AEE_01

Advanced Electrical Engineering

Michael E.Auer 02.05.2012 AEE01

Advanced Electrical Engineering Michael E. Auer

Basic Concepts



Chapter Content

• Introduction

• Complex Numbers and Phasors

• Circuit Theory Review

• Methods of Network Analysis

• Locus Diagrams

• Circuit Element Variations



Chapter Content

• Introduction • Complex Numbers and Phasors



• Locus Diagrams




This course: Fundamentals of Electro-Magnetism !!!



Fundamental Forces of Nature



The Electro-magnetic Spectrum

ac power network microprocessor



Example Lighting

Incandescence is the emission of light from a hot object due to its temperature.

Fluoresce means to emit radiation in consequence to incident radiation of a shorter wavelength

When a voltage is applied in a forward-biased direction across an LED diode, current flows through the junction and some of the streaming electrons are captured by positive charges (holes). Associated with each electron-hole recombining act is the release of energy in the form of a photon.

Scientific progress, but same fundamental laws

Light bulb Fluorescent lamp LED diode lamp



Chapter Content

• Introduction

• Complex Numbers and Phasors • Circuit Theory Review


• Locus Diagrams




Complex Numbers

Is useful to represent sinusoids as complex numbers.

jyxz +=θθ jezzz =∠=

1−=j

Rectangular coordinates

Polar coordinates

θθθ sincos je j ±=±

Relations based on Euler’s Identity

( )yzxz

==

)Im(Re



Relations for Complex Numbers



Phasor Domain

1. The phasor-analysis technique transforms equations from the time domain to the phasor domain. 2. Integro-differential equations get converted into linear equations with no sinusoidal functions. 3. After solving for the desired variable -such as a particular voltage or current - in the phasor domain, conversion back to the time domain provides the same solution that would have been obtained had the original integro-differential equations been solved entirely in the time domain.



Time Domain vs. Phasor Domain (1)

Phasor counterpart of

Rotating phasor Stationary phasor

φ∠ 0V

2 0πV ∠



It is much easier to deal with exponentials in the phasor domain than sinusoidal relations in the time domain. Just need to track magnitude/phase, knowing that everything is at frequency ω.

Time Domain vs. Phasor Domain (2)



Time Domain Frequency Domain

( )φωυ +== tRIiR cosm

φ∠= mRIV

Current through resistor

( )φω += tIi cosm

Time domain

Phasor Domain

Phasor Relation for Resistors



Time Domain

Time domain

Phasor Domain

Phasor Relation for Inductors



Time domain

Phasor Domain

Phasor Relation for Capacitors



Circuit Analysis in the Phasor Domain

Differential Equations

Solution of the Differential Equations

Solution

Time Domain Phasor Domain

Algebraic Equations complex !

Solution of the Algebraic Equations

Solution

Transformation

Retransformation



Basic Approach

1. Transform the circuit to the phasor or frequency domain. 2. Solve the problem using circuit techniques (nodal analysis, mesh

analysis, superposition, etc.). 3. Transform the resulting phasor to the time domain.

Time to Freq Solve

Equations in Freq Domain

Freq to Time



Phasor Analysis: General Procedure



Example: RL Circuit (1)



Example: RL Circuit (2)



Transformation Table

x means a travelling wave!



Traveling Waves in the Phasor Domain



Chapter Content

• Introduction


• Circuit Theory Review • Methods of Network Analysis

• Locus Diagrams




v1 = iiR1

v2 = iiR2

and

vi = v1 + v2 = ii(R1 + R2)

ii =vi

R1 + R2

v1 = viR1

R1 + R2

v2 = viR2

R1 + R2

Applying KVL to the loop,

Combining these yields the basic voltage division formula:

and

Voltage Division



ii = i1 + i2

and

i1 = iiR2

R1 + R2

Combining and solving for vi,

Combining these yields the basic current division formula:

where

i2 =vi

R2

i1 =vi

R1and

vi = ii1

1R1

+1R2

= iiR1R2

R1 + R2

= ii R1 || R2( )

i2 = iiR1

R1 + R2

Current Division



Thévenin

Norton

Thevenin and Norton Equivalent Circuits



βi1 =vo − vi

R1

+vo

RS

= G1 vo − vi( )+ GSvo

i1 = G1 vo − vi( )

