Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > The minimum number of 2-input NOR gates requ...
Start Learning for Free
The minimum number of 2-input NOR gates required to implement
Y = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)
Correct answer is '6'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
The minimum number of 2-input NOR gates required to implementY = ACE ...
Figure
Hence, the total number of 2-input NOR gates are 6.
Free Test
FREE
| Start Free Test |
Community Answer
The minimum number of 2-input NOR gates required to implementY = ACE ...
Minimum number of 2-input NOR gates required to implement the given expression
To find the minimum number of 2-input NOR gates required to implement the given expression, we need to analyze the expression and simplify it using Boolean algebra. We can break down the expression into smaller parts to identify the common terms that can be shared among the different sub-expressions.
Given Expression:
Y = ACE + AFC + ADE + ADF + BCE + BDF + BCF + BDE
Step 1: Simplifying the Expression
To simplify the given expression, we can use Boolean algebra rules such as the distributive law and De Morgan's theorem. By applying these rules, we can identify the common terms and reduce the number of gates required.
Step 2: Identifying the Common Terms
After simplifying the given expression, we can identify the common terms that can be shared among the sub-expressions, which helps in reducing the number of gates required. The common terms are:
- A
- C
- D
- E
- F
- B
Step 3: Implementing the Reduced Expression
Using the identified common terms, we can implement the reduced expression by sharing the common terms among the different sub-expressions. By doing this, we can eliminate the redundant gates and reduce the overall number of gates required.
Step 4: Counting the Number of Gates
After implementing the reduced expression, we count the number of 2-input NOR gates required. In this case, the minimum number of 2-input NOR gates required is 6.
Conclusion
By simplifying the given expression and identifying the common terms, we can implement the expression using a minimum number of 2-input NOR gates. In this case, the minimum number of gates required is 6.
To find the minimum number of 2-input NOR gates required to implement the given expression, we need to analyze the expression and simplify it using Boolean algebra. We can break down the expression into smaller parts to identify the common terms that can be shared among the different sub-expressions.
Given Expression:
Y = ACE + AFC + ADE + ADF + BCE + BDF + BCF + BDE
Step 1: Simplifying the Expression
To simplify the given expression, we can use Boolean algebra rules such as the distributive law and De Morgan's theorem. By applying these rules, we can identify the common terms and reduce the number of gates required.
Step 2: Identifying the Common Terms
After simplifying the given expression, we can identify the common terms that can be shared among the sub-expressions, which helps in reducing the number of gates required. The common terms are:
- A
- C
- D
- E
- F
- B
Step 3: Implementing the Reduced Expression
Using the identified common terms, we can implement the reduced expression by sharing the common terms among the different sub-expressions. By doing this, we can eliminate the redundant gates and reduce the overall number of gates required.
Step 4: Counting the Number of Gates
After implementing the reduced expression, we count the number of 2-input NOR gates required. In this case, the minimum number of 2-input NOR gates required is 6.
Conclusion
By simplifying the given expression and identifying the common terms, we can implement the expression using a minimum number of 2-input NOR gates. In this case, the minimum number of gates required is 6.
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Top Courses for Electrical Engineering (EE)View all
The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer?
Question Description
The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer?.
The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer?.
Solutions for The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer?, a detailed solution for The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? has been provided alongside types of The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The minimum number of 2-input NOR gates required to implementY = ACE + AFC + ADE + ADF + BCE + BDF + BCF+BDE is _____. (Correct upto nearest integer)Correct answer is '6'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup