Lectures IntroElectronics

Matt DOBBS Introduction to Signal Processing and Electronics for Scientists

.

Introduction to Signal Processing and Electronics forScientists

M. Dobbs

<[email protected]>

Department of Physics, McGill University

April 11, 2014

1 April 11, 2014


2 April 11, 2014

§ 1

Course Summary

The course will cover:

• DC circuits and networks

• Linear circuit elements: R, L, and C

• Charging C’s and L’s; energy storage

• Sinusoidal voltages and currents; phasors, and complex algebra techniques

• Filters: high-pass, low-pass, bandpass

• Resonant circuits

• Network theorems

• Fourier analysis of waveforms; frequency spectra

• Fourier transforms; the bandwidth theorem

• Semiconductor diodes

3


• Nonlinear circuit elements: diode circuits and rectification

• Laplace transforms

• Waves and pulses on transmission lines

In order to understand these concepts, we will frequently make use of the Audio Spectrum (40-20,000

Hz), where our ears can help us interpret the circuits and the output of our measurement devices.

There are five labs:

• The oscilloscope

• Wave shaping with RC, RL circuits

• Sinusoidal response of RC, RL circuits

• Resonance

• A mystery lab: use measurements to find the unknown elements in the box.

4 April 11, 2014

Contents

1 Course Summary 3

2 Introduction to Electronics in Physics and Elsewhere 11

2.1 Circuits come in all varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Static Circuits and Networks 17

3.1 Digital vs. Analog Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2 AC vs. DC circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3 Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4 Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.5 Current Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.5.1 Batteries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.6 Ohm’s Law & Review of Resistors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.6.1 Current Through a Resistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.6.2 Voltage Divider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.6.3 Resistor Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.6.4 Resistor Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5


3.6.5 Resistor Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4 Capacitors and Inductors 33

4.1 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.1.1 Capacitor Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.1.2 Charging a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.1.3 RC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.1.4 RC Filter Example from Chalk Board . . . . . . . . . . . . . . . . . . . . . . . . 39

4.1.5 Capacitor Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.2 Transformers and Mutual Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.2.1 Lenz’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.2.2 Transformers in AC circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.2.3 Transformer Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2.4 Example Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.3 Inductors and Self Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.3.1 Energy Storage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.3.2 Inductor Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.3.3 Inductor Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.3.4 Inductor Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5 Waterflow Analogy 59

6 Linear Network Theorems I 63

6.1 Superposition and Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6.2 Kirchoff’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

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6.2.1 Example: Method 1 total currents . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.2.2 Example: Method 2 circulating currents . . . . . . . . . . . . . . . . . . . . . . 66

7 Stepped Voltages with low and high-pass filters 69

8 AC Circuits, Sinusoidal Currents and Voltages, and Complex Representations 73

8.1 RLC AC-Circuit: Sinusoidal Representation . . . . . . . . . . . . . . . . . . . . . . . . 73

8.2 Representation with Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

8.3 RLC AC-Circuit: Complex Represntation . . . . . . . . . . . . . . . . . . . . . . . . . . 78

8.3.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

9 Linear Network Theorems II 81

9.1 Thevenin’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

9.1.1 DC Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

9.2 Norton’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

9.2.1 Conversion from Thevenin’s Theorem to Norton’s Theorem . . . . . . . . . . . . 85

10 Fourier Analysis 87

10.1 Time Domain Waveform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

10.1.1 Sine Wave timestream . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

10.2 Fourier Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

10.2.1 Application to Audio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

10.2.2 the Matlab function powerSpectrum.m . . . . . . . . . . . . . . . . . . . . . . 93

10.2.3 Example: Signal Buried in Noise . . . . . . . . . . . . . . . . . . . . . . . . . . 95

10.3 Example: Spatial Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

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10.4 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

10.4.1 Fourier Series Example: Square Wave (again) . . . . . . . . . . . . . . . . . . . . 99

10.5 Rectangular Pulse Train - Bandwidth Theorem . . . . . . . . . . . . . . . . . . . . . . . 103

10.5.1 Single Tone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

10.5.2 Three Tones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

10.5.3 Five Tones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

10.5.4 N →∞ Tones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

10.5.5 Bandwidth Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

10.6 System Frequency Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

10.6.1 Transfer Function Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

10.6.2 Bode Plot and Decibels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

11 Diodes 123

11.1 Diode Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

11.1.1 Half-wave Rectifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

11.1.2 Full-wave Rectifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

11.1.3 Diode Clipping Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

12 The S-plane and Laplace Transforms 143

12.1 Exponential solution to linear circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

12.1.1 S-plane Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

12.2 The S-plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

12.3 Using Poles and Zeros to Characterize Circuits . . . . . . . . . . . . . . . . . . . . . . . 152

12.3.1 Single Pole Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

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12.3.2 Multiple Pole Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

12.3.3 S-plane Zeros and Poles Calculation Examples . . . . . . . . . . . . . . . . . . . 154

12.4 Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

12.4.1 Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

12.4.2 Step Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

12.5 Laplace Transform Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

12.5.1 Example: RC circuit with step function input . . . . . . . . . . . . . . . . . . . 167

12.5.2 Example: RC circuit with a linearly rising input . . . . . . . . . . . . . . . . . . 168

13 Transmission Lines 171

13.1 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

13.2 Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

13.2.1 Characteristic Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

13.2.2 Reflections and Line Termination . . . . . . . . . . . . . . . . . . . . . . . . . . 179

13.2.3 Mis-matched Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . 180

13.2.4 Unterminated Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

13.2.5 Short-circuited Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

13.2.6 Reflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

13.3 Two-port S-parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

13.4 Dispersion and Lossy Cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

14 Op-amps 195

A Introduction to Circuit Simulation 205

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B Equations 213

C Waterflow Analogy 217

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§ 2

Introduction to Electronics in Physics and Elsewhere

Left: (source: NASA) The most cited scientific results to date come from WMAP, which measured the con-

situents of the universe and curvature of space time using observatons of the Cosmic Microwave Back-

ground radiation.

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Right: (source: M. Dobbs) The electronic readout system for the South Pole Telescope, built at McGill.

(source: http://en.wikipedia.org/wiki/Phoenix_lander) The Phoenix lander arrived on Mars in 2008 and included Cana-

dian instruments. It provided direct confirmation of the existence of water on mars.

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http://en.wikipedia.org/wiki/Phoenix_lander


Left: The Edge, guitarist for U2, made the sound from this VOX amp famous (listen to the Joshua Tree).

The characteristic sound comes from the inherent distortion (or non-linearity) of the circuits. For pure

sound, or for a good scientific instrument, this sort of non-linearity is generally not desireable.

Right: A very basic audio amplifier circuit, using an op-amp.

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2.1 Circuits come in all varieties

(source: http://gasstationwithoutpumps.wordpress.com/2012/06/24/temperature-lab-part-3-voltage-divider/) A simple low-pass RC fil-

ter implemented with leaded components on a breadboard. Students in this course will build many such

circuits.

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http://gasstationwithoutpumps.wordpress.com/2012/06/24/temperature-lab-part-3-voltage-divider/


(source: M. Dobbs) Circuit board developed at McGill for the CHIME telescope in 2014. The circuit board

is manufactured in Ontario and assembled by robots in Montreal with components that are sourced

internationally.

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(source: http://news.softpedia.com/news/Samsung-Proudly-Show-Its-32-nm-Manufacturing-Process-267745.shtml) Silicon wafer, manufac-

tured in Korea by Samsung, containing the A5 ARM processor for the Apple IPad2 tablet. This uses a

32 nm feature size.

16 April 11, 2014

http://news.softpedia.com/news/Samsung-Proudly-Show-Its-32-nm-Manufacturing-Process-267745.shtml

§ 3

Static Circuits and Networks

3.1 Digital vs. Analog Signals

CD vs. magnetic tape.

3.2 AC vs. DC circuits

• static circuits = voltages and currents are constant in time. Usually refered to as Direct Current

(DC), to distinguish them from circuits with time-varying currents, usually in the form of sine waves,

called alternating currents. A standard battery, like an AA cell, provides a static voltage of 1.5 V.

The outlets in your home provide an alternating voltage – usually a 120 V sine wave with 60 periods

per second (240V, 55 periods per second in Europe).

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• The SI unit Hertz refers to periods per second, [s−1], and is named for Heinrich Hertz who showed

electromagnetic waves exist.

(source: http://en.wikipedia.org/wiki/Alternating_current)

Often, we use the term DC to refer to a slowly varying signal, even though it is not, strictly speaking,

static. We also frequently say DC even when we are talking about voltages.

How do we measure the amplitude of a repetitive signal, like a sine wave? Consider the sine wave

Apsin(2πft)

• the peak amplitude is Ap

• the peak-to-peak amplitude (total excursion of the waveform) is 2 Ap.

• the root-mean-squared amplitude is√⟨A2p sin2 (2πft)

⟩= Ap/

√2

A detailed derivation of the RMS√

2 factor follows, using the angular frequency ω = 2πf .

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http://en.wikipedia.org/wiki/Alternating_current


(source: http://en.wikipedia.org/wiki/Root_mean_square)

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http://en.wikipedia.org/wiki/Root_mean_square


3.3 Electromagnetic Spectrum

• electricity distribution in north america, 60 Hz

• amplitude modulated (AM) radio 140-300 kHz

• frequency modulated (FM) radio 90-110 MHz

• WiFi 1.4 GHz

(source: http://en.wikipedia.org/wiki/Em_spectrum)

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http://en.wikipedia.org/wiki/Em_spectrum


3.4 Linear Systems

What is a linear system, and why do we like them?

Linear systems obey the principles of superposition and scaling. For two inputs

x1(t), x2(t)

with outputs

y1(t) = H [x1(t)], y2(t) = H [x2(t)]

then a linear system must satisfy

ytotal(t) = αy1(t) + βy2(t) = H [αx1(t) + βx2(t)]

for scalar α, β.

• superposition: H [x1(t) + x2(t)] = H [x1(t)] + H [x2(t)]

• scaling: H [αx(t)] = αH [x(t)]

A high quality audio recording system and/or audio playback system (your “hi-fi” stereo at home)

satisfies this. Given the audio signal x1(t) from a guitar and x2(t) from a human voice, the recording

system applies gains α, β to each, and the sound you hear when you playback the music is a faithful

representation of the sum of the two audio signals. In this case, the processing of the signal by the

recording and playback devices are represented by H , and ideally H = 1.

An important feature of a linear system is that information at two different frequencies do not mix.

This means that, for a linear piano, when the A-key and E-key are played together to strike a chord, the

only notes you hear are A (440 Hz) and E (659 Hz). They do not mix to form another note such as E#.

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The distortion effect, commonly used for electric guitars, is non-linear, and artificially creates frequency

mixing.

Linear systems are easy to understand, and easy to calculate the output of. THey are also very easy

to simulate in a computer.

3.5 Current Flow

The flow of electrons through a wire driven by a electric field potential energy difference is called a current,

defined as the rate of charge past a point in a circuit.

I [ampere] =Q[coulomb]

time[seconds]

One electron q is 1.602× 10−19 coulombs.

The voltage is the electric potential energy divided by the charge.

V =∆EElectric

q

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(source: http://cnx.org/content/m42324/latest/?collection=col11406/latest) The electron shown between two charged plates

has electric potential energy ∆EElectric = qV which will be transformed into kinetic energy (KE) when it

is ejected from between the plates.

Electrons will flow from the negative anode of a battery to its positive cathode. If positrons were

common, they would flow in the opposite direction.

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http://cnx.org/content/m42324/latest/?collection=col11406/latest


(source: http://en.wikipedia.org/wiki/Electric_current and http://www.physicsclassroom.com/class/circuits/u9l2c.cfm)

When we speak of current in a circuit, the arbitrary convention is that we are referring to the direction

of the flow of positive charges.

3.5.1 Batteries

Batteries harness chemical energy potential – the cathode is surrounded by postively charged ions, and the

anode by negatively charged ions. Electrons will travel through a circuit from the anode to the cathode,

releasing the stored chemical energy.

The amount of charge stored in a battery is usually specified in amp hours or mA hours. 1 amp hour

is 3600 Coulombs, or about 22.5 zeta-electrons (zeta = 1021).

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http://en.wikipedia.org/wiki/Electric_current

http://www.physicsclassroom.com/class/circuits/u9l2c.cfm


3.6 Ohm’s Law & Review of Resistors

The current through a conductor is proportional to the potential difference (or voltage) across it. We

define the resistance (actually, the 1/R) as the constant of proportionality and assign it the SI unit of

Ohms (Ω).

I [A] =V [V ]

R[Ω]

V = IR

R = V/I

A material is characterized by its resistivity

ρ[Ω ·m] = RArea

length

(source: http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity)

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http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity


which can be measured by placing an electric field E across the material, and measuring the magnitude

of the current density J (in amperes per square meter, A/m2) that flows.

ρ = E/J

.

The inverse of resistivity is conductivity

σ = 1/ρ

.

3.6.1 Current Through a Resistor

A 10 V battery sends current through a 100Ωresistor:

10 V

I

100 Ω

I = V/R = 10V/100Ω = 0.1A

This battery is doing work to push the current through the resistor. Where is this energy going? (1st

law of thermodynamics: energy is conserved in the universe)

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It is dissipated as heat:

P = IV = I × IR = I2 R

For this example, the 100Ωresistor dissipates P = I2 R = (0.1A)2 × 100Ω = 1 Watt.

3.6.2 Voltage Divider

10 V

I

80 ΩV1

20 ΩV2

I = V/R = 10V/(80 + 20)Ω = 0.1A

V1 = IR1 = 0.1A× 80Ω = 8V

V2 = IR2 = 0.1A× 20Ω = 2V

V = V1 + V2

3.6.3 Resistor Networks

Resistors in series add directly

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R1

R2

. . .

