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Robert Sobot Wireless Communication Electronics by Example Wireless Communication Electronics by Example Robert Sobot Wireless Communication Electronics by Example 123 Robert Sobot Electrical and Computer Engineering Western University London, ON Canada ISBN 978-3-319-02870-5 DOI 10.1007/978-3-319-02871-2 ISBN 978-3-319-02871-2 (eBook) Springer Cham Heidelberg New York Dordrecht London Library of Congress Control Number: 2013951765 Springer International Publishing Switzerland 2014 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com) To Allen Preface This tutorial book comes as a supplement to my ‘‘Wireless Communication Electronics, Introduction to RF Circuits and Design Techniques’’ textbook, which resulted from my lecture notes in ‘‘Communication Electronics I’’ undergraduate course that I was offering over the last 7 years to students at Western University in London, Ontario, Canada. My main inspiration to write this tutorial book again came from my students who would always ask ‘‘How to I practice for this course?’’ while being frustrated for having to browse through large number of books in order to find a few applicable examples for practice. In addition, most of the modern undergraduate textbooks follow the same approach of giving a number of solved examples, accompanied by a number of problems with no solutions. Feedback that I received from my students is that without being able to verify both the final results and the methodology, it is very discouraging to practice the unsolved problems from most textbooks. Consequently, many students become discouraged and never take that first most critical step. In this tutorial book, I choose to give not only the complete solutions to the given problems, but also to give detailed background information about the relevance of the problem and the underlying principles. By doing so, my hope is that my students will be able to make their first steps with my help, and then to develop their own confidence and understanding of modern electronics. In order to solve engineering problems, one must first understand the underlying principles, and one must know the basic set of methods of how to solve the problems. The intended audience of this book are primarily senior engineering undergraduate students who are just entering the field of wireless electronics after only first courses in electronics. At the same time, my hope is that graduate engineering students will find this book useful reference for some of the details that have been either only touched upon in the previous stages of their education, or explained from a different point of view. Finally, the practicing junior RF engineers may find this book handy source for quick answers that are routinely omitted in most textbooks. Paris, France, Winter, Spring and Summer, 2013 Robert Sobot vii Acknowledgments Writing a textbook is an undertaking that, by default, relies on ‘‘standing on the shoulders of giants.’’ I would like to acknowledge all those wonderful books that I used as the references and the source of my knowledge, and to say thank you to their Authors for providing me with the insights that otherwise I would not have been able to acquire. Under their influence, I was able to synthesize my own picture of reality, which is what acquiring of the knowledge is all about. My hope is that their guidance and shaping of my own understanding of the topics in this book are clearly visible, hence I do want to acknowledge their contributions, which are now being passed to the readers. In the professional life, one learns both from the written sources and from the ‘‘experience.’’ The experience comes from the interaction with people that we meet and projects that we work on. I am grateful to my former colleagues who I was fortunate to have as my technical mentors on really inspirational projects, first at the Institute of Microelectronic Technologies and Single Crystals, University of Belgrade, former Yugoslavia, then at PMC–Sierra Burnaby BC, Canada, where I gained most of my experiences about ‘‘the real world.’’ I would like to acknowledge contributions of Prof. John MacDougall who initialized and restructured the course into the form of ‘‘design and build’’ concept, and of Profs. Alan Webster, Zine Eddine Abid, and Serguei Primak who taught the course at various times. I would like to thank all of my former and current students who relentlessly keep asking ‘‘Why?’’ and ‘‘How did you get this?’’ with hope that material compiled in this book contains answers to at least some of those questions, and that it will encourage them to keep asking questions with unconstrained curiosity about all phenomena that surround us. Most of all, I want to thank my wife for her unconditional and continuous support, and our son for always being around my writing desk and for making me laugh. ix Contents Part I Problems 1 Introduction: Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2 Basic Terminology: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 3 Electrical Noise: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 4 Electronic Devices: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 5 Electrical Resonance: Problems . . . . . . . . . . . . . . . . . . . . . . . . . 25 6 Matching Networks: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7 RF and IF Amplifiers: Problems . . . . . . . . . . . . . . . . . . . . . . . . . 33 8 Sinusoidal Oscillators: Problems . . . . . . . . . . . . . . . . . . . . . . . . . 37 9 Frequency Shifting: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 10 Modulation: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 11 Signal Demodulation: Problems . . . . . . . . . . . . . . . . . . . . . . . . . 49 12 RF Receivers: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Part II Solutions 13 Introduction: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 14 Basic Terminology: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 15 Electrical Noise: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 xi xii Contents 16 Electronic Devices: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 17 Electrical Resonance: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 161 18 Matching Networks: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . 183 19 RF and IF Amplifiers: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 201 20 Sinusoidal Oscillators: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 215 21 Frequency Shifting: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 22 Modulation: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 23 Signal Demodulation: Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . 243 24 RF Receivers: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 Appendix A: Physical Constants and Engineering Prefixes . . . . . . . . . 267 Appendix B: Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Appendix C: Second Order Differential Equation. . . . . . . . . . . . . . . . 271 Appendix D: Complex Numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Appendix E: Basic Trigonometric Identities . . . . . . . . . . . . . . . . . . . . 275 Appendix F: Useful Algebra Equations . . . . . . . . . . . . . . . . . . . . . . . 277 Appendix G: Bessel Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Bibliographies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 Acronyms A/D AC ADC AF AFC AGC AM BiCMOS BJT BW, B CMOS CRTC CW D/A DAC dB dBm DC ELF EM eV FCC FET FFT FM GaAs GHz HBT HF Hz IC IF InGaAs InP Analog to digital Alternate current Analog to digital converter Audio frequency Automatic frequency control Automatic gain control Amplitude modulation Bipolar–CMOS Bipolar junction transistor Bandwidth Complementary metal–oxide semiconductor Canadian Radio–Television and Telecommunication Commission Continuous wave Digital to analog Digital to analog converter Decibel Decibel with respect to 1 mW Direct current Extremely low frequency Electromagnetic Electron volts Federal Communication Commission Field effect transistor Fast Fourier transform Frequency modulation Gallium arsenide Gigahertz Heterojunction bipolar transistor High frequency Hertz Integrated circuit Intermediate frequency Indium gallium arsenide Indium phosphide xiii xiv I/O IR JFET KCL KVL LC LF LNA LO MMIC MOS MOSFET NF PCB PLL PM pp ppm Q RADAR RF RMS SAW SHF SINAD S/N SNR SPICE TC THD UHF UV VCO V/F VHF V/I VLF VSWR Acronyms Input output Infrared Junction field–effect transistor Kirchhoff’s current law Kirchhoff’s voltage law Inductive–capacitive Low frequency Low noise amplifier Local oscillator Monolithic microwave integrated circuit Metal oxide semiconductor Metal oxide semiconductor field effect transistor Noise figure Printed circuit board Phase locked loop Phase modulation Peak to peak Parts per million Quality factor Radio detecting and ranging Radio frequency Root mean square Surface acoustic wave Super high frequency Signal to noise plus distortion Signal to noise Signal to noise ratio Simulation Program with Integrated Circuit Emphasis Temperature coefficient Total harmonic distortion Ultra high frequency Ultraviolet Voltage controlled oscillator Voltage to frequency Very high frequency Voltage current Very low frequency Voltage standing wave ratio Part I Problems Chapter 1 Introduction: Problems Electromagnetic (EM) waves and their transmission through space are possible due to several relatively simple but fundamental physical phenomena. In modern theoretical physics, EM wave is an abstract concept that epitomizes transfer of energy through the space, which is equivalent to say that information is transferred through the space. At the same time, the information itself is embedded into the wave in the form of the energy variation in time. In this chapter we review some of the fundamental concepts from physics related to energy, matter, EM waves, EM fields, propagation of energy through matter and space, and basic interaction between two waveforms. Problems: 1.1. An average sized snowflake consists of approximately n = 6.68559 × 1019 molecules. Assuming the complete matter of the snowflake is converted into energy, estimate for how long a laptop computer whose average power consumption is P = 25 W could be powered? 1.2. Our Sun produces energy in form of light by continuous nuclear fusion reaction in its core where hydrogen (H) atoms are fused together to form atoms of helium (He). Activity of the Sun is monitored by satellites positioned around Earth that measure the solar constant k S , i.e. average radiated power per square metre at the distance of one astronomical unit ( AU ) from the Sun. Estimated average value of the solar constant is k S = 1366 W/m2 . Knowing the average radius of the Earth r , the average distance between the Earth and the Sun l, and estimated mass of the Sun m S , estimate: 1. what is the total mass (M) of matter at rest that the Sun converts into energy per second? 2. what percentage of the total converted mass M at rest is used to supply energy to the Earth? 3. what is the total mass (M H ) at rest of hydrogen (H) used in the fusion process per second? 4. ignoring all other causes, and assuming that the Sun will collapse after converting 10 % of its current mass at rest, estimate how long the Sun will last? R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_1, © Springer International Publishing Switzerland 2014 3 4 1 Introduction: Problems 5. what is the total power radiated by the Sun and how much is received at the Earth? 1.3. Sketch an EM sinusoidal wave that propagates in x–direction with clearly marked all three axes, as well as the other relevant wave vectors. 1.4. Sketch a rough drawing of EM wave as being generated by a dipole antenna. 1.5. Show that expression E = E m sin(kz − ωt) describes a wave moving in the positive z direction. In addition, express k and ω parameters in terms of the corresponding wavelength λ, the frequency f , then find relationship to the wave velocity v = f (ω, k), i.e. as function of angular frequency ω and wave number k. 1.6. Starting with classical approximation of the wave equation of a sinusoidal EM wave E = E m sin(k z − ω t) derive expression and then calculate the speed of light in vacuum with no electric charges. 1.7. Find the ratio of electric and magnetic field amplitudes of a sinusoidal EM wave whose components are E = E m sin(kz − ωt) and B = Bm sin(kz − ωt). 1.8. We note that vector product of electric and magnetic fields E × B points in the same direction as the wave velocity vector. Find expression for magnitude of a vector S that is defined as S = 1/μ0 (E × B) and determine its measurement unit. Comment on the nature of these results. 1.9. Estimate peak values of electric E m and magnetic Bm components of EM wave generated by the Sun, as measured close to the Earth. 1.10. After learning that Poynting vector represents flow of energy per unit area, and recognizing that energy is related to work, that work is related to force, and the force is related to pressure, we realize that EM wave hitting a perfectly reflective surface must exert “radiation pressure” on the surface. Derive expression for average Poynting vector √S→ as function of the radiation pressure Pr (pressure is measured in N/m2 ). 1.11. A solar sail that is made of ideally reflective light material called Kapton is used to propel a probe away from the Sun by relying only on the pressure generated by the Sun light. Ignore influence of all planets in the solar system, and use results from the previous problems as needed. Assuming that the total mass of the sail plus payload is m = 2 kg, end the specific mass of the used Kapton is m k0 = 768.5 µg/m 2 , estimate mass of the payload. 1.12. The electrical and magnetic components of a composite electromagnetic wave are described as follows, E = E m cos(kz − ωt) y − E m cos(kz + ωt) y B = Bm cos(kz − ωt) z + Bm cos(kz + ωt) z (1.1) (1.2) Find expression for the average value of its Poynting vector S and comment on the result. 1 Introduction: Problems 5 Fig. 1.1 Problem 1.16: a long conductor (relative to the λ) is divided into infinitesimally short sections z ∞ λ, where each of the sections is modelled with distributed circuit elements R, L, C, and G 1.13. Calculate the intrinsic wave impedance, phase velocity, and wavelengths of an electromagnetic wave in free space oscillating at the following frequencies: f 1 = 10 MHz, f 2 = 100 MHz, f 1 = 10 GHz. 1.14. Plot the graph of the radial magnetic field H (r ) inside and outside of infinitely long wire in air of radius a = 5 mm aligned along the z–axis and carrying a DC current of I = 5 A. 1.15. Find the induced voltage of a thin wire loop of radius a = 5 mm in air subjected to a time–varying magnetic field H = H0 cos ω t, where H0 = 5 A/m and the operating frequency is f = 100 MHz. 1.16. Starting from an electrical line section model in Fig. 1.1 derive, first, expression for general characteristic line impedance, then the expression for lossless characteristic line impedance Z 0 . Calculate characteristic impedance of a two–wire transmission line that is characterized as L = 378 nH/m and C = 150 pF/m. 1.17. Instantaneous voltage of EM wave is described as v(t) = Vm cos (2π f t + φ0 ) where, ω = (2π 100) rad/s and φ0 = π/4. Calculate: (a) EM wave’s phase and the front–end’s distance from the starting point at t = 15 ms; (b) wavelength λ of this wave; and (c) amplitudes V of the instantaneous voltages at t = 0 s and t = 15 ms. 1.18. Instantaneous voltage of a waveform is described as v(t) = sin (2π t) + 1 1 1 sin (6π t) + sin (10π t) + sin (14π t) + · · · 3 5 7 Using plotting software of your choice, create the following plots (show the time axis over at least two periods of the slowest tone): 1. 2. 3. 4. on the same graph, plot v(t) and the first three terms terms; plot v(t) using the first ten terms; plot again the case 2. but this time remove the first two terms of the polynomial; try removing some other combinations of terms from the case 2. polynomial. 6 1 Introduction: Problems Fig. 1.2 Problems 1.21 and 1.22: two single tone signals and passive network (left), and time domain plot of an AM modulated waveform (right) Observe time–domain differences among waveforms in terms of their amplitude, rising and falling edges, period, especially relative to the ideal waveform that v(t) is expected to emulate. 1.19. Using the same time–domain square waveform function as in Problem 1.18 analyze the frequency content of waveforms in cases 2 and 3 by creating their respective frequency–domain plots, i.e. signal power versus frequency. 1.20. Given three waveforms, v1 = 1 + 0.5 sin(ω t), v2 = 1 + 0.5 sin(ωt − π ), and v3 = 0.5 + 0.5 sin(ω t − π ), without using any plotting software sketch by hands in correct scale the following plots: 1. 2. 3. 4. v1 , v1 , v1 , v1 , v2 , and v1 (t) − v2 (t); v2 , and v1 (t) + v2 (t); v3 , and v1 (t) − v3 (t); v3 , and v1 (t) + v3 (t); Comment on the created waveforms. 1.21. Two single tone signals, v1 = f (ω 1 , t) and v2 = f (ω 2 , t) are applied at the input nodes of a passive network bellow, Fig. 1.2 (right). Assuming that the two signals are drawn in scale, and v1 (t) has period of T1 = 1 µs, find the following: 1. expression for the output signal, vo (t) = f (v1 (t), v2 (t)), assume R1 = R2 ; 2. sketch to scale the signal frequency spectrum at the output node Vo when R1 = R2 , and clearly label frequencies on the horizontal axis; 3. expression for the output signal, vo (t) = f (v1 (t), v2 (t)), assume R2 = 2 R1 ; 4. sketch to scale the signal frequency spectrum at the output node Vo when R2 = 2 R1 , and clearly label frequencies on the horizontal axis. 1.22. For the given amplitude modulated (AM) signal vin , Fig. 1.2 (left), sketch shape of the envelope waveform. Chapter 2 Basic Terminology: Problems As electrical engineers we often take for granted scientific concepts related to potential and flow of current in conductive and semi–conductive materials. Nevertheless, our understanding of phenomena related to the flow of electricity inside electrical circuit is based on very few basic concepts studied in the previously taken courses. In this chapter we review some of the relevant elements from physics, terms and definitions, and techniques commonly used by electrical engineers while either analyzing or synthesizing electrical circuits. Intuitive understanding of classic concepts related to structure of an atom, internal forces, movement of charges, electrical field, and electric power is important and the following problems serve the purpose of refreshing the prior knowledge. Problems: 2.1. A hydrogen atom consists of one proton (m p , q p ) and one electron (m e , qe ) separated by distance r , where (m, q) pair corresponds to their respective mass and charge. Ignoring all other particles and forces and assuming the hydrogen atom to be at rest, find the ratio R of electrostatic FC and gravitational FG forces between these two particles. 2.2. A system of three charges, initially at rest, is located at three vertices of a square, Fig. 2.1. Sketch direction in which the charge q3 will begin to move assuming that charge q1 is a proton fixed in space, charge q2 is an electron fixed in space, and: (a) the charge q3 is a proton free to move in space; or, (b) the charge q3 is an electron free to move in space. 2.3. Derive expression for electric field around a single charged particle q. 2.4. In an experimental setup, two oil drops whose respective radiuses are measured as r1 = 1.670 µm and r2 = 1.775 µm while the oil’s mass density is ωoil = 837 kg/m3 . The two oil drops are first electrically charged and then suspended in the air by the means of electric field of E = 2.0 × 105 N/C that is acting in direction opposite to the gravity force. R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_2, © Springer International Publishing Switzerland 2014 7 8 2 Basic Terminology: Problems Fig. 2.1 Problem 2.2: a system of three charges located in space For the given data, calculate the difference in electric charges stored on these two oil drops. 2.5. Derive expressions for radial Er and the perpendicular Ep electric fields of an electric dipole at point r √ a, where distance between the two charges making the dipole is 2a. What happens if the derived results are applied for the case when r → 0? 2.6. Derive expressions for radial component of electric field Er at distance r from the centre of a very long metallic wire whose length is l and it is charged with a uniform linear charge density ql [C/m/m]. 2.7. Using a graphing tool of your choice, plot on the same graph various pairs of the following single tone signals at f = 10 MHz along with their respective sums: S1 = 2.0 sin (λt), S2 = 2.0 sin (λt + π/3), S3 = 2.0 sin (λt + π/2) S4 = 1.0 sin (λt + 3π/4), S5 = 2.0 sin (λt + 2π ), S6 = 3.0 sin (λt + 4π/3) For example, plot together S1 , S5 , and (S1 + S5 ) or S2 , S3 and (S2 + S3 ), and so on. In particular, observe their relationships in respect to their phase differences, and at the same time observe how the two amplitudes are related to each other at any given point in time, i.e. at some of the typical instantaneous phase points. For a given frequency, practice mental calculations of conversing various phase differences into the units of time. 2.8. Using a plotting software, create plot that includes the following five single tone signals (assume f = 10 MHz): S1 (t) = 2 sin (λ t), S2 (t) = − sin (2λ t), S3 (t) = 1 2 S4 (t) = − sin (4λ t), S5 (t) = sin (5λ t), · · · 2 5 and then plot their sum 2 sin (3λ t) 3 2 Basic Terminology: Problems 9 S(t) = 5 Sk k =1 Note that aside from the fundamental tone S1 at λ this particular waveform also contains both even and odd harmonics, i.e. both 2λ , 4λ , 6λ ,... and 3λ , 5λ , 7λ ,... terms. How the function Sk looks like? If number of the single tone signals Sk increases indefinitely, i.e. if k → ∞, what waveform shape S(t) is synthesized? 2.9. For a highly simplified case where the noise waveform is described as follows Sn (t) = 1 2 3 4 1 sin (λ t) − sin (2λ t) + sin (3λ t) + sin (5λ t) + sin (9λ t) 5 2 3 4 3 1. Find noise cancellation function Sm that should be used to remove this noise completely; 2. Using plotting software, plot together Sn , Sm and (Sn + Sm ). 3. Assuming realistic signal processing system, how much signal processing delay could you allow in the system, so that the cancellation is still satisfactory? (For the sake of argument, let us assume that “satisfactory” noise reduction is when the maximum noise amplitude is reduced about ten times.) Fig. 2.2 Problem 2.12: schematic diagram of circuit network Fig. 2.3 Problem 2.13: time domain plot of current flow 10 2 Basic Terminology: Problems 2.10. Calculate average power of a square pulse whose amplitude is v = 2V and the pulse width is t = 1 ms and its energy is dissipated in a resistor R = 100 π. What happens if the pulse duty cycle is changed to 25 or 80% ? 2.11. A current flowing in positive direction through an electronic component is defined as: −2t, if (t ≤ 0); i(t) = (2.1) +3t, if (t ≥ 0); 1. Calculate values of the current at the following two time instances i(−2.2 s) and i(+2.2 s); 2. Calculate the total charge g that has flown through the component within the time interval of (−2s ≤ t ≤ 3s); 3. and, find the average value of i(t) within the same time interval. 2.12. With reference to Fig. 2.2 calculate power absorbed/generated by each element in this circuit if: (vs , i s ) = (8 V, 7 A), v1 = 12 V, i2 = 2 A, i3 = 8 A. In this example, a resistor symbol is used to show an element capable of both absorbing and generating power. 2.13. Calculate the average power Pavg delivered/absorbed by a load R = 5 π and i rms for the current whose time domain waveform is shown in Fig. 2.3. Chapter 3 Electrical Noise: Problems Most humans do not even realize the existence of a phenomena known as a “cocktail party effect” where our brain is capable of filtering out sounds in a room full of people and tuning in to a single voice. From the perspective of wireless communication technology the pre twentieth century Earth was similar to an empty room, none to talk and none to listen. Except for the always existing cosmic radiation, the humans did not have a single device that is capable of using EM radiation for wireless communication. In the modern world, the air is always filled in with EM waves generated by countless wireless devices that we have. Now, the situation is similar to a room fool of people, all talking loud all the time. That constant din sets the minimum energy level of “noise”, thus any signal that is to be received must have energy level higher than the noise. In this section we review basic definitions, the theoretical background, and some of the basic techniques that help us design electronic equipment capable of extracting only the wanted signal from the background din. Problems: 3.1. Using plotting software of your choice, sketch two time–domain graphs: (a) plot an example of a voltage noise signal n(t, ω 1 , ω 2 , ω 3 , ...) next to an ideal single–tone voltage signal a(t, ω a ), so that the two voltage amplitudes are related as SNR = 20dB; and (b) plot a noisy single–tone signal A(t, ω) = a(t, ω a ) + n(t, ω), where the noise component is described as n(t, ω 1 , ω 2 , ω 3 , ...). For the purpose of this problem, experiment with the number of noise harmonics used to create the n(t, ω) and their frequencies relative to the single tone frequency so that the graphs are visually illustrative. Also, zoom–in the two graphs to one to two periods of the single–tone signal. Repeat this example for other SNR values (both positive and negative). 3.2. Using plotting software of your choice that is capable of performing FFT, sketch frequency domain plot of the noisy voltage signal from Problem 3.1. What is your observation? R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_3, © Springer International Publishing Switzerland 2014 11 12 3 Electrical Noise: Problems 3.3. Using plotting software of your choice, sketch a quantitative time–domain graph: (a) of two single–ended signals a(t) = VCM + A0 sin(ω t) + n(t) b(t) = VCM + A0 sin(ω t + λ ) + n(t) i.e. after each of the two single tone signals was affected by the same noise n(t); (b) of d(t) that is calculated as the difference of the two single–ended signals, i.e. d(t) = a(t) − b(t) Without affecting generality of the example, and for the simplicity of the plot, assume that the noise signal is of the form n(t) = ( A0/10) [sin(10ω t) + cos(20ω t)]. Choose values of the maximum amplitude A0 , the common mode DC voltage VCM , and frequency ω so that the graph is zoomed-in to one or two periods of the signal. Comment on the properties of signal d(t) relative to the signals a(t) and b(t). 3.4. For the given RLC network, Fig. 3.1, derive expression for its transfer function H ( jω) = vout /vin . Then, using plotting software of your choice, plot amplitude of the network transfer function |H ( jω)|, and find its bandwidth BW by using graphical method. Next, suggest the equivalent ideal (i.e. “brick–wall”) approximation of the transfer function and overlay the two transfer functions. For this exercise, use the following component values: R = 0.1 , 1 , 10 , L = 1 H, and C = 1F. What difference the value of the resistor makes? Explain. It is useful to explore influence of all the components on the centre frequency f 0 and bandwidth BW of the transfer function. For instance, how would you scale the component values so that the centre frequency of the BPF is set to f 0 = 10 MHz? To f 0 = 2.4 GHz? Explore how the behaviour of this RLC network changes if the components change their places inside the network by trying all other possible topologies. 3.5. Sketch a qualitative frequency–domain graph that shows: (a) a white noise power spectral density spectrum; (b) an ideal “brick–wall” bandpass (BP) filter transfer function; and (c) the white noise spectrum after being filtered by the BP filter. 3.6. Using the graph from Problem 3.5, show how the filtered white noise power is estimated graphically. Assume that the white noise PSD = 1 µW/ Hz, while the system is intended to amplify Ps = 100 mW electrical signal whose frequency Fig. 3.1 Problem 3.4: schematic diagram of RLC network 3 Electrical Noise: Problems 13 content is within the standard range of the human hearing. Calculate SNR for both cases, i.e. without and with BP filter applied. 3.7. Two random signal voltage generators are connected in series. The two voltage generators have maximal rms amplitudes e1 = 1 V and e2 = 10 V respectively. Derive the expression for the rms voltage amplitude at the terminals of this two source network. Comment on the result if the specific numerical values in this example are used. 3.8. Resistors R1 = 20 k and R2 = 50 k are at room temperature T = 290 K. For a given bandwidth of BW = 20 kHz find the thermal noise voltage for the three cases shown in Fig. 3.2, i.e.: (a) each resistor separately; (b) their serial combination; (c) their parallel combination; (d) the total available noise power Pn in the cases (a) to (c). 3.9.Two resistors R1 = 20 k at temperature T1 = 285.5K and R2 = 50 k at temperature T2 = 190K are connected in series. For a given bandwidth of BW = 20 kHz first find relevant expressions for the total output thermal noise voltage en , and the total available noise power at the output Pn of these two resistors, Fig. 3.2 (middle), and then, for the given data, calculate their respective numerical values. The term “available power” refers to the power delivered to the load if the load impedance were conjugately matched to the source impedance for the maximum power transfer. However, if the source and load impedances are not matched, then delivered power is obviously less than the available power, which still does not change the amount of the available power (i.e. in this case it would be only partially used). 