Linear circuit and superposition

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

Department of

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



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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

Technology

Linear circuit and superposition