RN

Rtotal = R1 + R2 + . . . + RN

10

20

30

40

Rtotal = 10Ω + 20Ω + 30Ω + 40Ω = 100Ω

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Resistors in parallel add inversely

V R1 R2

V = I1R1 = I2R2

Itotal = I1 + I2 =V

R1+V

R2

V =Itotal

1R1

+ 1R2

= Itotal ×Rtotal

1

Rtotal=

1

R1+

1

R2

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R1 R2. . . RN

1

Rtotal=

1

R1+

1

R2+ . . . +

1

RN

100 100 25 25

1

Rtotal=

1

100+

1

100+

1

25+

1

25=

10

100→ Rtotal = 10Ω

3.6.4 Resistor Examples

• Traditional leaded carbon resistor

(source: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/rescarb.html )

• surface mount resistors used on printed circuit boards

30 April 11, 2014

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/rescarb.html


(source: www.koaspeer.com)

• incandescent lightbulb

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www.koaspeer.com


• baseboard heater

3.6.5 Resistor Characteristics

Resistors are typically characterized by their resistance, precision (including variations with temperature),

ability to dissipate heat, and size.

The resistivity of a material typically has a temperature coefficient measured in parts per million

(ppm) per degree celsius. A standard resistor has a temperature coefficient of 100 pmm/C. Typically, the

resistance is specified at 20 C.

This means that, if you allow your circuit to run hot or cold, the component values will vary. A

100Ωresistor with 100 ppm/C temperature coefficient that is operating outdoors on the South Pole Tele-

scope on a cold winter day (-80 C) will have a resistance

100Ω× (1 + 100× 10−6ppm/C× (−80C− 20C)) = 99Ω

. In scientific applications, this will be particularly important for filters.

32 April 11, 2014

§ 4

Capacitors and Inductors

(source: Thanks to Prof. John Crawford for the use of his lecture notes, some of which are incorporated in this chapter.)

There are four basic passive linear circuit elements in electronics

• resistor

• capacitor

• inductor

• mutual inductance (transformer)

We have already seen that resistances dissipate energy (i.e., they get hot!). However capacitors and

inductors store energy. Capacitors store it in the electric field between the capacitor plates, while induc-

tances store it in the magnetic field in the neighbourhood of the wire.

4.1 Capacitors

The simplest type of capacitor is constructed from two parallel conducting plates separated by either

space, or some dielectric insulator.

33


(source: http://en.wikipedia.org/wiki/File:Capacitor_schematic_with_dielectric.svg)

If a battery of voltage V is connected to the two plates, the charge stored by the capacitor q will

be proportional to the voltage i.e., q = CV , where the constant of proportionality C is called the

capacitance.

The unit of capacitance is the Farad: if an applied voltage of 1 volt causes 1 coulomb of charge to be

stored, the capacitance is 1 Farad.

For the parallel plate capacitor above,

C =εA

dwhereA is the plate area and d is the plate separation. The constant ε is called the dielectric permittivity

and

ε = κεε0

where ε0 is called the permittivity of space, and κε is a dimensionless constant that depends on the

34 April 11, 2014

http://en.wikipedia.org/wiki/File:Capacitor_schematic_with_dielectric.svg


properties of the particular dielectric. The permittivity of space is ε0 = 8.85× 10−12F/m.

Note that if we construct a parallel plate capacitor with plate area A = 1cm2, and separation d = 1mm,

the capacitance will be C ≈ 1pF . To increase it, we could fill the space between the plates with a dielectric

with a high κε value.

Dielectric examples:

• Mica, κε ∼ 6 is an excellent insulator that resist breakdown at high electric field strengths.

• Air κε1.00059 at STP and 900 kHz. (κε can vary with frequency - a major challenge for modern

technology is achieving a good value at very high frequencies.)

• Teflon κε ∼ 2.1.

4.1.1 Capacitor Examples

(source: http://en.wikipedia.org/wiki/Capacitor)

35 April 11, 2014

http://en.wikipedia.org/wiki/Capacitor


4.1.2 Charging a Capacitor

Suppose we charge (i.e., store energy in) a capacitor by connecting it directly to a constant current source:

I

I

C

Since q = CV , and I = dq/dt, we find

I = CdV

dtor V =

1

C

∫Idt

I is constant. SodV

dt=I

Cand so the voltage simply ramps up at a constant rate.

(source: J. Crawford)

36 April 11, 2014


V0 is the constant of integration, V0 = 0 if the capactior is intially uncharged.

Applications

• function generators use this circuit to produce ramps and triangular waveforms.

• In particle physics, many detectors function by measuring the number of charged particles within a

“pixel” with a similar circuit.

(source: http://www.desy.de/~garutti/LECTURES/ParticleDetectorSS12/L6_SiliconLecture.pdf)

The energy stored by the electric field (work done by the battery providing the current) within the

capacitor is

W =∫ t0

(power)dt =∫ t0v · I dt =

∫ t0Cv

dv

dtdt = C

∫ V0vdv = 1

2CV2

We could have derived this for a voltage source instead of a current source (see http://hyperphysics.

phy-astr.gsu.edu/hbase/electric/capeng2.html)

37 April 11, 2014

http://www.desy.de/~garutti/LECTURES/ParticleDetectorSS12/L6_SiliconLecture.pdf

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html


and we would have arrived at the same answer, W = 12CV

2.

38 April 11, 2014


4.1.3 RC Circuit

4.1.4 RC Filter Example from Chalk Board

(source: Prof. M. Sutton 2013 Lectures) Examples on blackboard: RC circuit charging and discharging.

39 April 11, 2014


4.1.5 Capacitor Networks

Capacitors in parallel add directly

C1V1 C2

V2

the voltage across both capacitors is the same V1 = V2, so

Q1 = C1 V, Q2 = C2 V

Q1 + Q2 = (C1 + C2)V

Qtotal = CtotalV

C1 C2. . . CN

Ctotal = C1 + C2 + . . . + CNThis is intuitive, because it is equivalent to dividing the plates from one large capactior into many segments.

1µF 2µF 3µF 4µF

Ctotal = 1µF + 2µF + 3µF + 4µF = 10µF

44 April 11, 2014


Capacitors in series add inversely

V

I

C1V1

C2V2

VTotal = V1 + V2 =Q1

C1+Q2

C2

but Q1 = Q2 because there is no way for electrons to be created or destroyed between the two capacitors,

so

VTotal =Q

C1+Q

C2= Q(

1

C1+

1

C2) =

Q

Ctotal

1

Ctotal=

1

C1+

1

C2

45 April 11, 2014


C1

C2

. . .

CN1

Ctotal= 1

C1+ 1C2

+. . .+ 1CN

100 pF

100 pF

25 pF

25 pF1

Ctotal= 1

100+ 1100+ 1

25+ 125 = 10

100 → Ctotal = 10pF

46 April 11, 2014


4.2 Transformers and Mutual Inductance

(source: http://en.wikipedia.org/wiki/File:Faradays_transformer.png) “Drawing of the first transformer, built by British

scientist Michael Faraday in the 1830s. It consisted of an iron ring with two windings of insulated

wire around it.Faraday attached one winding to a sensitive galvanometer. When he touched the other

winding to a single cell battery, the winding created a changing magnetic field in the ring which induced

a momentary current in the second winding due to electromagnetic induction, which was registered by

the galvanometer.

Although this device looks remarkably like a modern transformer, Faraday is not considered the inventor

of the transformer. He only applied individual pulses of current to his device, not the alternating current

that allows modern transformers to work continuously. More importantly the device had the same number

of turns on both windings. Faraday did not discover the principle that makes transformers useful - that

it can be used to transform the voltage up or down by using different numbers of turns in the seconday

than in the primary winding.”

47 April 11, 2014

http://en.wikipedia.org/wiki/File:Faradays_transformer.png


A transformer can be made from two coils of wire wound on a cylinder or toroid. The cylinder can be

made of air or, better, a ferromagnetic material that will serve to better capture the magnetic field lines.

Current flowing in the ‘primary’ (input) coil produces a magnetic field ~B with magnetic induction

(units webers/m2), and this field passes through the n2 turns of the ‘secondary’ (output) coil.

Faraday showed that the rate of change of a quantity called the magnetic flux ΦB (defined as the surface

integral of B across the total area of the coil) is responsible for inducing the voltage across that coil. That

flux is proportional to | ~B|, which in turn is proportional to I1, the instantaneous current in the primary.

So,

ΦB ∝ | ~B| ∝ I1

and using Faraday’s law of induction |V2| ∝ ndΦBdt ,

|V2| ∝dI1

dt.

(source: Addison Wesley Longman via http://www.physics.sjsu.edu/becker/physics51/ac_circuits.htm)

48 April 11, 2014

http://www.physics.sjsu.edu/becker/physics51/ac_circuits.htm


The constant of proportionality is defined as the mutual inductanceM (units henry = volt second/amp).

1 henry means a constant of 1 volt across the secondary is induced by a change or 1 amp per second in

the primary.

Thus, for a transformer

V2 = MdI1

dt.

The direction of the output current flow I2 depends on the windings–it is easy to build a current

“inverter” by winding the transformer backwards.

4.2.1 Lenz’s Law

Lenz’s law: The current I2 induced int he secondary coil produces a field that opposes the change in the

flux ΦB.

(source: Left: J. Crawford)

In the simple cartoon, when the current in the lower coil is switched on, ΦB will increase in the upper

coil. The current in the upper coil tries to oppose the increasing flux by producing one with ~B in the

downward direction. This corresponds to the direction of current shown in the upper coil, and that current

would produce a voltage with the direction shown on the resistor.

(The resistor is actually not necessary the voltage will be induced even if the circuit is open.)

49 April 11, 2014


What happens if there is already a current in the lower coil, and we suddenly switch it off? In that

case, the upper coil will try to keep the field from collapsing, so the current flow in it would be opposite to

the direction shown. The voltage would then be opposite also, with the top wire negative, and the lower

positive.

4.2.2 Transformers in AC circuits

Here we have focused on changes in input current. What if the input I1 is a sine wave? The output

voltage, which is the derivative, will also be a sine wave with a phase offset of 90 degrees (cosine). For

AC circuits, transformers are very simple devices. For an ideal transformer (no loss), power is conserved.

This means:

V1 · I1 = V2 · I2

later, when we study AC circuits, we’ll find that

V1

V2=I2

I1=n1

n2

where n1n2

is the turns ratio.

50 April 11, 2014


4.2.3 Transformer Examples

(source: http://commons.wikimedia.org/wiki/File:Electronic_component_transformers.jpg www.electronicsarea.com http://en.wikipedia.org/

wiki/File:Polemount-singlephase-closeup.jpg )

4.2.4 Example Applications

• Electricity distribution: electricity is normally transported long distances over high voltage likes at

AC with very high (kilovolt) amplitude (why?). It is then transformed down to a few hundred volts

to bring it through city streets, and then transformed down again to the 220 VRMS that enters North

American homes.

• DC power supplies for home electronics are built by transforming 110 VRMS from wall outlets to a

lower voltage, rectifying it (taking the absolute value), then integrating it through a low pass filter

(like the RC filter introduced in this class).

51 April 11, 2014

http://commons.wikimedia.org/wiki/File:Electronic_component_transformers.jpg

www.electronicsarea.com

http://en.wikipedia.org/wiki/File:Polemount-singlephase-closeup.jpg

http://en.wikipedia.org/wiki/File:Polemount-singlephase-closeup.jpg


• voltage level translations inside analog electronics, like amplifiers

• isolation circuits which remove the DC level, but pass fast varying signals

• guitar pickup

(source: http://entertainment.howstuffworks.com/electric-guitar1.htm)

Discussion side note: humbucker pickups and the concept of differential cancellation as it applies to

science instruments.

4.3 Inductors and Self Inductance

An inductor is just half of a transformer. It makes use of Lenz’s law to store energy, just as a capacitor

does, as opposed to dissipating energy like a resistor.

52 April 11, 2014

http://entertainment.howstuffworks.com/electric-guitar1.htm


Consider the circuit below, at the moment it is “turned on”. If the inductor (coil) was replaced with a

short, a current V/R would flow immediately, with all of the voltage drop across the resistance R.

With the inductor in place, Faradays law of induction applies.

The current flow produces a flux through the coil, with in the direction shown. Lenzs law says that

the coil will attempt to produce a field that opposes the change in ~B. To do this, it induces a voltage

V (equal to the batterys voltage) across the coil, with the upper end positive and the lower negative.

Therefore, just at the instant the switch is closed, the current flow will be zero.

For a single coil, the voltage-current relationship is

V = Ldi

dt, i =

1

L

∫V · dt

L is called the self-inductance of the coil, or simply the inductance.

L depends on the number of turns of wire, and the magnetic properties of the material within the coil.

Any current-carrying conductor even a straight wire is surrounded by a magnetic field, and hence has

some self-inductance.

53 April 11, 2014


The inductance for a wire wrapped around a cylinder (called a solenoid) is

L =n2µA

l

(source: http://www.calctool.org/CALC/phys/electromagnetism/solenoid)

where l is the solenoid length, A = πr2 is its cross-section area, n is the number of turns, and µ is the

magnetic permeability of the solenoid material (µ0 = 4π × 10−7 for vacuum).

4.3.1 Energy Storage

Energy is stored in the magnetic field within the inductor coil.

If the voltage applied to the coil were constant, then the current in the coil would rise,


The intercept I0 is the constant of integration, representing a possible initial current in the coil.