3.10. Assuming that a realistic source resistor R1 at T1 drives a realistic load resistor R2 at T2 , find expression for noise figure NF at the output of the resistive networks, Fig. 3.2, in the following two cases: (a) the source resistance R1 and the load resistance R2 are connected in series, Fig. 3.2 (middle), and then comment on the conditions required to achieve minimum noise figure; (b) the source resistance R1 and the load resistance R2 are connected in parallel, Fig. 3.2 (right), and then comment on the conditions required to achieve minimum noise figure. 3.11. In the radio astronomy, the local calibration reference resistor is cooled to, for example, TR = 2.5K and its thermal noise power PR serves as the absolute Fig. 3.2 Problems 3.8, 3.9, and 3.10: schematic diagrams of a single realistic resistor R (left), series combination of resistors R1 , R2 (middle), and parallel combination of resistors R1 , R2 (right) 14 3 Electrical Noise: Problems Fig. 3.3 Problems 3.12, 3.13 and 3.14: Schematic diagram of an inverting operational amplifier reference level for the background radiation noise. Assuming an ideal amplifier and the overall measurement setup to be ideal, calculate the equivalent antenna temperatures Ta , if signal power levels measured at the antenna’s output terminals are PS = 3.01dB, 6.02dB, 9.03dB, ... above the background radiation power level? 3.12. Simplified schematic diagram of an inverting amplifier is shown in Fig. 3.3. For this problem assume that: (a) (b) (c) (d) voltage source v S is ideal; the operational amplifier has very high open–loop gain; no current flows into the two input nodes of operational amplifier and thermal noise is the only type of noise present in the system. Derive expression for rms value of the output noise voltage eout if the only noise source in the system is due to the total thermal noise voltage en at the input terminals of the operational amplifier (en is usually specified in the accompanying datasheets). The two resistors R1 , R2 are assumed ideal, i.e. noiseless. Then, calculate rms value of the amplifier’s noise voltage√eout , datasheets for the operational amplifier specify the input noise as en = 5 nV/ Hz, and the amplifier parameters are R1 = 1 k, and R2 = 10 k. 3.13. Use the same circuit and the same set of assumptions from Problem 3.12. However, this time the noisy operational amplifier is specified in terms of: (a) its total input noise voltage en ; and (b) its total input noise currents i n+ and i n− , where the two respective noise currents are associated with the positive and negative input nodes of the operational amplifier, (i n+ and i n− noise currents are also usually equal and specified in the accompanying datasheets). In addition, within the total frequency bandwidth of the system the two noise currents are assumed to be non–correlated. Again, the two resistors R1 , R2 are assumed ideal, i.e. noiseless. First, derive expression for rms value of the total output noise voltage eout , then calculate its numerical value. √ Datasheets for the operational amplifier specify the √ input noise as en = 5.082 nV/ Hz, the input noise current is i n = 5 pA/ Hz, and the amplifier parameters are R1 = 1 k, and R2 = 10 k. 3.14. Use the same circuit and the same set of assumptions from Problem 3.13. However, this time assume that the two resistors R1 , R2 are not ideal, i.e. they also contribute to the total noise power in the system. First, derive expression for rms value of the total output noise voltage eout . Second, calculate its numerical value. Datasheets for the operational amplifier specify the 3 Electrical Noise: Problems 15 Fig. 3.4 Problem 3.15: schematic diagram of an operational amplifier √ √ input noise as en = 5 nV/ Hz, the input noise current is i n = 5 pA/ Hz, and the amplifier parameters are R1 = 1 k, R2 = 10 k, and T = 3.75 ◦ C. Third, calculate the total noise voltage within frequency bandwidth BW = 20 Hz to 20 kHz. 3.15. An amplifier circuit, Fig. 3.4, is connected to the input signal source v S through node 1iso that the source resistance R S is matched to the amplifier’s input resistance R3 . The operational amplifier is assumed to have very high gain and infinite input impedance. Derive expression for the noise figure NF of this amplifier, then, calculate numerical value of the NF. Datasheets for the operational amplifier specify √ √ the total input noise as en = 5 nV/ Hz, the input noise current is i n = 5 pA/ Hz, and the amplifier parameters are R1 = 1 k, R2 = 10 k, R3 = 1 k, and T = 95 ◦ C. 3.16. An amplifier circuit, Fig. 3.5, is connected to the input signal source v S through node 1iso that the source resistance R S is matched to the amplifier’s input resistance. The operational amplifier is assumed to have very high gain and infinite input impedance. Derive expression for the noise figure NF of this amplifier, then, calculate numerical value of the NF. √ Datasheets for the operational amplifier specify √ the total input noise as en = 5 nV/ Hz, the input noise current is i n = 5 pA// Hz, and the amplifier parameters are R S = 800 , R1 = 1 k, R2 = 10 k, R3 = 1 k, R4 is adjusted so that the amplifier’s input resistance is matched with the source resistance, and temperature inside the amplifier box is measured as T = 80.8 ◦ C. Fig. 3.5 Problem 3.16: schematic diagram of an inverting operational amplifier 16 3 Electrical Noise: Problems 3.17. A realistic voltage source with the internal resistance is used to drive an ideal amplifier whose noise figure is NF and the power gain G. Derive expression for noise power No in dBm/Hz at the output terminal of the amplifier, and then calculate it by using the following numerical data: NF = 3dB, G = 12dB, and the environment temperature is T = 15 ◦ C. 3.18. Power spectrum graph of a signal and the noise floor power levels at the input side of an amplifier is shown in Fig. 3.6. What is the input side SNR I ? Assuming this signal is amplified by an amplifier whose gain is G = 10dB and NF = 3dB, redraw the power spectrum graph and comment on the result. What is the output side SNRo ? 3.19. Power levels of the space radiation routinely detected by Earth based radio telescopes are comparable, for instance, with power level of a signal generated by a cell phone placed on the Moon. In order to illustrate how weak this signal really is, relative to the surrounding EM radiation that is perceived as the noise, let us consider the following numerical example. Assume that an astronaut on the Moon is trying to contact the Earth by using his cellphone. At the same time, a bird that flies above the receiving radio antenna dish generates thermal noise that interferes with the phone signal, Fig. 3.7. A typical cell phone radiates signal at Pcell = 1W level over bandwidth of B = 5 MHz,the bird’s body temperature is T = 37 ◦ C, and average distance to the Moon is D = 384.4 × 106 m. Aside from the cell phone signal and the thermal noise generated by the bird, ignore all the other signal and noise sources, and assume that the signals are received uniformly over the whole surface of the antenna. Fig. 3.6 Problems 3.18: frequency domain of a signal spectrum relative to the noise floor Fig. 3.7 Problem 3.19: illustration for cellphone on the Moon example 3 Electrical Noise: Problems 17 Calculate: (a) Hight H above the antenna’s dish surface where the bird is located at the moment when SNR = 0dB, i.e. the cell phone and the thermal noise powers are equal? (b) SNR when the bird lands at centre of the dish cover, i.e. when H = h = 10m? 3.20. A signal whose amplitude is vin = 1 µV, bandwidth is B, and the source resistance Rin = 50 , passes through an ideal noiseless “brick-wall” type BP filter whose input impedance and bandwidth are always matched to the source. Derive expression and then calculate (a) SNR in dB at the BP filter output node, if bandwidth of the filter is B1 = 20 kHz and T1 = 88 ◦ C; (b) the environment temperature T2 in K if the same SNR as found in (a) needs to be maintained while using the input signal source whose bandwidth is B2 = 82.080 kHz. 3.21. Starting with the known expressions for thermal noise voltage generated by a resistor, for dynamic resistance R D of an R LC resonator at resonance ω 0 , and for effective bandwidth Beff of an R LC resonator, derive the well known expression for noise voltage squared en2 = kT/C . As a numerical example calculate the noise voltage en if C = 10 pF and T = 17 ◦ C. 3.22. An oscilloscope probe’s RC equivalent network model is specified as R = 1 M and C = 1 pF. Determine: (a) useful frequency bandwidth B of this probe; (b) the lower limit of voltage that can be measured by this probe at T = 33 ◦ C if all other effects and the probe load are ignored; 3.23. A television set consists of the following chain of five consecutive sub–blocks: two RF amplifiers with 20dB gain and 3dB noise figure each, a mixer with −6dB gain and 8dB noise figure, and two additional amplifiers with 20dB gain and noise figure of 10dB each. Calculate: (a) the system noise figure; (b) the system noise temperature Tn at T = 26 ◦ C. 3.24. An amplifier with the input signal power of 5 × 10−6 W, noise input power 1 × 10−6 W, has output signal power of 50 × 10−3 W and the output noise power 40 × 10−3 W. Calculate the noise factor F and the nose figure NF of this amplifier. 3.25. Calculate noise current and equivalent noise voltage for a diode biased with IDC = 1 mA at the room temperature 300K and within the bandwidth of 1 MHz. 3.26. The equivalent input resistance of an amplifier is Rin = 1 k, while its equivalent shot noise biasing current is I D = 1 µA at T = 70 ◦ C. The signal generator has internal resistance R S = 1 k and it provides a signal of at least v S = 10 µVrms . Calculate maximal bandwidth B of this amplifier. 18 3 Electrical Noise: Problems Fig. 3.8 Problem 3.28: noise frequency spectrum 3.27. A front end RF amplifier whose gain is A = 50dB and noise temperature TnA = 90 K provides signal to a receiver that has a noise figure of NF = 12dB. First, calculate the noise temperature TRx of the receiver itself, then calculate the overall noise temperature Tsys of the amplifier plus the receiver system at the room temperature T = 300 K. 3.28. A graph of a typical BJT input noise current power spectral density is shown in Fig. 3.8. Using approximative methods estimate the total input noise current of this BJT within the frequency bandwidth or 1 kHz to 100 MHz. Chapter 4 Electronic Devices: Problems Time domain electrical signals are processed using both linear and non–linear devices. Rapid changes of the voltage/current signal levels, however, cause wide range of different responses while being processed by electronic components and devices. In this chapter we review important practical aspects of setting up right conditions that enable active and passive devices to produce the desired response. Variations of variables external to the devices themselves, such as the power supply level and the environmental temperature, have great impact on the overall circuit behaviour. Thus, we practice to predict the overall circuit behaviour while accounting for the environment imperfections. Problems: 4.1. By definition, time domain voltage function v(t) across an inductor is related to the time varying current i(t) flow as v(t) ≡ L di(t) dt (4.1) where, L is the proportionality constant defined as inductance. If the current plot in time domain is shown in Fig. 4.1, sketch the voltage graph across the inductor. 4.2. A square pulse voltage source v(t) , capacitor C, and a resistor R are connected in a series circuit. At time t0 = 0 s the capacitor was completely discharged. Plot a graph of the voltage vC across the capacitor over the next 5 ms. Numerical data: pulse waveform frequency f = 1 kHz, its maximum amplitude is v(max) = 10 V, capacitor value C = 1 µF, and the resistor value is R = 1 kω. 4.3. Assuming VD D = 3.0 V power supply, design a Vref = 1V voltage reference using a resistive divider. Assume that there is no any other branching current and that the equivalent Thévenin resistance of this voltage reference is Rth ≤ 100 ω. How much the reference voltage changes if the power supply voltage variation is ±10 %? R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_4, © Springer International Publishing Switzerland 2014 19 20 4 Electronic Devices: Problems Fig. 4.1 Problem 4.1: time domain diagram of current waveform 4.4. A diode model is given as VD q VD VD − 1 = I S exp (4.2) − 1 ≈ I S exp I D = I S exp n VT nkT n VT where, I D – is current flowing through the diode I S – is the diode leakage current VD – is voltage across the diode, i.e. biasing voltage VT – is the thermal voltage (VT = kT /q) k – is Boltzmann constant T – is temperature in degrees Kelvin [K] q – is the elementary charge n – is the emission coefficient, usually between 1 and 2 Typical 1N4004/1A diode has the following parameters: I S = 76.9 nA and the emission coefficient n = 1.45. At the junction temperature of T = 28 ◦ C calculate the diode current I D both using the exact and approximated model, then find the calculation error in percents if: 1. the forward biasing voltage is VD = 600 mV ; and 2. the forward biasing voltage is VD = 50 mV. 4.5. The same 1N4004/1A diode from problem 4.4 is connected in series with an ideal current source I = 1 A. At junction temperature T = 28 ◦ C calculate voltage VD across the diode terminals as well as the voltage range if the current source varies by ±10 %. 4.6. For a BJT transistor whose I S = 5 × 10−15 A, and at room temperature VT = 25 mV, the biasing current is IC = 1 mA, and assume the emission coefficient n = 1. Calculate the base emitter voltage VB E . Now, for VB E = 0.50, 0.55, 0.60, 0.65, 0.70, 0.75, and 0.80 V calculate the collector current IC . Also, calculate gm of this BJT. 4.7. For the four resistive networks Fig. 4.2 calculate the equivalent impedances R AB at the following frequencies: DC, 1 Hz, 1, 10 kHz, 1, 100 MHz, and at ∞ frequency. When appropriate, instead of doing the exact calculations, estimate the results by using reasonable engineering approximation. Assume R = 1 kω, and C = 1 nF. 4 Electronic Devices: Problems 21 Fig. 4.2 Problem 4.7: schematic diagram of four resistive networks (a) Fig. 4.3 Problem 4.8: resistive networks (a) (b) (c) (d) (b) Fig. 4.4 Problems 4.9 (left) and 4.10 (right): schematic diagrams of networks 4.8. For the two networks given in Fig. 4.3, estimate maximum and minimum voltage gains V A,B /Vout,B . First assume impedance ratio of Z 1 /Z 2 = R1 /R2 = 10 : 1, and then assume Z 1 /Z 2 = R1 /R2 = 1 : 10. 4.9. For a network shown by schematic diagram in Fig. 4.4 (left), 1. assuming the base–emitter diode threshold voltage is Vth (B E) = 0 V, i.e. ideal BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential VC is required at the collector node C to maintain the saturation mode of operation? 2. assuming the base–emitter diode threshold voltage is Vth (B E) = 1 V, i.e. realistic BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential is required at the collector node VC to maintain the saturation mode of operation? 4.10. What is the required resistor ratio R1 /R2 for the network in Fig. 4.4 (right), so that the transistor Q 1 operates in saturation, if VCC = 10 V and R E = 1 kω: 22 4 Electronic Devices: Problems (a) (b) (c) Fig. 4.5 Problem 4.11: schematic of resitive network 1. assuming the base–emitter diode threshold voltage is Vth (B E) = 0 V, i.e. ideal BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential is required at the collector node VC to maintain the saturation mode of operation? 2. assuming the base–emitter diode threshold voltage is Vth (B E) = 1 V, i.e. realistic BE diode, find value(s) of R2 so that the transistor Q 1 is turned on. What potential is required at the collector node VC to maintain the saturation mode of operation? 4.11. Estimate impedances looking into networks (a), (b), and (c), Fig. 4.5. For the sake of argument, assume that the small signal internal base resistance rπ is very large, and the internal emitter resistance is re is a small non zero value. 4.12. Calculate thermal voltage VT under the following conditions: (a) T = −55 ◦ C, (b) T = 25 ◦ C, and (c) T = 125 ◦ C. As a side note, these three temperatures are commonly used to characterize military grade electronic equipment. 4.13. A simple voltage reference is built using a resistor and 1N4004 diode as in schematic Fig. 4.6 (left). Calculate voltage across the diode at room temperature 25 ◦ C, under the following conditions: VCC = 9 V, R = 1 kω, IS = 18.8 nA. Express the result using engineering number notification with three decimal places. 4.14. Find biasing voltage VB E at T = −55 ◦ C, T = 25 ◦ C, and T = 125 ◦ C, if BJT collector current is set to IC = 1 mA and I S = 100 fA, Fig. 4.6 (right). Repeat the calculations for I S = 200 fA. 4.15. For BJT transistor in Fig. 4.6 (right), assuming the base current to be negligible, estimate biasing voltage VB required at the base node so that the collector biasing current is set to IC = 1 mA ≈ IE . Data: I S = 100 fA, RE = 100 ω, T = 25 ◦ C. Fig. 4.6 Voltage reference network for problems 4.13 (left) and 4.14 (right) 4 Electronic Devices: Problems 23 4.16. Estimate the input impedance Z in looking into the base node, Fig. 4.6 (right), if emitter resistor is R E = 100 ω and the forward gain β F is assumed as: (a) β F = 99, and (b) β F → ∞. 4.17. Design preliminary resistive voltage divider to set the base biasing voltage for the circuit in Fig. 4.6 (right). Use your own engineering judgement and constrains for the design. Assume power supply voltage VCC = 3.3 V, emitter resistor R E = 100 ω, and β F = 99. 4.18. For a BJT transistor with I S = 100 fA, VBE = 591.6 mV at temperature T = 25 ◦ C calculate: (a) the collector current IC ; (b) the transistor’s transconductance gm ; (c) the intrinsic emitter resistance r E ; and (d) r E if IC = 2, 3 mA, .... What happens if the temperature changes to, for example, T = 30 ◦ C? Briefly state your observations. Chapter 5 Electrical Resonance: Problems Resonance in general is one of the fundamental phenomena in the universe. Electrical resonance in particular is so important for our telecommunication technology, that if by some weird coincidence all LC circuits were dissolved and if we lost the knowledge of its workings, our marvellous electronic technology would have become totally unusable. It is not difficult to imagine consequence of such unfortunate event for our technological civilization. And yet, the seemingly trivial parallel connection of two passive components, an inductor and capacitor, is all that is needed to create the electronic resonance. In this chapter we study various important aspects of this tiny circuit, both in frequency and time domain. Problems: 5.1. Derive expressions for impedance Z ab (ω) of a series RLC network and expression for admittance Yab (ω) of a parallel RLC network, Fig. 5.1. 5.2. Knowing that in case of an ideal RLC network its impedance Z ab (ω0 ) at resonance must be real, derive expressions for resonant frequencies f 0 of the series and parallel networks, Fig. 5.1. 5.3. Find Z ab (ω0 ) for ideal LC series and parallel networks, Fig. 5.2. 5.4. For a more realistic parallel LC network, Fig. 5.3, find expression for the resonant frequency ω p0 . The resistance R accounts for all thermal losses, i.e. it represents combined resistance of the inductor, the wires, and effective series resistance (ESR) of the capacitor. Then, using the result find expression for the resonant frequency ω p0 under condition that the “realistic” parallel LC networks degenerates into “ideal” parallel LC network, i.e. R → 0. What is the conclusion? 5.5. At the resonance ω p0 : (a) find expression for dynamic resistance R D = Rab (ω p0 ) of the LC network in Fig. 5.3. 5.6. One possible definitions for quality factor Q of a series resonant LC circuit is formulated as the ratio of the total energy stored the inductor (or the capacitor) versus the energy dissipated in the resistive part of the network, Figs. 5.1 and 5.3. R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_5, © Springer International Publishing Switzerland 2014 25 26 5 Electrical Resonance: Problems Fig. 5.1 Problems 5.1, 5.2: schematic diagrams of ideal series and parallel RLC networks Fig. 5.2 Problem 5.2: schematic diagram of ideal series and parallel LC network Fig. 5.3 Problems 5.4, 5.5, 5.6: schematic diagram of realistic parallel LC network Derive the three versions of expressions for Q s factor of a series RLC network, i.e.: (a) Q s (L , R), (b) Q s (R, C); and (c) Q s (R, L , C). 5.7. Derive relationships between series impedance Z s = Rs + j X s and its equivalent parallel admittance Y p = 1/Z s = 1/R p + 1/ j X p . What is the conclusion? 5.8. For a series RLC network, Fig. 5.1 (left), derive expression for bandwidth BW at the resonant frequency ω0 as a function of the Q. What is the conclusion? 5.9. Truly realistic model of an LC resonator must include losses of both the inductor and capacitor, modelled by effective series resistance ESR and resistor r1 , Fig. 5.4. Derive expressions for the resonant frequency ω0 and dynamic resistance R D of this circuit as the function of Q 1 and Q 2 factors of the inductor and capacitor. Finally, transform resonator in Fig. 5.4 into its equivalent parallel RLC network, assuming that the capacitor is lossless, i.e. ESR = 0. 5.10. For a given coil, L = 2 µH, Q = 200, f 0 = 10 MHz, calculate: 1. its equivalent series resistance, 2. its equivalent parallel resistance, 3. the value of the resonating capacitor C, 5 Electrical Resonance: Problems 27 Fig. 5.4 Problem 5.9: schematic diagram of realistic parallel LC network 4. resistance R which, when added in parallel expands the bandwidth to B = 500 kHz. 5.11. In RF circuits it is common practice to use RLC circuit to design a bandpass (BP) filter that is needed to isolate a single frequency carrier signal from the rest of EM radiation that is constantly present in the air. For the sake of argument, compare a series RLC resonating circuit with the combination of a lowpass (LP) and (HP) RC filters used one after another, which effectively also creates BP filter. Plot normalized transfer characteristics of RLC bandpass, LP, HP, and LPHP bandpass filters. In the plot, normalize resonant frequencies of both RLC and LPHP filters, as well as pole locations of LP and HP filters to 1Hz. Also, normalize all four output amplitudes to 1 V. In addition, try to estimate Q factor of RLC resonator, so that its bandwidth is equal to bandwidth of RC based LPHP bandpass filter. 5.12. For the given inductor L = 2.533 nH and a trimming capacitor whose range is C = 1.3 nF to 857.345 pF calculate the tuning range of this LC resonator. 5.13. Design an LC resonator whose resonant frequency is f 0 = 10 MHz if only L = 2.533 nH inductor and the following components are available: Fig. 5.5 Problem 5.17: Illustration of a BP transfer function’s frequency bandwidth 28 5 Electrical Resonance: Problems (a) C1 = 10 nF, C2 = 40 nF, and C3 = 50 nF. (b) C1 = 200 nF, C2 = 400 nF, C3 = 0.5 µF, and C4 = 2 µF. (c) C1 = 70 nF, C2 = 60 nF, C3 = 60 nF, and C4 = 100 pF. 5.14. Calculate the Q–factor of a serial RLC network if inductor L = 2.533 nH and the lumped wire resistance r = (λ )m, at: (a) f 1 = 10 MHz; and (b) f 2 = 100 MHz. 5.15. A 1mH inductive coil has wire resistance of R = 5 and self–capacitance of 5 pF. The inductor is used to create LC resonator at f 0 = 25 MHz. Calculate the effective inductance and effective Q factor. 5.16. Calculate resonant frequency of a serial RLC network with R = 30 , L = 3 mH, and C = 100 nF. Calculate its impedance at f = 10 kHz and f = 5 kHz. 5.17. A frequency response curve of an LC resonator looks as in Fig. 5.5. Assuming, f 1 = 450 kHz, f 2 = 460 kHz, and the resonance frequency f 0 = 455 kHz. Determine the resonator bandwidth, Q factor, inductance L if capacitance is C = 100 nF, the total internal circuit resistance R. 5.18. A series RC branch consists of R S = 10 and C S = 7.95 pF. Convert it into its equivalent parallel RC network form assuming operating frequency of f = 1GHz. Chapter 6 Matching Networks: Problems Inside electronic circuits, signals are processed either in the form of voltage or in the form of current. In the section related to voltage and current dividers we learned that the ratio between the signal source impedance and load impedances is extremely important because it directly affects how much the signal amplitude is reduced at the interface node between the source and load. In the ideal case, voltage amplitude stays unchanged if the source impedance is zero, and/or the load impedance is infinite (i.e. the load current is zero). Similarly, current amplitude stays unchanged if the load impedance is zero (i.e. the load voltage is zero) and/or source impedance is infinite. As a consequence, depending whether the goal is to pass voltage or current signal from one stage of the system to another, there are two opposing requirements. Most important conclusion, however, is that even if it were possible to achieve the ideal either voltage or current signal transmission from source to load, in both of these two ideal cases the transmitted signal power is zero because electric signal power is the product of voltage and current. In this section we practice Q–matching technique to design passive LC networks that serve the purpose of maximizing power transfer of voltage signals delivered by a source to the load, which implies that both voltage and current amplitudes of the signal must take non–zero values. We design the matching network circuits assuming complex–conjugate matching. Problems: 6.1. Design a single stage LC matching network in between source with R S = 5 ω and load R L = 50 ω termination at f = 10 MHz, Fig. 6.1 (left). Additional design constrain is that the matching network must maintain DC connection between the source and load. 6.2. Design a single stage LC matching network in between source with R S = 50 ω and load R L = 5 ω termination at f = 10 MHz, , Fig. 6.1 (right). Additional design constrain is that the matching network must maintain AC connection between the source and load. R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_6, © Springer International Publishing Switzerland 2014 29 30 6 Matching Networks: Problems Fig. 6.1 Problems 6.1 and 6.2: typical cases of non matched sources and load resistances, R S < R L (left), and R S > R L (right) Fig. 6.2 Problems 6.7 and 6.8: Inductive source driving resistive load (left), and resistive load driving capacitive RC load (right) 6.3. Using Q–matching technique, calculate the equivalent parallel network to series connection of R S = 5 ω and L S = 238.732 nH at f = 10 MHz. 6.4. Using results from problem 6.1, find reflection coefficient λ and mismatch loss ML at the interface between the serial and parallel sections of the matching network. 6.5. Using results from problem 6.1, estimate the 3 dB bandwidth, assuming that the overall matching network’s frequency characteristics is symmetrical. 6.6. Following up on results from problem 6.4, if the input signal occupies frequency bandwidth B from f = 8 MHz to f = 12 MHz then recalculate λ and ML. Briefly comment on the results. 6.7. Design single–stage LC matching network at 10 MHz when the source impedance is inductive, Z s = Rs + j L s , while the load is purely resistive, Fig. 6.2 (left). It is required that the matching network maintains DC connection between its input and output terminals. Data: R S = 5 ω, L S = 138.732 nH, R L = 50 ω. 6.8. Design single–stage LC matching network when parasitic capacitance C L exists in parallel with the load resistance R L , where, R S = 5 ω, R L = 50 ω, C L = 1 nF, f = 0.01 GHz. Assume DC connection inside the matching network. 6.9. Match 5 ω source resistance to 50 ω load resistance at 10 MHz by designing a general two stage matching network. In this example, the goal is to decrease the bandwidth relative to the single LC matching network solution. Make your own choice of the ghost resistance R I N T and briefly explain your reasoning. 6.10. Match 5 ω source resistance to 50 ω load resistance at 10 MHz. Design two stage LC matching network, where the first stage behaves as a LP filter, while the 6 Matching Networks: Problems 31 Fig. 6.3 Problem 6.12: Schematic diagram of RF amplifier connected to antenna through matching network second stage behaves as a HP filter. One of the goals of this matching network design is to increase the bandwidth relative to the equivalent single LC matching network solution. 6.11. An antenna impedance is assumed to be resistive R A = 50 ω. Tuneable RF amplifier is set to f 0 = 665 kHz and its input impedance is Rin = 2 kω. Design two possible matching networks using Q–matching technique and comment on differences between the two solutions. 6.12. For the given circuit, Fig. 6.3, RF antenna has internal resistance R A = 50 ω, frequency of the RF signal is f R F = 10 MHz, C2 = 11.213 pF, C0 → ∞, R1 = 5200 ω, and DC voltage at node 1iis set to V (1) = 1/4VD D . Design matching network that provides DC coupling to the RF antenna. In your solution use minimal number of components to design the matching network. 6.13. Using the parasitic absorption method at 10 MHz match source impedance of Z S = (5 + j10) ω to a load impedance of Z L = (50 − j30) ω (the capacitor is in parallel to the load resistance R L ). Chapter 7 RF and IF Ampliﬁers: Problems The transition from the classic linear low–frequency amplifiers that are found in literally in all undergraduate textbooks to RF amplifiers suitable for radio equipment can be made gradual if, before even moving into the higher frequency regions, first we review again important aspects of the circuit that become more protruding as the frequency increases. In the following examples we practice to recognize and mentally evaluate parameters such as input/output impedance, biasing points of BJT devices, Miller capacitance, and amplifier bandwidth. Instead of focusing at all details at once, and risking not to see the forest for the trees, we study each of the amplifier aspects in isolation while accepting numerical inaccuracies for the sake of developing intuition and sense for the circuit behaviour. Problems 7.1. Estimate impedances as seen by looking into the respective network nodes, as indicated in Figs 7.1b, c, and d. 7.2. Resistance seen by looking into a BJT emitter is Rout = 100 ω. Resistance looking into the base is Rin = 100 kω. For λ = 100, find reflected resistance at the base node R B and R E ? (Ignore base internal resistance and small emitter resistance re .) 