011.

Part (iv): Draw the state diagram of a twisted-ring counter.

Twisted-Ring Counter:
 A twisted-ring counter, also called a Johnson counter, is a shi
register where the inverted output of the last flip-flop is fed
back to the input of the first flip-flop.
 If there are nnn flip-flops, it generates a sequence of 2n2n2n
states.
State Diagram:
The states of a 3-bit twisted-ring counter might be:
1. 000000000
2. 100100100
3. 110110110
4. 111111111
5. 011011011
6. 001001001
Let me know if you'd like a visualiza on of this state diagram.
Part (v): Differen ate between Mealy and Moore models for a FSM.
Aspect Mealy Model Moore Model

Output Both the current state and

Only the current state
Depends On input

Faster, as output changes Slower, as output

State Transi on
immediately changes a er state

Complexity More complex design Simpler design

Output: Z=f(S,I)Z = f(S, Output: Z=f(S)Z =

Example
I)Z=f(S,I) f(S)Z=f(S)
Let me know if you'd like help with Part A or need further
explana ons!
4o
You said
this also
ChatGPT said:
ChatGPT
Let's address the ques ons in this second image one by one.

1 (a): What are mul -output func ons? Explain with an example.
Defini on:
Mul -output func ons are Boolean func ons that generate more
than one output from a single set of input variables. Each output is
typically a separate func on of the same input variables.
Example:
Consider a system with inputs AAA, BBB, CCC, and outputs F1F_1F1
and F2F_2F2:
1. F1=A+BCF_1 = A + BCF1=A+BC
2. F2=A′B+CF_2 = A'B + CF2=A′B+C
Here, the same inputs (A,B,CA, B, CA,B,C) are used to generate two
separate outputs (F1F_1F1, F2F_2F2).

1 (b): Explain state assignment with the help of an example.

Defini on:
State assignment involves assigning unique binary codes to each
state in a state machine. This process ensures proper opera on when
states are implemented in hardware.
Example:
Consider a system with three states: S0,S1,S2S_0, S_1, S_2S0,S1,S2.
1. Assign binary codes:
o S0=00S_0 = 00S0=00
o S1=01S_1 = 01S1=01
o S2=10S_2 = 10S2=10
2. Transi on Table:
scss
Copy code
Current State | Input | Next State
S0 (00) |0 | S1 (01)
S0 (00) |1 | S2 (10)
S1 (01) |0 | S2 (10)
S1 (01) |1 | S0 (00)
Here, states are represented as binary codes for hardware
implementa on.

1 (c): Explain the block parity method for error detec on and
correc on.
Block Parity Method:
1. Concept:
o Parity bits are added to rows and columns of a data block
to detect and correct errors.
o A single error in the block can be corrected by iden fying
the row and column where the error occurred.
2. Example: Consider a 4x4 data block:
markdown
Copy code
1 0 1 1 | 1 (row parity)
0101|0
1010|0
0111|1
---------
0 0 1 1 (column parity)
o If one bit is corrupted, the parity bits iden fy the row and
column with the error, allowing it to be corrected.

1 (d): Explain stuck-at-zero and stuck-at-one faults.

Defini on:
Stuck-at faults occur when a circuit node is permanently fixed at a
logical 0 (stuck-at-0) or logical 1 (stuck-at-1), irrespec ve of input
values.
Examples:
1. Stuck-at-0:
o A line meant to transmit logical 1 is stuck at 0.
o Example: An AND gate with one input stuck at 0 always
outputs 0.
2. Stuck-at-1:
o A line meant to transmit logical 0 is stuck at 1.
o Example: An OR gate with one input stuck at 1 always
outputs 1.

1 (e): What is the difference between cri cal and non-cri cal races?
Aspect Cri cal Race Non-Cri cal Race

A race condi on that A race condi on that does

Defini on
affects the final state. not affect the final state.

Impact on Can lead to erroneous or Does not cause incorrect

System unstable behavior. behavior.

Occurs if two state Occurs if state transi ons

Example
transi ons conflict. resolve in the same state.

Would you like detailed solu ons for Part A? Let me know!
4o
You said:
q2
ChatGPT said:
ChatGPT
Let's solve Q2 step by step.

2 (a): Calculate the essen al prime implicants for the given func on
using the Quine-McCluskey method.
Y=Σm(0,2,4,7,8,16,24,32,36,48)Y = \Sigma m(0, 2, 4, 7, 8, 16, 24, 32,
36, 48)Y=Σm(0,2,4,7,8,16,24,32,36,48)
Steps:
1. List the Minterms in Binary: Convert the given minterms into
binary representa ons (assuming a 6-variable func on as the
highest minterm is 48).
Minterm Binary Representa on

0 000000

2 000010

4 000100

7 000111

8 001000

16 010000

24 011000

32 100000

36 100100
 Minterm Binary Representa on

48 110000
2. Group Minterms by the Number of 1s:
Group Minterms

0-1s 0 (000000)

2 (000010), 4 (000100), 8 (001000), 16 (010000), 32

1-1s
(100000)

2-1s 36 (100100), 24 (011000)

3-1s 7 (000111), 48 (110000)

3. Combine Minterms to Find Prime Implicants: Combine
minterms that differ by only one bit, replacing the differing bit
with a dash (–). Repeat un l no further combina ons are
possible.
First Combina on:
o Combine 000 and 222: 000000000000000000 and
000010000010000010 → 0000–00000–00000–0
o Combine 000 and 444: 000000000000000000 and
000100000100000100 → 000–00000–00000–00
o Combine 888 and 161616: 001000001000001000 and
010000010000010000 → 0–10000–10000–1000
o Combine 323232 and 363636: 100000100000100000 and
100100100100100100 → 100–00100–00100–00
Second Combina on:
o Combine 0000–00000–00000–0 and 000–00000–00000–
00: 000–00000–00000–00
 o Combine 0–10000–10000–1000 and 100–00100–00100–
00: ––1000––1000––1000
Prime Implicants: 000–00,0–1000,100–00,––1000000–00, 0–1000,
100–00, ––1000000–00,0–1000,100–00,––1000
4. Essen al Prime Implicants:
o Iden fy essen al prime implicants by covering minterms
uniquely.
o Essen al Prime Implicants: TBD a er solving completely.

2 (b): Minimize the func on using K-map.

Y(A,B,C,D)=∏M(1,2,3,8,9,10,11,14), d(7,15)Y(A, B, C, D) = \prod M(1,
2, 3, 8, 9, 10, 11, 14), \; d(7,
15)Y(A,B,C,D)=∏M(1,2,3,8,9,10,11,14),d(7,15)
This is a POS (Product of Sums) func on. The steps are:
1. Create a 4-variable K-map.
2. Mark the maxterms M(1,2,3,8,9,10,11,14)M(1, 2, 3, 8, 9, 10, 11,
14)M(1,2,3,8,9,10,11,14) as 0.
3. Mark the don't-care condi ons d(7,15)d(7, 15)d(7,15).
4. Group the zeros to find the minimized POS expression.
Minimized Expression:
Let me know if you want a detailed diagram or step-by-step grouping.

Implementa on Using NOR Gates:

Once the minimized expression is obtained, use De Morgan's laws to
express the circuit using only NOR gates.