54 April 11, 2014

http://www.calctool.org/CALC/phys/electromagnetism/solenoid


The energy stored is

W =∫

Power · dt =∫ t0V · i dt =

∫Ldi

dti dt = L

∫ I0i · di = 1

2LI2

4.3.2 Inductor Examples

(source: Left: http://en.wikipedia.org/wiki/File:Electronic_component_inductors.jpg, Right: www.moelettra.it)

4.3.3 Inductor Networks

Inductors in series add directly. Inductors in parallel add inversely.

55 April 11, 2014

http://en.wikipedia.org/wiki/File:Electronic_component_inductors.jpg

www.moelettra.it


(source: http://en.wikipedia.org/wiki/Inductor)

56 April 11, 2014

http://en.wikipedia.org/wiki/Inductor


4.3.4 Inductor Applications

• filtering, e.g. pi-filter (on black board)

• power supply chokes

• tuned circuits, such as AM-radios. Consider a resistor-inductor-capacitor (RLC) in series (lab 5):

V = IR + LdI

dt+ 1/C

∫I · dt

for a pure AC current, I = Ip sinωt, the equation becomes

V = R · Ip sinωt + Lω · Ip cosωt− 1/(ωC) · Ip cosωt

with ω = 2πf . There is a frequency, ω = 1/√LC where the inductor and capacitor cancel, and are

invisible to the circuit. This is the resonant frequency. We see that

– the voltage across the Resistor is in phase with the current,

– the voltage across the inductor leads the current by 90 degrees,

– and the voltage across the capacitor lags by 90 degrees.

57 April 11, 2014


58 April 11, 2014

§ 5

Waterflow Analogy

• pressure ↔ voltage

• water volume per second flow ↔ current

• pump ↔ voltage source

• pipe constriction ↔ resistor

• flywheel ↔ inductor

• elastic membrane ↔ capacitor

59


(source: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.html)

(source: https://ece.uwaterloo.ca/~dwharder/Analogy/)

• see Section C for J. Crawford’s lecture notes on this

• Refer to https://ece.uwaterloo.ca/~dwharder/Analogy/ for helpful examples.

60 April 11, 2014

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.html

https://ece.uwaterloo.ca/~dwharder/Analogy/



Later we will study diodes, a non-linear circuit element.

(source: Left: http://en.wikipedia.org/wiki/Diode, right: https://ece.uwaterloo.ca/~dwharder/Analogy/)

61 April 11, 2014

http://en.wikipedia.org/wiki/Diode




62 April 11, 2014

§ 6

Linear Network Theorems I

Recall some basic properties of L,C:

• You can not instantaneously change the voltage across a capacitor.

• You can not instantaneously change the current in an inductance.

6.1 Superposition and Linearity

We are familiar with superposition and linearity in physics:

For any system described by linear equations (algebraic or differential), the response of the system

to a set of driving forces (Ftot = F1 + F2 + F3 + . . .) can be derived by separately calculating the

response to each force. The total response to Ftot will just be the sum of the individual solutions.

This may be familiar from mechanics, e.g. the harmonic oscillator.

For electric circuits, superposition and linearity imply that:

The current in any branch of a circuit can be found by calculating the current produced by each

voltage or current source separately with all other sources switched off. Then with all sources

63


operating, the current will be the algebraic sum of all the individual solutions.

Note carefully: switching off voltage sources implies imagining them replaced by a wire (i.e., a short

circuit), and switching off current sources implies replacing them by an open circuit.)

6.2 Kirchoff’s Laws

• energy conservation: The algebraic sum of the voltages around a closed loop is zero.

• charge conservation: The sum of the currents flowing into a node (junction point) is equal to the sum

of the currents leaving it.

6.2.1 Example: Method 1 total currents


64 April 11, 2014


Application of these laws produces a set of simultaneous equations that can be solved for all the currents.

The energy conservation equations are:

• +14− 1× I1 − 4× I2 − 6− 1× I3 = 0

• +4× I2 − 2× I5 + 3× I4 = 0

• +1× I3 + 6− 3× I4 − 5× I6 = 0

and the conservation of charge equations are:

• I1 = I2 + I5

• I2 = I3 + I4

• I4 + I5 = I6

65 April 11, 2014


6.2.2 Example: Method 2 circulating currents


We have automatically satisfied the current law. The energy conservation equations are:

• +14− 1× I1 − 4× (I1 − I2)− 6− 1× (I1 − I3) = 0

• +4× (I1 − I2)− 2× I2 + 3× (I3 − I2) = 0

• 1× (I1 − I3) + 6− 3× (I3 − I2)− 5× I3 = 0

and can be re-written

• −6× I1 + 4× I2 + 1× I3 = −8

66 April 11, 2014


• 4× I1 − 9× I2 + 3× I3 = 0

• 1× I1 + 3× I2 − 9× I3 = −6

which can be written in matrix notation as

In both cases, the end result is three equations with three unknowns, which can be easily solved.

• by using linear algebra and the ratio of determinants

• by using a program like matlab to write the soutions in matrix algebra,

67 April 11, 2014


68 April 11, 2014

§ 7

Stepped Voltages with low and high-pass filters

69


Draw the output voltage for the given (blue) input voltage:

70 April 11, 2014


Answers:

71 April 11, 2014


72 April 11, 2014

§ 8

AC Circuits, Sinusoidal Currents and Voltages, andComplex Representations

8.1 RLC AC-Circuit: Sinusoidal Representation

Consider a resistor-inductor-capacitor (RLC) in series (lab 5):

V = IR + LdI

dt+ 1/C

∫I · dt

73


for a pure AC current, I = Ip sinωt, the equation becomes

V = R · Ip sinωt + Lω · Ip cosωt− 1/(ωC) · Ip cosωt

with ω = 2πf = 2πT .

There is an angular frequency, ω = 1/√LC where the inductor and capacitor cancel, and are invisible

to the circuit. This is the resonant frequency.

The equation can be written in terms of sinusoids only:

V = R · Ip sinωt + Lω · Ip sin(ωt + 90) + 1/(ωC) · Ip sin(ωt− 90)

We see that

• the voltage across the Resistor is in phase with the current,

• the voltage across the inductor leads the current by 90 degrees,

• and the voltage across the capacitor lags by 90 degrees.

In this notation, we see that the amplitude of the voltage across each component is:for resistance for inductance for capacitance

|RI| |ωLI| |I/(ωC)|(in phase with I) (leads I by 90) (lags I by 90)

The impedance magnitude |Z| is just the ratio of the voltage amplitdue to the current amplitude

|V |/|I|for resistance for inductance for capacitance

|ZR| = R |ZL| = ωL |ZC| = 1/(ωC)and is easy to calculate for a particular ω.

74 April 11, 2014



The phasor diagram is: (source: J. Crawford)

and the total voltage is just the sum of the three vectors.

75 April 11, 2014


8.2 Representation with Complex Numbers

A complex number, such as rejφ = x + jy, is a natural representation for a number that has both an

amplitude and a phase.

(source: http://en.wikipedia.org/wiki/Complex_number)

In math, i (defined by i2 = −1) is normally used for the imaginary unit. Here j will be used, to avoid

confusion with current.

j2 = −1

x, y are real numbers representing the real and imaginary portion of the vector, respectively.

r =√x2 + y2

φ = arctany

x

• complex number addition represents vector addition

76 April 11, 2014

http://en.wikipedia.org/wiki/Complex_number


• Complex number subtraction represents vector subtraction.

• Multiplication by j rotates a vector by +90.

(a + jb)j = aj + bj2 = −b + ja

Division by j rotates a vector by -90.

• Euler’s formula A ejφ = A(cosφ + j sinφ)

• Multiplication

z1 × z2 = A1 ejφ1 × A2 e

jφ2 = A1 A2 ej(φ1+φ2)

• Divisionz1

z2=A1 e

jφ1

A2 ejφ2=A1

A2ej(φ1−φ2)

In problems involving sinusoidal relationships (e.g., simple harmonic motion in mechanics, or sinusoidal

A.C. in circuit problems), we think of the sine wave as the projection of a rotating vector. Therefore

77 April 11, 2014


y = A sinωt is the y projection of a vector represented by the complex number z = Aejωt. We are, in

a way, just ignoring the other projection. Note that in these problems, we rarely draw a picture of the

actual sine wave: all we care about are the amplitudes and relative phases of the sinusoidal quantities in

the problem, and the vector (or the complex number that represents it) contains just this information.

8.3 RLC AC-Circuit: Complex Represntation

For a resistor-inductor-capacitor (RLC) in series

V = IR + LdI

dt+ 1/C

∫I · dt

this time, the current is represented by I = Ipejωt (the imaginary projection gives the original Ip sinωt).

V = R · Ipejωt + jLω · Ipejωt +1

jωC· Ipejωt

78 April 11, 2014


and the total impedance of the loop is simply

Ztot = V/I = R + jωL +1

jωC

• R,ωL, 1/(ωC) all have units of ohms

• in the case of the inductance and capacitance, the j’s indicate phase relationships.

• jωL indicates that voltage leads current by 90 in an inductance;

• 1/(jωC) indicates that voltage lags current by 90 in a capacitance.

expressed in this way, networks of these impedances behave just like resistors

• impedances in series add

• impedances in parallel add inversely.

8.3.1 Examples

Calculate the voltage for this resonant circuit

79 April 11, 2014


Calculate the voltage for this circuit

• when ωL > 1ωC

• when ωL = 1ωC

• when ωL < 1ωC

and draw a qualitative network analysis on the black board.

80 April 11, 2014

§ 9

Linear Network Theorems II

The Thevenin and Norton Theorems are based on the superposition properties of linear theorems. They

provide a means for converting any circuit’s sources and impedances to a simple equivalent circuit.

81


9.1 Thevenin’s Theorem

For a linear electrical network

• the circuit at terminals A,B can be replaced by an equivalent voltage source Vth in series with an

equivalent resistance Rth.

• Vth is the voltage that would be obtained across terminals A,B of the circuit with terminals A-B open

circuited.

• Rth is the resistance that would be obtained looking into terminals A-B of the circuit with all its

current sources open circuited and all its voltage sources short circuited.

For DC circuits, the theorem works for voltage sources, currents sources, and resistances. For AC circuits,

it also works for circuits that include inductors and capacitors.

(source: http://en.wikipedia.org/wiki/Thevenin’s_theorem)

82 April 11, 2014

http://en.wikipedia.org/wiki/Thevenin's_theorem


9.1.1 DC Example


83 April 11, 2014



9.2 Norton’s Theorem

For a linear electrical network

• the circuit at terminals A,B can be replaced by an equivalent current source INo in parallel with an

equivalent resistance RNo.

• INo is the current that would be obtained from terminals A to B of the circuit if the terminals A-B

were short-circuited.

• RNo is the resistance that would be obtained looking into terminals A-B of the circuit with all its

current sources open circuited and all its voltage sources short circuited.

For DC circuits, the theorem works for voltage sources, currents sources, and resistances. For AC circuits,

it also works for circuits that include inductors and capacitors.


84 April 11, 2014



9.2.1 Conversion from Thevenin’s Theorem to Norton’s Theorem

Rth = RNo

Vth = INoRNo


85 April 11, 2014



86 April 11, 2014

§ 10

Fourier Analysis

10.1 Time Domain Waveform

A time domain waveform or “timestream” is just a series of data recorded versus time. Examples:

• music recorded on an LP or audio cassette

• strip-chart from a hospital heart monitor

• etc.

If the data is sampled (quantized) at regular time intervals, it is a discrete waveform, and has been

digitized.

• music on compact disks and ipod harddrives

87


10.1.1 Sine Wave timestream

Let’s create a timestream composed of a pure sine-wave. We’ll sample at the standard audio rate, 44,100

Hz,

Using matlab:

t = [0:1./44100:1]; % 1 second with 44,100 samples.

ts = sin(2.*pi*440.*t); % 440 Hz sine wave (A note)

plot(t,ts,’.’)

axis([0 0.01 -1 1])

xlabel(’Time (seconds)’)

sound(ts,44100); % play the sound on the computer speaker.

sound( sin(2.*pi*880.*t) ,44100); % play a high sound

sound( sin(2.*pi*80.*t) ,44100); % play a low sound

[mag, freq] = powerSpectrum(ts,44100 ) ;

semilogx(freq,mag,’-’);

xlabel(’Frequency (Hz)’)

88 April 11, 2014


The Fourier transform of a sine wave is a Dirac Delta function. It peaks at one specific frequency,

corresponding to the period of the sine wave.

The Fourier transform of a sine wave simply measures its period (or frequency). The amplitude of the

transform corresponds to the amplitude of the sine wave.

The Fourier transform produced a complex output. The complex phase corresponds to the phase of

the sine wave.

10.2 Fourier Theory

We can build up any waveform from a superposition of sine waves. For example, a square wave is just

the sum of a sine wave and its (properly weighted) harmonics.

89 April 11, 2014


(source: http://www.bores.com/courses/intro/freq/3_ft.htm)

90 April 11, 2014

http://www.bores.com/courses/intro/freq/3_ft.htm


A Fourier transform tells us which sine waves we need to add together to build up the waveform we

are analyzing. The sine waves are the basis set (just like the vectors x, y form the basis set for any 2-D

vector).

F (u) =∫ ∞−∞ f (x)e−j2πuxdx

remember, ejθ = cos θ + j sin θ, so the Fourier transform is really just a convolution with sine and cosine

waves.

• It is possible to construct any function as a sum of sine waves of various frequencies. The sine waves

form a basis set.

• The Fourier Transform power spectrum is the extent to which a function looks like a sine wave of a

particular frequency.

• It is fully reversable (preserves all information content). The inverse Fourier Transform is

f (x) =∫ ∞−∞ F (u)ej2πuxdu

• The Fourier transform for discrete waveforms is

F (u) =1

N

N−1∑x=0

f (x)e−j2πux/N

91 April 11, 2014


10.2.1 Application to Audio

Record an audio discrete timestream using a program like Audacity, and save it to a .wav file called

test.wav on the desktop. The timestream will be recorded at a rate of 44,100 Hz, the standard rate for

audio recordings such as CDs.