7.3. For network shown in schematic diagram, Fig. 7.1 a, 1. assuming the base-emitter diode threshold voltage is Vth (B E) = 0 V , i.e. ideal BE diode, find value(s) of R2 so that BJT transistor is turned on. What potential VC is required at the collector node C to maintain the saturation mode of operation? 2. assuming the base-emitter diode threshold voltage is Vth (B E) = 1 V , i.e. a realistic BE diode, find value(s) of R2 so that BJT transistor is turned on. What potential is required at the collector node VC to maintain the saturation mode of operation? 7.4. What is the required resistor ratio R1 /R2 for network in Fig. 7.1e, so that BJT transistor operates in saturation, if VCC = 10 V and R E = 1 kω: R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_7, © Springer International Publishing Switzerland 2014 33 34 7 RF and IF Amplifiers: Problems (a) (b) (c) (d) (e) Fig. 7.1 Problems 7.1, 7.3, and 7.4: network examples (a) (b) (c) Fig. 7.2 Problems 7.5, 7.6, 7.10, and 7.13: network schematics 1. assuming the base–emitter diode threshold voltage is Vth (B E) = 0 V, i.e. ideal BE diode, find value(s) of R2 so that the transistor is turned on. What potential is required at the collector node VC to maintain the saturation mode of operation? 2. assuming the base–emitter diode threshold voltage is Vth (B E) = 1 V, i.e. realistic BE diode, find value(s) of R2 so that the transistor is turned on. What potential is required at the collector node VC to maintain the saturation mode of operation? 7.5. If the only given data are RC = 10 kω and R E = 100 ω resistor values, estimate the upper bound of voltage gain Av for circuit in Fig. 7.2c and express the estimated gain in [dB]. 7.6. Given data for circuit in Fig. 7.2c is: RC = 10 kω, R E = 100 ω, I S = 100 fA, and VB E = 768.78 mV at temperature T = 25 √ C, and the input signal frequency is f = 10 MHz. In addition, there is also capacitor C connected in parallel with the emitter resistor R E that is not shown in the schematic. Estimate the circuit voltage gain Av if: (a) C = 1 µF, and (b) C → ∞. How large is the estimated gain difference between these two cases? How large is the gain difference in comparison with the gain calculated in Problem 7.5? Briefly comment on the results. 7.7. For grounded emitter amplifier powered from VCC = 10 V, assume VT = 25 mV, then estimate voltage gain A V in the following cases: (a) Vout = 7.5 V; (b) Vout = 5 V; and (c) Vout = 0.2 V. 7 RF and IF Amplifiers: Problems (a) 35 (b) (c) Fig. 7.3 Problems 7.9, 7.11, 7.12, and 7.14: network schematics 7.8. Assuming VT = 25 mV and that VB E ≤ VT , if VB E voltage of a BJT transistor changes by 18 mV how much is the correspond change of IC as expressed in dB? Repeat the calculations for VB E = 60 mV? 7.9. For a CE amplifying circuit in Fig. 7.3 estimate the Miller capacitance C M if RC = 9.9 kω, R E = 100 ω, CC B = 1 pF. 7.10. Estimate the inherent (i.e. due to the amplifier topology itself) input side bandwidth of CE amplifier in Fig. 7.2b if RC = 9.9 kω, R E = 100 ω, CC B = (1/π )pF (not shown in the schematic), λ = 190, R1 = 2 kω, R2 = 2 kω. 7.11. Estimate the range of frequencies where CE amplifier in Fig. 7.3b should be used, if the base side inductor L = 2.533 µH, RC = 9.9 kω, R E = 100 ω, CC B = 1 pF. 7.12. For amplifier in Fig. 7.3b , the small signal voltage gain is A V = −100. Estimate value of the inductor L so that the input stage resonates at f 0 = 15.915 MHz. Data: C = 1 pF. Assume the base current to be zero. 7.13. A signal generator with low internal resistance is coupled with the CE amplifier in Fig. 7.2c through a series capacitor C = 1 µF. Estimate the range of frequencies where the CE amplifier should be used, if RC = 9.9 kω, R E = 100 ω, CC B = 1 pF, R1 = 2 kω, R2 = 2 kω. 7.14. For circuit shown in Fig. 7.3c first suggest its application, and then estimate: (a) DC voltage at BJT collector node; (b) transconductance gain of BJT gm (Q 1 ); and (c) suggest the circuit’s input vi and output terminals vo , then estimate small signal AC voltage gain, if the gain is defined a A V = vo/vi . Data: assume λ = ∞, RC = 7.5 kω, I = 0.5 mA, C = ∞, VCC = 5 V, and VT = 25 mV. Chapter 8 Sinusoidal Oscillators: Problems Transmission of wireless signals is based on the idea of existence of a sinusoidal waveform, which in return is modulated in some way that encodes the intended message. In mathematical terms, a sinusoidal function is required for several key mathematical operations that we meet in the following chapters. Physical embodiment of a sinusoidal waveform is known as an oscillator, whose the only role is to deliver clean predefined either voltage or current waveform that is accurately described by a sinusoidal function. What is more, instead of designing a separate oscillator for each desired frequency, in practical realizations we prefer to design tuneable oscillators so that a single circuit delivers a sinusoidal waveform that can take a range of frequencies. In this chapter we review basic properties of closed loop systems, which are fundamental to oscillator operation. Instead of taking classical approach of studying specific oscillator topology, in this book we study oscillators from the perspective of closed loop systems, thus simplifying the analysis by looking at the forward path (i.e. amplifier) separately from the feedback path (i.e. passive feedback LC network). By doing so we can combine three separate concepts that we already know about, i.e. concept of a signal amplifier, concept of resonant LC circuit, and concept of feedback systems into the new concept of oscillators. Problems: 8.1. Derive expression for the general case of loop gain if the feedback loop consists of forward path amplifier with gain A and the feedback circuit path with gain β. 8.2. One of several versions of a phase oscillator, Fig. 8.1 (left), is based on a CE amplifier and three RC stages in the feedback loop. Derive expression for the minimal transistor gain factor βmin (not to be confused with the feedback loop parameter), and the resonant frequency ω0 , under the following assumptions: (a) the transistor’s output resistance ro is infinite; (b) all capacitors have the same value; (c) all resistors have the same values, while the transistor’s base resistance rb is absorbed in the left most resistor R; (d) for simplicity, details of biasing network are not shown; and (e) all elements are ideal, ignore the small emitter resistance re and base collector R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_8, © Springer International Publishing Switzerland 2014 37 38 8 Sinusoidal Oscillators: Problems Fig. 8.1 Problems 8.2, 8.3, and 8.6: simplified schematics of oscillator networks capacitance C BC . Then, calculate values for resistors R and capacitors C, if RC = 10 kω. 8.3. Estimate the resonant frequency ω0 of an oscillator whose feedback network is shown in Fig. 8.1(centre), if L 1 = 0.5 μH, L 2 = 1.5 μH and C = 126.65 pF. 8.4. Using the same example circuit and data as in Problem 8.3 estimate the feedback network’s gain factor β. 8.5. Assuming the same data as in Problems 8.3 and 8.4, estimate the effective resistance Reff that this feedback network presents to the output of the oscillator’s amplifier whose input impedance is Rin = 10 kω, if the effective inductor’s Q factor is Q(L eff ) = 50. 8.6. For the circuit shown Fig. 8.1 (right) derive: (a) expression for the resonant frequency ω0 ; and (b) expression for gm of BJT transistor. Then, use the following numerical data to calculate the resonant frequency ω0 and gm : RC = 10 kω, BJT output resistance rc = 10 kω, L = 2 μH, C1 = C2 = 253.30 pF, and Q L → ∞. Details of biasing network are omitted for simplicity. Repeat the same problem assuming finite Q L = 50, derive and recalculate the new equation(s) at the resonant frequency ω0 and gm . Fig. 8.2 Problem 8.7: simplified schematic of Clapp oscillator with varicap diode 8 Sinusoidal Oscillators: Problems 39 8.7. For the Clapp oscillator shown Fig. 8.2 calculate the oscillating frequency at: (a) zero bias VD = 0 V of the varicap diode, and (b) at VD = −7 V. Data: L = 100 μH, C1 = C2 = 300 pF, and C0 = 20 pF. Chapter 9 Frequency Shifting: Problems There are at least two fundamental reasons for the need of a practical technique to move (i.e. shift) frequency of a signal in the frequency domain while still preserving the embedded message. First, there are many radio transmitters in existence. Television, satellite, cell phone and other similar communication systems should be looked at as being nothing more than a sophisticated radio. Thus, from the perspective of wireless carrier signals, the space around us is equivalent to a room full of people, all talking loud and at the same time. As we easily imagine, it is difficult to hear message of a person standing on the other side of the room. In principle, there are two basic methods to enable a message to be transmitted between transmitter and the receiver. More obvious method is to apply the time sharing principle, i.e. to apply the rule that only one transmitter is active at any given moment. This is equivalent to a conference setup where one speaker after another gives a lecture, while all attendees quietly listen. Less obvious method is to apply the frequency sharing principle. Now, all transmitters are active all the time, however, each transmits its signal at different frequency bandwidth that equals to the resonant frequency of the receiver. This is equivalent to a room full of opera singers, all with perfect pitch, and all singing in non–overlapping frequency regions, e.g. bass voice is very easily distinguished from soprano voice. Visually, each frequency bandwidth serves role of a wire directly connecting transmitter and the receiver. As long as the receiver is “deaf” for all other frequencies except the one being used by the transmitter, the system works perfectly. In practice, the use of a narrowband LC resonator in the receiver’s front end sections makes the receiver able to “hear” only one transmitter, while all other active transmitters are effectively not heard. Second, conversion of an electrical signal into EM radiation happens in the antenna. In order to effectively produce EM radiation the antenna length should be at least as long as a few of the waveform’s wavelengths. As soon as we realize that, for example 3 kHz signal (which is in the range of a typical human voice) has wavelength of approximately 100 km it becomes obvious why direct wireless transmission R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2_9, © Springer International Publishing Switzerland 2014 41 42 9 Frequency Shifting: Problems of voice frequencies is not practical. Actually, we already use 100 km “antennas”, except that they are laid in trenches and called communication cables, phone lines. In this chapter we learn how to actually implement frequency sharing mechanism that has enabled our modern wireless communication systems. Problems: 9.1. There are four single–tone signals: S1 (t) = 3V sin(2π × 1 MHz × t) S2 (t) = 4V sin(2π × 20 MHz × t) S3 (t) = 12V cos(2π × 19 MHz × t) S4 (t) = 12V cos(2π × 21 MHz × t) For the given waveforms: (a) find expression for S(t) = S1 (t) × S2 (t). Then, using a graphing software plot waveforms of [S(t)], [S1 (T )], and [−S1 (t)] in the same window. Observe time domain relationships among these signals; and (b) then plot So (t) = 1/2 (S3 (t) − S4 (t)). Briefly comment on the results. 9.2. Starting from f L O = sin(2π × 10 MHz × t), find two other single–tones that could be used to generate a single tone at f L F = 1 kHz. Comment on the process and the result. 9.3. There is a large number of radio stations transmitting their respective programs at various carrier frequencies, all the same time. For example, a radio receiver is tuned to receive AM modulated wave transmitted at carrier frequency of f R F = 980 kHz. Local oscillator inside the receiver is set at f L O = 1435 kHz. For the given data: 1. 2. 3. 4. calculate frequencies at the output of the receiver’s mixer, which frequency should be chosen as the intermediate frequency (IF), at what frequency the ghost radio station would operate, and then plot frequency domain graph of the frequencies involved in this problem. 9.4. Tuned RF amplifier with LC tank whose Q = 20 is tuned at RF frequency f 0 . Estimate attenuation of the image signal, if the image frequency is 10 % higher then RF signal. 9.5. Two voltage signals v1 = V1 cos(ω1 t) v2 = V2 cos(ω2 t) are first added and then passed through an ideal diode, Fig. 9.1, whose voltage/current function is given as vD −1 (9.1) i D = I S exp Vt 9 Frequency Shifting: Problems 43 Fig. 9.1 Problem 9.5: Simplified schematic of a diode mixer First, derive expression for the current signal at the output of the diode as a function of the two input voltages i D = f (v1 , v2 ), and comment on its frequency spectrum; second, assume that V1 = V2 = 1 V, f 1 = 10.000 MHz, f 2 = 10.001 MHz, Vt = 25 mV, and I S = 1 µA, then write expression for the low– frequency current tone coming out of this network. 9.6. Two voltage signals v1 = V1 cos(ω1 t) v2 = V2 cos(ω2 t) are either first added and then applied to the gate of an ideal BJT transistor Q 1 , Fig. 9.2 (left), or v1 is applied to the gate of Q 2 and v2 is applied to the emitter node by means of a 1:1 ratio transformer, Fig. 9.2 (right). Find expressions for frequency spectrum of the collector current in both cases. Assume ideal transistors with the current gain of β. 9.7. Two voltage signals v1 = V1 cos(ω1 t) v2 = V2 cos(ω2 t) are applied to JFET transistors J1 and J2 that are used instead of Q 1 and Q 2 respectively, in the same topology as in Fig. 9.2. Find expression for the square term in the output frequency spectrum. Fig. 9.2 Problem 9.6 and 9.7: Simplified schematic of BJT mixer 44 9 Frequency Shifting: Problems Fig. 9.3 Problem 9.8: Simplified schematic of a dual–gate FET mixer and its equivalent circuit diagram, where M1 and M2 are assumed identical 9.8. Two voltage signals v1 = V1 sin(ω1 t) v2 = VDC2 + V2 sin(ω2 t) are applied to dual–gate FET transistor used as a mixer, Fig. 9.3 (left). Note that a dual gate FET transistor M1,2 is equivalent to cascode configuration of two stacked transistors M1 and M2 , Fig. 9.3 (right). Assume an ideal FET transistor. Find expression for the square term in the output frequency spectrum. Chapter 10 Modulation: Problems For the purpose of creating wireless communication system, a simple high frequency sinusoidal waveform is actually not much useful. By itself, the only information it carries is that there is an active source in existence. In order to carry any more complicated message, a music or speech for example, the single tone waveform must have some time variation that can be controlled and used to embed the message. By inspection of a simple sinusoidal function S(t) = A sin(ω t + λ) we conclude that there are three possible parameters available for manipulation in time, amplitude A, frequency ω , and phase λ. Naturally, amplitude A is the first obvious choice, simply because we can choose to use a switch and turn on and off the sinusoidal source. That switching action translates into a binary type of modulation (i.e. time change) of the amplitude. If periods of the silence and the signal activity are timed to “sound” as series of long and short beeps, then transmission technique known as Morse code is created. Or, if the instantaneous amplitude of the sinusoidal waveform is somehow controlled by the instantaneous amplitude of a speech, then we say that the sinusoidal waveform is amplitude modulated (AM). Second choice would be to somehow control the instantaneous frequency ω of the high frequency sinusoidal waveform by the instantaneous amplitude of a speech, then we say that the sinusoidal waveform is frequency modulated (FM). In this scenario, deviation of the instantaneous frequency relative to a no–signal frequency f 0 is proportional to the speech amplitude. Third choice would be to control the instantaneous phase λ and make it proportional to the speech amplitude, in which case we say that phase modulation (PM) has been implemented. Each of those there modulation techniques has its advantages and disadvantages depending on the intended application. In this chapter we practice some of the basic principles related to signal modulation. R. Sobot, Wireless Communication Electronics by Example, 45 DOI: 10.1007/978-3-319-02871-2_10, © Springer International Publishing Switzerland 2014 46 10 Modulation: Problems Problems: 10.1. For given audio signal and its carrier waveform Aa (t) = 3 + 1.5 sin (2π × 1500 × t) Ac (t) = 6.0 sin (2π × 50, 000 × t) do the following: 1. 2. 3. 4. 5. Plot both these two waveforms in a single plot; Construct AM modulated wave; Determine the modulation factor and percent modulation; What are the frequencies of the audio signal and the carrier? What tones are visible in frequency spectrum of the modulated wave? 10.2. How many AM broadcasting stations can be accommodated in a 100 kHz bandwidth if the highest frequency used to modulate these carriers is 5 kHz? 10.3. Determine the power content of each of the sidebands and of the carrier of an AM signal whose total power is 1,200 W while the modulation index is set at 85 %. 10.4. An AM signal whose modulation index is 70 % contains 1500 W at the carrier frequency. Determine the power content of the upper and lower sidebands for this percent modulation. Then, calculate the power at the carrier and the power of each of the sidebands when the percent modulation drops to 50 %. 10.5. For FM waveform whose frequency modulation index of m f = 1.5 and the modulation signal’s frequency is f b = 10 kHz: 1. estimate the required bandwidth BFM (using Carson’s rule); 2. calculate ratio of the total power PT relative to the power in the FM unmodulated waveform; 3. find which harmonics has the highest amplitude. 10.6. Recommend a design parameters of an AM standard broadcast receiver that is to operate with an intermediate frequency (IF) of 455 kHz, thus 1. Calculate the required frequency for the local oscillator f LO when the receiver is tuned to f c = 540 kHz, assuming that frequency of the local oscillator is above the frequency of the received signal. 2. Calculate the required frequency for the local oscillator f LO when the receiver is tuned to f c = 540 kHz, assuming that frequency of the local oscillator is below the frequency of the received signal. 10.7. A carrier waveform whose frequency is f c = 107.6 MHz is frequency modulated by a baseband sine wave whose frequency is f m = 7 kHz. The output FM signal has frequency deviation of πf s = 50 kHz. For the given data, 10 Modulation: Problems 47 Fig. 10.1 Problem 10.1: schematic of reactance modulator Fig. 10.2 Problem 10.2: illustration of a phase modulation circuit with varicap diode 1. Calculate frequency swing of the carrier in this FM signal; 2. Determine the highest and the lowest frequencies attained by the modulated signal; and 3. Calculate the modulation index of the output FM wave? 10.8. The total signal power of FM transmitter is PT = 100 W while modulation index is m f = 2.0. 1. Calculate power levels contained in all frequency components; and 2. if frequency of the used modulation signal is f m = 1.0 kHz estimate the FM signal bandwidth requirement. 10.9. For the given data in circuit Fig. 10.1 find value of capacitor C so that frequency of the output waveform is f out = 3.5 MHz. Data: C T = 83.4 nF, L T = 20 nH, R = 100 φ, gm (M1 ) = 10 mS. 10.10. For phase modulation circuit, Fig. 10.2, that consists of an inductor in parallel with series combination of capacitor C and varicap diode Cd0 , find the phase deviation constant K . Data: Q = 70, C = 10 Cd0 , V0 = 15 V. Chapter 11 Signal Demodulation: Problems Modulated waveform must be received and the message recovered. The message recovery in AM communication systems requires a circuit that is capable to extract the carrier envelope. A diode–capacitor based circuit known as envelope detector (or, peak detector) is essential part of both AM and FM receivers, as well as many other analog signal processing systems. In this chapter we familiarize with the envelope detector operational principle and operation. Problems: 11.1. With reference to the simplified schematic Fig. 11.1 and if R = 2 kω, estimate: 1. 2. 3. 4. 5. Detector input impedance; Total signal power delivered to the detector; Voltages v0 (max), v0 (min), and V0 (DC); Average output current I0 (DC); and Appropriate capacitor value C to prevent diagonal clipping distortion for maximal modulation frequency f m (max) = 5 kHz and maximal modulation index m a = 0.9. 11.2. Assume that the input signal Vin of an AM diode detector, Fig. 11.2, is a 665 kHz IF carrier modulated with a 5 kHz tone. Component values are: C1 = 220 pF, C2 = 22 pF, R1 = 470 ω, R2 = 4.7 kω, R L = 50 kω. Diode I D versus VD characteristics is shown in the graph. 1. Sketch qualitatively the detector output tones along an λ axis showing relative amplitudes of the tones; 2. Sketch AM waveform shape at nodes 1–5; 3. Sketch equivalent circuit at 5 kHz. Calculate amplitude ratio of the input signal and signal at node 3; 4. Sketch equivalent circuit at 665 kHz. Calculate amplitude ratio of the input signal and signal at node 3; 5. Briefly comment on the above results. R. Sobot, Wireless Communication Electronics by Example, 49 DOI: 10.1007/978-3-319-02871-2_11, © Springer International Publishing Switzerland 2014 50 11 Signal Demodulation: Problems Fig. 11.1 Problem 11.1: simplified schematic of peak detector Fig. 11.2 Problem 11.2: simplified schematic of peak detector, and voltage current characteristics the diode Fig. 11.3 Block diagram for Problem 11.3 11.3. Voltage signal received by a 50 ω antenna has amplitude of 10 µV. Gain contributions are noted next to each block of the system (see Fig. 11.3). Estimate: 1. input signal power in W and dBm units, 2. power delivered to the speaker in dBm and W. 11.4. A signal waveform used to modulate RF carrier has symmetrical triangular shape with zero DC = 1 V component and amplitude of Vb = 2 Vpp while the carrier wave has amplitude of VC = 2 Vp . Sketch these two waveforms before and after the mixer circuit. Estimate modulation index m by inspection of the plots. 11.5. For unmodulated signal, AM current in the antenna is I0 = 1 A, while sinusoidal modulation wave causes the antenna current to be Im = 1.1 A. Calculate the modulation index m if the antenna impedance is R. 11 Signal Demodulation: Problems 11.6. An FM signal S FM (t) is described as S FM (t) = 2,000 sin 2π × 108 + 2 × π × 104 cos (π × 104 t) t as it arrives at a 50 ω antenna. Determine: 1. 2. 3. 4. 5. 6. the carrier frequency, the transmitted power, modulation index m f , the message signal frequency f b , required signal bandwidth B by using Bessel functions (not Carson’s rule), power in the largest and smallest sidebands as predicted by Bessel functions. 51 Chapter 12 RF Receivers: Problems System level analysis of an RF receiver includes characterization of its components, the internal blocks, tuneability range, image frequencies, signal to noise ratio, and dynamic range. In this chapter we practice some of the key system level concepts. By all means, the reader is advised to take other textbooks and, with the acquired introductory knowledge from this book, to keep developing knowledge of RF communication circuits and high frequency concepts and design techniques. Problems: 12.1. Sketch a block diagram of a heterodyne AM receiver, and illustrate waveform shapes at output of each stage. The transmitted message waveform is shown in Fig. 12.1. 12.2. An AM receiver is designed to receive RF signals in the in the 500 kHz to 1600 kHz frequency range with the required bandwidth of B = 10 kHz at the resonant frequency f 0 = 1050 kHz. The RF amplifier uses inductor L = 1 µH. 1. calculate bandwidth at f max = 1600 kHz and capacitance C; 2. calculate bandwidth at f min = 500 kHz and capacitance C; and 3. comment on the results. 12.3. An AM receiver is designed to receive RF signals in the f min = 500 kHz to f max = 1600 kHz frequency range. All incoming RF signals are shifted to intermediate frequency f I F = 465 kHz. AM receiver tuning is commonly done by a knob that simultaneously tunes resonating capacitors in the RF amplifier and LO oscillator sections. For the receiver architecture in Fig. 12.2 (matching network not shown), 1. calculate tuning ratio C R F (max)/C R F (min) of the resonator capacitor in the RF amplifier, 2. calculate tuning ratio C L O (max)/C L O (min) of the resonator capacitor in the local oscillator LO, 3. recommend the resonating frequency for the local oscillator. R. Sobot, Wireless Communication Electronics by Example, 53 DOI: 10.1007/978-3-319-02871-2_12, © Springer International Publishing Switzerland 2014 54 12 RF Receivers: Problems Fig. 12.1 Problem 12.1: time domain plot of baseband modulating signal Fig. 12.2 Problem 12.3: AM receiver block diagram Fig. 12.3 Problem 12.5: input–output characteristics of an RF amplifier 12.4. The LO oscillator frequency is 11 MHz, and RF signal frequency is 10 MHz. What is the image frequency? 12.5. Input–output power characteristics of an amplifier is given in Fig. 12.3. Estimate the gain, 1dB compression point, and the third order intercept point IIP3. 12.6. A receiver whose IF frequency is f I F = 455 kHz is tuned to an RF signal with f R F = 950 kHz. No other transmission frequency is allowed within the RF band f R F = 950 kHz ± 10 kHz. However, for the sake of argument, let us imagine existence of a nearby non–linear transmitter whose emitting frequency spectrum consists of its both first and the second harmonics. Aside from the obvious 950 kHz ±10 kHz frequency range, what other frequency range(s) should also be prohibited for the external transmitter? 12 RF Receivers: Problems 55 Fig. 12.4 Problem 12.9: block diagram of an FM transmitter 12.7. Medium wave AM transmitters operate in the 540–1610 kHz range with 10 kHz spacing while using 455 kHz IF frequency. Estimate the range of the local oscillator frequencies and suggest bandpass filter(s) suitable for use with AM medium wave receiver? 12.8. A double conversion receiver architecture is based on two IF frequencies, I F1 = 10.7 MHz and I F2 = 455 kHz. If the receiver is tuned to a f R F = 20 MHz signal, find frequencies of the local oscillators and the internal frequencies. 12.9. An FM transmitter, Fig. 12.4, used the local VCO controlled by audio signal to generate waveform whose frequency is centred at f0 = 3.5 MHz. The oscillator shifts its frequency by ωf 0 = ±1.6 kHz when the audio signal amplitude is Vin = 3.6 V pp . Each FM station is assigned a frequency channel that is B = 150 kHz wide. At the point of antenna find: (a) the carrier rest frequency f c ; (b) the carrier frequency deviation ωf c ; (c) the FM modulation percentage; and (d) peak–to–peak voltage of the message signal Vin needed to cause 100 % of modulation. 12.10. An FM signal whose centre frequency is f 0 = 200 kHz and its frequency deviation is ωf 0 = ±200Hz is passed through two frequency multipliers, Fig. 12.5. The two newly created waveforms are then added together by a summing block. Signal coming out of the summing block is passed through LC resonator whose centre frequency and bandwidth are aligned with the lower–side–band (LSB) part of the output spectrum. At the output of the LC resonator, power of the LSB waveform is measured as Pout = 1 mW at the edges of its bandwidth. Assuming an ideal system: 1. calculate f 1 ± ωf 1 ; 2. calculate f 2 ± ωf 2 ; Fig. 12.5 Problem 12.10: block diagram of an RF section 56 12 RF Receivers: Problems Fig. 12.6 Problem 12.11: simplified schematic diagram of a three stage RF amplifier 3. calculate value of Q factor in the LC resonator; 4. sketch detailed plot of the output power spectrum. For the vertical axis use [dBm] units, and for the horizontal axis use [Hz] units. Clearly show power levels of all relevant tones. 12.11. Conceptual and simplified schematic diagram of a three stage RF amplifier is in Fig. 12.6. Power level Pin of the RF source signal that corresponds to the 1dB compression point, signal–to–noise ratio SNR measured at the internal node 1i, signal power level P2 measured at the internal node 2i, and the current i out of the signal delivered to the load R L are shown explicitly. In addition, bandwidth of the LC resonator is B is set by the design of RF amplifier’s LC resonator. 1. Data: R L = 100 λ, temperature is T = 26.85 √ C, power gain of the first stage is A1 = 10 dB, noise figure of the second stage is N F2 = 6.02 dB, and noise figure of the third stage is N F3 = 9.0309 dB, 2. Assumptions: the total input side noise is generated only by the equivalent input resistance Rin while each gain stage generates its own noise, the load resistor is ideal, all base currents are ignored. Using the given data and assumptions, estimate: a. b. c. d. power gain of the second stage only A2 in [dB]; noise voltage vn at the output; noise figure NF of this amplifier; dynamic range DR of this amplifier. 