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B = 02 =) X, Y, Z has to during off le not dike LS But m when 'A' = 0 ㅕ A = 3 using Sanu, A filled 긋 = efficient usage of all op will be heater. refrigerator, be on au Peak A = on, is These TR only rali. A-3 3 essential 9 17 is Power. Switched off. ( Je = 3 all op. Statement, Town table bastery. = 0. only healey can be on. can be banery only Ref. Olp = 0] X =)0 2 J → 4 S 6 7 8 9 10 )) 12 13 ly 15 16 D J O D --- 8 - O oo 1 0 1 O 1 1 1 O 1 1 J O Z= PO r S 1 D O 0 C 1 0 O B 1 1 J 0 O 0 0 J 0 1 1 0 0 J 1 ) J O O O 010 0 1 1 O 11 P D O 6 1 I O 0 0 O 1 00 0 1 O 0. 0 0 0 0 0 1 I ) 1 X = m (4, 6, 12, 14) Y = m (12, 13, 14, 15) O O 0 0 0 YZ 0 0 0 D O 1 00- 1 I 1 0 1 -0 0 O 1 1 0 Σ ml 2,3,6,7, 10, 11, 14, 15)

B = 02 =) X, Y, Z has to during off le not dike LS But m when 'A' = 0 ㅕ A = 3 using Sanu, A filled 긋 = efficient usage of all op will be heater. refrigerator, be on au Peak A = on, is These TR only rali. A-3 3 essential 9 17 is Power. Switched off. ( Je = 3 all op. Statement, Town table bastery. = 0. only healey can be on. can be banery only Ref. Olp = 0] X =)0 2 J → 4 S 6 7 8 9 10 )) 12 13 ly 15 16 D J O D --- 8 - O oo 1 0 1 O 1 1 1 O 1 1 J O Z= PO r S 1 D O 0 C 1 0 O B 1 1 J 0 O 0 0 J 0 1 1 0 0 J 1 ) J O O O 010 0 1 1 O 11 P D O 6 1 I O 0 0 O 1 00 0 1 O 0. 0 0 0 0 0 1 I ) 1 X = m (4, 6, 12, 14) Y = m (12, 13, 14, 15) O O 0 0 0 YZ 0 0 0 D O 1 00- 1 I 1 0 1 -0 0 O 1 1 0 Σ ml 2,3,6,7, 10, 11, 14, 15)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Please create schematic and truth table so that it matches given truth table. Will upvote !
### Karnaugh Map Analysis and Digital Circuit Illustration #### Karnaugh Maps (K-Maps) **K-Map for Output X:** - **Variables:** X, Y̅, Z - **Configuration:** - Rows labeled as DE: 00, 01, 11, 10 - Columns labeled as BA: 00, 01, 11, 10 - Map Values: Contains '1's at (01, 01), (01, 11), (11, 00). - **Expression Derived:** \( X = \overline{C} \cdot \overline{A} \) **K-Map for Output Y:** - **Variables:** X, Y̅, Z - **Configuration:** - Rows labeled as DC: 00, 01, 11, 10 - Columns labeled as BA: 00, 01, 11, 10 - Map Values: Contains '1's in the entire last row. - **Expression Derived:** \( Y = DC \) **K-Map for Output Z:** - **Variables:** X, Y̅, Z - **Configuration:** - Rows labeled as: 00, 01, 11, 10 - Columns labeled as: 00, 01, 11, 10 - Map Values: Contains '1's in the last two columns. - **Expression Derived:** \( Z = B \) #### Circuit Diagram - **Inputs:** A, B, C, D - Components: - Two NOT gates (inverters) used for inputs A and C. - Two OR gates are used to combine inputs for outputs X and Y. - Direct connections represent input variables passing through. - **Outputs:** - \( X \): Produced by combining A̅ with an OR gate. - \( Y \): Produced by combining results via an OR gate. - \( Z \): Directly connected from input B. **Note:** - The circuit is marked as incorrect for the provided truth table. The arrow points out a section where the circuit connections do not accurately represent the intended logic expressions derived from the Karnaugh Maps.