Using matlab:

cd ./Desktop

[ts, sampleRate] = wavread(’test.wav’); % read .wav file as a timestream ts

sampleRate % print the sample rate to the screen

plot( ts(1:1000) )

% display the Fourier content of this timestream


semilogx(freq,mag,’-’);

axis([10 25000 0 0.01]); xlabel(’Frequency (Hz)’)

92 April 11, 2014


10.2.2 the Matlab function powerSpectrum.m

function [mag, freq] = powerSpectrum( ts, sampleRate )

% This function returns the absolute magnitude power spectrum

% in a vector mag which corresponds to x-axis frequency freq (units = Hz).

% plot(freq, mag) will display the power spectrum

%

% Inputs ts: a column vector, of arbitrary length containing the

% input timestream data.

% sampleRate: the sampling rate, in Hz, for the ts data.

%

N = length(ts) ;

halfN = floor(N/2) ;

rawFFT = fft(ts) ;

mag = abs( rawFFT(1:halfN ) ) *2./N ;

freq = [0:1:halfN-1] ;

freq = freq/ halfN * sampleRate/2 ;

93 April 11, 2014


end

94 April 11, 2014


10.2.3 Example: Signal Buried in Noise

Using matlab:

t = [0:1./44100:1]; % 1 second with 44,100 samples.

ts = 0.1 * sin(2.*pi*440.*t); % 440 Hz sine wave (A note)

plot(t,ts,’.’)

axis([0 0.01 -1 1]); xlabel(’Time (seconds)’)


ts_noise = randn(1,44101) ;

plot( t, ts_noise, ’.’ );

axis([0 0.01 -1 1])

sound(ts_noise,44100); % play the sound on the computer speaker.

% Let’s hide our A-note in the white noise.

ts = ts + ts_noise;

plot(t,ts,’.’); axis([0 0.01 -1 1])


% but, in fourier space, this signal is loud and clear


plot(freq,mag,’-’);

axis([0 1000 0 0.2]); xlabel(’Frequency (Hz)’)

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Notice that the Fourier Transform of a Gaussian random timestream (random noise) is WHITE (equal

power at all frequencies).

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10.3 Example: Spatial Fourier Transform

(source: http://lambda.gsfc.nasa.gov/)

97 April 11, 2014

http://lambda.gsfc.nasa.gov/


10.4 Fourier Series

Let’s step back now, and look at the mathematics of the Fourier transform.

Joseph Fourier (1768-1830), in his paper “On the Propagation of Heat in Solid Bodies” demonstrated

an expansion of methematical functions in a series of trigonometric functions.

Essentially, Fourier has realized that Any periodic waveform can be represented as the sum of a

suitably chosen harmonic set of properly phased sine waves. i.e.

y(x) = b0 +∞∑1an sin(nx) +

∞∑1bn cos(nx)

This was controversial– even his colleagues Laplace and Lagrange were not convinced.

We already have seen one such example, a square wave, and found that even abrupt discontinuities can

be built up from smooth sine waves.

Finding the coefficients is simple: multiply both sides by sin(mx) and cos(mx) and integrate over the

period. ∫ 2π

0sin(mx) sin(nx)dx = 0 for m 6= n,= π for m = n.∫ 2π

0cos(mx) cos(nx)dx = 0 for m 6= n,= π for m = n.∫ 2π

0sin(mx) cos(nx)dx = 0 for all m,n.

(said differently, these sine and cosine waves form an orthogonal set.)

b0 is found by taking the averagevalue of the function over the period. Thus the Fourier Series is:

an =1

π

∫ 2π

0sin(nx)y(x) · dx

bn =1

π

∫ 2π

0cos(nx)y(x) · dx

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b0 =1

2π

∫ 2π

0y(x) · dx

10.4.1 Fourier Series Example: Square Wave (again)


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(source: http://en.wikipedia.org/wiki/Fourier_series)

see http://upload.wikimedia.org/wikipedia/commons/2/2b/Fourier_series_and_transform.

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http://en.wikipedia.org/wiki/Fourier_series

http://upload.wikimedia.org/wikipedia/commons/2/2b/Fourier_series_and_transform.gif




gif for a nice animated GIF.

For Fourier Series representing periodic functions

• the fundamental frequency f0 is the same as the frequency 1/T of the original waveform.

• in electro-speak b0 is the “DC component of the waveform

• Even functions (symmetric for reflection about the y-axis) contain only cosine waves.

• Odd functions contain only sine waves.

(not surprizing as cosine/sine waves are even/odd functions).

• The mix of sine and cosine waves will change as we alter the start time (move the x-axis) of the wave.

A true square wave could never be amplified and displayed on any instrument (such as the scopes

in the lab, which have analog amplifiers at their front end), because all real-world devices half limited

bandwidth.

• your ears have a bandwidth of about 22 kHz

• the oscilloscopes in the lab have a bandwidth of about 50 MHz

This limited bandwidth means that the full Fourier series of the waveform cannot be represented by

the instrument.

An RC filter has a time constant (time for voltage to fall to V0/e) τ = RC and a bandwidth ω =

2πf = 1RC .

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This means a device with a finite bandwidth, such as the 50 MHz lab scopes, will have a finite time-

constant τ = 12πf = 3.2 ns.

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10.5 Rectangular Pulse Train - Bandwidth Theorem

Previously, we studied the periodic functions which can be “built up” by super-imposingsinusoids that are harmonics (an integer multiple) of a fundamental frequency f0.→ Is it possible to create a single pulse?

In this section, we’ll explore the super-position of sinusoids that fall in some narrowbandwidth range, ∆ω.

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10.5.1 Single Tone

Consider the timestream for a single tone at angular frequency ω0 = 2πf0, and usingthe complex notation

s(t) = Aejω0t

This is an infinite sinusoid timestream, which is represented in Fourier space by a deltafunction at ω0.

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We will use f0 = 50 Hz and A = 10 as an example.

10.5.2 Three Tones

Let’s add two additional tones, dividing the amplitude by the number of tones (3) andnormalizing by the bandwidth.

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We will use f0 = 50 Hz, 48 Hz and 52 Hz, such that ∆f = 4 Hz.

s(t) =A∆ω

3(ejω0t + ej(ω0+∆ω

2 )t + ej(ω0−∆ω2 )t

s(t) =A∆ω

3ejω0t(1 + ej

∆ω2 t + e−j

∆ω2 t)

s(t) =A∆ω

3

1 + 2 cos(∆ω

2t) ejω0t

we have written this in a way that shows us these three tones are indistinguishable fromthe amplitude of the fundamental tone at f0 being modulated.

A∆ω3

[1 + 2 cos(∆ω

2 t)]

is the modulation envelope.

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• The period of the “fundamental” or carrier is T = 1/f0.

• The period of the “modulation envelope is TMOD = 4π∆ω = 2

∆f = N∆f .

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10.5.3 Five Tones

Let’s put five tones in the same bandwidth of 4 Hz.

s(t) =A∆ω

5(ejω0t + ej(ω0+∆ω

2 )t + ej(ω0−∆ω2 )t + ej(ω0+∆ω

4 )t + ej(ω0−∆ω4 )t

s(t) =A∆ω

5ejω0t(1 + ej

∆ω2 t + e−j

∆ω2 t + ej

∆ω4 t + e−j

∆ω4 t)

s(t) =A∆ω

5

1 + 2 cos(∆ω

2t) + 2 cos(

∆ω

4t) ejω0t

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we now have a modulation period of 4/∆f , or 1 second.

• The packet width (width of main bump) is still ∆t = 1/∆f ∼ 0.25 s.

• The modulation period is TMOD = 5∆f .

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10.5.4 N →∞ Tones

A pattern is emerging:

• packet width is ∆t = 1/∆f .

• The modulation period is TMOD = N∆f

If we take N →∞, can we get a one-time only pulse?

Consider N tones within the same bandwidth.


Define ∆ω′ = ∆ω/N

s(t) =A∆ω

N∆ω′ejω0t

N2∑

n=−N2ejn∆ω′t∆ω′

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taking N →∞, the summation becomes an integral

s(t) =A∆ω

∆ωejω0t

∫ ∆ω2

−∆ω2ejω

′tδω′

This is easy to integrate

s(t) =2A∆ω

∆ω

sin(∆ω2 t)

tejω0t = A∆ω

sin(∆ω2 t)

(∆ω2 t

ejω0t)

which is the “sync” function sin(x)/x.

And so the superposition of an infinite number of sinusoids in the limited bandwidthof 48-52 Hz produces the waveform:

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This is a pulse that is not periodic, it happens only once (the modulation period isinfinite!).

The packet width is still ∆t = 1/∆f = 0.25 s.

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10.5.5 Bandwidth Theorem

(also sometimes called the “classical uncertainty theorem”)

∆f∆t ' 1

∆ω∆t ' 2π

(since ω = 2πf ).Any phenomena that occurs in a limited time interval ∆t has a frequency spread

of ∆f = 1/∆t Hz. If the time interval is small, then the frequency spread will be large.Question:

what if you produce a pure sine wave, e.g. at 1 kHz, that lasts for 1 second?We already learned that the Fourier transform of a sine wave is a delta function– doesn’tthis have zero width?

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Question:what if you produce a pure sine wave, e.g. at 1 kHz, that lasts for 1 second?We already learned that the Fourier transform of a sine wave is a delta function– doesn’tthis have zero width?

Answer:No!, the function you produced is a sine-wave, convolved with a 1 second long boxcar.This produces signal at 1 Hz and all its harmonics.

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10.6 System Frequency Response

Consider any circuit, such as an R/L low pass filter

Vin Vout

The system frequency response or transfer function can be characterized mathe-matically by the function h(t), wherein the output y(t) can be calculated from the inputx(t) timestream by convolving

y(t) = h(t)⊗ x(t)

(⊗ denotes a convolution) as is often the case, many systems are simpler when workingin Fourier space,

Y (f ) = H(f ) ·X(f )

BlackBox Networkh(t)H(ω)

vouty(t)Y (ω)

vinx(t)X(ω)

For the R/L low pass filter, the output is a simple voltage divider

Vout = Vin ×R

R + jωL

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and H(f ) is given by the ratio

H(ω) =VoutVin

=R

R + jωL

where ω = 2πf .

10.6.1 Transfer Function Example

Knowing the Fourier space transfer function can make it very easy to calculate theresponse of the circuit to an input waveform.

Consider the R/L low-pass filterVin Vout

response to a square wave.

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In Section 10.4.1, the Fourier series representation of a square wave was calculated,allowing the square-wave timestream to be written as a sum of sine waves:

Vin = x(t) = Asinω0t +

1

3sin 3ω0t +

1

5sin 5ω0t + . . .

using

H(ω) = Vout/Vin =R

R + jωL

the attenuation of each Fourier series coefficient can be calculated.

Let’s consider ω0 = R2L, then plugging ω0, 3ω0, 5ω0, into H(ω) to get the attenuation

factor

|H(ω0)| = | R

R + j R2LL| ' 0.89

|H(3ω0)| ' 0.55

|H(5ω0)| ' 0.37

So the the filter has removed much of the amplitude from the higher harmonics at 3ω0

and 5ω0, rending the output more like a single tone (more like a sine wave, which is aDirac delta function in frequency space).

Similar analyses can be applied to transfer functions for other “physical circuits”,

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(source: http://en.wikipedia.org/wiki/File:Frequency_response_example.png) Here, the vibrational resonances of a struc-ture are characterized by a Fourier space transfer function, and applied to a square waveinput force.

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http://en.wikipedia.org/wiki/File:Frequency_response_example.png


10.6.2 Bode Plot and Decibels

The transfer function of a linear, time in-variant system is most often displayed inthe form of a Bode Plot.

The Bode plot for the R/L low-pass filter

Vin Voutwhich has a single pole

at f = R/(2πL) ≡ 100 Hz is

(source: http://en.wikipedia.org/wiki/File:Bode_Low-Pass.PNG) whichis independent of the input waveform.

A Bode plot provides both the magnitudeand phase of the transfer function H(f ).

The y-axis for the phase plot is normallypresented in degrees or radians. The Bodeplot shows that the filter has a |∆φ| = 45

phase shift at the pole frequency. This is a

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http://en.wikipedia.org/wiki/File:Bode_Low-Pass.PNG


generic property of a single pole filter suchas an RC, L/R, or LC filter.

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The y-axis for the magnitude plot is most often presented in decibels (dB). This is thelogarithmic ratio of output power to input power, multipled by 10.

Magnitude Ratio[dB] = 10 log10(PoutPin

).

Most often, people choose to work in voltage units rather than power. Since P ∝ V 2,

Magnitude Ratio[dB] = 10× log10(PoutPin

) = 10 log10(VoutVin

)2

= 20 log10(VoutVin

).

(we have used logAb = b logA).The decibel scale is logarithmic.

A 20 dB output is 10 times bigger than the input.A 6 dB output is 2 times bigger than the input.A 3 dB output is

√2 times bigger than the input.

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§ 11

Diodes

(source: Adam Gilbert’s lectures on Diodes.)

(source: Wikipedia)

123


Diodes are the first non-linear circuit element we have encountered.Adam Gilbert’s (excellent!) slides on diodes are uploaded to MyCourses in the usefulinfo folder.Let’s explore some common circuits that use diodes.

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Diodes

What is a Diode?

• Semi-conductor device

• Conducts current in only one direction

• Very little forward voltage drop

Figure 1 – Ideal diode characteristics

Non-ideal diodes have a:

• small forward voltage drop – typically 0.7V for Silicon, 0.3V for Germanium, 0.4V for Schottky.