12.12. A cell phone receiver operates at room temperature, T = 16.85 √ C, and it has the following specifications: noise figure NF = 20 dB, bandwidth B = 1 MHz, signal to noise ratio SNR = 0 dB. Also, non–linearity of the receiver circuit can be described by the following output signal y(x) function: y(x) = 2x − 0.267x 3 12 RF Receivers: Problems 57 which applies to the desired incoming signal x1 = 0.1 V sin ω1 t. At the same time when the x1 signal is being received, there is another cell phone nearby transmitting a signal whose function is x2 = A2 sin ω2 t, i.e. (A2 → 0.1 V). For the described situation calculate: 1. 2. 3. 4. the 1dB compression point in [dB]; the input intercept point IIP3 in [dB]; receiver sensitivity S in [dBm]; estimate maximum dynamic range DR in [dB], and effective dynamic range DR in [dB] (assuming 2/3 of the maximum DR); 5. amplitude of the second signal A2 in [V] that would cause the “signal blocking” effect. In that case it is assumed that output of the receiver, when both x1 and x2 signals are present, can be described as 3 2 y(x1 , x2 ) ∞ a1 + a3 A2 A1 cos ω1 t. 2 Part II Solutions Chapter 13 Introduction: Solutions Examples of physical phenomena given in Chap. 1 are related to fundamental concepts in physics related to energy, matter, EM waves, EM fields, propagation of energy through matter and space, and basic interaction between two waveforms. In this chapter we find detailed, tutorial like numerical solutions to the given problems as well as comments related to historical developments of the theory and some of the relevant practical applications. Solutions: 1.1. Einstein discovered the mass–energy equivalence concept that is famously formulated as E = (mc2 )2 + ( pc)2 (13.1) where, m is the mass of object whose equivalent energy is calculated, c is the velocity of light in vacuum, and p is momentum of the mass m in motion. However, if the mass is not moving then p = 0 and (13.1) reduces into the well known form E = mc2 (13.2) This equitation is one of the fundamental natural principles, and it is used to describe, for example, the hydrogen fusion process in the starts as well as the energy conversion process in nuclear reactors. In this example we quantify amounts of energy potentially available to us through exploits of this principle. A snowflake consists of water molecules (H2 O), thus, first we find its total mass (m S ) by adding masses of all n individual molecules that constitute the snowflake. We already know that each water molecule consists of two hydrogen atoms (atomic weight H = 1.00794 g/mol) and one atom of oxygen (atomic weight O = 15.9994 g/mol). Thus, a single water molecule has atomic weight of: R. Sobot, Wireless Communication Electronics by Example, 61 DOI: 10.1007/978-3-319-02871-2_13, © Springer International Publishing Switzerland 2014 62 13 Introduction: Solutions H2 O = 2 × H + O = 18.01528 g/mol (13.3) or, if the mass is expressed in g then H2 O = H2 O 18.01528 g/mol = = 2.99151 × 10−23 g NA 6.0221415 × 1023 g/mol (13.4) then the complete snowflake has mass m S of m S = n × H2 O = 6.68559 × 1019 × 2.99151 × 10−23 g = 2 mg = 2 × 10−6 kg (13.5) Velocity of the snowflake is negligible relative to the speed of light, which leads into the equivalent energy of: E = m c2 = 2 × 10−6 kg × 2.99792458 × 108 m/s = 1.79751036 × 1011 J (13.6) Power is defined as the energy transfer rate, therefore in order to provide the average power of P = 25W the total energy E must be distributed over the following time t= 1.79751036 × 1011 J E = = 7.190 × 109 s √ 228 years P 25 W (13.7) We conclude that if we were able to completely convert, for instance, only 2 mg of any matter (including a snowflake) into energy we would be able to provide power to our hand-held electronic equipment for many years. Following this line of thought, we already developed nuclear power plants that are based on this fundamental principle and used to provide power to whole cities. However, we still need to learn how (and if) this principle could be scaled down and used to provide energy to our smaller everyday’s gadgets, e.g. electronic equipment, houses, and cars. 1.2. For all practical purposes the Sun is a nuclear reactor that, from our perspective, represents an infinite source of energy. The Sun energy is radiated continuously into the space, and even though it enabled the life on Earth, huge percentage of this freely available energy is still not harnessed by the humans. In this example we find how much of this energy is really available and for how long it is expected to last. We keep in mind that the solar constant k S is measured at the Earth distance from the Sun, and that by definition it represents the radiated energy flux per time, i.e. flow of energy per second as measured over a square metre surface. For purposes of this problem, it us handy to first express the solar constant in terms of the converted mass, as k S = 1366 J/s kg m/s2 kg W Nm = 1366 3 = 1366 2 = 1366 2 = 1366 2 m m m s ms s (13.8) 13 Introduction: Solutions 63 and to use it to find the equivalent amount of “mass flow” m → per square metre per second, which is calculated from E → = m → c2 as m→ = E→ 1366 kg/s3 −14 kg = 2 = 1.51988 × 10 2 8 c m2 s m 2.997924580 × 10 /s (13.9) where m → and E → are respective mass and energy flow per metre square per second. The calculated mass m → is equivalent to the radiated energy by the Sun per square metre per second in all directions, as measured at 1AU distance (i.e. at the Earth, because that is how k S is measured). 1. To find the total mass per second (M) that is being converted into energy, we integrate m → over the spherical surface A whose radius is l = 1 AU , thus M = A × m → = 4ω l 2 × m → 2 kg = 4ω × 1.495978707 × 1011 m × 1.51988 × 10−14 2 m s kg (13.10) = 4.27435 × 109 s 2. This total mass M is converted into energy that is being dissipated from the Sun in all directions over every second. However, by looking from the Sun the Earth is visible only as a circular “shield” whose diameter is r and whose area S is 2 S = ω × r 2 = ω × 6.3568 × 106 m = 1.26948 × 1014 m2 (13.11) therefore, the energy reaching the Earth is equivalent to mass flow m E , m E = m → × S = 1.51988 × 10−14 kg kg × 1.26948 × 1014 m2 = 1.92946 2 m s s ∴ mE √ 2 kg s (13.12) which, in comparison with the total converted mass M = 4.27435 × 109 kg/s, represents only a minor amount of about 4.5 × 10−8 % , i.e. for all practical purposes it is negligible relative to the total converted mass being dissipated in all directions of the space. 3. The hydrogen fusion process occurring in the stars takes several steps, however in principle it involves conversion of four hydrogen atoms (or, equivalently, two deuterium atoms) into one helium-4 atom plus some amount of energy. This extra energy is direct consequence of the mass difference between the starting hydrogen mass and the resulting helium mass. By experiments we found that atomic mass of the four hydrogen atoms and one helium atom are respectively 64 13 Introduction: Solutions H = 4 × 1.00794 g/mol ∞ 6.694896 × 10−27 kg He = 4.002602 g/mol ∞ 6.646476 × 10−27 kg therefore, the mass λm that is actually converted into energy is difference between the ending and the starting mass, i.e. λm = H − H e = 4.842 × 10−29 kg (13.13) which is relative to the mass of four H atoms as R= 6.694896 × 10−27 kg √ 138 4.842 × 10−29 kg (13.14) where, R is the ratio of the mass of 4H atoms initially available before the fusion starts and the equivalent mass converted during the fusion process. In other words, only 1/138 of the initial H mass is actually converted into the radiated energy while the rest of the initial mass is still preserved in the form of helium, which means that the Sun must start the fusion process with the total of M H = R × M = 138 × 4.27435 × 109 kg/s √ 5.9 × 1011 kg/s (13.15) of hydrogen in order to sustain the solar constant. 4. The Sun’s current mass is estimated as m S = 1.9891 × 1030 kg, thus to convert 10% of m S at the estimated mass conversion rate M H it will take time t that is found as t= 0.1 × 1.9891 × 1030 kg 0.1 × m S = = 3.371 × 1017 s MH 5.9 × 1011 kg/s ∴ t √ 10.7 × 109 years (13.16) 5. The total power PS generated by the Sun is calculated by using the complete surface A of the sphere surrounding the Sun with diameter 1 UI and the solar constant, i.e. PS = k S × 4ω × UI 2 2 = 1366W/m2 × 4ω × 1.495978707 × 1011 m = 3.842 × 1026 W It is interesting to see how power generated by the Sun compares to the World energy consumption. According to publicly available statistical data, in the year 2012 the World energy consumption reached E = 550 × 1018 J, which is approximately equivalent to 13 Introduction: Solutions P= 65 550 × 1018 J E = √ 17.3 × 1012 W t 365 × 24 × 3600 s (13.17) In comparison, power from the Sun that is reaching the Earth (that its, only the Sun facing side), after using (13.11), is found as P = k S × S = 1366 W × 1.26948 × 1014 m2 √ 173 × 1015 W m2 (13.18) which is to say that, if we could capture all power delivered by the Sun we would have had approximately 10,000 times more power than what we currently need. 1.3. Maxwell formalized symmetrical relationship between spatially varying electric and magnetic fields. That is, a changing electric filed is always accompanied with the changing magnetic field. Through his work Maxwell also calculated the speed of light and, therefore, concluded that light must be a form of electromagnetic radiation that propagates through space with finite velocity. Important note about these two fields is that they propagate together through space, always being in phase, and most importantly once the energy started moving through the space it will keep moving even if the energy source is removed. For instance, that is why we receive light from the distant stars even though they may not be there anymore. A common way to illustrate an EM wave is to show two sinusoidal waveforms as being perpendicular and propagating in the same direction, one for its electrical and one for its magnetic component, Fig. 13.1. If the waveform propagates in the x-direction, then the electric vector E and magnetic vector B must follow rules for the vector product E × B in the right-handed coordinate system, which then dictates that the electric vector must be in y-direction, while the magnetic vector B must point in the z-direction. It is important to notice that in this interpretation, Fig. 13.1, even though only a single sinusoidal functions are shown in the x and y-directions, the respective electric and magnetic wave components extend infinitely in these two directions. 1.4. A dipole antenna is simplest geometrical form of antenna that is derived from a capacitive structure. Therefore, it consists of two symmetrical conductive elements, for example two simple metallic rods. Once the antenna is stimulated with an AC Fig. 13.1 Problem 1.3: illustration of an EM sinusoidal wave propagating in the time-direction 66 13 Introduction: Solutions Fig. 13.2 Problem 1.4: illustration of a three step capacitor morphing into dipole antenna Fig. 13.3 Problem 1.4: illustration of an EM wave generated by a dipole antenna signal generator source, it creates EM filed that keeps propagating through the space at the speed of light even when the source is turned off. In order to demonstrate how a dipole antenna structure is derived, let us take a look at Fig. 13.2. We already know that a simple capacitor stimulated by AC signal generator creates changing electric field in between the two capacitive electrodes, Fig. 13.2 (left). However, even if the two electrodes are physically separated at the end that is further away from the source, the electric field does not break, Fig. 13.2 (centre). Instead, once the two electrodes are vertically aligned with each other, the field morphs into radial shape, Fig. 13.2 (right). In other words, we can imagine that a dipole antenna is only a fully “open” capacitor. As the AM current is continuously changing its direction along the antenna electrodes (as shown by green arrow), the electric and magnetic fields also keep changing their respective directions, Fig. 13.3. In this graphical quasi 3D representation of EM field moving radially away from the dipole antenna, we should actually imagine a torus shaped electrical field lines orthogonally wrapped around circular magnetic field lines. In reality, the whole space is filled with the alternating fields, not only the indicated vector lines. In order to better visualize this dynamic process, watch some of very good 3D animations available on the web that show how this radiation pattern is created and how it propagates through the space. 1.5. In principle, there are at least two ways to estimate number of participants in a parade marching along the street. We either take a fixed position at the street side and watch the parade passing in front of our eyes while counting the participants in 13 Introduction: Solutions 67 Fig. 13.4 Problem 1.5: illustration of an EM sinusoidal wave and its space–time parameters time, or we make aerial photo of the whole street and count the participants in space. In the former case we keep fixed position of observation, while in the latter case we keep fixed time of observation. We take similar approach if we want to observe a wave propagation. From a strictly mathematical point of view, equation E = E m sin(k z − ω t) is function of both time and space coordinates, i.e. E = f (z, t). That is to say, by setting one of the two coordinates constant we are able to observe how E changes relative to the other coordinate. In the given equation, both the wave number k and angular frequency ω can be also seen as “scaling” or “gain” parameters for their respective coordinates. Let us first consider spatial wave movement by “freezing” the time, i.e. by setting t = const, and by observing only E = f (z). If that is the case, and if without loosing generality we set t = 0, then E = E m sin(k z) (13.19) which is a simple sinusoidal function in z direction, Fig. 13.4 (top). As for any other sinusoidal function, it is easy to identify its period T . We only must keep in mind the physical meaning of the coordinate z and visualize this sinusoidal function in space. If that is the case then the spatial period T must be measured in units of meters, thus to emphasize this detail instead of using the general letter T we use letter π when referring to the spatial period. This graph is equivalent to a photo of the sinusoidal function, i.e. it shows amplitude at various spatial points at the same moment in time. It is useful to take a closer look at the spatial “gain” factor k. At the point in space where one full spatial period π is measured, i.e when z = π (keep in mind that time is constant and set to t = 0, and that a sinusoidal function has period of 2ω ), we write 68 13 Introduction: Solutions kz − ωt = 2ω ∴ kπ − 0 = 2ω ∴ k= 2ω π (13.20) where we used the fact that a full period of a sinusoidal function is equivalent to 2ω angular measure. In literature k is often referred to as the “wave number.” Similarly, we can set the space coordinate to a constant, e.g. z = 0, and observe time variant function E = E m sin(−ω t) (13.21) Again, (13.21) represents a simple sinusoidal function whose initial phase is negative, Fig. 13.4 (bottom). Because the point of observation is fixed, this graph shows how amplitude of the sinusoidal function changes in time, i.e. it shows the amplitude at various time points as seen from the same point in space. As we are familiar with using time variable when describing sinusoidal functions, we keep using letter T do denote period of this function and we measure it in seconds. Again, it is useful to take a closer look at the time “gain” factor ω. At the moment in time when one full time period T is measured, i.e when t = T (keep in mind that space is constant and set to z = 0), we write kz − ωt = 2ω ∴ 0 − ωT = 2ω ∴ |ω| = 2ω = 2ω f T (13.22) where the “–” sign indicates the initial phase. In both cases shown in Fig. 13.4 the wave propagates in positive directions of its respective coordinates. Now, it is very useful to determine velocity of the wave’s front-end propagation through space by using the well known classical Newtonian equation for the relationship among distance z, time t, and velocity v, i.e. z = vt (13.23) Because it takes time t = T to travel distance in space of one wavelength π, by definition and after substituting (13.20) and (13.22) we write v∞ π 2ω ω ω λz = = πf = = λt T k 2ω k (13.24) With (13.24) we have a tool to calculate time–space coordinates of a propagating wave whose frequency is f and the wave number k. 1.6. Theoretical and independent derivation of the speed of light from Maxwell’s equations was one of the most convincing confirmations that light is indeed a form of EM radiation. By following same basic idea used in Example 1.5 that gave us connection between time and space propagation of waves, we can derive and exactly calculate the speed of light starting from Maxwell’s classical equations assuming vacuum and no free electric charges (φ = 0 and J = 0). 13 Introduction: Solutions 69 Starting with three-dimensional Maxwell’s equations in form of ≤ ×E=− ∂B ∂t ≤ × B = μ0 ε0 (13.25) ∂E ∂t (13.26) First, we find second derivatives of E and B by finding space derivative ∂/∂z of (13.25) and then after substituting the space derivative into (13.26) we write ≤(≤ × E) = − ∂(≤ × E) ∂ 2E = −μ0 ε0 2 ∂t ∂t (13.27) At this point we need a theorem from calculus that states that ≤(≤ × E) = −≤ 2 E + ≤ · (≤ · E) (13.28) where, if φ = 0 then ≤ · E = 0, which further leads into one dimensional form of the wave equation as ≤ 2 E = μ0 ε0 ∂2 E ∂t 2 ∴ ∂2 E ∂2 E = μ0 ε0 2 2 ∂z ∂t (13.29) Now we can find one dimensional second derivatives in space and time as ∂E = k E m cos(kz − ωt) ∂z ∂2 E = −k 2 E m sin(kz − ωt) = −k 2 E ∂z 2 (13.30) ∴ ∂E = −ω E m cos(kz − ωt) ∂t ∴ ∂2 E = −ω2 E m sin(kz − ωt) = −ω2 E ∂t 2 (13.31) therefore, the wave equation (13.29) becomes −k 2 E = −μ0 ε0 ω2 E ∴ v∞ ω 1 =≥ k μ0 ε0 = (13.32) 1 4ω × 10−7 N/A2 × 8.854, 187, 817 × 10−12 F/m ∴ v = c = 299, 792, 458 m s (13.33) 70 13 Introduction: Solutions where, value of vacuum permittivity ε0 constant was independently established experimentally by measuring electrostatic force between capacitor plates. However, value of magnetic constant (vacuum permeability) in SI system was postulated as μ0 = 4ω × 10−7 N/m to help define the unit for electric current . Therefore, in SI units, the speed of light is c = 299, 792, 458 m/s. 1.7. As a side note, starting with one dimensional form of (13.25), which is simply ∂B ∂E =− ∂z ∂t (13.34) after substituting the two derivatives into (13.34), we write k E m cos(kz − ωt) = ω Bm cos(kz − ωt) ∴ Em ω = =c Bm k (13.35) which gives us another interesting relationship that involves EM field components and the speed of light. Thus, knowing amplitude of one EM wave component we easily calculate amplitude of the second. 1.8. We continue discussion started in Problem 1.7, by saying that magnitude of the S = 1/μ0 (E × B) vector product is found as |S| = 1 1 1 1 E m2 |E × B| = = E m Bm = c Bm2 μ0 μ0 μ0 c μ0 ∴ S= 1 N A2 V 2 W2 V2 A 2 s m = = m N m3 m s (13.36) that is, this vector product has the unit of power per square metre, thus it represents density of the energy being transferred. In addition, by applying the right hand rule for the vector product, we find that the resulting vector S points in the same direction as the wave velocity. In other words, it points in direction of the energy flow. We already encountered the special case of this vector in Problem 1.2 when we called it the solar constant. The energy flow vector is very important parameter because it not only quantifies the amount of energy but also contains information about the flow direction. In the literature, vector S is referred to as Poynting vector and it is explicitly used for the purpose of estimating an EM wave energy flow. 1.9. Now that we are familiar with the nature of Poynting vector and its relationship with the solar constant, we can perform various calculations, including calculations of EM wave component amplitudes as the wave arrives to the Earth. Electrical and magnetic field components of EM radiation generated by the Sun are given as E = E m sin(ky −ωt) y, and B = Bm sin(kz −ωt) z. Therefore, average 13 Introduction: Solutions 71 value of Poynting vector ∗S∓ is calculated as ∗S∓ = 1 c Bm2 1 c Bm2 ∗E × B∓ = ∗sin2 (kx − ωt)∓ x = x μ0 μ0 2 μ0 ∴ c Bm2 ⇒ 2 μ0 2 μ0 k S 2 × 4 ω × 10−7 N/A2 NA2 × 1366 W/m2 = Bm = = 3.384 µT c 2.99792458 × 108 m/s kS = ∴ E m = c Bm = 2.99792458 × 108 m/s × 3.384 × 10−6 T = 1.014 × 103 V m 1.10. Afer realizing that energy density can be also expressed as a function of pressure, the whole discussion about EM waves becomes really interesting. Electromagnetic waves are assumed to be a non-material form of existence, which is defined through very abstract concept of energy. At the same time, one of the implications of Poynting vector is that interaction of EM wave with material object results in physical pressure. This statement by itself sounds almost as science fiction. However, as we will see shortly, this concept of non-material with material interaction opens completely new ways of thinking about, for instance, propulsion systems in space. To see how the pressure comes into the equation, let us take another look at Problems 1.8 and 1.8, where we already established the nature of Poynting vector ∗S∓, and rearrange the measuring units as ∗S∓ = c Bm2 2 μ0 (13.37) ∴ ∗S∓ ∼ ∴ P∼ W J s−1 N m s−1 N m = = = 2 2 2 m m m2 m s N W m −1 = m2 m2 s (13.38) By simple rearrangement of (13.37) to match the units in (13.38) we write expression for EM wave pressure as P= B2 ∗S∓ = m c 2 μ0 72 13 Introduction: Solutions Fig. 13.5 Solar sail concept which can be then interpreted as the pressure on both sides of a surface immersed in the EM field. However, if the EM wave is perfectly reflected, then the wave pressure is distributed over only one side of the surface, that is, the pressure on that single side increases by the factor of two (because area of two sides of a surface is twice the area of one side). Thus, we conclude that radiation pressure Pr on one side of a perfectly reflective surface must be Pr = 2 P = B2 2 ∗S∓ = m c μ0 N/m2 which is to say that by knowing either of the two components of EM wave we are able to calculate mechanical pressure caused by the light. Practical consequences of this conclusion we illustrate in the following examples. 1.11. First, Maxwell’s theory predicted the light pressure, then in 1899 Pyotr Lebedev experimentally demonstrated the concept. Thereafter, a number of science fiction writers used the concept, most notably Arthur Clarke in his 1964 novel “Sunjammer”, which inspired naming of the next solar sail experimental space flight demonstration planned for 2014. In this example we try to numerically illustrate feasibility of the concept (Fig. 13.5). In order to neutralize gravitational pull from the Sun, force produced by solar pressure must be at least as equal. From the basic physics we know that pressure P is defined as force F over the surface A, thus radiational force Fr is expressed as Fr = Pr A. In addition, gravitational force between two masses is defined by the Newton’s gravitational law. In addition, assuming a space probe of mass m, the Sun mass m S , area of the solar sail A, and the distance of the probe from the Sun a, we write expression for the balance between radiational and gravitational forces as G 2S mS m A ≤ Fr = Pr A = 2 a c We already know that Poynting vector ∗S∓ represents power per surface, therefore the total power of the Sun PS is distributed over spherical surface 4 ω a 2 . Thus, 13 Introduction: Solutions 73 G mS m 2 PS ≤ A 2 a c 4 ω a2 which, for the given numerical data, leads into c 4 ω G mS m 2 PS 3 × 108 m s−1 × 4 ω × 6.674 × 10−11 N m2 kg−2 × 1.989 × 1030 kg × 2 kg = 2 × 3.842 × 1026 W 3 2 A ≥ 1.302 × 10 m A≥ This solar sail can be made of a square shaped area whose side is 36.079 m. Sail of this size must weight m k = m k0 A = 768.5 µg/m2 × 1.302 × 103 m2 = 1 kg (13.39) therefore, mass of the useful payload is 2 kg − 1 kg = 1 kg. 1.12. Waves are described by ordinary trigonometry equations, therefore in our analysis of multi-tone waves we can assume that they have been created by superposition (i.e. interference) of two or more single-tone waves. Consequently, the overall analysis is simplified because more complicated equations are derived by following the elementary trigonometry algebra rules. While keeping in mind the two space– time variables, there is a number of interesting situations that arise that depend on the phase relationship between the two super-positioning waves. In this example introduce the concept of “standing waves”, which is mathematically described as the superposition of two waves traveling in opposite directions. In this case, the electric filed is described as if it were composed of the sum of two waves, one traveling in positive z-direction, and one traveling in negative z-direction. Therefore, we write E = E m (cos(kz − ωt) − cos(kz + ωt)) y = 2E m sin(kz) sin(ωt) y B = Bm (cos(kz − ωt) + cos(kz + ωt)) z = 2Bm cos(kz) cos(ωt) z (13.40) (13.41) Then, by definition, 1 E×B μ0 2E m 1 cos(kz) cos(ωt) z = (2E m sin(kz) sin(ωt) y) × μ0 c S= = E m2 sin(2kz) sin(2ωt) x c μ0 and now we can calculate average value ∗S∓ by definition, (13.42) 74 13 Introduction: Solutions T 1 ∗S∓ = S dt T 0 T E m2 1 = sin(2kz) sin(2ωt) dt x c μ0 T 0 E m2 T =2ω/ω sin(2kz) cos(2ωt)|0 =− x c μ0 2ω T =0 (13.43) Let us take a closer look at the above results. Equation (13.40) shows that the electric field becomes zero when sinkz = 0, in other works when kz = nω ∴ z= nω π nω = =n 2ω k 2 π (13.44) where, n = 0, 1, 2, . . .. As the time advances, these points in space are not moving. Similarly, the points of maximum amplitude E max = 2E m are also not moving in space. We conclude that at distances equal to multiples of the half–wavelength the total amplitude of the electrical component is zero. At the same time, from (13.41) we see that Bm = 0 if coskz = 0, i.e. 1 kz = n + ω 2 ∴ z= n 1 + 2 4 π (13.45) where, n = 0, 1, 2, . . .. Again, as the time advances, at these points in space the corresponding total amplitude of the magnetic field is always zero, which coincides with the maximum amplitudes of the total electric field. Result found in (13.43) is also very interesting. We already concluded that amplitude of Poynting vector represents rate of energy flow in the vector’s direction, thus saying the ∗S∓ = 0 is to say that the energy of the wave used in this example does not propagate in space. This case is known as a “standing wave”, it is very important and it can be created, for example, by using two ideal mirrors positioned exactly at the distance equal to an integer number of wavelengths that effectively create a resonant cavity, Fig. 13.6. The wave energy is oscillating, however it is confined within the cavity. For example, this phenomena forms the basis for design of lasers. It is important to us to recognize that because standing waves do not propagate in space, they can not be used transmission of radio signals. Therefore, we must design transmission circuits so that reflection, and therefore creation of standing EM waves, is avoided. 1.13. We have already learned that SI unit for electric field E is V/m, while SI unit for magnetic filed H is A/m. It is then easy to realize that the ratio between the two fundamental fields is 13 Introduction: Solutions 75 Fig. 13.6 Problem 1.12: example of a standing wave formed between two perfectly reflective surfaces at distance of d = 3/2π E ∼ H V m A m = V =α A (13.46) which is, naturally, known as the free space impedance Z 0 . We keep in mind that in free space B = μ0 H (i.e. for all practical purposes they are interchangeable), thus after substituting (13.32) and (13.35), we derive several alternative expressions of the free space impedance. Relative permeability μr and permittivity εr of free space are defined as unity, while free space permeability μ0 and permittivity ε0 are defined relative to the free space, resulting in the intrinsic impedance as E =c B ∴ ∴ Z 0 = μ0 c = E =c μ0 H 1 ε0 c = ∴ E = μ0 c H μ0 = 4ω c × 10−7 m/s √ 120 ω √ 377 α ε0 (13.47) By definition, phase velocity was derived from (13.34) in (13.35), Therefore, the EM wave phase velocity in free space is, obviously, the velocity of light νp = c = π 1 ω = =π f = ≥ k T μ0 ε0 ∴ ν p = 2.99792458 × 108 m/s √ 300, 000 km/s The wavelength then is calculated as (13.48) 76 13 Introduction: Solutions ⎧ ⎪ ⎨30 m for f 1 = 10 MHz νp 2ω ν p π= = = 3m for f 2 = 100 MHz ⎪ f ω ⎩ 3 mm for f 3 = 10 GHz (13.49) As this example shows, wavelength of EM waves oscillating at some of the commonly used frequencies are comparable with physical size of our equipment. That fact is important for design engineers when considering, for instance, the antenna size (e.g. the dipole length), or the length of conductive wires on printed circuit boards (PCB). As the EM wave frequency increases, its wavelength becomes comparable first with the size of PCB, then with the sizes of discrete components themselves. We note that at typical RF signal frequencies used in this book (i.e. f = 10 MHz), size of PCB (which is in order of several centimetres per side) is much less then the estimated 30 m calculated wavelength of a 10 MHz signal. Therefore, the use of approximated Maxwell’s equations is justified. On the other hand, if the signal frequency is in order of multiples of GHz, even integrated circuit (IC) designs (which typically occupy only a few millimetres per side) must account for the phase differences along the signal path and use the transmission line theory in the design circuit calculations. It is now good time to comment on the terms of “high frequency” (HF) and “low frequency” (LF). As we see, the high frequency term is relative to the length of the conductive medium (e.g. wire). If along the conductive path we can measure significant variation of potential, or even full positive–zero–negative swing, then we say that the signal frequency is high. However, if along the conductor length we measure only negligible variations in the potential, we say that the signal frequency is low and that the conductor represents good approximation of an equipotential surface. It should be obvious that even 60 Hz signal (π =) is considered “high frequency” if transmitted over distances of several thousands kilometres. Naturally, there is wide “grey area” between HF and LF regions, thus it is up the designer to decide what approximation is acceptable for the particular design. 