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Linear circuit and superposition
- 1. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering EE211 Circuit 1 Week 6 Module 1 By Samson Cheung ([email protected]) 1
- 2. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Circuit theorems Linear circuits Superposition 2
- 3. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Linear circuit is an important class of circuits – all circuits introduced in this course are linear circuits. Linear circuits are characterized by additivity and homogenity. Linear circuits imply the principle of superposition, which simplifies circuit analysis by separating the contributions from different independent sources 3
- 4. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering A large Simplify complex circuits circuit analysis Circuit Theorems ‧Linear Systems ‧ Superposition ‧Source Transformation ‧ Thevenin’s Theorem ‧Norton’s Theorem ‧ Max. power transfer 4
- 5. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering A system takes an input and creates an output Human body is a system Input: Food Output: Any human activity US Government is a system Input: Tax Dollars Output: Infrastructure, National security, etc. 5
- 6. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Mathematical Notation: Input H(vs) = i Circuit as a system H Example: A speaker Input: all the independent voltage and current sources ◦ Example: input voltage vs from the amplifier (changes over time) Output: current/voltage across a load resistor ◦ Example: current i driving the diaphragm What about external power supply? ◦ Part of a dependent source whose value is based on the input ◦ It is part of the system 6
- 7. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering A linear circuit (system) is a special system whose output is linearly related (or directly proportional) to its input. Output i H Input vs Linear circuits are very useful in modeling devices and very well understood 7
- 8. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Additivity: ◦ the output to a sum of inputs is Output i the sum of the corresponding outputs H(v1+v2) = H(v1) + H(v2) + + Input vs Homogeneity: ◦ the output to a scaled input is Output i the scaled output H(Kv1) = KH(v1) K Input vs K 8
- 9. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Additivity: ◦ New Input: Vs’ = V1+V2 ◦ New Output: Vo’ = H(Vs’) = (V1+V2)/2 = V1/2 + V2/2 = H(V1)+H(V2) System Description: Homogenity: Vo H (Vs ) ◦ New Input: 4 // 4 Vs Vs’ = KVs Vs ◦ New Output: 2 4 // 4 2 Vo’ = H(Vs’) = (KVs)/2 = KH(Vs) 9
- 10. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering io 2io Dependent sources are not input! They are controlled by signals internal to the system and are considered as part of the system. 10
- 11. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Ii Linear Io System H + Vi Vo - Vector input and output Vector addition Vector-scalar multiplication 11
- 12. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Vt +- System Description: KCL: 12
- 13. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Why is it a good idea? With only one source, the rest of the circuit can be simplified by combining all resistors. Circuit with voltage Circuit with voltage source Is turned off source Vs turned off 13
- 14. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering 1. Turn off all but one independent source. ◦ Voltage source = 0V (closed) ◦ Current source = 0A (open) 2. Find the voltage or current of interest (the output) using only that source. 3. Repeat steps 1&2 for each source. 4. Add the contributions of the individual sources to find the final answer. 14
- 15. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Find I0 12V 2k 4mA – + 2mA 1k 2k I0 15
- 16. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Step 1: Keep the 2mA current source Current source killed Voltage source killed 2k -2mA 2mA 1k 2k i1 parallel 16
- 17. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Step 2: Keep the 4mA current source Voltage source killed Current source killed 2k 4mA v 4mA 1k 2k i2 17
- 18. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Step 3: Keep the 12V voltage source Current source killed 12V Current source killed 2k – + 1k 2k i3 i3 18
- 19. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Step 4: Add all three parts together 12V 2k 4mA – + 2mA 1k 2k I0 19
- 20. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering Find v. 20
- 21. E l ec t r i c a l E l e c t r iCao m C o u ptuee r & c l & pm t r Department of Engineering v0 Part 1: Keep the first 5V source Ohm’s Law: i0 = v0 -5 + i0 + i0 + 2(i0-i1) = 0 i1 2(-i0+i1) +2i1 +2v0 = 0 i0 2v0 v0 = 5/4 V v1 Part 2: Keep the 2nd 5V source Ohm’s Law: i3 = v1 5 + i3 + i3 + 2(i3-i4) = 0 -5 + 2(-i3+i4) +2i4 +2v1= 0 i3 v1 =-5/8 V i4 2v1 Part 1+ Part 2: v= v0+v1 = 5/8 V 21