B- D Figure 1- Boolean Circuit You are given the Boolean circuit in the figure above. 1) Convert this circuit to a truth table 2) Convert the truth table to the Sum of Products expression 3) Using Boolean algebra, simplify the circuit to its minimum form 4) Using a Karnaugh map, simplify the circuit to its minimum form 5) Infer the number of basic logic gates needed for the representations in questions 2-4. Which representation would use the minimum number of logic gates in implementation?

Electric Motor Control
10th Edition
ISBN:9781133702818
Author:Herman
Publisher:Herman
Chapter22: Sequence Control
Section: Chapter Questions
Problem 6SQ: Draw a symbol for a solid-state logic element AND.
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can you please help to answer question number 5 based on the answer that i give for question number 1 to 4.. Thank you in advance.
B
D
Figure 1- Boolean Circuit
You are given the Boolean circuit in the figure above.
1) Convert this circuit to a truth table
2) Convert the truth table to the Sum of Products expression
3) Using Boolean algebra, simplify the circuit to its minimum form
4) Using a Karnaugh map, simplify the circuit to its minimum form
5) Infer the number of basic logic gates needed for the representations in questions 2-4.
Which representation would use the minimum number of logic gates in implementation?
Transcribed Image Text:B D Figure 1- Boolean Circuit You are given the Boolean circuit in the figure above. 1) Convert this circuit to a truth table 2) Convert the truth table to the Sum of Products expression 3) Using Boolean algebra, simplify the circuit to its minimum form 4) Using a Karnaugh map, simplify the circuit to its minimum form 5) Infer the number of basic logic gates needed for the representations in questions 2-4. Which representation would use the minimum number of logic gates in implementation?
A
Do
Do
Do
A+L
Twith Ioble
A
(ĀB+č ) (A+c) D= (ÃB+ē) (A+ c)
1
1
1
2)
D= Eml1,5,7)
D: ABC+ ABC+ ABC
3)
D: ĀBC+ABc + ABC
- BC + AC
00.01
10
4) A
D= BC+ AC
O DO - o –
O O 0 - -
Transcribed Image Text:A Do Do Do A+L Twith Ioble A (ĀB+č ) (A+c) D= (ÃB+ē) (A+ c) 1 1 1 2) D= Eml1,5,7) D: ABC+ ABC+ ABC 3) D: ĀBC+ABc + ABC - BC + AC 00.01 10 4) A D= BC+ AC O DO - o – O O 0 - -
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