1.14. In this example we revisit applications of Ampère’s law by looking at magnetic field distribution within and outside of a cylindrical conductive wire. Straightforward implementation of Ampère’s law, d H · d l = Ifree,enc + dt L ∂D ∇ × H = Jfree + ∂t D · ds integral form (13.50) differential form (13.51) S for case of constant current I (t) = const and all the charges contributing to the current flow, the current density J is uniform through any cross-section area of radius r inside the conductor, up to the radius a at the conductor’s surface. Hence, portion of the total current Ir flowing thorough any area with radius r (0 ≤ r ≤ a) inside the conductor is determined by the ratio between the full cross-section area ωa 2 and the inside area ωr 2 , i.e. Ampère’s law can be written as 13 Introduction: Solutions 77 180 160 H, A/m 140 120 100 80 60 40 20 0 0 5 10 15 20 25 30 r, mm Fig. 13.7 Problem 1.14: Magnetic field distribution inside and outside of an infinitely long wire or radius a = 5 mm carrying a current of 5 A H · 2ωr = I ωr 2 ωa 2 ∴ H= Ir 2ωa 2 (13.52) where 0 ≤ r ≤ a. Outside of the conductor the current density is equal to zero, which simplifies (13.50) (or (13.51)) so that the magnetic filed H outside of the conductor is calculated as H · 2ωr = I ∴ H= I 2ωr (13.53) where r ≥ a. The total magnetic field outside and inside of the infinitely long conductive wire is ⎧ Ir ⎪ ⎪ = 31.810 × 103 r A/m r ≤ 5 mm ⎪ ⎨ 2ωa 2 H (r ) = ⎪ ⎪ ⎪ ⎩ I = 0.798 A/m r ≥ 5 mm 2ωr r It is important to notice that the magnetic field is affected by the surrounding material, it increases linearly inside the conductor because more current contributes to the magnetic field, while outside of the wire the magnetic field strength is inversely proportional to the distance because the whole current has been accounted for and there is no external contributors to the field, Fig. 13.7. 1.15. Free charge inside a time varying magnetic field B start to move in space, which is by definition an electric current. If the charge is contained inside a metallic conductor, then the moving charges create induced electrical field, which by definition creates voltage potential along a path. The newly induced potential is equal 78 13 Introduction: Solutions Fig. 13.8 Problem 1.15: the time rate of change of the magnetic flux density induces a voltage to the loop line integral of the electrical filed E around the conductor. It the same time, this potential is also connected with the B and its flux calculated across the associated surface. Straightforward implementation of Faraday’s law d E · dl = − B · ds integral form dt S L ∂B ∇×E=− differential form ∂t (13.54) (13.55) results in expression for potential V as d d V =− E · dl = B · ds = μ0 H · ds dt S dt S L d d = μ0 H0 cos(ω t)n · ds = μ0 H0 cos(ω t) ωa 2 n dt S dt = −ωa 2 ωμ0 H0 sin(ω t) = −0.031 sin(2ω × 108 t) V where, B = μ0 H is the magnetic flux density, n is the unity vector in the same direction as magnetic field H vector. Maximum amplitude of the induced potential is approximately 31 mV when measured at the ring terminals. This problem example is a typical demonstration of Maxwell’s equation in a form of Faraday’s law, also known as the transformer law, where the time-varying magnetic field induces a voltage response in a conductive loop located nearby, Fig. 13.8. The voltage induction principle is one of the most important discoveries that enabled radio communication, electric motors, and generation of electric energy. 1.16. Low frequency approximation is a very handy tool in cases when wavelength of of a signal is much longer than length of the conductive media (e.g. a wire) that is used to transmit the said signal. In the low frequency approximation the assumption is that the conductive surface is equipotential, i.e. at any given moment in time the potential is equal at all points in space. Or, in other words the conductor is ideal, that is without any parasitic RLC components. However, when the signal wavelength is comparable with the conductor’s length non equipotential surfaces become reality. In Problem 1.13 we could see a range of wavelengths used in telecommunications, 13 Introduction: Solutions 79 Fig. 13.9 Problem 1.16: section of a long conductor (relative to the π) located between z and (z + λz) while in this example we take a first look at what is known as transmission line model. A long connector is cut into infinitesimally short pieces, and each section is modelled as RLC network, Fig. 13.9. We keep in mind that in this case units are normalized per length. With the help of Kirchhoff’s voltage law around circuit loop that includes V (z) and V (z + λz), Fig. 13.9, V (z + λz) − V (z) = −(R + jωL) I (z) λz (13.56) at the same time, by definition, potential difference across the conductor section λz is written as V (z + λz) − V (z) d V (z) = lim − (13.57) − λz∼0 dz λz which, after substitution of (13.56) and after applying the limit λz ∼ 0 leads into − d V (z) = (R + jωL) I (z) dz (13.58) a we derive expression Similarly, applying Kirchhoff’s current law to the node for change of the current as function of the wire length as I (z + λz) = I (z) − V (z + λz)(G + jωC) λz (13.59) and, by definition, we write I (z + λz) − I (z) d I (z) = lim λz∼0 dz λz ∴ d I (z) = −(G + jωC) V (z) dz (13.60) Equations (13.58) and (13.60) are coupled first-order differential equations. Solution to this system of equation is found by decoupling the two equations, which is accomplished by differentiating both sides, first (13.58) and then (13.60). By doing 80 13 Introduction: Solutions so, explicit solutions for V (z) and I (z) are found. Therefore, starting with spatial differentiation of (13.58) and substituting (13.60), − d 2 V (z) d I (z) = (R + jωL) dz dz ∴ d 2 V (z) = (R + jωL)(G + jωC) V (z) dz d 2 V (z) − (R + jωL)(G + jωC) V (z) = 0 dz d 2 V (z) − k 2 V (z) = 0 dz (13.61) where, a complex propagation constant k is defined as k = (k) + j(k) = (R + jωL)(G + jωC) (13.62) and it is function of the transmission line geometry (keep in mind that the R, L, C, and G are all distributed parameters, i.e. functions of length, calculated separately for specific shape of the conductor). Repeating the same procedure, while starting with spatial differential of (13.60) instead, similar solution to (13.61) is found for the current spatial dependance, i.e. d 2 I (z) − k 2 I (z) = 0 dz (13.63) Equations (13.61) and (13.63) show explicitly, for a given moment in time, the voltage/current spatial dependance along the transmission line. Solutions to these two decoupled equations are already well known to be in the following form (assuming, of course, that the transmission line is aligned with the z-axis): V (z) = V + e−kz + V − e+kz + −kz I (z) = I e − +kz +I e (13.64) (13.65) As a convention, each of these two equations is interpreted as combination of two waveforms, one propagating in the positive z-direction, while the second one propagates in the negative z-direction. The two equations (13.64) and (13.65) are correlated because they describe the same waveform, which means that they are connected through the transmission line impedance. For example, substituting (13.64) back into (13.58) results in the explicit relationship between the voltage V (z) and current I (z) as 13 Introduction: Solutions − 81 d V (z) = (R + jωL) I (z) dz ∴ 1 d V (z) I (z) = − (R + jωL) dz d(V + e−kz + V − e+kz ) 1 =− (R + jωL) dz k + −kz − V − e+kz ) = (V e (R + jωL) k V (z) = (R + jωL) (13.66) which is to say that the expression connecting current I (z) and voltage V (z) must be impedance, and because it is important parameter of a transmission line it is named characteristic line impedance Z 0 , i.e. after substituting (13.62) (R + jωL) V (z) = = Z0 = I (z) k (R + jωL) (G + jωC) (13.67) which, in ideal lossless case, i.e. no thermal dissipation R = G = 0, simplifies as Z0 = L C (13.68) Characteristic impedance of a lossless transmission line (13.68) is not function of frequency. That fact should be contrasted to the general definition (13.67) that is a complex quantity and does take into account always present thermal losses (but not inter-component interference). However, the characteristic impedance is very strong function of the line geometry (through the distributed values L and C), hence must be calculated for each type of transmission line, for example for two-wire line, coaxial line, parallel-plate line, etc. Transmission line model emphasizes the fact that waves propagate through the media with finite velocity, which causes different potentials between points where the wave has “arrived” and the points that are still “waiting” for the waves to arrive. Thus, the importance of a transmission line characteristic impedance is in the fact that, from strictly electrical perspective, each section of the line represents resistance. Therefore, any subsequent section or component that is connected to a particular section should have the same impedance for maximum power transfer, otherwise, at the interfacing point some of the transmitted energy will be “reflected back” and perceived as the signal loss. Calculation of the reflection coefficient at the interface of two impedances gives us very useful design parameter. For the given numerical example, we calculate 82 13 Introduction: Solutions Z0 = L = C 378 nH/m = 50.2 α 150 pF/m (13.69) which implies that this two-wire connection it is a good choice as the interface to a 50 α antenna, or a 50 α input of an oscilloscope. 1.17. Instantaneous values of a sinusoidal wave are very often expressed in their respective interchangeable units. For example, inquiring about a wave’s instantaneous phase in radians is equivalent to asking about location of the wave’s front end in time. A fine point to note, though, is that it is customary that phase of a periodic waveform is folded back into the first period, while the wave’s front end position is expressed in absolute time. Similar comment applies when comparing the waves’s wavelength and the wave’s front end position in space. In this example we illustrate these relations by using example of a waveform whose frequency is f = 100 Hz. By observation of the given waveform expression v(t) = Vm cos (ω t + φ0 ) (13.70) where, radial frequency was given as ω = (2ω 100) rad/s and the initial phase φ0 = then by definition it follows that ω/4, ω = (2ω 100) rad/s ⇒ f = 100 Hz ∴ 1 T = = 10 ms f ∴ π = c T = 2.99792458 × 108 m/s × 10 ms √ 3, 000 km where correlation between frequency and the wavelength is achieved through velocity of light c and the wave period T . Similarly, correlation between phase and the absolute time is established by using relationship for radial period T = 2ω ∴ 2ω ∞ 10 ms ∴ 15 ms ∞ 3ω = 12ω 4 which is phase of a waveform whose initial phase is φ0 = 0, for example light grey sinusoidal waveform in Fig. 13.10. However, in this example the zero moment in time is associated with the initial phase of φ0 = ω/4. Thus, at the moment t = 15 ms the phase is shifted to φ= ω 13ω 12ω + = 4 4 4 13 Introduction: Solutions 83 Fig. 13.10 Problem 1.17: unit conversion As already mentioned, it is customary to express all phase results relative to the period, i.e. after removing integer multiples n × 2ω , (n = 0, 1, 2 . . .), which implies that phase takes only values within 0 ≤ φ ≤ 2ω range, Fig. 13.10, which in this example means that φ = 5ω/4. Instantaneous voltage amplitude relative to the maximum amplitude Vm at t = 0 s is calculated directly from (13.70) as Vm V = Vm sin φ0 = ≥ 2 while at t = 15 ms we find, for example, by inspection of Fig. 13.10 that V = Vm sin Vm 13ω = −≥ 4 2 Familiarity with various views and units related to sinusoidal waveforms are very important for practicing engineers. As a side note, we observe huge value of wavelength for a 100 Hz waveform. Indeed, waveforms of LF signal are comparable with the Earth’ dimensions. Knowing that practical length d of an antenna intended for reception/transmission of a signal should be comparable with the signal’s wavelength (for example, d = π/4), it should be obvious that wireless transmission of LF signals is simply not practical. That is, we already use long-distance cables as telephone lines laid in trenches, instead of using them as the antennas. 1.18. Fourier gave us a tool that is equivalent to X-ray generator used by medical professionals. The X-ray machine enables us to see inside of a living being, without the need to physically open the body. Similarly, Fourier transform and polynomials enable engineers to “see” inside a complicated waveform and to find what is was made from. Picture that shows content of a waveform is called “frequency spectrum” 13 Introduction: Solutions 1.0 1.0 vΣ10(t), vHP(t) vΣ3(t), v1(t), v3(t), v5(t) 84 0.5 0 -0.5 vΣ3 v1 v3 v5 -1.0 0 0.5 0 -0.5 vHP vΣ10 -1.0 0.5 1.0 time 1.5 2.0 0 0.5 1.0 1.5 2.0 time Fig. 13.11 Problem 1.18: a complicated “squarish” looking waveform v3 (t) on the left, synthesized by a simple addition of three single-tone functions v1 (t), v3 (t), v5 (t). Better approximation of a square waveform v10 (t) (right) is achieved by adding first ten terms of (13.71). Waveform v H P (t) is created by removing the first two terms from v10 (t), which is equivalent to applying a HP filter. Observe that sharp edges are preserved due to HF content (horizontal axis is in units of period T ) and it serves the same purpose as X-ray pictures to doctors. In this example we learn how to synthesize a square pulse waveform by using addition of several sinusoidal functions. In that respect, Fourier polynomials are our recipes for how to make complicated periodic waveforms. Function v(t) in (13.71) represents Fourier polynomial of a square wave function (actually it should have been 4/ω v(t), to be mathematically correct). v(t) = sin (2ω t) + 1 1 1 sin (6ω t) + sin (10ω t) + sin (14ω t) + · · · (13.71) 3 5 7 By increasing the number of terms the square wave approximations becomes better, with the ideal square function requiring infinite number of sinusoidal terms. However, if some of the terms are missing, the shape of square waveform becomes distorted when we observe it in time domain, Fig. 13.11 (right). In electrical engineering terms, the process of removing for instance low-frequency terms (e.g the terms associated with ω and 2 ω, i.e. low-frequency terms) is equivalent of applying high-pass (HP) filter. Similarly, when only high-frequency terms are removed, in electrical terms it is equivalent to application of a low-pass (LP) filter. If both lowfrequency and high-frequency terms are removed and only the “middle” terms are left in the equation, it is equivalent of applying a band-pass (BP) filter. It is useful exercise to remove one or more neighbouring terms from the middle of the polynomial and observe the time domain distortion. In addition, find in literature similar Fourier polynomials that describe triangular, sawtooth, or any other “standard” waveform and repeat the exercise. 1.19. While time domain plot in Fig. 13.11 is very useful to compare signal nonlinearities, i.e. distortion, it still does not give us clear picture of what v10 (t) is made from. For that purpose we need to use Fourier transform and convert v10 (t) 13 Introduction: Solutions 85 Signal Power, dB 0 HPF -20 -40 -60 -80 -100 -120 0 5 10 15 20 25 frequency, kHz Fig. 13.12 Problem 1.19: frequency spectrum of square waveform v10 , Fig. 13.11 (right) (the vertical axis is scaled relative to the first harmonic’s power) into its equivalent frequency domain plot P( f ) to show power content of v10 (t) at various frequencies. In the frequency spectrum plot, Fig. 13.12, the ten single-tone signal terms that are used in the synthesis of v10 (t) with their respective frequencies and relative powers are clearly visible. Red dash-dot line (HPF) shows profile of a HP filter that would be used if the first two terms are to be removed, v H P (t), for example if only crossover points of the waveform need to be detected. Of course, in this example we started from a signal whose frequency content was already known, however, this example illustrates usefulness of Fourier transform to examine content of an unknown signal, and it is indispensable to circuit designers. 1.20. It is very often needed to remove unwanted components in the frequency spectrum of a signal. For example, if a noise becomes embedded into the information carrying tone becomes, then some form of filtering must be applied in order to recover the information. Application of LP, HP, or BP filters is obvious and most often used solution to the problem. With this approach all tones outside of the filter’s pass-band will be indiscriminately suppressed. However, it is not difficult to imagine a case where a single component in the signal frequency spectrum needs to be removed while leaving the rest of the spectrum intact. Fortunately, with a bit of understanding how sinusoidal waveforms interact with each other, we can design circuits to perform mathematical computations and to achieve the goal. In this example we learn how two or more sinusoidal signals can be added together, so that the resulting waveform is derived either by “constructive” or “destructive” action. In other words, instead of trying to suppress a certain tone by the filtering action, it is always possible to add another tone that is “negative” and therefore cancels the targeted term. For example, with reference to (13.71), if the goal is to produce a waveform y(t) that is identical to v(t), except for only the third term v3 (t) = 1/5 sin(10ω t), then we 86 13 Introduction: Solutions can write the following: y(t) = v(t) + (−v3 (t)) (13.72) 1 1 1 1 = sin (2ω t) + sin (6ω t) + sin (10ω t) + sin (14ω t) · · · − sin (10ω t) 3 5 7 5 1 1 1 = sin (2ω t) + sin (6ω t) + sin (14ω t) + sin (18ω t) + · · · 3 7 9 As trivial as it may seem, (13.72) gives us direction of how to design a system that can perform cancelation of v3 (t) tone. Aside from physical connection to v(t), we need to design signal generator whose output equals −v3 (t), which is a sinusoidal waveform. In the following chapters we will learn how to design a circuit that generates sinusoidal output with specific amplitude, phase and frequency. For the time being we assume that we have such circuit. Lastly, we need an adding circuit (keep in mind that the operation of subtraction may be looked at as the addition of negative number). Depending whether the information signal is in form of current or voltage, then the addition is performed either by connecting two wires together (and therefore summing the two currents), or by using an operational amplifier whose two inputs perform subtraction of the two voltage signals. In this example we learn a special very important case of “constructive” addition of two waveforms, that is, we find out how to create “differential” signal out of two single-ended signals. 1. Before we proceed to performing signal operations, let us first take a closer look at the given signals v1 (t) = 1 + 0.5 sin(ω t) v2 (t) = 1 + 0.5 sin(ω t − ω ) Both of these two sinusoidal signals contain a constant term equal to one. This is important observation, because in electrical engineering, a constant current or voltage is referred to as DC. That means, value of every point of the sinusoidal function is increased by, in this case, one. Therefore, this DC value takes the role of average value for the sinusoidal signal (because without it, average value of zero is assumed in all sinusoidal waveforms). To emphasize this role, we refer to the average value of sinusoidal signal as the common mode (CM). We quickly realize that in this case CM = 1 is greater than maximum amplitude Vm = 0.5, therefore v(t) is always positive, which is important information for the circuit designer. Next observation we make is that maximum amplitudes of v1 (t) and v2 (t) are equal. Following that, the same comment applies to the two frequencies ω 1 and ω 2 , their values are also equal. Finally, because the two frequencies are equal, we can look at phase difference. In this case, phase difference λφ = ω . Let us now see significance of these observations. Subtraction of v1 (t) and v2 (t) leads into 13 Introduction: Solutions 87 1.5 v1(t), v2(t), v(t) 1.0 0.5 0 -0.5 v v1 v2 -1.0 -1.5 0 π 2π 3π 4π time Fig. 13.13 Problem 1.20: a differential waveform v(t) (black line), created by subtracting of two waveforms v1 (t) and v2 (t) with the same common mode voltage VCM = 1V , same frequency, and the opposite phase 1 1 v(t) = v1 (t) − v2 (t) = 1 + sin(ω t) − 1 + sin(ω t − ω ) 2 2 1 = [sin(ω t) − sin(ω t − ω )] 2 1 = [sin(ω t) + sin(ω t)] 2 = sin (ω t) (13.73) The resulting sinusoidal waveform v(t) has CM = CM1 = CM2 = 1 − 1 = 0, which is logical because the two initial CM values were equal. Maximum amplitude of the resulting sinusoid is the sum Vm = Vm1 −(−Vm2 ) = 0.5−(−0.5) = 1. It is important to understand that the sign of a sinusoidal amplitude is assigned relative to the CM (just remember a simple sinusoid where CM = 0). In other words, the two sinusoids v1 (t) and v2 (t) always have the opposite signs, which is caused by the phase difference of exactly λφ = ω . To summarize, when two sinusoids have same CM, same frequency, same amplitude, and phase difference of ω (i.e. they are inverted relatively to each other), their difference is a sinusoid with the same frequency whose amplitude is doubled and CM is zero, Fig. 13.13. Important to notice, by doing the mathematical operation of subtraction we achieved factor of two gain. Differential signal processing is most important in modern electronics and we will see it again in the following chapters. 2. As counterintuitive it is, the subtraction of two inverted sinusoids results in amplified sinusoid, let us now see what is the result if the same two sinusoids are added, i.e. 88 13 Introduction: Solutions 1 1 v(t) = v1 (t) + v2 (t) = 1 + sin(ω t) + 1 + sin(ω t − ω ) 2 2 1 = 2 + [sin(ω t) + sin(ω t − ω )] 2 1 = 2 + [sin(ω t) − sin(ω t)] 2 =2 (13.74) This time, two time varying components (AC) canceled each other because they are inverted (we can imagine that each waveform was pulling in the opposite direction with the same force, thus going nowhere), and we are left only with a DC signal the is result of the sum of two CM levels. 3. When two waveforms are not perfectly symmetrical (which is more realistic case), then resulting signal differs from the previous two cases, for example, in this exaggerated case of CM misalignment 1 1 1 + sin(ω t − ω ) v(t) = v1 (t) − v2 (t) = 1 + sin(ω t) − 2 2 2 1 1 = + [sin(ω t) − sin(ω t − ω )] 2 2 1 1 = + [sin(ω t) + sin(ω t)] 2 2 1 = + sin (ω t) (13.75) 2 the two CM levels did not cancel, therefore, the resulting differential signal does not have zero CM and the sinusoid takes asymetrically positive and negative values, Fig. 13.14, which may or may not be what we wanted to achieve. 4. Addition of two inverted sinusoids with misaligned CM results in 1 1 1 + sin(ω t − ω ) v(t) = v1 (t) + v2 (t) = 1 + sin(ω t) + 2 2 2 3 1 = + [sin(ω t) + sin(ω t − ω )] 2 2 3 1 = + [sin(ω t) − sin(ω t)] 2 2 3 = (13.76) 2 which means that two inverted sinusoids cancel even if their CM are different, while the outcome is DC level that is the sum of two initial CM levels. As already mentioned, differential signal processing is extremely important in modern signal processing, and it is, therefore, very important to quantify all possible 13 Introduction: Solutions 89 1.5 v1(t), v2(t), v(t) 1.0 0.5 0 -0.5 v v1 v3 -1.0 -1.5 0 π 2π 3π 4π time Fig. 13.14 Problem 1.20: a differential waveform v(t) (black line), created by subtracting of two waveforms v1 (t) and v3 (t) with non-equal common mode voltage, same frequency, and the opposite phase imperfections due to the signal misalignments, or departures of any kind from the ideal symmetrical case. This example also illustrates how important it is for engineers to develop “mental math” skills that enable us to imagine that a sinusoid is created by adding two components, namely the varying AC and the static DC, and to separately manipulate the two of them. 1.21. Analog signal processing is nothing more than direct implementation of mathematical equations by means of transfer functions that are used to describe electronic components, i.e. R, L, and C (and from now on M, a memristor). That is, each component by itself embodies one or more specific mathematical operations, for example operation of addition is implemented by using a resistive divider for two voltage signals, while operation of the first derivative is implemented by knowing that first derivative of a current signal is proportional to the voltage across inductor whose inductance value serves as the proportionality constant. Therefore, it is possible to piece together any equations by means of combining various analog components. The equation synthesis is achieved either as direct one-to-one matching, for example exponential function is embedded in a BJT transistor’s operation, or by close approximation, for example nth root function can be implemented by using Taylor expansion and exponential function, and so on. In this example we learn how simple operation of linear addition is implemented by using a voltage divider. 1. Because the two resistors are equal R = R1 = R2 = 1 kα the two input voltages are added and divided equally. Thus, after passing through the voltage divider network, the two signals are attenuated in the same proportion, which follows from Kirchhoff’s current law (regardless whether the input voltage signal is AC or DC) as 90 13 Introduction: Solutions Fig. 13.15 Problem 1.21: block diagram of an analog voltage adder circuit v2 − vo v1 − vo + =0 R R ⇒ vo = v1 + v2 = 2vo 1 (v1 + v2 ) 2 ∴ vsum = v1 + v2 = 2 vo (13.77) which further illustrates operation of addition. Side effect of this approach is that the addition operation is accompanied with the 1/2 scaling factor, which implies that correct implementation of voltage signal addition as in (13.77) requires two equal resistors and an amplifier whose gain is G = 2, Fig. 13.15. 2. From given data T1 = 1 µs it follows that f 1 = 1/T1 1 = 1MHz. Further, by inspection of Fig. 1.2 we write T2 = 1/2 T1 = 0.5 µs, therefore f 2 = 1/T2 = 2MHz. Additional two pieces of information we can extract from Fig. 1.2 is that the two maximum amplitudes are related as V2m = 2 V1m and that the phase difference is zero (even though one frequency is double of the other, the two are synchronized at all times and have the same zero-amplitude crossing points in time). Therefore, v1 (t) = 1 sin[(2ω 1MHz) t] v2 (t) = 2 sin[(2ω 2MHz) t] ∴ vsum = v1 (t) + v2 (t) = sin[(2ω 1MHz) t] + 2 sin[(2ω 2MHz) t] (13.78) From the graph given in Fig. 1.2 and (13.78) we conclude that the linear addition of two single tone signals does not produce any new frequency tones in the output spectrum (we still have the 1 and 2 MHz single tones at the output), and that the output amplitude of v2 is still twice the amplitude of v1 . Knowing the two voltages (v1m = 1), (v2m = 2) and resistance R, we can correlate the two corresponding powers (P1 , P2 ) as 13 Introduction: Solutions 91 Fig. 13.16 Problem 1.21: relative power spectrum of two waveforms v1 (t) and v2 (t) P1avg = 2 v2 v2 v1rms v2 1 4 = 1m = and P2avg = 2rms = 2m = R 2R 2R R 2R 2R ∴ P1avg 1 = P2avg 4 (13.79) which is to say that 1 : 4 is relative power ratio between the two signals, both at the input and the output nodes. Compact way of presenting normalized power vs. frequency relationship is to use a frequency spectrum graph, Fig. 13.16 (left). 3. When the two resistors are not equal, i.e. R2 = 2 R1 = 2R, the two input voltages are added and divided in the same ratio as the two resistors, thus we write v1 − vo v2 − vo 1 + = 0 ⇒ 2v1 − 2vo + v2 − vo = 0 ∴ vo = (2 v1 + v2 ) R 2R 3 Therefore, in this case the output voltage vo is not just a simple sum of the two input signals, instead voltage v1 is multiplied by factor two before it is added to v2 , which is then alltogether multiplied by 1/3 factor, thus we write vo = 1 {2 sin[(2ω 1 MHz) t] + 2 sin[(2ω 2 MHz) t]} 3 (13.80) We observe that the output frequency spectrum still did not change, while the relative amplitudes of the two output tones are now equal. 4. In this case, after passing through the voltage divider network, the two signals are not attenuated in the same proportion. Signal v2 passes through twice the resistance relative to the resistance of the path that v1 takes. Consequently, at the output side of the voltage divider network the proportion between v1 and v2 changed by factor two relative to the input side, and therefore became equal. Now, the two signal powers contribute equally to the power of the output signal vo , Fig. 13.16 (right). This example illustrates linear operation of addition, which may change relative powers of the internal tones in the output signal, however it does not change frequency content. This is very important observation, because in the following chapters we 92 13 Introduction: Solutions Fig. 13.17 Problem 1.22: signal envelope sketch will realize that design of wireless circuits is based on using non-linear operations, such as multiplication, to produce output tones at different frequency relative to the input tones. This “frequency shifting” operation is fundamental for wireless RF communication systems and must be recognized as different than linear addition. 