• leakage current when reverse biased

• reverse breakdown

Figure 2 – Diode I/V curve – image taken from http://en.wikipedia.org/wiki/Diode

I=Is ( e Vd/Vt -1 ) , Vt=KT/qK is the Boltzman constant, T is the temperature in Kelvin, Vd is the voltage across the diodeq is the charge on an electron, Is is the reverse leakage current

How do they work?A piece of semiconductor material (e.gSilicon/Germanimum – group 4 elements) isdoped to have an n-doped region adjacent to ap-doped region – the silicon is actuallydeposited with the dopants mixed in (see vapor-phase epitaxy )

P-doping is achieved by infusing the materialwith group 3 acceptors elements e.g Boron,Gallium, this adds 'holes' to the material

N-doping is achieved by infusing the materialwith donors (group 5 elements e.g Arsenic,Phosphorous) – this adds free electrons to thematerial

At the P-N junction boundary diffusion takesplace, this diffusion uncovers ions and resultsin a net negative charge in the P-doped sectionand a net positive charge in the N-dopedsection. The region in the device withuncovered ions is known as the depletionregion. There is an electric field present acrossthe depletion region which slows down thediffusion process.

Can anyone guess which way we'd need toapply a voltage to reduce the depletion region?

Figure 3 – Diode I/V curve – image taken from http://en.wikipedia.org/wiki/P-n_junction

What do diodes look like?

Diodes come in many sized packages, from devices that require tweezers to pick up to devices that can be bolted in place.

Figure 4 – Some example diodes – image taken from http://en.wikipedia.org/wiki/Diode

There are many types of diodes, to name a few:

Laser diodes

Photo diodes

Light emitting diodes (LEDs)

Zener diodes

Simplest diode circuit – lighting up an LED

In the following figure you can see some of the information that manufacturer has provided in the component data sheet.

Figure 5 – Part of data sheet for a typical LED (LTST-C190EKT)

Lets try building a circuit to run the LED at its recommended current:

• LEDs are special and require much higher forward voltage levels.

• The relative luminous intensity is 1 in the data sheet at 10mA of current.

• The data sheet indicates that a forward voltage of 1.9V is required for 10mA

What resistor value is required in this circuit to light the LED?

Figure 6 – Simplest diode circuit

R = (5V – 1.9V ) / 10mA = 310 Ohms

How much power is this circuit using up?

P= 5V * 10mA =50mW

P=10mA^2 *310 + 1.9V * 10mA = 50mW

What would happen to the power consumption in this circuit if we designed the LED to run from a 12V supply?

What about the LED power consumption?

Circuit analysis with diodes

Analyzing circuits with diodes can be a tricky. We often don't know if the diode is conducting or not.

1) Start your analysis by assuming the diode is non-conducting

2) Calculate what forward voltage would be across the diode

3) If this voltage is less than the diode model threshold you are using (Ideal vs Non ideal diodes) then the diode is non-conducting and your analysis is over

4) If the voltage is greater than the threshold you have specified, you need to replace the diode with an appropriate model i.e short for an Ideal diode or a voltage source for non-ideal diode and recalculate the circuit voltages/current

5) You're done! - Note that if you want to model the I/V characteristics of the diode you're better off using SPICE!

Lets work out this circuit on theblackboard:

Lets say we're interested in thecurrent through the diode, andthe voltage at the anode.

Figure 7 – A more complicated diode circuit

Diode followed by capacitor

Note that there isno mechanism forthe capacitor todischarge so onceit finds a peak itjust sits therewaiting for ahigher one.

The differencebetween the peakand the capacitorvoltage is the 0.7Vforward drop ofthe diode

The most basic peak detector

Bandwidth governed bythe RC time constant.

Smaller RC results infaster decay

When does the diodeconduct again after itfinds a peak?

Usual approach is toapproximate the sinusoidby a triangular waveform

Maybe try this problem athome.....

The full wave rectifier

We now conducttwo times persinusoid.

Commonmethod toconvert fromAC to DC

The Zener diode

Can you seesomething funnyabout this circuit?

We are running thediode around itsreverse break downvoltage!

Often used as aprecision voltagereference, circuitprotection, cheappower supplies


11.1 Diode Circuits

11.1.1 Half-wave Rectifier

(source: http://www.brighthubengineering.com/consumer-appliances-electronics/96645-efficiency-of-ac-rectifiers/)

Another version of the half-wave rectifier is:

Vin Vout

which will be used as a building block for a full-wave rectifier.

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http://www.brighthubengineering.com/consumer-appliances-electronics/96645-efficiency-of-ac-rectifiers/


11.1.2 Full-wave Rectifier

A full-wave rectifier acts like the mathematical operation of applying the absolute value.Using the building blocks from the previous example, it can be drawn:

Vin Vout

This is more commonly drawn:

A full-wave rectifier is the first element in common power supplies, like the “bricks”that power most laptop computers.

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(source: http://www.electronics-tutorials.ws/diode/diode_6.html)

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http://www.electronics-tutorials.ws/diode/diode_6.html


A transformer can be added at the input to change the output voltage amplitude.

(source: http://www.brighthubengineering.com/consumer-appliances-electronics/96645-efficiency-of-ac-rectifiers/)

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http://www.brighthubengineering.com/consumer-appliances-electronics/96645-efficiency-of-ac-rectifiers/


11.1.3 Diode Clipping Circuits

(source: http://www.electronic-basics.com/2009/05/diode-limiter.html)

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http://www.electronic-basics.com/2009/05/diode-limiter.html


(source: http://www.daenotes.com/electronics/digital-electronics/clipper-circuits)

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http://www.daenotes.com/electronics/digital-electronics/clipper-circuits


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§ 12

The S-plane and Laplace Transforms

(source: Many examples and ideas for this chapter are taken from John Crawford’s lecture notes.)

Warning: this chapter builds on what has been learned about Fourier transforms anduses that as a foundation. Students should ensure they have a solid understanding ofFourier transforms before diving in to this chapter.

12.1 Exponential solution to linear circuits

We have studied the transfer function for AC signals transmitted through a circuitnetwork consisting of linear circuit elements (L,C,Rs).

BlackBox Networkh(t)H(ω)

vouty(t)Y (ω)

vinx(t)X(ω)

143


We found that certain inputs give the same form of output:

• (1) DC inputs: once the output reaches a steady state, it will be proportional to theinput.

• (2) AC (sinusoidal) inputs: we found that a sinusoidal current, i = ~Iejωt was asolution to the differential equation involving any combination of L,C,Rs. (we have

written ~I to remind ourselves that it is a complex number, with its phase denotingthe phase of the sinusoid).i.e. for a resistor-inductor-capacitor (RLC) in series

V = I(t) R + LdI(t)

dt+

1

C

∫I(t) · dt

v =

R + jωL +1

jωC

~Iejωt = ~V ejωt

or ~V = ~Zjω~I with ~Zjω = ~V /~I

Using math jargon, we might say that ejωt is an eigenfunction of the operator

~Z = R + Ld

dt+

1

C

∫dt.

If this is the case for sine waves (imaginary exponentials), might it also be true for realexponentials? (i.e., is i = Ieσt a solution to our differential equation?)

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• (3) Exponential inputs: i = Ieσt

v =R + σL +

1

σC

Ieσt = V eσt

or V = ZσI

(the vector notation is not needed above, because there is no phase information). Thismeans ordinary exponentials are also eigenfunctions. Putting an exponential in, we getan output that is proportional to the same exponential. σ can be positive or negative,denoting exponential growth or decay.

We can combine inputs (2) and (3) into a single complex solution,

• (4) Complex exponential inputs: i = ~Iest = ~Ieσtejωt

v =R + sL +

1

sC

~Iest = ~V est

or ~V = ~Zs~I

where both ~I and s = σ+jω are complex numbers. This solution represents a sinusoidalfunction with an amplitude that is exponentially growing or shrinking with time.

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(source: John Crawford)

146 April 11, 2014


It is a representation of the general caseencompasing all of the previous solutions(eigenfucntions).

• DC case: σ = 0, ω = 0→ s = 0

• AC case: σ = 0, ω 6= 0

• ordinary exponential: σ 6= 0, ω = 0

In the s-plane, the complex impedance forour R,L,C circuit elements become

ZR = R

ZL = sL

ZC =1

sCand these impedances can be used with Kir-choff’s laws, Th’evenin’s theorem, Norton’stheorem, and superpositoin. All of theseprevious ideas still apply, just as they didfor the complex impedances we derives forAC sinusoids.

(source: http://www.dsprelated.com/dspbooks/mdft/Comparing_Analog_Digital_

Complex.html)

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http://www.dsprelated.com/dspbooks/mdft/Comparing_Analog_Digital_Complex.html

http://www.dsprelated.com/dspbooks/mdft/Comparing_Analog_Digital_Complex.html


12.1.1 S-plane Example

Consider the following circuit with an exponentially decaying input voltage

vin = 20e−2te4jt

vin

2 H 0.1 H

1 Ω

the complex s-impedances are

ZL = sL = (−2 + j4) · 2H = (−4 + j8) Ω

ZC =1

sC=

10

(−2 + j4)Ω

ZR = 1Ω

and so the current is

~I =~V

R + sL + 1sC

=~V

1 + (−4 + j8) + 10−2+j4

= ~V1− j28 + j14

(12.1)

and the current magnitude is

|I| = V

√12 + 22

√82 + 142

= 20

√5√

260= 2.8A

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and the phase of ~I/~V represents the phase by which the current leads the voltage (usingEquation 12.1)

φ = arctan(−2)− arctan(14/8) = −123.5.

The final solution for the current is

i = ~Iest = 2.8e−j(123.5)e−2tej4t

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12.2 The S-plane

Various functions–voltages, currents, impedances, transfer functions–can be plotted onthe s-plane.

Let’s use a simple RC filter example to explore this

vin

C

R

vout

vin is any signal in the s-plane. The s-impedance is Zs = R + 1sC and the transfer

function is

H(s) =voutvin

=R

R + 1sC

=sCR

sCR + 1=

s

s + 1/τ

where τ = RC.

• H(s) = 0 at s = 0, called a “zero”.

• H(s) =∞ at s = −1/τ , called a “pole”.

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• zero at s = 0: this is a DC voltage. With a blocking capacitor, there is no steady-state output voltage, Vout = 0.

• pole at s = −1/τ : for an input voltage vin = V et/RC the ratio of the output to inputvoltage is infinite. This seems non-sensical, but by solving the differential equationsfor this input, you would find the output is vout = V (1 − t/τ )et/RC which gives atransfer function of vout/vin = (1− t/τ ) which grows to to −∞ at t =∞.

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12.3 Using Poles and Zeros to Characterize Circuits

The positions of poles and zeros in a circuit’s transfer function allows us to infer itsbehaviour for various inputs.

12.3.1 Single Pole Example

Consider the circuit

vin

C

R

vout

1. draw the poles and zeros in the σ versus jω plane:


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2. draw a vector from each pole and each zero to the signal s of interest. We willconsider only sinusoidal signals s = jω here:


3. The transfer function H(jω) is the ratio of the zero-vector over the pole vector.

H(s) =vector(zero→ s)

vector(pole→ s)

4. The magnitude |H(jω)| is just the ratio of the lengths of these two vectors (recallthat the magnitude of a complex number is equal to the magnitude of the numeratorover the magnitude of the denominator).The distance from the zero to s is just ω, while the distance from the pole to s is√ω2 + 1/τ 2, so |H(jω)| = ω√

ω2+1/τ2.

5. phase(H(jω)) Recall that the phase of a complex number is the difference of the

153 April 11, 2014


phase of the numerator and the phase of the denominator.90 − arctan(ωτ ), which is 90 at low frequency and 0 at very high frequency.

12.3.2 Multiple Pole Example

The above formalism works for more complicated circuits that have many poles andzeros. In general, transfer functions can be factorized into polynomials in s and writtenin the following form

H(s) = k(s− a1)(s− a2)(s− a3) . . . (s− am)

(s− b1)(s− b2)(s− b3) . . . (s− bm)

The procedure becomes

1. draw a vector from each pole and each zero to the signal s

2. |H(jω)| = product of all the zero-vector lengthsproduct of all the pole-vector lengths

3. φ(H(jω)) = (sum of all the zero-vector angles)−(sum of all the zero-vector angles)

12.3.3 S-plane Zeros and Poles Calculation Examples

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S-plane Examples (source: John Crawford Lectures) 1. The Twin-Tee Network

C

R

C/2 C/2

2R 2Rvi

vo

Fig. 9 The Twin-Tee network. We want to find the transfer function of the network above. To do this, we replace all C's by 1/sC (so the C/2's become 1/2sC) and solve for vo/vi. This requires some lengthy algebra, but with patience we find

H(s) = s2 +ω02

s2 + 4ω0s +ω02 , where ω0 =

1RC

(9)

We can factor the numerator to give H(s) = (s + jω0 )(s − jω0 )

(s+.26ω0 )(s + 3.73ω0 ) (10)

The s-plane therefore has two zeros at ±jω0 and two poles at -.26ω0 and -3.73ω0. If we want to find the frequency response for input sinusoidal signals, we draw the following vectors:

XX

jω0

–jω0

s

–.26ω0–3.73ω0

σ

jω

2

Fig. 10 Sinusoidal response for the Twin-Tee network At ω = 0 (DC), the product of the numerator vector lengths will be ω02 as will the denominator vectors (3.73ω0 x .26ω0 = ω02), so the output voltage will equal the input. The radian frequency indicated in the figure above is just below ω0. We can see that at this point the magnitude of H(jω) is close to zero, because the vector to the upper zero point is very short. Right at ω = ω0, |H(jω)| will be exactly zero; there will be no output voltage. As ω increases above ω0 the output will again increase, and at very high frequencies all vectors will have nearly the same lengths. Therefore the output voltage amplitude will again be nearly equal to the input. We can also see the phase behaviour. At very low ω, the two numerator vectors add to phase 0˚, and the two pole vectors also add to 0˚. The output voltage will therefore be in phase with the input. For ω just below ω0, the zero vectors still add to 0˚, and the two pole vectors add to arctan(1/.26)+arctan(1/3.73) = 75 + 15˚ = 90˚, so the angle by which the output voltage leads the input will be –90˚ (i.e., the output voltage will lag the input). Just above ω =ω0 the upper zero vector changes phase from -90˚ to +90˚, so the output will lead the input by 90˚. At extremely high frequencies, all vectors will point nearly straight up, so the output will again be in phase with the input. Fig. 11 shows the variation of both vo/vi and φ as a function of ω. Here the value of ω0 has been taken to be 20 rad/s.