Transcribed Image Text:### Karnaugh Map Analysis and Digital Circuit Illustration #### Karnaugh Maps (K-Maps) **K-Map for Output X:** - **Variables:** X, Y̅, Z - **Configuration:** - Rows labeled as DE: 00, 01, 11, 10 - Columns labeled as BA: 00, 01, 11, 10 - Map Values: Contains '1's at (01, 01), (01, 11), (11, 00). - **Expression Derived:** \( X = \overline{C} \cdot \overline{A} \) **K-Map for Output Y:** - **Variables:** X, Y̅, Z - **Configuration:** - Rows labeled as DC: 00, 01, 11, 10 - Columns labeled as BA: 00, 01, 11, 10 - Map Values: Contains '1's in the entire last row. - **Expression Derived:** \( Y = DC \) **K-Map for Output Z:** - **Variables:** X, Y̅, Z - **Configuration:** - Rows labeled as: 00, 01, 11, 10 - Columns labeled as: 00, 01, 11, 10 - Map Values: Contains '1's in the last two columns. - **Expression Derived:** \( Z = B \) #### Circuit Diagram - **Inputs:** A, B, C, D - Components: - Two NOT gates (inverters) used for inputs A and C. - Two OR gates are used to combine inputs for outputs X and Y. - Direct connections represent input variables passing through. - **Outputs:** - \( X \): Produced by combining A̅ with an OR gate. - \( Y \): Produced by combining results via an OR gate. - \( Z \): Directly connected from input B. **Note:** - The circuit is marked as incorrect for the provided truth table. The arrow points out a section where the circuit connections do not accurately represent the intended logic expressions derived from the Karnaugh Maps.
## Explanation of Text and Diagram The image contains handwritten notes explaining logic operations using a truth table and set expressions. Here's a transcription and explanation suitable for an educational context. ### Transcription --- **Logic Operations Explanation** 1. **B = 0 => \(\bar{Z} = 0\)** 2. **X, Y, Z** have to be 'on' only during off-peak rates. Let \(\Theta = \Theta_{\text{off}}\). i.e., when \(A = 0 => \eta\) is not efficient usage of power. So, all \(\varphi_p\) will be switched off, like refrigerator, heater, battery. \( => A = 0\) ⇒ all \(\varphi_p = 0\). 3. Using all these statements, the truth table is filled. 4. But \(A = 0\), only heater can be on since it is essential and battery can be on. \(A = 0\) ⇒ only Refrigerator \(\varphi_p = 0\) 5. **X => 0** --- ### Truth Table | D | C | B | A | X | Y | Z | |---|---|---|---|---|---|---| | 1 | 0 | 0 | 0 | 0 | 1 | 0 | | 1 | 0 | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 1 | 0 | | 0 | 0 | 0 | 1 | 0 | 0 | 1 | | 1 | 1 | 0 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 0 | 0 | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 0 | 1 | 0 | 1 | 1 | | 1 | 0 |
Transcribed Image Text:## Explanation of Text and Diagram The image contains handwritten notes explaining logic operations using a truth table and set expressions. Here's a transcription and explanation suitable for an educational context. ### Transcription --- **Logic Operations Explanation** 1. **B = 0 => \(\bar{Z} = 0\)** 2. **X, Y, Z** have to be 'on' only during off-peak rates. Let \(\Theta = \Theta_{\text{off}}\). i.e., when \(A = 0 => \eta\) is not efficient usage of power. So, all \(\varphi_p\) will be switched off, like refrigerator, heater, battery. \( => A = 0\) ⇒ all \(\varphi_p = 0\). 3. Using all these statements, the truth table is filled. 4. But \(A = 0\), only heater can be on since it is essential and battery can be on. \(A = 0\) ⇒ only Refrigerator \(\varphi_p = 0\) 5. **X => 0** --- ### Truth Table | D | C | B | A | X | Y | Z | |---|---|---|---|---|---|---| | 1 | 0 | 0 | 0 | 0 | 1 | 0 | | 1 | 0 | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 1 | 0 | | 0 | 0 | 0 | 1 | 0 | 0 | 1 | | 1 | 1 | 0 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 0 | 0 | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 0 | 1 | 0 | 1 | 1 | | 1 | 0 |
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