1.22. A continuous single tone waveform by itself does not convey any useful information, aside from revealing its own existence. Nevertheless, even that single bit of information, i.e. existence of a waveform, is used to design first forms of communication systems for transmitting information over distances greater than the distance achievable by a human voice. If time multiplexed, combination of “short” and “long” beeps enables creation of Morse code where each letter of alphabet is encoded as a unique combination of beeps. For instance, “RF” would be encoded as “· − · · · − ·” and transmitted wirelessly by well timed turning ON and OFF power button of a transmitter that emits a single tone. In this example we learn about a bit more sophisticated encoding principle, where we exploit fact that a single tone sinusoidal waveform can change either its amplitude, frequency, or phase. If amplitude of the waveform is changed in accordance with the information signal, we refer to this technique as amplitude modulation. The time varying maximum amplitude points generate a curve known as the “envelope waveform” and its shape is exact replica of the information being transmitted, Fig. 13.17. Because a sinusoid is symmetrical waveform, its amplitude change is also symmetrical. Consequently, we note existence of two identical half-envelopes venv+ = |venv− | whose absolute values are equal (which is always the case). In this particular case only, the information is a low-frequency sinusoid, thus each half-envelope has a shape of a sinusoidal signal. Also, the two half-envelopes have common modes at CM+ for the positive, and CM− for the negative half-envelope. By using appropriate signal processing electronics, either of the two half-envelopes can be extracted, which is to say that the information can be extracted from any of the half-envelopes. Chapter 14 Basic Terminology: Solutions Concept of an electric field and forces among charged particles at the atomic level help us model macro-behaviour of materials and devices. The macro-models based on these concepts enable us to predict outcome of an experiment, or to design a device to perform a certain (mathematical) function. Duality of matter and waves turns out to be fundamental for macro operation of these devices, thus in this section we review some of the mathematical techniques commonly used in the wireless electronic circuit design. Solutions: 2.1. In classical physics model we use concept of the four fundamental forces in the universe that are responsible for all non-contact interactions. Namely the four forces are gravity, electromagnetism, strong nuclear force, and weak nuclear force. The two nuclear forces are assumed to be “short distance” forces responsible for holding the matter together. At the same time, gravity and electromagnetism are assumed to be “long distance” forces that greatly influence bio-chemical, electrical and electronic processes. In this example we review basic definitions and compare electrostatic and gravitational forces. Direct application of Coulomb’s and gravitational law gives q q ke e 2 p k e qe q p FC R= = m erm p = FG Gm em p G 2 r 2 (1/4ω λ0 ) Nm2/C2 1.602176565 × 10−19 C = 6.67384 × 10−11 N m2/kg2 9.10938291 × 10−31 kg 1.672621777 × 10−27 kg = 2.26882 × 1039 That is, for all practical purposes force between an electron and proton is determined only by the electrostatic attraction, while the gravitation force is safely neglected because it is approximately 1039 times smaller. Therefore, although we think about R. Sobot, Wireless Communication Electronics by Example, 93 DOI: 10.1007/978-3-319-02871-2_14, © Springer International Publishing Switzerland 2014 94 14 Basic Terminology: Solutions the charges as material particles, in electrical engineering we neglect the effects of gravity. 2.2. As we already know, scalar variables are determined only by a number that quantifies their magnitude (e.g. temperature), while vector variables are the ones that need both magnitude and direction for the complete description (e.g. force). Electrostatic force is responsible for the movement of free electrons, which by definition is electrical current. In this example, we review vector arithmetics of forces among electrical charges. r23 = a, while distance from q1 to q3 equals to Distance from q2 to q3 equals √ the square diagonal r13 = a 2. Due to the fact that only charge q3 is mobile, there is no need to calculate force between the two fixed charges (q1 , q2 ). Indeed, charge q3 is pulled by (q1 , q2 ), each pooling in its respective direction, with Couloumb’s law stating that electrostatic pulling/pushing force is inversely proportional to the distance. Therefore, magnitude of force F12 relative to F13 stands as the ratio ke F23 = F13 ke q2 q3 2 r23 q1 q3 2 r13 = r13 r23 2 ∴ F23 √ 2 a 2 = F13 = 2 F13 a (14.1) which helps us to sketch correctly scaled force vector diagram. Direction of the vectors is determined by polarity of the there charges. 1. when q3 is a proton, i.e. positive charge, it is pushed away from q1 and pulled by q2 . Thus, vector plot is constructed as in Fig. 14.1 (left). From 14.1 and the simple geometric rules of similar and right-angled triangles, Pythagoras’ theorem, and rules for squares we conclude (after normalizing F13 = 1) that √ 2 √ 2 ⎧ √ 2 2 F3 = + 2− = 5−2 2 2 2 Fig. 14.1 Solution 2.2: Electrostatic forces among three electric charges in space (14.2) 14 Basic Terminology: Solutions 95 while, horizontal and vertical projections of F3 are √ √ 2 2 F3x = 2 − and F3y = 2 √ 2 F3y 2 tan φ = = → φ = 28.675∞ √ F3x 4− 2 (14.3) which, as referenced in Fig. 14.1 (left), fully defines vector F3 . 2. when q3 is an electron, i.e. negative charge, it is pulled by q1 and pushed away by q2 . Thus, vector plot is constructed as in Fig. 14.1 (right). Following exactly the same reasoning as in the previous case we find magnitude and direction of F13 as ⎧ F3 = √ 5 − 2 2 and π = 28.675∞ (14.4) This exercise should serve a purpose of refreshing vector algebra, basic geometry, and basic physics. 2.3. Using a “test charge” q0 at distance r from charge q we measure Coulomb’s force Fc acting at the test charge, and we use it to define electric field around the charge q as F q 1 q E = lim = ke 2 r = r q0 ≤0 q0 r 4ω λ0 r 2 We observe that strength of electric field around a point charge is not function of the used test charge, and the field decreases with square of the distance. 2.4. Developing ability to connect seemingly unrelated events and concepts in a logical order is essential skill. In this example, we practice how to put together a few very basic concepts and how to derive very important scientific conclusion. The elementary electric charge constant was determined by famous Millikan’s oil-drop experiment. In the experiment, Fig. 14.2, a large number of very fine drops of oil whose mass density is known with high precision were electrically charged and suspended in the air by the means of electric field acting in direction opposite to the gravity force. By repeating the measurement many times for various sizes of the oil-drops and magnitude of the electric field, Millikan noticed that all calculated charges stored at the oil drops were multiples of a certain “unity charge”. He therefore concluded that the total charge comes in discrete quanta, thus, the minimal possible charge must be the one of a single electron. Knowing mass density and radius of the two given oil drops and by assuming that the oil drops take ideal spherical shape (which is reasonable assumption if the drops are in the suspended, i.e.weightless, state) it is straightforward to calculate their respective masses m 1 and m 2 as 96 14 Basic Terminology: Solutions Fig. 14.2 Solution 2.4: Simplified diagram of Millikan’s experimental setup m 1 = φoil V = φoil 4ωr 3 kg 4ω(1.670µm)3 = 837 3 = 1.6329146 × 10−14 kg 3 m 3 and m 2 = 1.9606909 × 10−14 kg These two masses are pulled down by gravitational force and at the same time pulled up by the electrical force. When the oil drops are suspended, the two forces are in balance, i.e. mg mg + q E = 0 ∴ q = − E which leads into m 1.6329146 × 10−14 kg × 9.806650 2 m1g s q1 = − = −8.0067111 × 10−19 C =− E 2.0 × 105 CN and q2 = −9.6139048 × 10−19 C Therefore, the difference in electrical charge is q = q2 − q1 = −1.6071937 × 10−19 C (14.5) which is about 0.3 % higher then the accepted value of of electron charge qe = 1.602176565 × 10−19 C. As we already know, the electric charge is one of the fundamental constants in physics and it is important that it is determined by high accuracy. As a side note, in his real experiment, Millikan apparently measured the elementary charge with less than 2 % measuring error. 2.5. In reality, all natural processes are fundamentally non-linear. That is, strictly speaking, a linear function is just a mathematical idealization that does not exist in nature. However, within a relatively small enough window of observation any nonlinear function can be linearized to a various degree. Therefore, by approximating a nonlinear function around the point of interest with its small linear section, we consciously trade accuracy for simplicity. Typical example of this approach is that 14 Basic Terminology: Solutions 97 Fig. 14.3 Solution 2.5: Geometrical view of an electic dipole while we walk we do not think about the Earth as a sphere, instead within our close proximity we perceive it as a plain. Over long period of time, mathematicians worked out linearization methods for many non-linear functions even before they became commonly used to model various physical phenomena. For example some of commonly used mathematical approximations used by engineers are based on Taylor expansion series. In this example, we use Taylor expansion of inverse square root function, which replaces the nonlinear inverse square root with a “nice” polynomial function that is much easier to deal with. Similarly to the Fourier polynomials, the first two polynomial terms are the linear part (i.e. the constant and the first order variable (a + bx)), while the higher order terms add more and more non-linearity while at the same time reducing the approximation error. By definition (14.6), the electric dipole moment is a vector quantity whose magnitude equals to the product of positive and negative electrical charges, while the vector direction is from the negative to the positive charge. In simpler words, it is a measure of the separation between positive and negative electrical charges or its polarity. We already know that a potential V at any point in space is calculated using the superposition principle, thus we write V = k=n ⎪ k=1 1 Vk = 4ω λ0 q q − r1 r2 Using Pythagoras’ theorem, Fig. 14.3, we write r12 = (r sin ∂ )2 + (r cos ∂ − a)2 ∴ ⎨ ⎨ r1 = r 2 + a 2 − 2ra cos ∂ and r2 = r 2 + a 2 + 2ra cos ∂ 98 14 Basic Terminology: Solutions Assuming r ≥ a, we use only the linear terms of the Taylor series1 to simplify expressions for r1 and r2 as 1 1 = ⎩ r1 r 1+ a 2 r 1 −2 a r a cos ∂ 1 a 2 a 1 ∗ r 1 − 2 r + r cos ∂ 1 a 2 1 1 cos ∂ 1− ∗ − r2 r 2 r r which leads into 1 q 1 V = − 4ω λ0 r1 r2 1 a 2 a 1 a 2 a q cos ∂ − 1 + cos ∂ 1− ∗ + + 4ω λ0 r 2 r r 2 r r p cos ∂ p·r 2aq cos ∂ = = (14.6) V = 4ω λ0 r 2 4ω λ0 r 2 4ω λ0 r 2 where, p = 2 pq is known as the electric dipole moment. Electric dipole moment concept enables us to treat the dipole as a single charge. Using E = −∓V in spherical coordinates we write p cos ∂ εV = Er = − εr 2ω λ0 r 3 p sin ∂ 1 εV = E∂ = − r ε∂ 4ω λ0 r 3 For small distances r the electric field becomes infinite, which is mathematical consequence of the initial assumption that r ≥ a. However, from large enough distance, dipole appears as a simple charge, which simplifies further analysis. In engineering jargon, in most cases results within 10 % of the correct result are considered “good enough”. In the case of mental calculations sometimes even being within the order of magnitude of the correct result is close enough because the estimate is achieved fast and without the use of calculating equipment. We keep in mind that mental calculations skills are extremely useful to practicing engineers because, even with all involved inaccuracies, the engineer develops intuitive feeling for the problem at hand. 2.6. In order to practice new principles, it is natural to start with simplest possible cases and progressively work out more complicated ones. For instance, in this example instead of considering only point charges, we introduce multiple point charges distributed over a cylindrical object. Geometry of the charged objects determines shape of the surrounding electric field, thus in our work we progress from a point, to 1 √ Taylor’s series around x = 0 is: 1/ 1+x ∗ 1 − (1/2) x + (3/8) x 2 − (5/16) x 3 + · · · 14 Basic Terminology: Solutions 99 a line, a cylinder, a sphere, and then to non-regular shapes. It should be obvious by now that non-regular objects may be “dissected” and considered simply as a sum of basic regular objects, which in itself can be seen as being equivalent to the non-liner to linear function approximation. The total charge uniformly distributed on the given wire is found simply as q = ql l, thus direct implementation of Gauss’s Law in cylindrical coordinates leads into D · ds = qfree,enc → S λ0 E · ds = qfree,enc S ∴ Er · 2ωrl = ql l λ0 → Er = ql r 2ωr λ0 (14.7) where, r is the unity vector perpendicular to the wire surface. 2.7. Duality between matter and energy is still one of the greatest mysteries that we are facing in modern science. Conclusion that particles and waves are interchangeable came from the famous double–slit experiment. Basic property of waves, i.e. interference, is perfectly well modelled by interaction of sinusoidal functions. Thus, mathematics of classic waveforms is based on trigonometry algebra. In this example we review some of basic definitions related to sinusoidal functions and their relationships, namely addition of two sinusoidal function and their phase difference. It is a simple matter to reconstruct a sinusoidal function if its zero–crossing points (i.e. its period) and the amplitude are known in advance. For example, phase difference between S1 and S2 waveforms, Fig. 14.4, is ω/3. In the time–domain, for a 10 MHz waveform, the ω/3 phase difference is expressed in units of time as follows. First, the waveform period is calculated as T = 1/ f = 100 ns ⇒ 2ω , then from a Fig. 14.4 Solutoin 2.7: Illustration of a relationship between sinusoidal waveforms 100 14 Basic Terminology: Solutions simple proportions between ω/3, 2ω , and 100 ns it follows that ω/3 is equivalent to 16.667 ns. Similarly, every time when S1 and S2 cross their respective amplitudes are equal, thus amplitude of S1 + S2 at the same instance must be twice the amplitude of either waveform. Because their phase difference is ω/3, the two forms cross for the first time 16.667 ns. At that moment S1 = 2 sin (ω/3) = 1.732, thus the instantaneous sum of the two sinusoids is 3.464. Similarly, when S2 = 0 then S1 = 1.732 the sum must be 1.732 as well. We observe that addition of these two sinusoids produced another sinusoid whose amplitude is greater than any of the two individual amplitude. In this case we say that this was “constructive interference” because the resulting sinusoid is “stronger” then each of the two sinusoids. By careful engineering, that is, by understanding how the phase difference and amplitude of two or more sinusoids add together, we are able to manipulate waveforms in many ways. Some of the practical uses of wave interference are shown in the following examples. 2.8. Very important application of constructive addition principle is to use a carefully chosen set of sinusoids to synthesize other periodic functions. In this example we use a set of five sinusoidal functions to synthesize a sawtooth waveform. If the given set of sinusoids is written as 2 S1 (t) = (−1) 1 2 S2 (t) = (−1)2+1 2 3+1 2 S3 (t) = (−1) 3 2 S4 (t) = (−1)4+1 4 2 S5 (t) = (−1)5+1 5 1+1 sin (1 α t) sin (2 α t) sin (3 α t) sin (4 α t) sin (5 α t) then it is easy to see that all functions have the same general form, thus their sum is S(t) = 5 ⎪ k=1 Sk = 5 ⎪ k=1 (−1)k+1 2 sin (k α t) k where, term (−1)k+1 alternates between −1 and +1 and therefore controls the initial phase, term 2/k controls amplitude of its corresponding term, and term (k α t) sets the harmonic frequency. When only the first five terms are used, then the sawtooth function is still visibly distorted, Fig. 14.5 (left) while, for instance, 25 terms produce much more realistic sawtooth function, Fig. 14.5 (right). Needles to say, an infinite number of terms is required to synthesize mathematically perfect sawtooth function. 101 4 3 2 1 0 -1 -2 -3 -4 S1 S2 S3 S4 S5 S 0 0.5 1.0 time 1.5 amplitude amplitude 14 Basic Terminology: Solutions 2.0 4 3 2 1 0 -1 -2 -3 -4 S(t) 0 0.5 1.0 1.5 2.0 time Fig. 14.5 Soluton 2.8: Illustration of sawtooth waveform synthesis using five sinusoids (left), and twenty five (right) 2.9. Noise cancellation concept is based on the idea that a multitone noise waveform Sn is added to its “mirror image” waveform Sm so that Sn + Sm = 0, and is therefore cancelled. The cancelation action is much easier executed in frequency domain, by simply adding negative version of each component in the noise frequency spectrum. We duly note that even though we used term “noise” in this example, any other signal tone can be cancelled using the same principle. In an ideal case, the cancelation function’s spectrum must contain inverted, i.e. “mirrored” images of all single tone components found in the noise function. Thus, each single tone noise component would be added with its mirrored tone, therefore each of the sums would be zero. In other words, the mirrored tone is shifted by T /2 (i.e. by ω ) relative to the original tone. Therefore, 1. Noise Sn (t) and the corresponding cancelation function Sm (t) must be related as follows, 1 2 3 4 1 sin (α t) − sin (2α t) + sin (3α t) + sin (5α t) + sin (9α t) 5 2 3 4 3 1 2 3 4 1 Sm (t) = − sin (α t) + sin (2α t) − sin (3α t) − sin (5α t) − sin (9α t) 5 2 3 4 3 Sn (t) = or, equivalently Sm (t) = 1 2 1 sin (α t + ω ) − sin (2α t + ω ) + sin (3α t + ω )+ 5 2 3 3 4 sin (5α t + ω ) + sin (9α t + ω ) 4 3 2. In this idealized case, the cancelation is perfect, Fig. 14.6 (left) because each tone in the spectrum is perfectly cancelled in each instance of real time. 3. However, in a realistic case there is always a finite delay time, i.e. “processing time”, between the moment when the noise value presents itself and the cancelation value is created. That is, the sequence of events in time is as follows. 14 Basic Terminology: Solutions 4 3 2 1 0 -1 -2 -3 -4 0 Sn(t) Sm(t) Sn(t)+Sm(t) amplitude amplitude 102 0.5 1.0 time 1.5 4 3 2 1 0 -1 -2 -3 -4 Sn(t) T/10 T/20 T/100 0 0.5 1.0 1.5 time Fig. 14.6 Solution 2.9: One of the possible illustrations of destructive interference First, amplitude of the incoming noise signal is measure at a certain instance in time. Then, signal whose amplitude is same but with the opposite sign is created. Finally, the two signals are added so that their sum is a zero value signal. Naturally, execution of these actions takes finite amount of time, which means that by the moment the correct cancelation value is created, the original instantaneous value of the noise signal has already changed and the sum is not zero (except by a chance, of course). In practice, the delay is measured relative to period T of the first harmonic (whose period is the longest, of course) because the absolute value of the period is not relevant at all. In Fig. 14.6 (right) there are three cases shown, i.e. when the cancelation signal was delayed relative to the noise signal by: (a) T/10 or 10 % of the period; (b) T/20 or 5 % of the period; and (c) T/100 or 1 % of the period. A rough comparison of the noise and the produced “cancelled” signal’s amplitudes, Fig. 14.6 (right), shows that in this particular case we must finish the complete signal processing in less then 1% relative to the period T , if we are to suppress the noise signal by factor of ten. By choosing higher operating frequencies we effectively reduce the allowed processing time, i.e. the intended electronics must operate under much harder time constrains. That is why, for example, it is not difficult to design noise cancellation headphones by using the state of the art technology that enables the design of electronics operating in multi gigahertz range, while the audio signal is in kilohertz range. For example, a 1 kHz signal’s period is T = 1 ms, thus one percent of T is 10 µs. By the standards of modern technology, within a 10 µs window it is possible to take millions of samples of the signal and perform large number of numerical operation while the signal itself barely changed its amplitude. However, the situation is is completely different if we try to cancel a 10 GHz tone. One percent of its period is only 1 ps, therefore even the fastest processors of the day can not to much within this time window. 2.10. By definition, electrical power P is product of the two fundamental electrical variables, voltage and the current, measured at the same time instance. Due to Ohm’s law, there are three equivalent forms that describe instantaneous electric power as 14 Basic Terminology: Solutions 103 p = vi = v2 = i2 R R (14.8) where, for simplicity the resistance R is assumed constant in time. A square pulse waveform is characterized by only two amplitude levels; first half of its period the amplitude takes value of zero, while over the second half of its periods the amplitude constantly holds its maximum value. In accordance to (14.8), if at any given moment either voltage or current takes zero value, the power must take zero value too. A periodic function is often described by duration of its duty cycle. By definition, a square wave is said to work on 50 % duty cycle, which means that half of its period the square wave holds its high value, and half of the period it holds zero. Because the pulse amplitude is constant over its corresponding half–period, in this example given as 1ms (i.e. the full period is T = 2ms), we simply write 2 (2 V)2 v (t) = = 40 mW for 0 ∼ t ∼ 1 ms R 100 ν p(t) = 2 2 v (t) = (0 V) = 0 mW for 1 ms ∼ t ∼ 2 ms R 100 ν hence, by definition, average power Pavg over the full period T = 2 ms is Pavg = 1 T 0 T p(t) dt = 1 2 ms 1 ms 40 mW dt + 0 1 × 40 mW × 1 ms = 20 mW = 2 ms 2 ms 0 mW dt 1 ms (14.9) which is the constant power level that needs to be continuously held over the full time period so that it dissipates as same amount of energy as power “burst” of the square wave that lasts only half–period but has double amplitude. In other words, the total energy dissipated within the pulse period is due to the energy only within the high voltage half–period, i.e. W = Pt = 40 mW × 1 ms = 40 mJ which is equivalent to the total energy dissipated by constant power of 20 mW continuously delivered over the time period of 2 ms. In order to be able to mentally evaluate or estimate integrals, it is very useful to keep in mind that a definite integral represents area of the region bounded by the function (where area below the horizontal axis is assumed negative and taken away from the sum). 2.11. Whether an element serves the role of a power source or a power load depends on the flow of current in and out of that element. Positive power implies that the energy flows into the element where it is dissipated, while the negative power value 104 14 Basic Terminology: Solutions Fig. 14.7 Solution 2.11: Graph of a piecewise linear function in current-time coordinating system that is easily integrated geometrically implies that the energy is leaving the element while being delivered to its load. In this example we review simple definitions related to the current flow and charges. 1. After straightforward implementation of (2.1) and with reference to Fig. 14.7, we write i(−2s) = −2A/s (−2s) = 4A; and i(3s) = 3A/s (3s) = 9A; 2. By direct implementation of definition for current, and in the case of linear current change in time, definite integral is trivial to solve from the graph in Fig. 14.7 by adding areas of right angled triangles bounded by the i(t) functions, thus, i(t) ⇒ dq dt ∴ q= 3s −2 s i(t)dt = 0 −2 s 3s (−2t)dt + (3 t) dt = 17.5 C 0 In this trivial case of linear i(t) function the same result is derived by inspection of Fig. 14.7 and calculation of surface under the function, i.e. by adding the two triangular areas, as (4 × 2)/2 + (9 × 3)/2 = 17.5 C, where the two catheti of each right angled triangle are in the units of amperes and time respectively, therefore the product (i.e. area) is in coulombs. 3. Knowing that the total charge moved in within the time interval of (−2 s, 3 s) equals 17.5 C, the average current is easily found by calculating the hight of a rectangle whose area is 17.5 C and horizontal side is 5 s long in current-time coordinating system, therefore the constant current level that would move the same amount of charge is found as i avg = 17.5 C/5 s = 3.5 A, marked as red dash–dot line. 2.12. Understanding that definition of power source is based on the flow of current relative to potential at the device’s terminal helps us to track the total energy in the system. As we remember, the law of conservation of energy must apply to closed systems. By inspection of Fig. 2.2 and from the given data, it is straightforward to write: (vs , i s ) = (8 V, 7 A) with the current leaving positing terminal of the element, therefore Ps = −56 W. Voltage v2 = vs = 8 V, the current entering R2 , therefore, P2 = 16 W. In accordance to Kirchhoff’s current law (KCL), i s = i 1 + i 2 , therefore, i 1 = 7 A − 2 A = 5 A leaving R1 , which leads into P1 = −60 W. This result is followed 14 Basic Terminology: Solutions 105 by conclusion that v3 = vs + v1 = 20 V, which means that P3 = 160 W. Voltage controlled current source gm forces current of i gm = 0.25A/V × 12 V = 3 A leaving the device, while v(gm ) = v3 = 20 V. Thus, we reach conclusion that P(gm ) = −60 W. It is simple to verify the conservation of energy law by adding all powers in the system as Psystem = Ps + P1 + P2 + P3 + P(gm ) = −56 W − 60 W + 16 W + 160 W − 60 W = 0 which illustrates behaviour of ideal voltage/current sources in a multi-device closed system. As ideal elements, they are capable of absorbing/generating infinite amount of power in order to hold the required voltage/current at its terminals. 2.13. By inspection of Fig. 2.3, and from definition of electric power P = i 2 R, we conclude that during the first time period power absorbed by the load is P1 = 500 W, during the second time period dissipated power is P2 = −125 W, during the third time period net power is P3 = 0 W, and during fourth time period absorbed power is P4 = 500 W. Therefore, the average power is found as Pavg = P1 + P2 + P3 + P4 = 218.75 W 4 (14.10) In this case of constant current sections it is easy to geometrically solve the definition integral, thus RMS current is ⎩ i rms = (10 A)2 × 1 ms + (−5 A)2 × 1 ms + (0 A)2 × 1 ms + (10 A)2 × 1 ms 4 ms = 7.5 A Chapter 15 Electrical Noise: Solutions Signal to noise ratio (SNR) is one of the most basic specifications used to characterize quality of the received signal, as well as (implicitly) the complexity of the required electronics. In the following examples, we develop a feeling for relationships between the noise and signal, we learn about differential signals, thermal noise, and we practice to use frequency spectrum plots to help us in the design process. Solutions: 3.1. Saying that two voltage amplitudes are related as SNR = 20 dB is just another way of saying that amplitude of the one assumed to be “signal” is ten times greater than amplitude of the one assumed to be “noise”. For instance, for two given voltage amplitudes A1 = A0 (i.e. “signal”) and A2 = A0/10 (i.e. “noise”), we write SNR := 10 log = 10 log = 20 log ∴ = 20 log P1 P2 (15.1) A21/R A2/R A1 A2 A0 = 10 log A1 A2 2 (15.2) A0 10 = 20 log 10 = 20 dB (15.3) We keep in mind that SNR is defined as ratio of two powers (which is unitless quantity, e.g. P1/P2 ). Often, it is more practical to use units of [dB] by using log function as in (15.1), while SNR of the two respective voltage (or current) amplitudes is derived and expressed in units of [dB] as in (15.2). Therefore, if ratio of two powers is, for example, P1/P2 = 10 we can express it as power SNR P = 10 dB by using (15.1), or equivalently we can use the respective two voltage amplitudes and say that voltage R. Sobot, Wireless Communication Electronics by Example, 107 DOI: 10.1007/978-3-319-02871-2_15, © Springer International Publishing Switzerland 2014 108 15 Electrical Noise: Solutions 15 15 a(t) n(t) a(t)+n(t) 10 amplitude, V amplitude, V 10 5 0 -5 -10 -15 5 0 -5 -10 0 50 100 150 200 -15 time, µs 0 50 100 150 200 time, µs Fig. 15.1 Solution 3.1: time domain waveforms SNR V = 20 dB as in (15.3). All three expressions are equivalent. Main reason for using dB units is due to properties of the log function,1 which enable us to use the simple operations of addition and subtraction when comparing two signals that are already expressed in decibels. At the same time, by using the log function we can plot both weak and strong signals in the same plot, even if their powers are orders of magnitudes apart (which is, again, the case for transmitted and received EM signals in radio communications). In addition, it turns out that B is too large unit for modern wireless communication applications, thus we use dB. For the sake of creating meaningful visualization of an ideal single tone signal a(t, ω a ) and a noise signal n(t, ω 1 , ω 2 , ω 3 ) we arbitrary choose three random tones and add them to the main tone as a(t) = A0 sin ω a t n(t) = n 1 sin(ω 1 t + λ1 ) + n 2 sin(ω 2 t + λ2 ) + n 3 sin(ω 3 t + λ3 ) where A0 = 10(a1 +a2 +a3 ) so that we set the required SNR = 20 dB (Fig. 15.1). For purposes of this illustrative plot, the quasi–random “noise” function n(t) is created simply by adding three random sine/cosine terms with various frequencies. Also, frequencies of the noise harmonics are set a bit higher than the signal frequency (which is not to say that it can not be the other way around). It is clearly visible from the graph that the two signals, a(t) and n(t), have the maximum amplitudes ratio of approximately ten. However, when noise affects the wanted signal, mathematically speaking, that means the two functions are simply added together in time–domain as a(t) + n(t) (Fig. 15.1, right). This graph helps us visualize how an example of a single–tone voltage signal that contains high frequency noise with SNR = 20 dB may look like. As the additional exercise, plot more examples with various SNR to develop your own sense of noisy signals and to practice estimating the associated SNR from the resulting time–domain plots (which is what we do when we use an oscilloscope). For 1 log(x y) = log x + log y and log(x/y) = log x − log y 15 Electrical Noise: Solutions 109 example, find what is SNR in [dB] if amplitude of the noise function is |n(t)| = A0 while amplitude of the signal function is |a(t)| = A0/10 ? 