G1 β +1( )vi = G1 β +1( )+ GS[ ]vo

vo =G1 β +1( )

G1 β +1( )+ GS

vi ×R1RS

R1RS

=β +1( )RS

β +1( )RS + R1

vi

Applying KCL at the output node,

Current i1 can be written as:

Combining the previous equations

Thevenin Equivalent Voltage (1)

open circuit



vo =β +1( )RS

β +1( )RS + R1

vi =50 +1( )1 kΩ

50 +1( )1 kΩ +1 kΩvi = 0.718vi

Using the given component values:

and

v th = 0.718vi

Thevenin Equivalent Voltage (2)



ix = −i1 − βi1 + GSvx

= G1vx + βG1vx + GSvx

= G1 β +1( )+ GS[ ]vx

Rth =vxix

=1

G1 β +1( )+ GS= RS

R1

β +1

Applying KCL,

Rth = RSR1

β +1= 1 kΩ

20 kΩ50 +1

= 1 kΩ 392 Ω = 282 Ω

Thevenin Equivalent Resistance



in = i1 + βi1= G1vi + βG1vi

= G1 β +1( )vi

=vi β +1( )

R1

Applying KCL,

in =50 +120 kΩ

vi =vi

392 Ω= (2.55 mS)vi

Short circuit at the output causes zero current to flow through RS. Rth is equal to Rth found earlier.

short circuit

Norton Equivalent Circuit



Check of Results: Note that vth = inRth and this can be used to check the calculations: inRth=(2.55 mS)vi(282 Ω) = 0.719vi, accurate within round-off error. While the two circuits are identical in terms of voltages and currents at the output terminals, there is one difference between the two circuits. With no load connected, the Norton circuit still dissipates power!

Final Thevenin and Norton Equivalent Circuits



Find Thevenin and Norton Equivalent Circuits!

nthth ivR ,,



• An equivalent circuit is one whose v-i characteristics are identical with the original circuit.

• It is the process of replacing a voltage source vS in series with a resistor R

by a current source iS in parallel with a resistor R, or vice versa.

Source Transformation (1)



(a) Independent source transform

(b) Dependent source transform

• The arrow of the current source is directed toward the positive terminal of the voltage source.

• The source transformation is not possible when R = 0 for voltage source and R = ∞ for current source.

+ +

+ +

- -

- -

Source Transformation (2)



Find io in the circuit shown below using source transformation.

Source Transformation Example



It states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to EACH independent source acting alone.

The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately.

Superposition Theorem (1)



We consider the effects of 8A and 20V one by one, then add the two effects together for final vo.




Steps to apply superposition principle 1. Turn off all independent sources except one source. Find

the output (voltage or current) due to that active source using nodal or mesh analysis.

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to the independent sources.




Two things have to be keep in mind: 1. When we say turn off all other independent sources:

Independent voltage sources are replaced by 0 V (short circuit) and

Independent current sources are replaced by 0 A (open circuit).

2. Dependent sources are left intact because they are controlled by circuit variables.




Example 2 Use the superposition theorem to find v in the circuit shown below.

3A is discarded by open-circuit

6V is discarded by short-circuit

Superposition Theorem Example 1



Use superposition to find vx in the circuit..

2A is discarded by open-circuit

20 Ω v1

4 Ω 10 V

+ −

(a)

0.1v1 4 Ω 2 A

(b)

20 Ω

0.1v2

v2

10V is discarded by open-circuit Dependant source

keep unchanged !!!