0

0.2

0.4

0.6

0.8

1

-80

-40

0

40

80

0 50 100 150 200

|H(jw)|

Phas

e (d

eg)

ω (rad/s) Fig. 11 Variation of |H(jω)| (solid curve) and φ (vo re vi; dotted curve) as a function of ω.

3

The circuit is called a "notch filter" because just at the radian frequency ω0 it rejects the input, so that no signal arrives at the output. One of its uses is to reject unwanted signals (e.g., 60 Hz noise) that would otherwise interfere with some detected signal. Some of the characteristics of this circuit can be understood from simple considerations. At very low frequencies the lower resistively coupled branch transfers all of the input signal to the output. At very high frequencies the upper capacitively coupled network behaves like a short circuit and similarly passes the whole input signal to the output. At the notch frequency ω0 there is a phase lead (vo re vi) of 90˚ in the upper branch, and a phase lag of 90˚ in the lower branch. At this frequency the magnitudes of the two signals are the same, so the two cancel. 2. A Series Resonant Circuit

vi

L C

R vo

Fig. 12 Series LCR circuit

The transfer function of this circuit is

€

H(s) =vovi

=R

R + sL +1sC

=RL

s

s2 + s RL

+1LC

(11) We can factor the denominator and get

€

H(s) =RL

s

s+R2L

+R2

4L2−1LC

#

$ %

&

' ( s +

R2L

−R2

4L2−1LC

#

$ %

&

' (

(12)

If

€

R2

4L2<1LC

the root will be imaginary, so we would write

€

H(s) =RL

s

s+R2L

+ j 1LC

−R2

4L2#

$ %

&

' ( s +

R2L

− j 1LC

−R2

4L2#

$ %

&

' (

(13)

4

What does the pole-zero plot look like? Suppose we have imaginary roots, giving us eq. 13. The numerator tells us that there is a zero of H(s) at the origin, and there are two

poles at

€

s = −R2L

±1LC

−R2

4L2

The pole-zero plot looks like this:

x

x

σ

jω

Fig. 13 Pole-zero plot with one zero at origin and two complex conjugate poles The poles are at positions on the s-plane that show how the circuit would behave if it were excited by an impulse: this would produce a ‘ringing’ signal at the frequency

€

ω =1LC

−R2

4L2 and decaying exponentially with the factor

€

e−R2Lt .

To qualitatively see the frequency response for an input sine wave draw a signal point along the ordinate at position jω, and imagine changing ω by moving the point upward along the axis. Draw a ‘zero-vector’ from the origin to the signal, and two pole vectors from the poles to the signal, like this:

5

x

x

σ

jωSignal

Fig. 14 One zero-vector and two pole-vectors drawn to a signal point The magnitude of the transfer function (i.e. output/input) is given by the length of the zero-vector divided by the product of the lengths of the two pole vectors. You can see that this quantity will be a maximum when the signal point is closest to the pole: at that point the upper pole vector will be shortest, so the denominator will be a minimum. If the signal point is at zero (DC) the output will be zero, and if ω is very high the three vectors

will be almost the same length, so |H(s)| will approach

€

RLω

ω2 =RωL

.

We can also see what happens to the phase of the output with respect to the input. The phase is the (phase of the zero-vector) – (sum of the phases of the pole vectors). With the signal point shown the sum of the pole vector phases would be something like (75-45)˚ degrees and the zero vector has a phase of 90˚ degrees. So the output voltage would lead the input by about (90-30) = 60˚ . If the frequency were raised to a point where the signal passes the upper pole, the phase shift will pass through zero degrees. At a very high frequency the output will lag the input by 90˚. The case discussed above is for the solution of H(s) with imaginary roots (eq. (13). If the product LC is increased the resonant frequency will decrease and each of the poles in fig, 13. will approach the σ axis. When the square root is zero, the two poles will merge on the σ-axis, forming a double-pole. This is the case called critical damping – with an impulse the response is no longer no longer a damped sine wave, but is exponential with the most rapid asymptotic approach to zero. If the square root becomes positive, eq. (12) shows that there will be two poles on the negative σ-axis. As LC is decreased, one pole will move towards the origin, and the other will move away. 3. Compensated Attenuator

6

A resistive voltage divider is a simple example of a circuit that produces a reduced output voltage for some input. Often, though circuit capacitances ‘spoil’ the output – they may for example distort an input pulse. An oscilloscope probe is a good example: often there is a ‘x10’ setting on the probe to reduce the voltage fed to the scope input, but the scope’s input capacitance provides an extra unwanted element like C1 in fig. 15A below:

vi

C1R1

R2vo

vi

R2

R1 C1

C2

vo

A B

Fig. 15 A. Uncompensated attenuator, B. Compensated attenuator We would like the output to be smaller than the input by just the divider ratio R1/(R1+R2). But the capacitor C1 (which could be the scope’s input capacitance) will reduce the output at high frequencies. If the input voltage is a step or pulse, the capacitance will spoil the rise time. If we can somehow modify the divider so that it reduces sine waves of all frequencies by the same factor, then the divider becomes ‘perfect’. Output pulses also will be perfect (reduced) copies of the input if all their Fourier components are reduced by the same factor. This is accomplished by adding capacitor C2 (fig. 15B) to ‘compensate’ the divider. It is easy to calculate its capacitance. The s-impedance of a parallel RC pair is

€

R 1sC

R +1sC

=R

1+ sτ, where τ is the RC time constant. We can then write the transfer function

of circuit B as

€

H(s) =

R11+ sτ1

R11+ sτ1

+R2

1+ sτ2

(14)

We can immediately see that if τ1=τ2 the ratio is just R1/(R1+R2), independent of frequency.

7

What are the poles and zeros for this circuit? With some algebra we can manipulate eq. (14) to get an s in the numerator and another in the denominator, like this:

€

H(s) =R1(1+ sτ2 )

(R1 + R2) + s(R1τ2 + R2τ1)

This says that there is a single zero at s=–1/τ2 and a single pole at

€

s = −R1 + R2

R1τ2 + R2τ1

This can be written as

€

s = −1

R C= −

1τ

where R|| is the parallel combination of R1 and

R2 and C|| is the parallel combination of C1 and C2.

€

We can draw the two poles this way:

jω

σx-1/τ1 -1/τ||

zero pole

If τ1≠τ|| there will be phase shifts at different frequencies. But if τ1 = τ|| (which implies τ1=τ2 ) the pole merges with the zero. The two annihilate! This gives our frequency independent transfer function. The process of finding such a circuit is called “pole-zero cancellation”.


12.4 Laplace Transform

Recall: we found that we could represent any physically realistic waveform as an infiniteset of sine waves using the Fourier Transform:

F (ω) =∫ ∞−∞ f (t)e−jωtdt

This constitutes an integral only along the imaginary axis of the s-plane. We couldconstruct functions elsewhere in the s-plane, which amounts to allowing for an exponen-tial rise or fall of the signal amplitude.

The Laplace transform allows us to decompose any time domain function as compo-nents in the s-plane.

L(f (t)) = F (s) =∫ ∞0 f (t)e−stdt

• Note that the integration range has been restricted to t = 0→∞, so we call it the“one-sided Laplace transform”.

• Since it represents only signals starting at t = 0, there are waveforms it cannotreproduce.

• e.g. An infinitely long sine wave cannot be synthesized by Laplace transform com-ponents, but one beginning at t = 0 and extending to t =∞ can.

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12.4.1 Inverse Laplace Transform

The inverse Laplace Transform is

f (t) = L−1(F (s)) =1

2πjlimT→∞

∫ γ+jTγ−jT e

stF (s)ds

where γ is a real number chosen such that the contour path of integration is in the regionof convergence of F (s). Often γ = 0 and we can write

f (t) =1

2πj

∫ jω=+∞jω=−∞ estF (ω)dω

The techniques for calculating the iLT are treated in detail in texts on complex variables.A large number of practical cases can easily be looked up in tables, which will be themethod we use here.

163 April 11, 2014


12.4.2 Step Function

A step function,is perhaps the simplest Laplace transform

F (s) =∫ ∞0 1× e−stdt =

1

swhich produces a “spike” around a pole at s = 0.

Laplace transforms are very convenient for handling pulses (recall that Fourier trans-forms were not!). Pulses can be constructed by adding step functions,


and the Laplace transform is

F (s) =1

s(1− e−sτ).

164 April 11, 2014


12.5 Laplace Transform Applications

Like the Fourier Transform, the Laplace transform changes a differential equation suchas

V = I(t) R + LdI(t)

dt+

1

C

∫I(t) · dt

into an algebraic equation

v(s) =R + sL +

1

sC

~Iesti(s)Time Domain Frequency Domainf (t) F (s)dfdt sF (s)− f (0)∫ t0 f (t)dt F (s)

s

Calculating circuits in the s-plane using Laplace transforms

1. Calculate the circuit impedances exactly as you would for a group of resistors oper-ating at DC, (or complex impedances at AC), replacing

ZR = R, ZC = 1/(sC), ZL = sL

2. For the signal voltage calculate (or look up) the Laplace transform of the time domainsignal v(t).

165 April 11, 2014


3. The result will be the Laplace transform of some quantity (e.g. i(s) for the current,vo(s) for the voltage).

4. Calculate (or look up) the inverse LT to get back to the time domain.

All the usual network techniques, including Thevenin’s theorem, Norton’s theorem,Kirchoff’s laws, and superposition apply.

166 April 11, 2014


12.5.1 Example: RC circuit with step function input

Consider the RC circuit

vin

C

R

vout

and let vi(t) be a voltage step vi = 0→ V at t = 0The input waveform in Laplace space is vin(s) = V/s.We calculated the transfer function in Sec. 12.2,

H(s) =voutvin

=R

R + 1sC

=sCR

sCR + 1=

s

s + 1/τ

and so the output is

vout(s) =V

s× s

s + 1/τ=

V

s + 1/τ.

Referring to the table of Laplace transforms provided in the equation sheet appendix,we find that L(e−at) = 1/(s + a), to the time domain answer is

vout(t) = V e−t/τ

167 April 11, 2014


12.5.2 Example: RC circuit with a linearly rising input

Now consider the same circuit,

vin

C

R

vout

but with a linearly rising input vin(t) = at.The LT gives us

vin(s) = a∫ ∞0 te−stdt = a/s2

so the output is

vout(s) = vin(s)×H(s) =a

s2× s

s + 1/τ=

a

s(s + 1/τ ).

To find the iLT, lets factorize this into two terms

vout(s) =a

s(s + 1/τ )=aτ

s− aτ

s + 1/τ.

and look-up the iLT for each term

vout(t) = aτ (1− e−t/τ).

The result is a differentiated signal vout = RC dvin/ dt, that becomes “correct” after the

intial rise time (dictated by the time constant RC).

168 April 11, 2014



169 April 11, 2014


170 April 11, 2014

§ 13

Transmission Lines

Electromagnetic waves (light!) travels at c ' 3 × 108 m/s in vacuum. The typicaltransmission speed for electrical signals on a wire is about 3/4c. As a rule of thumb, thepropagaton rate is about 1 foot per nanosecond.

171


13.1 The Wave Equation

Consider a wave of arbitrary shape f (x) that begins is initially (t = 0) at the origin andis travelling in the +x direction with speed c.

An observer moving along with the wave would describe events using the coordinatex′ = x−ct, and would simply observet he unchanged function f (x′). In the (x, t) frame,the equation of this right-moving wave is

y = f (x− ct) or y = f (t− x/c)

and a left-moving wave would be described

y = f (x + ct) or y = f (t + x/c).

We are often interested in sinusoids (especially since any function can be expressed asa sum of sinusoids). A right moving sinusoidal wave is

y = A sinωt′ = A sinω(t− x/c) = A sin(ωt− kx)

(and a left moving sinusoid is y = A sin(ωt+ kx).) Using the complex phasor notation,the right-moving sinusoidal wave becomes

y = Aej(ωt−kx)

this represents a rotating phasor–the sine wave is the projection of this function ontothe real or imaginary axis.

172 April 11, 2014


For y = f (x− ct),∂y

∂x= f ′(x− ct) and

∂2y

∂x2= f ′′(x− ct)

with ’ denoting the derivative with respect to (x-ct). Similarly

∂y

∂t= −cf ′(x− ct) and

∂2y

∂t2= c2f ′′(x− ct).

Combining these partial derivatives, we arrive at the wave equation

∂2y

∂x2=

1

c2

∂2y

∂t2

A system which satisfies this equation describes a wave travelling along the x-axis withspeed or propagation c.

173 April 11, 2014


13.2 Transmission Lines

Information (in the form of waves) is transmitted from one place to another (such asacross a circuit board or from one instrument to another) on transmission lines.