3.2. In these two examples the “noisy” signal s(t) is synthesized as sum of a 10 kHz single tone and three randomly chosen tones as s(t) = 10V sin(2π × 10 kHz × t) + 0.3V sin(2π × 601 kHz × t + 13) + 0.5V sin(2π × 713 kHz × t − 42) + 0.2V sin(2π × 907 kHz × t + 87) (15.4) Note that the maximum amplitude of the three noise components added together is one, which is ten times smaller relative to the main 10 kHz tone amplitude. We observe that time domain plot (i.e. by using an oscilloscope), Fig. 15.1 (right), shows 10 kHz noisy signal with approximately SNR = 20 dB, however, we have no way of determining the frequency content of the noise itself, and subsequently how to design a low-pass (LP) filter to remove the noise. Here is where the frequency plot (i.e. the use of spectrum analyzer) becomes indispensable. In the frequency plot, Fig. 15.2, it becomes obvious that only three single tones are present in the signal. On the horizontal axis we clearly read frequency values of the frequencies, while the vertical axis shows powers relative to the 10 kHz tone that was normalized to valued of 0 dB. By zooming on this plot we can read relative signal powers in units dB, which we already know from (15.4). We keep in mind that in this simulated plot the noise floor (measured as below −100 dB) is due to numerical noise of the simulator tool. It goes without saying that it is extremely important for an engineer to become proficient in using both time domain and frequency domain graphs to evaluate signals. For example, from Fig. 15.2 we easily conclude that if we design a LP filter whose frequency slope starts at just above 10 kHz and extends all the way 0 Signal Power, dB -20 -40 LP filter -60 -80 -100 -120 -140 0 0.2 0.4 0.6 frequency, MHz Fig. 15.2 Solution 3.2: frequency spectrum plot 0.8 1.0 110 15 Electrical Noise: Solutions Fig. 15.3 Solution 3.3: time domain differential signal to 600 kHz (dash–dot red line) it will be sufficient to suppress all three noise tones down into the noise floor. Naturally, this is much easier design requirement relative to the case where the first noise tone is somewhere much closer to the 10 kHz signal (because steeper filter slope requires higher order, i.e. more complicated, filter). 3.3. When two single–ended signals a(t) and b(t) that have their amplitudes equal, their common–modes equal, their frequencies equal, and their phases inverted, are subtracted from each other we refer to the resulting signal d(t) = a(t) − b(t) as differential signal. It is important to realize that the two waveforms are two separate real physical signals, while the d(t) is only measured result of the applied mathematical operation. One way of looking at single–ended versus differential signals is that amplitude of a single–ended signal is measured relative to a stable reference level (to which we arbitrary assign value of “zero” and for the convenience we name it GND), while “differential” measurement implies that the desired signal and its reference level are phase inverted versions of each other, for instance a(t) and b(t) in Fig. 15.3 (left). Mathematically speaking, both measurements are relative, thus we can always declare instantaneous value of, for instance, b(t) = 0 and imagine that at every moment a(t) is measured relative to the “instantaneous zero level”. Also, one can always imagine to be “walking along” the a(t) path while measuring the “distance” to b(t). If one needs to “look down” (Fig. 15.3, green arrow (1)) then the distance is positive, and “looking up” (Fig. 15.3, green arrow (2)) means the negative result (Fig. 15.3, right). Let us see again how this case looks from strictly mathematical point of view, where the two complementary general signals a(t) and b(t) have common mode levels CMa and CMb , have same frequency ω a = ω b = ω , and are subjected to the interfering noise signals n a (t) and n b (t), i.e. 15 Electrical Noise: Solutions 111 a(t) = CMa + Aa sin ω t + n a (t) b(t) = CMb + Ab sin(ω t + π ) + n b (t) ∴ a(t) = CMa + Aa sin ω t + n a (t) b(t) = CMb − Ab sin ω t + n b (t) (15.5) Obviously, the difference is d(t) = a(t) − b(t) = [CMa − CMb ] + [(Aa + Ab ) sin ω t] + [n a (t) − n b (t)] (15.6) It is interesting to note that (15.6) implies that if n a (t) = n b (t) = n(t), i.e. both signals a(t) and b(t) are exposed to the same interference n(t), then the common noise is canceled. Similarly, if CMa = CMb = CM, then the resulting signal is at zero common mode level, i.e. CMd = 0. Finally, if the two amplitudes are equal, i.e. Aa = Ab = A, then this special case of (15.6) reduces to d(t) = 2 A sin ω t = D sin ω t (15.7) where, D = 2 A. From Figs. 15.3 (right) and (15.7) we observe two important consequences of deriving the differential signal. First, both the common noise and the common mode DC level are cancelled. Second, the amplitude of the resulting amplitude of the differential signal is twice the amplitude of either a(t) or b(t) signals, i.e. D = 2 A. Thus, mathematical operation of subtracting these two signals results in the new signal that is effectively amplified by the factor of two, i.e. there is inherent 6 dB voltage gain in this operation. These two properties of a differential signal, i.e. the noise cancelation and the inherent gain, are attractive enough to justify the additional effort required to actually implement the subtraction function and to operate with two signals instead of one. Therefore, it is valid to ask the following question, how realistic is to expect that both signals a(t) and b(t) are subjected to the same noise, and what is the implementation cost of the differential signal processing system? When two wires carrying the two signals, a(t) and b(t), are physically located close and parallel to each other along their full lengths, it is reasonable approximation that through either inductive or capacitive coupling any external signal (thus perceived as the “noise”) will be equally injected into both wires. Therefore, we establish an engineering rule for designing a “differential signal path” by specifying that the two wires must be close and parallel to each other, as much as possible, along the full length of the path. For instance, that is one of the reasons why “twisted pair” cable is used in phone connections. The cost of benefiting from the noise cancellation and the achieved 6 dB gain is that two wires and two signals must be used instead of one. In addition, an operational amplifier is needed to perform the subtraction of the two signals. Thus, 112 15 Electrical Noise: Solutions processing differential signals necessitates the use of more wires, and the use of both single–ended–to–differential and differential–to–differential operational amplifiers. That is, the overall hardware complexity is drastically increased. Then one may also ask, why do not we use only single–ended signal processing instead, which is simpler do design and implement? To answer this question, we should keep in mind the main purpose of communication electronics, which is to transmit a signal as far as possible while using as little as possible amount of energy. One possible analogy could be with the need to drive a car as far as possible with the minimum amount of fuel. After travelling a long distance, the signal inevitably becomes attenuated and distorted by the noise. At this moment, the removal of noise and the 6 dB gain make the whole difference in the world. Therefore, to make efficient and elegant systems, whenever possible we must eventually take advantage of differential signal processing. Of course, in order to achieve full benefits of differential design, we must make sure that the two paths and all associated components are perfectly matched so that the desired cancelations are achieved. As further exercise, plot more graphs similar to Fig. 15.3, where for example the noise is injected only into one of the two signals, or where the two CM levels are not identical, or any other mismatch combination that you could think of. In all cases, try to estimate the resulting SNR. 3.4. By noting that the RLC network in Fig. 3.1 consists of a resistor R in series with the parallel impedance Z LC = Z L ||Z C , we write 1 ZC = Z L = jωL (15.8) jωC ∴ ZC Z L jωL (15.9) Z LC = = ZC + Z L 1 − ω2 LC Then, we simply apply the voltage divider rule for R and Z LC , Fig. 3.1, and directly write expression for the voltage transfer function H ( jω) as2 H ( jω) = vout Z LC = = vin R + Z LC 1 R Z LC +1 = jωL R(1 − ω2 LC) + jωL ∴ |H ( jω)| = H (ω) = ωL R 2 (1 − ω 2 LC)2 + ω 2 L 2 = R 1 1 ω 2 L2 − 2 CL + ω 2 C 2 + 1 (15.10) 2 |z| = √(z)2 + →(z)2 15 Electrical Noise: Solutions 113 Fig. 15.4 Solution 3.4: frequency domain plots of a resonator transfer function H (s) After choosing component values R = 1 π, 10 π, 100 π, L = 1 H , and C = 1 F, it is now trivial to plot either of the two forms of function (15.10) and create Fig. 15.4 for the three resistor values. However, let us stop here for a moment and have the following discussion. First, let us take a closer look at Fig. 3.1. From strictly electrical point of view we observe that the node aiis shorted to the ground if either L = 0 H or C = ∞ F, i.e. if in accordance to (15.8) if either Z L = 0 or Z C = 0. That is to say, in either of these two cases the output voltage becomes Vout = 0 V , which leads into |H (ω )| = 0. But, we also know that Z L (ω ) = 0 if ω = 0 (i.e. at DC), and Z C (ω ) = 0 if ω = ∞ (as shown by (15.8)) even if the two components (L , C) have non–zero finite values. Same conclusion may be reached from strictly mathematical perspective, by finding the following limits of function (15.10) as lim H (ω ) = ω ≤0 R lim H (ω ) = ω ≤∞ R 1 1 0 − 2 CL + 0 + 1 = 1 1 ∞ − 2 CL + ∞ + 1 1 =0 ∞ = 1 =0 ∞ As expected, our conclusion based on engineering reasoning is consistent with the mathematical behaviour of function (15.10), i.e. gain of its equivalent network in Fig. 3.1, for the two extreme frequencies. Similarly, we make additional observation that the maximum value of H (ω ) = 1, simply because that is the maximum gain of a passive RLC network. In this particular case the maximum gain happens when Z LC becomes an open connection. Obviously, open connection implies infinite impedance, which from (15.9) leads into conclusion that Z LC = ∞ when ≥ ω = 1/ LC . Note that the same conclusion is reached by using strictly calculus methods and by finding maximum of the first derivative of (15.10). It is really matter of convenience which method is used. For instance, from the first form of |H (ω )| in (15.10) we easily find that the maximum is achieved when 114 15 Electrical Noise: Solutions lim ω ≤ ≥1 LC H (ω ) = R 2 (1 − L C 1 LC LC)2 + L C = L C L C =1 (15.11) These observations reveal to us that gain of the RLC network, as described by H (ω ), has very interesting property that, as the frequency ω changes from zero to infinity, its value starts at zero, rises to one, and falls again to zero. Intuitively, we can say that having a function whose gain changes with frequency must be useful for some engineering purpose. In the case of this particular function, we note that if H (ω ) is multiplied by another function G(ω ), then G(ω ) H (ω ) = G(ω ) if, and only if ω = ω 0 = ≥ 1 LC (15.12) where, ω 0 is referred to as the resonant frequency and is arguably the most important variable in wireless communications. Here is an example of how we interpret and use this seemingly trivial property of the RLC network. If instead of being a single tone signal, i.e. G(ω 1 ), the function G(ω ) consist of several harmonics, i.e. G(ω 1 , ω 2 , . . . , ω 0 , . . . , ω n , . . .), then by applying the RLC network function we can “extract” the signal whose frequency is equal to the resonant frequency (because that tone is multiplied by one), and “remove” all the others (because they those tones are multiplied either by less then one or by zero). For instance, if we set ω 0 = ω 2 then G(ω 1 , ω 2 , . . . , ω 0 , . . . , ω n , . . .) H (ω 2 ) = G(ω 2 ) if ω 2 = ω 0 = ≥ 1 LC (15.13) In this scenario, G(ω 1 , ω 2 , . . . , ω 0 , . . . , ω n , . . .) may represent the mix of all various frequencies of the signals present around us due to transmission of all radio and TV stations, cell phones, satellites…. However, if the RLC network is incorporated into our receiver, we effectively create the “entry door” into the system that rejects all other frequencies except the ω 0 , which means that the receiver is tuned into, i.e. it hears only the transmitter working on the ω 0 frequency and completely rejects all the other transmitters working at different frequencies. What is more, by changing values of either L or C we can change value of ω 0 and tune the receiver to some other transmitting frequency. On the other hand, value of the real resistance R is directly responsible for the width of the frequency bandwidth BW. We will come back to this topic again, but for the moment we keep in mind that design of narrowband RLC filters assumes that we use an ideal inductor made of ideal wire with zero resistance, i.e. there are no internal thermal losses in the system and the corresponding energy loss. A brick–wall type bandpass (BP) filter transfer function is defined as being centered around frequency ω 0 , to have gain A = 1 within the frequency range BW , and gain A = 0 for any other frequency (Fig. 15.4, right). The bandwidth BW 15 Electrical Noise: Solutions (a) 115 (b) (c) Fig. 15.5 Solution 3.5: frequency domain noise plots itself is defined by two points on the transfer curve that are 3 dB below the function’s maximum value. Therefore, the consequence of applying BP filter to any multi– tone signal (i.e. using multiplication operation) is that all frequencies outside of the BW frequency range are completely suppressed (because amplitudes of those tones are multiplied by zero gain), while the power level inside the BW frequency range stays unchanged (because amplitudes of those tones are multiplied by the gain of one). Obviously, the “brick–wall” filter is only a crude approximation of the real BPF transfer function, nevertheless it is very practical and drastically simplifies our system analysis. This approximation is even more justified if we keep in mind that −3 dB is equivalent to 1/2 of the original power level. From the engineering perspective, when the power level drops by half or more, we justify the BPF approximation and neglect those low power levels (while keeping in mind that the calculated SNR will be overly optimistic). 3.5. To the first approximation, power spectral density (PSD) function of a white noise type signal is assumed flat across all frequencies in the (DC to f ≤ ∞) range, that is the PSD function is at the constant power level P. In other words, when comparing any two 1 Hz wide frequency windows, as illustrated by red boxes in Fig. 15.5a, the measured power levels are always the same. For example, the same power is contained in (10–20 Hz) as is in (1,000–1,010 Hz) frequency window. A brick–wall type bandpass (BP) filter transfer function is defined as being centered around frequency ω 0 , to have gain A = 1 within the frequency range BW , and gain A = 0 for any other frequency (Fig. 15.5b). Therefore, the consequence of applying BP filter to the white noise (i.e. using multiplication operation) is that all frequencies outside of the BW frequency range are completely suppressed (because amplitudes of those tones are multiplied by zero gain), while the power level inside the BW frequency range stays unchanged (because amplitudes of those tones are multiplied by the gain of one) (Fig. 15.5c). We dully note that the total amount of energy contained within white noise is, by definition, infinite. Direct consequence is that if the complete white noise is allowed to enter an amplifier, then the amplifier would need to draw an infinite amount of energy from the energy source (e.g. the battery) in order to process the noise signal. And that is even before the wanted signal is introduced into the system. Naturally, this is not realistic engineering solution. Therefore, we must apply BP filtering operation before the the amplification, so that BW of the BP filter is wide 116 15 Electrical Noise: Solutions enough to let through only the signal itself, and at the same time to allow only limited amount of white noise to “sneak by” and enter the system. With this approach, both the signal and noise are amplified by using finite energy source (even though we never wanted to amplify the noise in the first place, but that is unavoidable). By using BP filter, we now calculate SNR by comparing complete signal energy with the amount of noise energy that was allowed to enter (which is now finite). 3.6. By definition, for the known P S D, the total power P contained within a given frequency ( f 1 , f 2 ) window is calculated by solving integral P= f2 PSD df (15.14) f1 which, for the trivial case of P S D = const and ( f 1 , f 2 ) = (0, ∞), reduces to P = PSD ∞ d f = P S D × ∞ = ∞ [W] (15.15) 0 for any positive value of (P S D > 0). Or, in geometrical sense, the integral (15.14) is just another way of calculating rectangular area under the P S D function, i.e. greyed areas in Fig. 15.6. Obviously, if at least one side of a rectangle is infinitely long then its area must be infinite. Similarly, we calculate noise power within bandwidth of the human voice (i.e. BW = 20 Hz, 20 kHz) while holding the same P S D as Pn = P S D 20×103 d f = 1 µW/Hz × 19, 980 Hz = 19, 980 µW (15.16) 20 within the given BW. Thus, the total noise power is, simply, the area under P S D function in Fig. 15.6 (right). This example clearly illustrates the need for using BP filtering function in front of the amplification function. For instance, for a given Ps = 100 mW voice signal, we calculate the two SNR levels as SNR = 100 mW Ps = −∞ dB = Pn ∞ and SNR = Ps 100 mW 7 dB = Pn 19, 980 µW In other words, without the application of a BPF to help us limit power of the surrounding noise, the wanted signal is completely swamped by the noise and it would be impossible to recover it again. In that case, the noise power is infinitely greater than the signal power. Thus, we always try to design BPF whose BW is wide just enough to let the wanted signal through, which results in the maximum possible SNR for the given design. 15 Electrical Noise: Solutions 117 Fig. 15.6 Solution 3.6: power spectrum density (PSD) functions As a side note, we keep in mind that any realistic electronic system (including amplifiers, of course) has its own inherent BW due to the fact that the realistic components used to manufacture the electronic system are frequency limited on their own. We keep this point in mind throughout the rest of the book. 3.7. By definition, instantaneous amplitude of a random noise source is in accordance with Gaussian probability distribution function, thus, variance φ 2 is equivalent to the average mean–square variation of the amplitude around its average value. In other words, the average mean-square current or voltage variation around its average value, i 2 or e2 , is equivalent to the variance φ 2 , which leads into conclusion that the rms current/voltage value equals to the standard deviation φ . For instance, in the case of two random voltage sources acting in series, this statistical property of random signals results in the following conclusion for the total average mean–square voltage e2 2 2 2 2 2 2 2 2 2 2 ettr = e1rms + 2 e1rms e2rms + e2rms = e1rms + e2rms ms = e1rms + e2rms (15.17) because, the average value of product of two independent random values is zero, i.e. 2 2 e1rms e2rms = 0. Thus, we generalize (15.17) for n thermal voltage sources 2 2 2 + e2rms + · · · + enrms e2 = e1rms (15.18) Therefore, we keep in mind that the noise powers of random signals add up, not the noise voltages. For simplicity, in the rest of this book we assume the average and rms values and simply write that e= e12 + e22 + · · · + en2 (15.19) For the specific numbers used in this example we write, e= e12 + e22 = 12 + 102 V = 10.05 V (15.20) 118 15 Electrical Noise: Solutions The above result clearly illustrate that within a network that contains more than one source, main contributor to the noise voltage is the source with the highest amplitude. In this particular case, by simply ignoring the ten times smaller source the overall error is only 5 % relative to the exact value. By recognizing this fact, we can quickly estimate the total voltage contribution without using exact numerical calculations. 3.8. Thermal noise generated by a resistor is modelled as a simple ideal voltage source en whose voltage amplitude is function of the internal resistance R [π], the environment temperature T [K], and frequency bandwidth B = B [Hz] as en (R) = ≥ 4kT R B V (15.21) where, R represents equivalent resistance of the resistive network, and k is Boltzmann constant. Following up on discussion in Problem 3.7, and while keeping in mind that the noise powers add up, not the noise voltages, first, we find the equivalent network resistance, and then finding the equivalent thermal noise generator is straightforward by using (15.21). When two resistors R1 and R2 are connected in series, then the equivalent resistance is R S = R1 + R2 , and their equivalent mean–square noise voltage of the Thévenin voltage source 2 2 2 = en1 + en2 = 4kT B(R1 + R2 ) = 4kT B R S en12 V2 (15.22) Similarly, when two resistances are connected in parallel, 1/R P = 1/R1 + 1/R2 , we write expression for their equivalent mean–square noise current of the Norton current source by using equivalent noise conductances G = 1/R as 2 2 2 = i n1 + i n2 = 4kT B(G 1 + G 2 ) = 4kT B(1/R1 + 1/R2 ) i n12 ∴ 2 = 4kT B(R1 ||R2 ) = 4kT B R P en12 V2 (15.23) Therefore, we find, (a) Each of the two stand–alone separate resistors generates thermal noise as follows: en2 (R1 ) = 4 × 20 kπ × 1.3806488 × 10−23 J/K × 290 K × 20 kHz = 6.406 × 10−12 V2 en2 (R2 ) = 4 × 50 kπ × 1.3806488 × 10−23 J/K × 290 K × 20 kHz = 16.016 × 10−12 V2 ∴ en (R1 ) = 2.531 µV en (R2 ) = 4.002 µV 15 Electrical Noise: Solutions 119 Fig. 15.7 Problem 3.8: model of a realistic resistor (left), and illustration for available noise power (right) (b) Equivalent resistance Rs of the two resistors in series is R S = (20 kπ + 50 kπ) = 70 kπ ∴ en (R S ) = 4.735µV. (c) Equivalent resistance R p of the two resistors in parallel is R P = (20 kπ||50 kπ) = 14.286 kπ ∴ en (R P ) = 2.139 µV. (d) With the given square value of the noise voltage en2 and its associated resistance R, it is straightforward to calculate available noise power. Available, i.e. maximum possible power, is delivered to the load if the load impedance equals the source impedance. Under the matched impedances condition Z S = Z ∗L , Fig. 15.7 (right), only one half of the source voltage is available to the load, therefore for a single resistor R S we write the expression for available noise power as Pn = (en/2)2 e2 4R S kT B = n = = kT B RS 4 RS 4 RS (15.24) For two resistors R1 and R2 in series, the equivalent resistance is R S = R1 + R2 , which after using (15.22) gives expression for the noise power as Pn (R1 + R2 ) = (en/2)2 4kT B(R1 + R2 ) = kT B = R1 + R2 4(R1 + R2 ) (15.25) Finally, when two resistances are in parallel, we write expression for their available noise power as Pn (R1 ||R2 ) = (en/2)2 4kT B(R1 ||R2 ) = kT B = R1 ||R2 4(R1 ||R2 ) (15.26) Therefore, the resulting available noise power is always the same, Pn = kT B = 8.008 ×10−17 W. We dully note that passive networks do not have power amplification. Finally, we note that in order to reduce the thermal noise, everything else being equal, the equivalent resistance needs to be reduced. Thus, we have one guideline for design of low–noise circuits: the internal circuit nodes should have low–resistance as much as possible, which is often contradictory to other requirements that we try to achieve with the same circuit. Thus, in the circuit design process we always have to make some engineering compromises. 120 15 Electrical Noise: Solutions 3.9. For the series resistor combination in Fig. 3.2 (middle), for the two resistors at two different temperatures, we write en2 = e12 + e22 = 4k∂f R1 T1 + 4k∂f R2 T2 = 4k∂f (R1 T1 + R2 T2 + R1 T2 − R1T 2) = 4k∂f [T2 (R1 + R2 ) + R1 (T1 − T2 )] = 1.680 × 10−11 V2 (15.27) where all of the above expressions are identical and they result in the same numerical answer. Expression for the available noise power Pn is found by definition, where the noise power is calculated across the series connection of the two resistors, as (en/2)2 4k∂f [T2 (R1 + R2 ) + R1 (T1 − T2 )] = R1 + R2 4(R1 + R2 ) R1 = kT2 ∂f + k∂f (T1 − T2 ) R1 + R2 = 6 × 10−17 W Pn = (15.28) We observe that if (T1 = T2 = T ) the second term in (15.28) cancels itself and the noise power expression collapses to the already known expression Pn = kT ∂f . However, when two resistors are at different temperatures then, through the voltage “gain” factor R1 /(R1 + R2 ) in the second term, the resistors create voltage divider effect for the noise voltage sources. 3.10. By definition, noise factor F is calculated as the ratio of the output noise power Po and the input noise power Pin . Therefore, by using the same reasoning and results from problems 3.8 and 3.9, we write (a) Expression for the output noise power of serial resistor network is already found as in (15.28). The “input” noise power is, however, only due to thermal noise of resistor R1 distributed over the series resistance R S = R1 + R2 . Thus, we write Pin = (en (R1 )/2)2 4kT1 B(R1 ) R1 = kT1 B = R1 + R2 4(R1 + R2 ) R1 + R2 (15.29) We can now easily find expression for the noise figure NF by substituting (15.28) and (15.29) as 15 Electrical Noise: Solutions F= Po = Pin 121 kT2 B + k B(T1 − T2 ) R1 R1 + R2 R1 R1 + R2 T2 R1 + R2 =1+ −1 T1 R1 ∴ T2 R1 + R2 −1 NF = 10 log F = 10 log 1 + T1 R1 kT1 B dB (15.30) Obviously, if R1 ∓ R2 , then [(R1 + R2 )/R1 ] ≤ 1, which leads into F ≤ 1 and NF ≤ 0 dB. In all other cases NF becomes greater than 0 dB. Equation (15.30) implies that in the case when the load is connected in series to the source, in order to reduce NF the source impedance should be as high as possible, i.e. much higher than the load impedance. (b) In the case of parallel resistive network, but this time we use G = 1/R terms, the output noise power is found as Po = kT2 B + k B(T2 − T1 ) R2 R1 + R2 (15.31) while the input available noise power is found as Pin = kT1 B R2 R1 + R2 (15.32) which leads directly into Po F= = Pin kT2 B + k B(T2 − T1 ) R2 R1 + R2 R2 R1 + R2 T2 R1 + R2 =1+ −1 T1 R2 ∴ T2 R1 + R2 −1 NF = 10 log F = 10 log 1 + T1 R2 kT1 B dB (15.33) Obviously, this time, if R2 ∓ R1 , then [(R1 + R2 )/R2 ] ≤ 1, which leads into F ≤ 1. In all other cases NF becomes greater than 0 dB. Equation (15.33) implies that in the case when the load is connected in parallel to the source, in order to reduce NF the load impedance should be as high as possible, i.e. much higher than the source impedance. 122 15 Electrical Noise: Solutions 3.11. In (15.24) we already confirmed that power and temperature are directly proportional, thus the equivalent temperature is yet another unit of expressing power of the incoming signal within the given frequency bandwidth B. For the case of a signal that is present at the antenna being calibrated against the local thermal etalon, we can derive the following relationship. PR = kTR B W ⇒ PR = 10 log(kTR B) dB ∴ PS = PR + 3.01 dB = 10 log(kTR B) + 3.01 dB = 10 log(kTR B) + 10 log(2) = 10 log(2kTR B) dB ⇒ PS = 2kTR B W (15.34) where, PS is the measured power of the antenna signal. To find the equivalent antenna temperature Ta , from (15.24) we write Ta = 2kTR B PS = = 2TR = 5 K kB kB (15.35) Similarly, after setting PS = 6.02 dB = 10 log(4) W, PS = 9.03 dB = 10 log(8) W, etc., by repeating as same calculation as above, we respectively calculate Ta = 5 K, 10 K, 20 K, . . ., which illustrates the relationship between noise power and temperature. Additionally, we note that the noise temperature is also used in the radio astronomy to determine physical temperatures of the celestial bodies (once some other variables are also known). 