Superposition Theorem Example 2



Chapter Content

• Introduction



• Methods of Network Analysis • Locus Diagrams




Things we need to know in solving any resistive circuit with current and voltage sources only:

Number of equations

• Ohm’s Law b

• Kirchhoff’s Current Laws (KCL) n-1

• Kirchhoff’s Voltage Laws (KVL) b – (n-1)

Introduction

Number of branch currents and branch voltages = 2b (variables)

Problem: Number of equations!

mesh = independend loop



Mesh Analysis (1) 1. Mesh analysis provides a general procedure for analyzing

circuits using mesh currents as the circuit variables.

2. Mesh analysis applies KVL to find unknown currents.

3. A mesh is a loop which does not contain any other loops within it (independent loop).



Mesh Analysis (2)



Mesh Analysis (3) Example – circuit with independent voltage sources

Note:

i1 and i2 are mesh current (imaginative, not measurable directly)

I1, I2 and I3 are branch current (real, measurable directly)

I1 = i1; I2 = i2; I3 = i1 - i2

Equations:

R1∙i1 + (i1 – i2) ∙ R3 = V1 R2 ∙ i2 + R3 ∙(i2 – i1) = -V2

reordered:

(R1+ R3) ∙ i1 - i2 ∙ R3 = V1 - R3 ∙ i1 + (R2 + R3)∙i2 = -V2



Mesh Analysis (4)

Formalization: Network equations by inspection.

−

=

⋅

+−

−+

2

1

2

1

323

331

)()(

VV

ii

RRRRRR

General rules: 1. Main diagonal: ring resistance of mesh n 2. Other elements: connection resistance between meshes n and m

• Sign depends on direction of mesh currents!

Impedance matrix Mesh currents

Excitation



Mesh Analysis (5)

Example: By inspection, write the mesh-current equations in matrix form for the circuit below.



Mesh Analysis Special Cases

dependent source

ideal voltage source



Mesh Analysis



Nodal Analysis (1) It provides a general procedure for analyzing circuits using node voltages as the circuit variables.

Example

3



Nodal Analysis (2)



Nodal Analysis (3)

v1 v2

Example Apply KCL at node 1 and 2

G1

G3

G2

G1∙v1 + (v1 – v2) ∙ G3 = 1A G2 ∙ v2 + G3 ∙(v2 – v1) = - 4A

reordered:

(G1+ G3) ∙ v1 - v2 ∙ G3 = 1A - G3 ∙ v1 + (G2 + G3)∙v2 = - 4A



Nodal Analysis (4)

Formalization: Network equations by inspection.

−

=

⋅

+−

−+A2

A1)(

)(

2

1

323

331

vv

GGGGGG

General rules: 1. Main diagonal: sum of connected admittances at node n 2. Other elements: connection admittances between nodes n and m

• Sign: negative!

Admittance matrix Node voltages

Excitation



Example: By inspection, write the node-voltage equations in matrix form for the circuit below.

Nodal Analysis (5)



Nodal Analysis Special Case

dependent source



AC Network Nodal Analysis



Chapter Content

• Introduction




• Locus Diagrams • Circuit Element Variations



Phasor Inversion Example



R-C-Circuit Locus



1. The inversion of a straight line through the origin is again a line through the origin.

2. The inversion of a straight line not through the origin is a circle through the origin and vice versa.

3. The inversion of a circle that does not pass through the origin is again a circle that does not pass through the zero point

Locus Inversion Theorems



through the origin not through the origin

Inversion of a Straight Line



Inversion of a Circle not Through the Origin



TUU

a

e g( )j

j mit ω

ωω

= =+

=1

1 ΩΩ

Locus of a Low-Pass Transfer Function



Locus of a Parallel-Series Circuit (1)



Locus of a Parallel-Series Circuit (2)



Equivalent Networks



Dual Networks (1)

In other words, both circuits are described by the same pair of equations:



Dual Networks (2)



Dual Networks (3)

Rules for obtaining the dual of a planar circuit, regardless of wether or not it is a series-parallel network: Rule1:

Insight of each mesh, including the infinite region surrounding the circuit, place a node. Rule2:

Suppose two of this nodes, for example a and b, are in adjacent meshes. Then there is at least one element in the boundary common to these two meshes. Place the dual of each common element between nodes a and b.