A transmission line might consist of a simple wire, a coaxial cable, or a more compli-cated system such as a stripline, twisted pair cable, or optical fibre.

Thus far, we have approximated the wires in our circuits as having zero resistance,capacitance, and inductance. We have not yet considered the transmission time. Theseapproximations are far from the reality.

A common transmission line is a coaxial cable, like the RG-59 cables often used inlabs to connect to measurement devices.

(source: RG-59 coaxial cable. A: Plastic outer insulation, B: Copper-clad aluminium braid shield conductor, C: Dielectric, D: Copper-clad steel central conductor.

http://en.wikipedia.org/wiki/File:RG-59.jpg)

174 April 11, 2014

http://en.wikipedia.org/wiki/File:RG-59.jpg


• The space between the conductor and shield forms a tubular capacitor, with a di-electric in between. This creates a capacitance per unit length C ′.

• The conductor is surrounded by a magnetic field, creating an inductance per unitlength L′.

• We will assume resistance per unit length R′ is small and can be neglected, as isoften the case. This is a “loss-less” transmission line.

The cable can be modelled as an infinite number of infinitely short sections, each withcapacitance C ′ · dx and inductance L′ · dx,


• Since a current flows in each capacitance, there will be a change in the current fromi to i′ due to C ′, ∆i = C dv

dt .

• Since a changing current causes a voltage drop across the inductance L′, so therewill be a difference in the voltage from v to v′, ∆v = Ldi

dt.

175 April 11, 2014



Assuming a linear variation of v and i across each element, we find

v − (v +∂v

∂xdx) = L′dx

∂i

∂t→ ∂v

∂x= −L′∂i

∂t(13.1)

differentiating with respect to x

∂2v

∂x2= −L′ ∂

2i

∂t∂x(13.2)

And similarly,

i− (i +∂i

∂xdx) = C ′dx

∂v

∂t→ ∂i

∂x= −C ′∂v

∂t(13.3)

differentiating with respect to t

∂2i

∂x∂t= −C ′∂

2v

∂t2(13.4)

Combining equations 13.2 & 13.4, we arrive at the wave equation

∂2v

∂x2= L′C ′

∂2v

∂t2

176 April 11, 2014


with a velocity of propagation c = 1/√L′C ′.

A typical coaxial cable (such as RG62U) has L′ = 0.38µ H/m and C ′ = 44 pF/m,giving c = 2.45 × 108 m/s (about 80% of the speed of EM waves in free space). Thisreduces velocity is due to the dielectric medium between the conductor and shield.

See this animation http://en.wikipedia.org/wiki/File:Transmission_line_

animation.gif.

13.2.1 Characteristic Impedance

Our equation 13.1 for the voltage variation

∂v

∂x= −L′∂i

∂tprovides a means of calculating the characteristic impedance Z0 of the transmissionline. Beginning with v = f (x− ct), the derivative is

∂v

∂x= f ′(x− ct) = −L′∂i

∂t

∂i

∂t=f ′(x− ct)

L′

which we can integrate over t

i =f (x− ct)

cL′+ Const =

v

cL′+ Const

177 April 11, 2014

http://en.wikipedia.org/wiki/File:Transmission_line_animation.gif.

http://en.wikipedia.org/wiki/File:Transmission_line_animation.gif.


the constant is a DC current and is not relevant for the wave description.The ratio of the amplitude of a single voltage wave to its current wave is

v

i= cL′ =

L′√L′C ′

=

√√√√√√L′C ′

= Z0.

This new quantity, the characteristic impedance Z0, has units of ohms and appearsas a pure resistance, which may seem surprizing since our model contained no resistiveelements.

The RG62U cable has Z0 = 93Ω.Naively, one may have guessed that a transmission line would behave like a simple

parallel plate capacitor (created by the conductor and ground). However, if we send apulse down the input end of the line it will not all charge up instantaneously, since thewave moves down the line at a finite speed. Therefore the current flowing into the inputend will be constant (at least until it is affected by reflections on the line).

For a wave sent through a transmission line, normally we wish to have as much poweras possible is absorbed by the load at the far end of the line and as little power aspossible will be reflected back towards the source. This happens when the transmissionline is matched, meaning the load impedance is equal to Z0.

178 April 11, 2014


13.2.2 Reflections and Line Termination

Consider a signal sent down a transmission line (schematically, we could suddenly con-nect a battery to the end of a transmission line, such that this step function in voltagesends a pulse down the cable).

The line has characteristic impedance Z0.The signal propagates at speed c = s 1√

L′C ′, charging the inductances and capacitances

as it moves along the line.Pictorially, a right-going signal produces a V+, I+ moving together down the line:


the ratio of V/I is Z0 everywhere. A “+” subscript is used to denote a right-goingvoltage or current.

Placing the signal at the opposite end of the line:


179 April 11, 2014


Since the I+ and I− currents are moving in opposite directions, |I+| = −|I−|.For right-going waves, V+/I+ = +Z0 (current is moving in the +x direction)For left-going waves, V−/I− = −Z0 (current is moving in the −x direction).

If the transmission line is lossless (no resistance along the wires or between the wires),the Z0 is pure real and the transmission line can be terminated by placing a resistorRL = |Z0| at the end of the line.


Since the current to voltage ratio is V/I = |Z0| = RL when the signal reaches the endof the line, Ohm’s law is satisfied, and all the current can flow through R0, depositingthe entire signal power there (with no reflected power).

In this scenario, the load is properly “matched” to the cable impedance. In essence,the signal behaves as if it is flowing down a lossless transmission line, that is connectedto another identical transmission line of infinite length.

13.2.3 Mis-matched Transmission Lines

What happens if the line is terminated with an impedance ZL 6= Z0 ??

The situation can be resolved if there is more than one wave travelling on the line. A

180 April 11, 2014


left-going reflected wave is introduced,

V+ + V− = VL (13.5)

I+ + I− = IL

and since I− = V−/R0

V+

R0− V−R0

=VLRL→ V+ − V− = VL

R0

RL

and adding & subtracting the above with Eq. 13.5, the voltages of the incident andreflected waves can be found

V+ =VL2

1 +R0

RL

, V− =VL2

1− R0

RL

The reflection coefficient is the ratio of the reflected to incident waves

ρV =V−V+

=RL −R0

RL + R0

The current reflection coefficient is

ρI =I−I+

= −ρV

181 April 11, 2014


13.2.4 Unterminated Cable

This is a cable with nothing at the far end. In this case

RL =∞, ρV = +1, ρI = −1

and all the power is reflected. This makes sense, because power needs to be dissipatedsomewhere, and there is no resistance in our idealized lossless cable.

The superposition of the incident and reflected waves looks like:


the two waves add coherently to produce double the voltage and no current at theoutput. The extra voltage is coming from the charge stored in the inductors L′ beingtransmitted to the capacitors C ′.

182 April 11, 2014


13.2.5 Short-circuited Cable

This is a cable with nothing at the far end. In this case

RL = 0, ρV = −1, ρI = +1

and all the power is reflected. This makes sense, because power needs to be dissipatedsomewhere, and there is no resistance in our idealized lossless cable.

The superposition of the incident and reflected waves looks like:


there is no voltage at the output, as expected, since the voltage is flowing across a short.

13.2.6 Reflections

Reflections on transmission lines can be problematic, as they corrupt the informationcontent of the signal. Consider a digital transmission


183 April 11, 2014


and how a single pulse might look after superimposing several reflections


13.3 Two-port S-parameters

(source: http://en.wikipedia.org/wiki/Scattering_parameters)

13.4 Dispersion and Lossy Cables


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http://en.wikipedia.org/wiki/Scattering_parameters

Telegraphy and Cable Transmission

1798!! De Silva (Spain) sends message 40 km -sparks from Leyden jars1839: Cooke and Wheatstone – England: needle telegraph patent1844: Samuel Morse USA: recording telegraph patentBy 1850: nearly all continents had extensive telegraph land-lines1850: First submarine cable laid from Dover to Cap Grisnez1858: 1st transatlantic cable: insulation blown out by 2kV pulses!1866: Cable laid by the Great Eastern ship from Ireland to Newfoundland

– success!By 1900: nearly all industrialized countries linked by undersea and

overland cables2004: Fibre-optic cables and wireless have largely superseded copper

DispersionBUT: Transmission speed low, ~ 40 characters/minuteAnd speech transmission over long distances was impossibleThe culprit: Dispersion- velocity is frequency dependentThe reason: Line resistances

Equations

)1()( RitiL

xv

+∂∂

−=∂∂

)2()( GvtvC

xi

+∂∂

−=∂∂

Equations of the realistic lineFor sine waves, with , these equations become

)3()( iRLjxv

+−=∂∂ ω

)4()( vGCjxi

+−=∂∂ ω

Differentiate (3) by x and substitute from (4)

(5)k where))(( 22

2

jvvGCjRLjxv

+==++=∂∂ αγγωω

Solution: )6(][ tjjkxxjkxx eeBeeAev ωαα += −−

Phase velocity of the wave

)7(884

1 22

2

22

2

2 ⎥⎦

⎤⎢⎣

⎡++−=

LR

CG

LCRGLCk

ωωωω

If we solve for k in (5) and keep only terms to 2ndorder in the binomial expansion, we get

This says that ω/k (the velocity) is frequency-dependent - The higher ω waves move faster, messing up the received signal.

Solution: Oliver Heaviside (1887): ADD inductances to theline to make

L/C=R/G (8)

The bracketed term then becomes = 1!

Then LCk ω= as in the zero-resistance cable

and LCvp

1=

Undersea Fibre-Optic Cables in 2004


194 April 11, 2014

§ 14

Op-amps

195

Operational Amplifier Lecture

What is an operational amplifier (opamp)?Vout = Ga * ( V + - V -

)

Vs+ and Vs- represent the power supplies to the amplifier

Figure 1 – Symbol for operational amplifier – taken from http://en.wikipedia.org/wiki/Operational_amplifier

Ideally:

• Gain Ga is infinitely large

• Infinite input impedance - No current flows into the V+ (Vp) or V- (Vn) terminals

• Infinite Gain Bandwidth Product – Can work well at all frequencies

What can we build with opamps?

Opamps are extremely versatile see: http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf

We can use opamps to build:

• Filters

• Summers / Subtracters

• Integrators / Differentiators

• Current / Voltage sources

• Rectifiers / Peak detectors

• Oscillators

Basic Opamp analysis technique:

1) Assume that no current flows into the input terminals of the opamp

2) Assume that the V+ and V- terminals are at the same voltage (result of very large gain)

Non-inverting amplifier

(assumed no current flowing into amplifier V- terminal)

What is the gain in the above circuit?

Inverting amplifier

Iin= (Vin – 0V ) / R1

Vout = 0V – R2 * Iin = - ( R2 / R1) * Vin

Note that Iin must also flow through R2

What happens if the gain of the opamp is not infinite?

Vout = Ga * (Vp – Vn) = - Ga *Vn (1)Iin=(Vin-Vout)/(R1+R2) (2)Vn= Iin*R2 + Vout (3)

Vout= - Ga * R2*(Vin-Vout)/(R1+R2) – Ga *Vout

Vout (1+Ga-Ga*R2/(R1+R2)) = -Ga * R2 *Vin /(R1+R2)

Vout/Vin = -R2 / ( (1/Ga+1)*(R1+R2) – R2)

So you can work it out with non infinite gains but in general its not needed (typically an off the shelf opamp will have 100k gain – gain will drop with increased frequency though)

What do you think this circuit does?

Low pass filter:

This is the inverting circuit topology

What do you think the cut off (-3dB) frequency is?If we remove R2 what do you think the circuit will do?

For fun what is Vout1 and Vout2 with respect to Vin1?

Remember the diode equation: , where Can you use this type of circuit for anything useful?


204 April 11, 2014

Appendix A

Introduction to Circuit Simulation

In class we will use LTSpice version IV (download from http://www.linear.com/

designtools/software/), which is a repackaging of the open-source circuit simulationtool called SPICE that was originally developed at UC Berkeley. We choose it becauseit is available free, and has versions for mac and windows.

Adam Gilbert has prepared a list of tips for using LTSpice, appended on the followingpages.

205

http://www.linear.com/designtools/software/

http://www.linear.com/designtools/software/

LTspice Tips and Tricks

HotkeysHotkeys can be very useful, especially on a Mac since the tool bar isn't present.

The LTSpice defaults are shown in the right most panel in the image below. These defaults can be changed from the 'Control Panel' 'Drafting Options' 'Hot Keys' section, see left most panel.

Grounds and global nets

You must have at least 1 ground in your circuit – this is used as a reference point for all your voltage measurements. Remember that your multimeter has two wires – in LTSpice for a voltage measurement one of these wires is always touching ground. Note that you should carefully think about where you want to place your ground connection – putting it in a funny place will cause confusion i.e your voltagemeasurements could all be shifted around by a fixed quantity.

Labelled nets with useful names makes probing them more intuitive.

Types of Simulation

In this course you will be using the Transient and AC analysis.

Transient analysis

To perform a Transient analysis on the toolbar go to “Simulate” and then “Edit Simulation command”. For Mac users you have to press 'S' and then right click the mouse to find the appropriate tool

Once there, for a transient analysis you should set the settings to the following. You must either tick thebox “Start external DC supply voltages at 0V” or “Skip initial operating point solution”. Both ticked is also fine.

If you do not tick one of these boxes, the simulation tool will calculate the DC bias point before startingthe transient simulation so you may end up seeing a simulation that looks like the following: i.e you missed all the fun stuff at the start – the capacitor was already charged to 5V before any points are plotted.

After clicking OK on the “Edit simulation command” window, you must left click the schematic to place the simulation directive.

