linear circuit and superposition
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E l e c t r i c a l & C o m p u t e r E l e c t r i c a l & C o m p u t e r Engineering
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Linear Circuit and Superposition
EE211 Circuit 1 Week 6 Module 1By Samson Cheung ([email protected])
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Circuit theorems
Linear circuits
Superposition
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Overview
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Big picture
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
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Circuit Theorems
A largecomplex circuits
A largecomplex circuits
‧Linear Systems ‧ Superposition‧Source Transformation ‧ Thevenin’s Theorem
‧Norton’s Theorem ‧ Max. power transfer
‧Linear Systems ‧ Superposition‧Source Transformation ‧ Thevenin’s Theorem
‧Norton’s Theorem ‧ Max. power transfer
Simplifycircuit analysis
Simplifycircuit analysis
Circuit TheoremsCircuit Theorems
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Concept of a system
A system takes an input and creates an output
Human body is a systemInput: Food
Output: Any human activity US Government is a systemInput: Tax Dollars
Output: Infrastructure, National security, etc.
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Circuit as a system
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
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Input
Mathematical Notation:H(vs) = i
Circuit as a system H
Example: A speaker
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Linear Circuit
A linear circuit (system) is a special system whose output is linearly related (or directly proportional) to its input.
Linear circuits are very useful in modeling devices and very well understood
Input vs
Output i
H
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Key Properties of Linear circuits Additivity:
◦ the output to a sum of inputs is the sum of the corresponding outputs
H(v1+v2) = H(v1) + H(v2)
Homogeneity: ◦ the output to a scaled input is
the scaled output
H(Kv1) = KH(v1)
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Input vs
Output i
Input vs
Output i
+
+
K
K
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Example 1
24//42
4//4
)(
ss
so
VV
VHV
Additivity: ◦ New Input:
Vs’ = V1+V2
◦ New Output: Vo’ = H(Vs’)
= (V1+V2)/2
= V1/2 + V2/2
= H(V1)+H(V2) Homogenity:
◦ New Input: Vs’ = KVs
◦ New Output: Vo’ = H(Vs’)
= (KVs)/2
= KH(Vs) 9
System Description:
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Example 2: w/ dependent source
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2io
𝑉 𝑜−𝑉 𝑠
2𝐾
𝑉 𝑜
4 𝐾
io
System Description:
Input =, Output,
KCL @ top node:
It is a linear circuit as the output is just a scaled version of the input.
Dependent sources are not input!
They are controlled by signals internal to the system and are
considered as part of the system.
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A linear circuit (system) can have multiple inputs and multiple outputs.
Additivity: =
Homogenity: =
More than one input & output
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Linear System
H
Ii
Vi
Io
+Vo
-
=
Vector-scalar multiplication
Vector addition
Vector input and output
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Example Additivity:
= +
=+= + Homogenity:
= +
=
Input = , Output,
Vt+ -
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System Description:
𝑉 𝑜−𝑉 𝑠
2𝐾
𝑉 𝑜
4 𝐾
𝑉 𝑜−𝑉 𝑡
4 𝐾
KCL:
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The Principle of Superposition
A simple idea from the last example:
Principle of SuperpositionFind the output of a linear circuit based on the algebraic sum of the output due to each independent source.
Circuit with voltage source Is turned off
Circuit with voltage source Vs turned off
Why is it a good idea? With only one source, the rest of the circuit can be simplified by combining all resistors.
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Procedure
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.
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Example 1
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2kW1kW
2kW12V
I0
2mA
4mA
– +
Find I0
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Example 1
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2kW1kW
2kW
i1
2mA
Step 1: Keep the 2mA current source
Current source killed
Voltage source killed
-2mA
parallel
𝑖1=−2𝑚𝐴 ∙1𝑘Ω∥2𝑘Ω
1𝑘Ω=−
43𝑚𝐴
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2kW1kW
2kW
i2
4mA
Step 2: Keep the 4mA current source
Voltage source killedCurrent source
killed
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Example 1
4mA
v
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2kW1kW
2kW12V
i3
– +
Step 3: Keep the 12V voltage source
Current source killed
Current source killed
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Example 1
i3
𝑖3=−12𝑉
1𝑘Ω+2𝑘Ω=−4𝑚𝐴
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2kW1kW
2kW12V
I0
2mA
4mA
– +
Step 4: Add all three parts together
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Example 1
𝐼 0=𝑖1+𝑖2+𝑖3=−43+0−4𝑚𝐴=−5
13𝑚𝐴
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Find v.
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Example 2
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v1
2v1
Part 1: Keep the first 5V sourceOhm’s Law: i0 = v0
-5 + i0 + i0 + 2(i0-i1) = 0 2(-i0+i1) +2i1 +2v0 = 0 v0 = 5/4 V
Part 2: Keep the 2nd 5V sourceOhm’s Law: i3 = v1
5 + i3 + i3 + 2(i3-i4) = 0 -5 + 2(-i3+i4) +2i4 +2v1= 0 v1 =-5/8 V
Part 1+ Part 2:v= v0+v1 = 5/8 V
v0
2v0i0 i1
i3i4
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Example 2