00 und '

ZviiZv =′⋅=

′ = ′ = ′ = ′ ⋅ ′ ′ = ′′

′ = ′′

′ = ′ ′ = ′

∑ ∑u i i G u i Cut

u Lit

i i u uq q

µµ

νν

0 0 dd

dd

qq iivvtvCi

tiLviRvvi

==

==⋅=== ∑∑

dd

dd 0 0

νν

µµ Given Network

Dual Network

qqq

q iZvZv

iZCLZLC

ZRG ⋅==′⋅=′=′=′ 0

0

202

020

'

i-v-Duality

Duality relations for the basic network elements:

For example Z0 = 1Ω

I-V-Relations in Dual Nezworks



Chapter Content

• Introduction




• Locus Diagrams




• All electronic components have manufacturing tolerances. o Resistors can be purchased with ± 10%, ± 5%, and

± 1% tolerance. (IC resistors are often ± 10% and more.) o Capacitors can have asymmetrical tolerances such as +20%/-50%. o Power supply voltages typically vary from 1% to 10%.

• Device parameters will also vary with temperature and age. • Circuits must be designed to accommodate these variations. • Mainly Worst-case Analysis and Monte Carlo Analysis

(statistical) are used to examine the effects of component parameter variations.

Circuit Element Variations



• For symmetrical parameter variations

Pnom(1 - ε) ≤ P ≤ Pnom(1 + ε) • For example, a 10 kΩ resistor with ±5% percent tolerance could take on

the following range of values:

10kΩ(1 - 0.05) ≤ R ≤ 10kΩ(1 + 0.05) 9500 Ω ≤ R ≤ 10500 Ω

Tolerance Modeling



• Most circuit parameters vary from less than ± 1 % to greater than ± 50%.

• As a consequence, more than three significant digits is meaningless.

• Results should be represented with three significant digits: 2.03 mA, 5.72 V, 0.0436 µA, and so on.

Numeric Precision



• Worst-case analysis – Parameters are manipulated to produce the worst-case min and max

values of desired quantities. – This can lead to over design since the worst-case combination of

parameters is rare. – It may be less expensive to discard a rare failure than to design for 100%

yield.

• Monte-Carlo analysis – Parameters are randomly varied to generate a set of statistics for desired

outputs. – The design can be optimized so that failures due to parameter variation

are less frequent than failures due to other mechanisms. – In this way, the design difficulty is better managed than a worst-case

approach.

Circuit Analysis with Tolerances



Problem: Find the nominal and worst-case values for output voltage and source current.

Solution: • Known Information and Given

Data: Circuit topology and values in figure.

• Unknowns:

Nominal voltage solution:

VOnom = VI

nom R1nom

R1nom + R2

nom

=15V18kΩ

18kΩ + 36kΩ= 5V

VOnom, VO

min , VOmax , II

nom , IImin , II

max

Worst Case Analysis Example



• Parameters are varied randomly and output statistics are gathered. • We use programs like MATLAB, Mathcad, SPICE, or a spreadsheet to

complete a statistically significant set of calculations. • For example, with Excel, a resistor with a nominal value Rnom and

tolerance ε can be expressed as:

R = Rnom(1+ 2ε(RAND() − 0.5))

The RAND() function returns random numbers uniformly distributed between 0 and 1.

Monte Carlo Analysis



Histogram of output voltage from 1000 case Monte Carlo simulation.

VO (V)

Average 4.96

Nominal 5.00

Standard Deviation 0.30

Maximum 5.70

W/C Maximum 5.87

Minimum 4.37

W/C Minimum 4.20

Monte Carlo Analysis Results

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