Note that for the “stop time” you can enter times in the format: 1ms , 2us instead of 1e-3, 2e-6

AC analysis

To perform an AC analysis , go to the Edit simulation panel and enter simiar details. Make sure to placethe directive on the schematic. You may only have one type of simulation running at a time so if you have a transient analysis directive on the schematic make sure you either delete it or that it has a semi- colon at the front of it.

In the following image you can see that we are performing an AC analysis and that the previous transient analysis command has been commented out with a ; . Note that this is automatic on the windows version of Ltspice.

Note also that for a you must have an AC source for the ac analysis to work. To achieve this you shouldright click on the source and put a 1 in the box for AC amplitude.

Probing the circuit

You can add multiple cursors to the nets you are probing by right clicking on the net name in the scope window. With two cursors you can calculate voltage/current and time differences.

See the following image:

Note also that you can edit the expression that the LTSpice will plot. An an example, you may wish to calculate the voltage difference across a single component - by labelling both sides it becomes easy. The expression would be “ V(netlabel1) – V(netlabel2)”


(source: http://cds.linear.com/docs/en/product-info/LTspiceIV_flyer.pdf)

212 April 11, 2014

http://cds.linear.com/docs/en/product-info/LTspiceIV_flyer.pdf

Appendix B

Equations

SineWave : A(t) = Apsin(2πft), ARMS = Ap/√

2

ω = 2πf =2π

T

P = IV

V =∆EElectric

q

Impedances in series add direclty.Impedances in parallel add inversely.

Resistors:V = IR

Resistivity : V =∆EElectric

q

ZR = R(Fourier space)

ZR = R(Laplace space)

Capacitors:

C =εA

d, ε = κεε0

V =1

C

∫Idt or I = C

dV

dt

Qstored = C × V

EnergyStored : W = 12CV 2

ZC = 1/(jωC)(Fourier space)

ZC = 1/(sC)(Laplace space)

Inductor:

V = Ldi

dt, i =

1

L

∫V · dt

L =n2µA

l

EnergyStored : W = 12LI2

ZL = jωL(Fourier space)

ZL = sL(Laplace space)

Transformer:

Transformer : V = MdI

dt

V1V2

=I2I1

=n1

n2

Complex Numbers:

rejφ = x+ jy, j2 = −1

r =√x2 + y2, φ = arctan

y

x

ejθ = cos θ + j sin θ

213


Fourier Series:

y(x) = b0 +∞∑1

an sin(nx) +∞∑1

bn cos(nx)

an =1

π

∫ 2π

0sin(nx)y(x) · dx

bn =1

π

∫ 2π

0cos(nx)y(x) · dx

b0 =1

2π

∫ 2π

0y(x) · dx

Fourier Transform:

F (f) =∫ ∞−∞

f(t)e−j2πftdt

f(t) =∫ ∞−∞

F (f)ej2πftdt

F (ω) =∫ ∞−∞

f(t)e−jωtdt

f(t) =1

2π

∫ ∞−∞

F (ω)ejωtdω

(notice that, when written in terms of the angular frequency ω, thesymmetry is affected such that there is a funny normalization for theinverse Fourier Transform.)

y(t) = h(t)⊗ x(t)

Y (f) = H(f) ·X(f)

(⊗ denotes a convolution)Bandwidth Theorem:

∆f∆t ' 1

∆ω∆t ' 2π

The Wave Equation:

∂2y

∂x2=

1

c2∂2y

∂t2

Transmission line Characteristic Impedance:

Z0 =v

i=

√L′

C ′

Other:

Magnitude Ratio[dB] = 10 log10(PoutPin

).

214 April 11, 2014


(source: http://www.stanford.edu/~boyd/ee102/laplace-table.pdf)

215 April 11, 2014

http://www.stanford.edu/~boyd/ee102/laplace-table.pdf


216 April 11, 2014

Appendix C

Waterflow Analogy

John Crawford’s notes on the waterflowanalogy to electric circuits.

217

Current in Circuits—Fluid in PipesAn Analogy

Early in electronic history, investigators thought of electricity as a flow of "something",analogous to the flow of some sort of fluid. Eaarly names for electrical quantities reflectthese ideas, and emphasize the link with mechanical quantities. For example we read ofelectromotive force or electrical pressure for voltage, current for charge flow, andaccumulator for capacitance. Certain aspects of the analogy between the electrical andmechanical quantities are very close indeed. The relationships can be understood byconsidering a simple mass-spring system capable of storing mechanical potential andkinetic energy, and a circuit that can store both electrostatic and magnetic energy.

A Mass-Spring System

m

F

kxx

γx

If we apply a force F to this mass, making it move in the x direction, we can calculate itssubsequent motion. We assume that the spring force kx acts in the direction shown, asdoes air resistance. Here we assume that this viscous resistance is simply proportional tovelocity ˙ x ; the coefficient of viscosity is γ.

The equation of motion for the mass will then be

m˙ x = F − ˙ x − kx (1)

An LRC Circuit

v

LR

C

i

The relation between voltage and current for this circuit is usually written

v = Ldi

dt+ Ri +

q

C(2)

where i is the current flowing in the circuit, and q is the charge on the condenser.

We can express this in terms of the single variable q by remembering that i =dq

dt:

v = L˙ q + R ˙ q +q

C(3)

Rearranging (1) F = m˙ x + ˙ x + kx (1')

We can now compare the 2 equations to pick out the analogous quantities:

v ↔ F

L ↔ mR ↔ γC ↔ 1/k

This says that a voltage is analogous to a force, and inductance to some massive quantity,a resistance to a viscosity, and a spring to "inverse k". This last quantity appearsfrequently in discussions of elasticity, and is defined as compliance. A very compliantspring is one with a low value of k; a small force would produce a large spring extension.

This still doesn't look much like a current flow. However, if we think about the elementsof a plumbing 'circuit', we can construct a loop with elements that have mass, viscousresistance, and springiness. It might look like this:

Pump

Flywheel

Constriction

Accumulator

Reservoir

elasticmembrane

v

LR

C

ivc

The corresponding electrical elements are shown to indicate analogous objects. The pumpprovides a pressure (the force quantity), the moving massive element is a flywheel, whichspins as fluid passes it, and the resistive element is a constriction in the pipe. The deviceon the right is a cylinder with an elastic membrane stretched across it. This acts as anenergy and fluid storage device; in hydraulic engineering it is referred to as anaccumulator.

To get a feeling for the way that the fluid circuit might behave, imagine that the pumpproduces a sudden step of pressure, like this:

pressure

t

P

Initially, the inertia of the flywheel would prevent a sudden current flow in the loop—thefull pressure change would appear in the pipe between the pump and the flywheel.However, in time the flywheel would start to spin, and pressure would then build upbetween the flywheel and the resistance. With current flowing, there would be a pressuredrop across the constriction in the pipe proportional to the speed of the current flow. Theflowing current would distort the membrane in the accumulator, and the pressure in thispart of the pipe would steadily increase. Since the membrane prevents continuous flowback to the reservoir, in its final state the circuit would simply contain fluid at rest, withpressure everywhere equal to P. However, before this final equilibrium state, thebehaviour of the current would be oscillatory because of the inertial behaviour of theflywheel. We might expect to see the pressure in the accumulator act like this:

pressure

t

P

In the electric circuit, the behaviour is very similar. When a voltage step V is applied, theinductance initially produces a 'back emf' equal to the voltage step; this back emfgradually decreases as current builds up, and as di/dt begins to decrease. The risingcurrent produces a voltage drop across the resistor. The current flow begins to charge thecapacitor, and the voltage across the capacitor increases.

Charge still continues to flow into the capacitor even after vc has reached the applied stepvoltage V. At this point, the current starts to decrease, but does not reverse—with di/dtnow negative, the inductive voltage is in the same direction as the generator voltage, andcurrent keeps flowing into the capacitor. With low resistance, the maximum voltage willapproach 2V when the instantaneous current i finally drops to zero. The subsequentbehaviour will be a damped oscillation. When this finally stops, the voltage throughoutthe circuit will be V, and the current flow will be zero. Note that the capacitor does notallow any steady DC flow in this final state.

Energies

The potential energy stored in a stretched spring is 1

2kx2 and the electrostatic energy

stored by a charged capacitance is 1

2CV2 =

1

2C

q

C

2

=1

2

q2

C. Here again, the capacitance

behaves like a stretched spring, with C playing the rôle of a compliance.

The kinetic energy stored by a mass moving at velocity ˙ x is 1

2m˙ x 2 , and the magenetic

energy stored by an inductance carrying current i is 1

2Li2 , or

1

2L˙ q 2 . So again, L is a

“mass-like quantity.

There are simple and important energy arguments that tell us how C’s and L’s should actin circuits. Each of these elements stores some energy W, and that energy cannot be

changed instantaneously. If it were possible to do so, dW

dt would be infinite. This would

mean that if we dumped the energy from an L or C instantaneously, its pulse amplitudewould be infinite. If we tried to pump energy into an element in zero time, the powersupply would have to supply energy at an infinite rate. In any real circuit, then, it takestime to change the energy stored.

In simple and explicit terms:

1. You can’t instantaneously change the voltage across a capacitor.2. You can’t instantaneously change the current in an inductance.

We can illustrate the usefulness of rule 1 by considering the behaviour of simple seriesand parallel combinations of R and C.

Blocking Capacitors

Here is an RC combination often used to connect successive stages of circuits. We showit with its fluid equivalent. Imagine that a pulse (of pressure in the fluid case, and ofvoltage for the electrical circuit) is applied. What happens to the output pressure po(voltage vo) on the resistive element?

In the fluid circuit the pressure change on both sides of the membrane at the instant of thepulse step will be the same. Since there is now a pressure drop across the resistiveelement, current will begin to flow. The membrane will start to stretch, storing elasticenergy. If the pulse is sufficiently long, the current will eventually drop to zero, since themembrane is not supposed to leak.

p

C

v R

vopo

Now, if the pressure from the pump suddenly drops to zero, the membrane will try toreturn to its ‘relaxed’ position. As it does this, the pressure po will become negative (i.e.,lower than the reservoir pressure) and fluid will flow backwards. Depending on the pulselength, the pressure po will look like one of the curves below:

The form of the pulse depends on the actual accumulator and resistive elements used andon the pulse length. But note that in all cases both the positive and negative steps of theoutput are exactly the same height as the amplitude of the applied pressure step P. Thedecay during current flow will be exponential.

We will have exactly the same behaviour in the RC circuit. The output pulse vo will havea step amplitude V for both the positive and negative transitions — our rule 1 requiresthat this be true. The decaying part of the curve will be exponential in form with timeconstant RC. The curves are identical to the diagram above, with vo replacing po.

The capacitor prevents any constant DC (unidirectional) current flow. We can calculate

the current flowing in the resistor at any time: by Ohm’s law it will simply be vo

R. Since

no continuous DC can flow — only current change — this implies that the area under thepositive portion of the vo curve (which represents the charge flowing) must be equal tothe area under the negative portion.

Because the capacitor blocks DC current flow (hence the name ‘blocking capacitor’), itisolates the DC voltages on its left and right side but allows voltage variations to pass.The general rule for designing such a blocking capacitor in signal applications is tochoose a time constant RC>>the time for signal variation (e.g., the length of an inputpulse). The output pulse in this situation would look like the picture (c) above.

Often we deal with sinusoidal waveforms (or Fourier superposition of sines). In suchcases, if we want most of the generator voltage to appear across the resistor, we shouldchoose the capacitive impedance Zc <<R. The capacitor still blocks DC effectively, butallows the AC signal (i.e., the variation) to pass.

Oscilloscopes are provided with switchable input blocking capacitors. If the scope inputswitch is set to DC there is a direct connection from the circuit being probed to thescope’s input amplifier, with some high resistance to ground (ususally ~ 1MΩ).However, if the input switch is set to AC, a capacitor is connected in series to the input.This allows you to see variations in the input signal, but it blocks DC voltages.

P

P

PP

P P

(a) (b) (c)

Charging a Capacitor Through a Resistor

If we turn the circuit around so the resistor is on the pump (generator) side, we have thisarrangement:

With an input pulse, fluid (charge) fills the accumulator (capacitor), building up pressure(voltage). Note that rule 1 still tells us what happens to the capacitor: when the inputpulse is applied, the voltage across the C can’t change instantaneously, so vc must start atzero. The initial current flow is therefore limited by the resistor. A solution to the simpledifferential equation for this circuit tells us the form of vo after the generator voltage stepV is applied. It is:

vo = V 1 − e− t

RC

A Bypass Capacitor

In many circuits we see parallel combinations of R and C. Imaging that such a pair isconnected to some constant current generator. This arrangement and its fluid equivalentare shown below:

pv

vopo

R

C

vov

Note that if we want to find the voltage variation vo in response to a step of current I, wecan transform the circuit to its Thévenin equivalent. The circuit would look exactly likethe charging circuit of the previous section, and the output voltage after the initial currentstep would just be:

vo = IR 1 − e− t

RC

Here, at t=0, the output voltage starts at zero; again rule 1 says that this must be true.Initially all current flows into the capacitor, and it is only when voltage builds up that theresistor starts to share some of the current. (In the fluid circuit the accumulator takes allthe initial current flow, and none flows into the restricted branch.) It is only when voltagehas built up to IR that the resistor takes all the current, and the capacitor (which cannotpass DC) takes none. The initial current in the capacitor in this case bypasses the resistor— hence the name.

If the current generator in this case produces not a voltage step but sinusoidal AC, and thecapacitance is very large, essentially all the current will flow through the capacitivebranch and very little will flow through the resistor. This is the condition correspondingto ZC = R .

Flow

Pump

i

vo

CR

I

po

Gauge

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