3.12. Although noise analysis of amplifiers may be involved, it follows already known principles: (a) each noisy component is replaced with its equivalent noiseless component model plus the ideal voltage/current source that models the associated noise voltage/current itself, and (b) networks containing multiple sources are solved by applying the superposition principle. Thus, the noisy operational amplifier is replaced with its equivalent noiseless version plus the thermal input–referred voltage noise source en at the positive input node, Fig. 15.8. Doted line represents boundaries of the original noisy operational amplifier. We note that polarity of ideal thermal noise sources is meaningless because the signal is random. In addition, in the noise analysis we use the squared voltage values for the statistical addition of multiple noise signals (which simplifies the analysis because the signal power is proportional to the squared voltage). Contribution of voltage noise source en : is found by shorting the input signal source v S , thus for the non–inverting configuration input–output relationship we write eout (en ) = i R1 (R1 + R2 ) = en (R1 + R2 ) R1 (15.36) 15 Electrical Noise: Solutions 123 Fig. 15.8 Solution 3.12: schematic diagram of a noisy operational amplifier Fig. 15.9 Solution 3.13: noise calculation schematic diagram for operational amplifier Therefore, for the given numerical data we write ≥ ≥ 10 kπ 5 nV/ Hz = 55 nV/ Hz eout = 1 + 1 kπ (15.37) It would be now trivial to calculate ≥ the total noise voltage within a certain BW, simply by multiplying the last result by BW . For example, if the amplifier bandwidth were specified as BW = 400 Hz–19 kHz, the total measured noise voltage at the output terminal would have been ≥ (15.38) eout (total) = 55 nV/ Hz × (19 k − 400)Hz = 7.5 µV 3.13. The noisy operational amplifier is replaced with its equivalent noiseless version plus the thermal voltage noise source en at the positive input node, and the thermal current noise sources i n− and i n+ associated with its input terminals (Fig. 15.9). Doted line represents boundaries of the original noisy operational amplifier. Now, there is more than one noise source, and we have to apply the superposition principle. Each of the noise sources is considered on its own as follows. Contribution of voltage noise source en is already found in (15.36), therefore for the given numerical data 124 15 Electrical Noise: Solutions Fig. 15.10 Solution 3.13: schematic diagram for noise calculations ≥ ≥ 10 k 5.082 nV/ Hz = 55.902 nV/ Hz eout (en ) = 1 + 1k ∴ (eout (en ))2 = 3.125 × 10−15 V2/Hz (15.39) Contribution of current noise source i n− : is found after shorting the noise voltage source, and after the current noise source i n+ is left open (Fig. 15.10). Under these circumstances, we can reason that there is no noise current flowing through R1 because both of its nodes are at the same potential (i.e. gnd). Therefore, the complete noise current i n− is forced through R2 , which generates noise voltage at the output node (keep in mind that one node of R2 is at the gnd potential) ein− = i n− R2 ∴ 2 ein− = (i n− R2 )2 (15.40) Therefore, for the given numerical data we write ≥ ≥ ein− = i n− R2 = 5pA/ Hz × 10 kπ = 50 nV/ Hz ∴ 2 ein− = 2.5 × 10−15 V2 /Hz (15.41) Contribution of current noise source i n+ : is found after shorting the noise voltage source, and after the current noise source i n− is left open, Fig. 15.11. We reason that there is no resistor for i n+ to develop voltage, i.e. (i n+ × 0 π) = 0, because the current source is shorted by ideal wire (of course, in this case we are not discussing the theoretical dissipated power by the ideal source). Therefore, total noise contribution of the two combined sources is found by statistical addition (i.e. by adding the noise powers) of (15.39) and (15.41) as 15 Electrical Noise: Solutions 125 Fig. 15.11 Solution 3.13: schematic diagram for noise calculaitons Fig. 15.12 Solution 3.14: schematic diagram for noise calculations 2 eout = 2 een 2 + ein− 2 R2 en + (i n− R2 )2 = 1+ R1 = (3.125 + 2.5) × 10−15 V2 /Hz = 5.625 × 10−15 V2 /Hz ∴ ≥ eout = 75 nV/ Hz (15.42) 3.14. Now, both the noisy operational amplifier and noisy resistors are replaced with their equivalent noiseless versions plus the respective thermal voltage noise sources (Fig. 15.12). Important point to keep in mind is that the thermal noise voltage appears across the noisy resistor terminal. That is, in our model that consists of both noiseless resistor and the associated thermal voltage generator, the thermal noise appears between the same circuit nodes where the noisy resistor resistor was connected. For example, in Fig. 15.12 the noise voltage e R2 is assumed to be between the negative input terminal and the output terminal of the operational amplifier. That is exactly where the noisy resistor R2 was connected, and therefore in this noise model we do not account for the voltage drop through the corresponding noiseless R2 element. Thus, relative to the already found solution (15.42) in Problem 3.13, we need to add contribution of the two noisy resistors. Contribution due to noise source e R1 : is found from the equivalent circuit network (Fig. 15.13, left) as follows. The analysis can be done in a number of ways, here we use engineering reasoning to say that voltage e R1 appears at the negative input of the 126 15 Electrical Noise: Solutions operational amplifier (remember: e R1 voltage is generated across the noisy version of R1 , which was connected between the ground node and the negative input node). Therefore, the noise current through R1 must be i R1 = e R1 /R1 . The same current must flow through R2 , thus we write eout (R1 ) = i R1 R2 = ∴ 2 eour (R1 ) = e2R1 R2 R1 e R1 R2 R1 2 = 4kT R1 ∂f = 4 × 1.3806488 × 10 −23 R2 R1 2 V2 = 4kT R1 × 276.9 × 1 k × 10 k 1k R2 R1 2 2 V2 /Hz V2 /Hz = 1.529 × 10−15 V2 /Hz (15.43) where, in order to simplify the rest of calculations, for the moment we prefer to use units. Contribution due to noise source e R2 : is found from the equivalent circuit network (Fig. 15.13, right) as follows. The noise voltage is is due to the noisy R2 resistor that was connected between the amplifier output node and virtual ground, therefore we simply write V2/Hz 2 (R2 ) = 4kT R2 ∂f V2 = 4kT R2 V2 /Hz eout (R2 ) = e R2 ∴ eour 2 eour (R2 ) = 4 × 1.3806488 × 10−23 × 276.9 × 10 k V2 /Hz = 0.153 × 10−15 V2 /Hz (15.44) Therefore, after combining (15.42) to (15.44), the complete solution is 2 eout = R2 1+ R1 2 en + (i n− R2 ) + 4kT R1 2 R2 R1 2 + 4kT R2 V2 /Hz ∴ ≥ (3.025 + 2.5 + 1.529 + 0.153) × 10−15 V/ Hz ≥ = 84.895 nV/ Hz eout = (15.45) In the above analysis it is useful to keep the same order of magnitude for the intermediate results, so that it becomes obvious which component is the main contributor to the noise. Also, by keeping the B information away, simplifies the calculations. Therefore, it is now convenient to calculate the total noise rms voltage by performing the B calculation only once as 15 Electrical Noise: Solutions 127 Fig. 15.13 Solution 3.14: schematic diagram for noise calculaions Fig. 15.14 Solution 3.15: schematic diagram for noise calculations eout (total) = eout × = 12µV ≥ ≥ B = 84.895 nV/ Hz × (20 kHz − 20Hz) (15.46) The above analysis is valid under assumption that the noise spectrum density function is constant within the given bandwidth. Otherwise, the exact definition for a resistor’s thermal noise that includes the frequency dependent integral must be used. 3.15. Main purpose of calculating noise figure NF of a device is to quantify how much noise power is actually created internally within the device itself, which makes NF one of the device’s very useful quality measures. We keep in mind that square voltage is equivalent to normalized power (because V2 /1 π ∼ W), and also that division of a square voltage by the unit of frequency (i.e. V2 /Hz) is equivalent to power spectral density en2 . Complete schematic diagram suitable for amplifier noise analysis includes all thermal noise sources, Fig. 15.14, along with their respective noiseless components. By definition, noise factor F is a ratio of SNRi at the input of a device, and Signal– to–Noise–Ratio at its output SNRo . Thus, in order to calculate F we have to find ratio of the noise power PI that was delivered to the amplifier by the source (i.e. presented at the input node 1i), and the total noise power eout at the output of the 128 15 Electrical Noise: Solutions amplifier. The fact that eout is measured at the amplifier’s output node complicates the analysis because the source noise power is measured at the input node. Additionally, the output noise power eout is the sum of the amplified input noise power plus the internally generated noise power. It is important to realize, therefore, that in order to compare the input noise power PI , as delivered by the source to the input node, with the internally generated noise power, the amplifier noise contribution needs to be referenced relative to the same node. Thus, we introduce the term input–referred voltage/current noise power PA . In accordance with this definition, PA is equivalent to the complete internally generated noise power that is integrated at the input node, i.e. before the amplification. Naturally, if an amplifier is ideal (i.e. noiseless) the input–referred noise power PA = 0, which is equivalent to F = 1. In that case, the output noise power is due to only the source noise power amplified by the amplifier. We see how any noise internally generated by the amplifier is easily quantified by the noise factor. Assuming the amplifier gain is A, the total output noise power is then A(PI + PA ), the input signal power is PS and, with this in mind, it is convenient to rewrite expression for noise figure F as PS SNRi PA PI = (15.47) F= =1+ A PS SNRo PI A (PI + PA ) which clearly illustrates the point. We now derive input–referred noise and the input noise powers as follows. Contribution due to noise source e R S : thermal noise voltage generated by the source resistance R S is delivered to the input node through the resistive voltage divider (R S , R3 ), thus at node 1iwe write 2 R3 R S + R3 (15.48) When the source resistance is matched to the amplifier input impedance, i.e. R S = R3 from (15.48) we have eI = R3 eRS R3 = 4kT R S R S + R3 R S + R3 ∴ PI = e2I = kT R3 V2 Hz e2I = 4kT R S ∴ PI = 5.083 × 10−18 V2 /Hz (15.49) Contribution due to noise source e R3 : thermal noise voltage generated by the source resistance R3 is delivered to the input node through the resistive voltage divider (R3 , R S ), thus at node 1iwe write 15 Electrical Noise: Solutions 129 2 RS ∴ = 4kT R3 R S + R3 (15.50) When the source resistance is matched to the amplifier input impedance, i.e. R S = R3 from (15.51) we have RS e R3 e3 = R S = 4kT R3 R S + R3 R S + R3 e32 = kT R3 e32 V2 Hz (15.51) Contribution due to noise source e R1 : thermal noise voltage generated by the source resistance R1 is delivered to the input node through the resistive voltage divider (R1 , R2 ), thus at node 1iwe write 2 R2 R1 + R2 (15.52) (Noise voltage source e R2 is shorted (its internal resistance is zero), and we remember that an operational amplifier is also a voltage source, thus its output impedance is zero. That is, for purposes of this analysis R2 is effectively connected to the gnd node through the operational amplifier’s output.) Contribution due to noise source e R2 : thermal noise voltage generated by the source resistance R2 is delivered to the input node through the resistive voltage divider (R2 , R1 ), thus at node 1iwe write e1 = R2 e R1 R2 = 4kT R1 R1 + R2 R1 + R2 ∴ e12 = 4kT R1 2 R1 R1 + R2 (15.53) Contribution due to noise voltage source en : the internal thermal noise voltage generated by operational amplifier is directly connected to the input node, thus at node 1iwe write 2 = en2 (15.54) en1 = en ∴ en1 e1 = R1 e R2 R1 = 4kT R2 R1 + R2 R1 + R2 ∴ e12 = 4kT R2 Contribution due to noise current source i n+ : this current source forces noise current through parallel connection R3 ||R S , which generates voltage directly at the input node, ein+ = i n+ R3 ||R S ∴ 2 R3 R S R3 2 2 ein+ = i n+ = i n+ R3 + R S 2 (15.55) Contribution due to noise current source i n− : this current source forces noise current through parallel connection R1 ||R2 , which generates voltage directly at the negative input node of operational amplifier, which appears also at the positive input node, thus 2 R1 R2 2 (15.56) ein− = i n− R1 ||R2 ∴ ein− = i n− R1 + R2 130 15 Electrical Noise: Solutions Therefore, after combining the six amplifier thermal noise power terms (15.51) to (15.56), the complete solution for the internal noise power spectrum density is PA = 2 eout 2 2 R2 R1 + 4kT R2 + R1 + R2 R1 + R2 R3 R S 2 R1 R2 2 2 + i n− V2 /Hz R3 + R S R1 + R2 = kT R3 + 4kT R1 2 en2 + i n+ ∴ PA = (5.083 + 16.803 + 1.680 + 25 + 20.661 + 6.25) × 10−18 V2 /Hz = 75.447 × 10−18 V2 /Hz (15.57) It is now straightforward to substitute (15.57) and (15.49) into (15.47) and to calculate F =1+ 75.477 × 10−18 V2 /Hz = 15.849 5.083 × 10−18 V2 /Hz ∴ NF = 10 log(15.849) = 12 dB (15.58) 3.16. Using the same methodology from the previous problems, while paying attention to the subtle differences between gains of non–inverting and inverting amplifiers, with reference to Fig. 15.15 we derive NF as follows. Contribution due to noise source e R S : thermal noise voltage generated by resistance R S is delivered to the input node through the resistive voltage divider (R S , R1 ||R4 ), thus at node 1iwe write eR S = R1 ||R4 eRS (R1 ||R4 ) = 4kT R S R S + R1 ||R4 R S + R1 ||R4 ∴ (eR S )2 = 4kT R S R1 ||R4 R S + R1 ||R4 2 V2/Hz (15.59) In order to match the source resistance to the amplifier input impedance first we calculated R4 so that R S = R1 ||R4 , i.e. R4 = R1 R S /(R1 − R S ), then from (15.59) we write PI = (eR S )2 = kT R S V2 Hz ∴ PI = 3.909 × 10−18 V2 /Hz (15.60) 15 Electrical Noise: Solutions 131 Fig. 15.15 Solution 3.16: schematic diagram for noise calculations Fig. 15.16 Solution 3.16: schematic diagrams for noise calculations Contribution due to noise source e R3 : thermal noise voltage due to resistance R3 is directly to the positive input node of the operational amplifier. Thus, ≥ connected ≥ e R3 = 4kT R3 V/ Hz, is also measured at the negative input node of the operational amplifier (Fig. 15.16, left). Our reasoning goes as following. We know that thermal voltage e R3 is amplified and is therefore measured as eout (e R3 ). This output voltage due to e R3 is then easily found by following current i 0 through the (R2 + R1 + R S ||R4 ) resistive chain. The feedback current i 0 is controlled by e R3 potential at the negative input terminal of the operational amplifier that forces i 0 down through (R1 + R S ||R4 ) resistor chain, i.e. eout (e R3 ) = i 0 (R2 + R1 + R S ||R4 ) = e R3 (R2 + R1 + R S ||R4 ) (15.61) R1 + R S ||R4 Nevertheless, our goal is to find an equivalent voltage source eR3 that, if connected to 1i, would also generate eout (e R3 ). We note that under conditions of this particular setup where eR3 is the only signal source there is no current flow through R3 , which implies that the positive node of the operational amplifier is at ground potential, Fig. 15.16 (right). Thereafter, we write 132 15 Electrical Noise: Solutions eout (e R3 ) = −i 0 e R2 = − R3 R2 R1 ∴ (eR3 )2 = R1 eout (e R3 ) R2 2 (15.62) From (15.61) and (15.62) it follows that (eR3 )2 2 R1 R2 + R1 + R S ||R4 = e R3 R2 R1 + R S ||R4 R1 R2 + R1 + R S ||R4 2 = 4kT R3 R2 R1 + R S ||R4 V2/Hz (15.63) which, the given numerical data results in R4 = 4 kπ, and (eR3 )2 = 1 k 10 k + 1 k + 666.667 2 −23 × 4 × 1.3806488 × 10 × 353.95 K × 1 k V2/Hz 10 k 1 k + 666.667 = 9.578 × 10−18 V2/Hz (15.64) Contribution due to noise source e R1 : in order to derive thermal noise voltage generated by resistance R1 , as seen from the input node 1ithrough the resistive voltage divider (R1 , R4 ||R S ), we use similar reasoning as for eR3 above, thus we write eR1 = i R1 R1 = ∴ (eR1 )2 = 4kT R1 R1 e R1 R1 = 4kT R1 R1 + R4 ||R S R1 + R4 ||R S R1 R1 + R4 ||R S 2 (15.65) which, the given numerical data results in (eR1 )2 = 4 × 1.3806488 × 10−23 × 353.95 K × 1 k = 7.037 × 10−18 V2/Hz 1k 1 k + 666.667 2 V2/Hz (15.66) Contribution due to noise source e R2 : thermal noise voltage generated by resistance R2 is delivered to the input node through the resistive network (R2 , R1 ). Current through R1 and R2 is set by e R2 , i.e. i R1 = i R2 , thus at node 1iwe write 15 Electrical Noise: Solutions 133 Fig. 15.17 Solution 3.16: schematic diagrams for noise calcualtions R1 e R2 R1 = 4kT R2 R2 R2 eR2 = i R1 R1 = ∴ (eR2 )2 = 4kT R2 R1 R2 2 V2/Hz (15.67) which, the given numerical data results in (eR2 )2 = 4 × 1.3806488 × 10−23 × 353.95 K × 10 k = 1.955 × 10−18 1k 10 k 2 V2/Hz V2/Hz (15.68) Contribution due to noise source e R4 : thermal noise voltage generated by resistance R4 is delivered to the input node through the resistive divider (R4 + R1 ||R S ). Thus, with reference to Fig. 15.17 at node 1iwe write eR4 = i R4 R S ||R1 = ∴ (eR4 )2 = 4kT R4 e R4 R S ||R1 R4 + R S ||R1 R S ||R1 R4 + R S ||R1 2 V2/Hz (15.69) which, the given numerical data results in (eR4 )2 = 4 × 1.3806488 × 10 = 0.782 × 10−18 −23 V2/Hz × 353.95 K × 4 k 444.444 4 k + 444.444 2 V2/Hz (15.70) Contribution due to noise voltage source en : the internal thermal noise voltage generated by operational amplifier is directly connected to the positive input node of operational amplifier, thus it is at the same position as e R3 in the network (Fig. 15.16, left). Thus, we reuse (15.63) and write 134 15 Electrical Noise: Solutions Fig. 15.18 Solution 3.16: schematic diagram for noise calculations (en )2 = R1 R2 + R1 + R S ||R4 R2 R1 + R S ||R4 en2 V2/Hz (15.71) which, the given numerical data results in (en )2 = 17.5 × 10−18 V2/Hz (15.72) Contribution due to noise current source i n+ : this current source forces noise current through resistance R3 , which generates voltage directly at the positive input node of the operational amplifier ein+ = i n+ R3 . Again, we reuse (15.63) and write )2 = (ein+ R1 R2 + R1 + R S ||R4 R2 R1 + R S ||R4 2 (i n+ R3 )2 V2/Hz (15.73) which, the given numerical data results in )2 = 12.25 (ein+ V2/Hz (15.74) Contribution due to noise current source i n− : this noise source forces a current through R2 , because due to the virtual ground both terminals of (R1 + R S ||R4 ) resistive network is at ground potential, which implies that i n− flows through R2 , Fig. 15.18. With that in mind, eout (i n− ) = i n− R2 is seen at the output node. Thus, referencing the output voltage back to node 1ithrough voltage divider R1 , R2 we write ein− = eout (i n− ) i n− R2 R1 == R1 R2 R2 ∴ (ein− )2 = (i n− R1 )2 (15.75) which, the given numerical data results in (ein− )2 = (5p × 1 k)2 V2/Hz = 25 × 10−18 V2/Hz (15.76) Therefore, after combining the seven amplifier thermal noise power terms (15.63) to (15.76), the complete solution for the internal noise power spectrum density is 15 Electrical Noise: Solutions 135 2 = (e )2 + (e )2 + (e )2 + (e )2 + (e )2 + (e )2 + (e )2 V2/Hz PA = eout n in+ in− R1 R2 R3 R4 ∴ PA = (7.037 + 1.955 + 9.578 + 0.782 + 17.5 + 12.25 + 25) × 10−18 V2 /Hz = 74.102 × 10−18 V2 /Hz (15.77) It is now straightforward to substitute (15.77) and (15.60) into (15.47) and to calculate 74.102 = 19.955 3.909 NF = 10 log(19.955) = 13 dB F =1+ ∴ (15.78) 3.17. The input noise power is calculated as N I = kT = 1.3806488 × 10−23 J/K × 288.15K = 3.978 × 10−21 W/Hz ∴ N I = −174 dB m/Hz (15.79) We note that the −174 dBm/Hz is is often taken as the minimal noise power reference at room temperature. Then, this thermal noise power is amplified by gain G, i.e. No = N I + G (15.80) However, the amplifier itself contributes the additional NF, which adds up to the output noise power. That means the total noise power at the output node is No = N I + G + NF = −174 dBm + 12 dB + 3 dB = −159 dBm (15.81) 3.18. Within the given bandwidth B for the signal before amplification (narrow red arrow), Fig. 15.19, we estimate SNR I as the difference between the signal peak and the noise level, i.e. SNR I = −80 dBm − (−100 dBm) = 20 dB (15.82) The amplifier provides G = 10 dB of power gain for both noise and the signal. However, it also provides internally generated noise power N F = 3 dB, which is added only to the amplified noise level. Therefore, the two amplified power levels are −80 dBm + 10 dB = −70 dBm for the signal (wide blue arrow), and −100 dBm + 10 dB + 3 dB = −87 dBm for the nose floor (blue dots). Therefore, after the amplification we have SNRo = −70 dBm − (−87 dBm) = 17 dB which reflects the amplifier’s noise figure NF = 3 dB as expected. (15.83) 136 15 Electrical Noise: Solutions Fig. 15.19 Solution 3.25: signal to noise frequency spectrum 3.19. It is convenient to express both the total cell phone power and the thermal noise power radiated by the bird in units of W/Hz, so that PS = Pcell B W/Hz PN = kT and W/Hz (15.84) At the average lunar distance D the total cell phone power is distributed over surface of a sphere with radius D. Similarly, the total thermal noise power generated by the bird is distributed over surface of a sphere with radius H . Thus, we find the two respective power fluxes as PS f = PS 4π D 2 PN f = and PS 4π H 2 (15.85) The two signals are received by the whole area A of the antenna dish, which gives us expressions for the total received signal and noise powers as S= PS A 4π D 2 N= and PS A 4π H 2 (15.86) We are now ready to actually write explicit expression for SNR as PS A 2 S PS H 2 Pcell H 2 SNR = = 4π D = = PS N PN D 2 kT B D 2 A 2 4π H (15.87) which, for the given numerical data works as: (a) From (15.87) when signal–to–noise–ratio is set to SNR = 0 dB, that is to say SNR = 1, we write Pcell H 2 =1 kT B D 2 ∴ H= kT B D 2 = 56.247m Pcell (b) After setting H = h from (15.87) we write (15.88) 15 Electrical Noise: Solutions 137 SNR = Pcell h 2 1 = −15 dB = kT B D 2 31.623 (15.89) which clearly illustrates how difficult is to receive a week signal coming from the space. 3.20. We already found expression for a signal power delivered to the matched load, for example in (15.24). Here, BPF is matched to the signal source both in frequency bandwidth and resistance, while the source resistance generates the thermal noise. (a) Straightforward implementation of (15.24) results in 2 vin 2 vin S 4Rin SNR = = = = 50.138 = 17 dB N kT1 B1 4Rin kT1 B1 (15.90) (b) After the signal and BP filter bandwidths change to B2 , then from (15.90) we write 2 vin = 88 K (15.91) T2 = 4Rin k B2 S N R 3.21. In the textbook(s), we already showed expressions for dynamic resistance R D of an R LC resonator at resonance ω 0 , for thermal noise voltage, and for effective bandwidth to be RD = Q ω0C en2 = 4kT R D Beff and and Beff = π B-3 dB (15.92) 2 which, after combining the three of them together leads into en2 = 4 k T ∴ en = Q f0 kT Q π B = 2π k T = ω0C 2 ω0C Q C 1.3806488 × 10−23 J/K × 290.15 K = 20 µV 10pF (15.93) 3.22. Thermal noise of a resistor R at temperature T within a bandwidth B that is controlled by a capacitor C leads into en2 = 4kT R B and en2 = kT C (15.94) ∴ B= 1 = 25 kHz 4RC (15.95) 138 15 Electrical Noise: Solutions Now, the total integrated thermal voltage within the bandwidth B is calculated by using either of the two expressions in (15.94), e.g. en = kT = 65 µV C (15.96) 3.23. First, all given gain and noise figures specifications have to be expressed as −6 dB ≡ 1/4 10 dB ≡ 10 3 dB ≡ 2 8 dB ≡ 6.31 20 dB ≡ 100 Then, straightforward implementation of Friss formula for a five–stages system gives F2 − 1 F3 − 1 F4 − 1 F5 − 1 + + + A1 A1 A2 A1 A2 A3 A1 A2 A3 A4 6.31 − 1 10 − 1 2−1 10 − 1 + + =2+ + 100 100 × 100 100 × 100 × 1/4 100 × 100 × 1/4 + 100 = 2.270 (15.97) F(tot) = F1 + NF = 10 log F = 3.561 dB (15.98) Tn = (F − 1) T = (2.270 − 1) 299.15 K = 380 K (15.99) 3.24. By definition (15.47) we find PI 5 × 10−6 −6 SNR I N F= = I = 1 × 10 −3 = 4 PO SNR O 50 × 10 NO 40 × 10−3 ∴ NF = 10 log(4) = 6 dB (15.100) 3.25. A simple model developed in solid state physics describes a diode noise manly as a “shot–noise” i sn , which is function of the diode’s biasing current I D . It is very practical to remember approximate values of diode parameters at 1 mA biasing current at room temperature 27◦ over 1 MHz bandwidth, i.e. 2 = 2 q ID B i sn ∴ i sn = 2 × 1.602176565 × 10−19 C × 1 mA × 1 × 106 Hz = 17.90 nA At the same time, voltage across a pn junction and dynamic diode resistance r D are 15 Electrical Noise: Solutions 139 VT = kT = 25.865 mV ≈ 26 mV q ∴ rD = VT = 25.865 π ≈ 26 π I DC Therefore, the shot–noise current i sn flowing through the diode resistance r D generates noise voltage en = i sn r D = kT 2 q ID B = kT q ID = 1.3806488 × 10 −23 J 2B q ID /K × 300.15 K × (15.101) 2 × 1 MHz 1.602176565 × 10−19 C × 1 mA = 463 nV In order to reduce a diode noise, (15.101) implies that we need to reduce threshold voltage VT , thus to lover the internal resistance r D , which is achieved by lowering temperature and by increasing the biasing current, or by decreasing bandwidth. For example, by doubling the biasing current to I D = 2 mA, everything else being equal, the dynamic resistance r D is halved and the diode shot–noise voltage reduced to en = 327 nV. Or, equivalently, in order to achieve the same noise reduction by controlling the temperature while keeping the same I D = 1 mA biasing current we wold have to cool the diode down to −61◦ c. Concurrent biasing current increase and temperature reduction results in en = 231 nV. We could go even further and create, for example, a plot (15.101) to show how the noise voltage changes relative to the biasing current I D = 100µA to 10 mA, and relative to temperature T = 50–400 K(in steps of 50 K) (Fig. 15.20). (Note that both axis use logarithmic scale.) Within this temperature range for a given biasing current, e.g. I D = 1 mA, the noise voltage changes over order of magnitude, where the temperature dependance is more pronounced at lower temperatures. Similar plot could be created if we wanted to see how the noise voltage changes with, for instance, bandwidth and the biasing current. 3.26. We already know that wider amplifier’s bandwidth allows more noise to enter the system, thus to increase the integrated noise power level. As soon as the noise power level brings SNR of this amplifier down to zero, we state that the maximum allowed bandwidth of this amplifier is reached, because beyond that particular value of bandwidth B the input signal is completely swamped with the noise without any chance of being recovered latter. In this problem, we identify three sources of noise at the input terminals of the amplifier and apply the superposition principle, Fig. 15.21. (a) The shot–noise current i sn passing through the source resistor R S (while v S is shorted) causes voltage 140 15 Electrical Noise: Solutions -6 10 en [V] 400K 10-7 50K B=1MHz 10-8 -4 10 -3 -2 10 10 ID [A] Fig. 15.20 Solution 3.25: noise vs. drain current parametric plot Fig. 15.21 Solution 3.26: illustration diagram 2 i sn = 2q I D B ∴ 2 esn (R S ) = (i sn × R S )2 = 2q I D B R S2 (15.102) This voltage is forced by the current through R S resistor and it is therefore directly connected to the input node of the amplifier. (b) the thermal noise voltage generated by the source resistance R S en (R S ) = 4kT R S B (15.103) is spread across (Rin , R S ) divider, therefore at the input terminals ≥ 4kT R S B Rin en (R Sin ) = R S + Rin ∴ 2 Rin 2 en (R Sin ) = 4kT R S B R S + Rin = 4kT B R S where, R S = R S Rin R S + Rin 2 (15.104) 15 Electrical Noise: Solutions 141 (c) Similarly, thermal noise generated by the amplifier input resistance Rin en2 (Rin ) = 4kT Rin B = 4kT B Rin where, = Rin Rin RS R S + Rin 2 (15.105) RS R S + Rin 2 Therefore, from (15.102) (15.104) and (15.105) the total noise at the input of the amplifier is 2 (R S ) + en2 (R Sin ) + en2 (Rin ) (15.106) en2 = esn while the SNR(in) definition leads into SNR(in) = 20 log vS =0 en ⇒ v S = en ∴ v2S = en2 2 = esn (R S ) + en2 (R Sin ) + en2 (Rin ) B = 2q I D B R S2 + 4kT R S B + 4kT Rin ∴ B= v2S ) 2q I D R S2 + 4kT (R S + Rin = 10.208 MHz (15.107) 3.27. In order to compare and manipulate various units, it is much easier if all data is converted to the common unit, in this case a plain numbers, therefore NF = 12 dB ∴ F = 15.85 and A = 50 dB = 1 × 105 (15.108) which leads into TRx = (F − 1) T = (15.85 − 1) × 300 = 4, 455 K and 4, 455 K Tsys = 90 K + ≈ 90 K 1 × 105 (15.109) This example illustrates dominance of the first stage in the overall system level noise calculations. 142 15 Electrical Noise: Solutions Fig. 15.22 Solution 3.28: noise spectrum 3.28. In order to estimate the total input noise current, the input current power spectral density plot is approximated with piecewise linear sections (Fig. 15.22). By doing so, the process of integration is reduced to simple additions of regular surfaces under the PSD curve. We easily recognize the following three geometrical shapes. 1. Rectangular area (the noise ground level is at 0 A2/Hz, thus : i t2 = 1 × 10−25 A2/Hz × (100 × 106 − 1 × 103 )Hz ≈ 10 × 10−18 A2 (15.110) 2. LF triangular area: 1 (1 × 10−23 − 1 × 10−25 )A2/Hz × (2 × 104 − 1 × 103 )Hz 2 (15.111) = 94.05 × 10−21 A2 i L2 F = 3. HF triangular area: 1 (1 × 10−24 − 1 × 10−25 )A2/Hz × (1 × 108 − 3 × 107 )Hz 2 (15.112) = 31.5 × 10−18 A2 2 iH F = Now we estimated the total noise current as 2 i n = i t2 + i L2 F + i H F = 6.45 nA (15.113) Chapter 16 Electronic Devices: Solutions Complex electronic systems are designed by following the modular, similar to Lego Bricks, approach. Each of the basic electronic building blocks is first designed and studied separately, and thereafter it becomes only the designers imagination and creativity that limits how complicated or how unusual the new circuit design will be. Nevertheless, thorough understanding the basic building blocks behaviour is the first step on the road to great designs and inventions. In this section we review important voltage/current divider based devices, frequency limitations of fast pulse waveforms, and the elements of non-linear device behaviour, which is fundamental for the overall wireless circuit functionality. Solutions: 4.1. Rapid amplitude changes of a current flowing through an inductive element generate induced voltage at terminals of the inductive element. As per (4.1), amplitude of the induced instantaneous voltage is directly proportional to the slope (i.e. rate of change) of current that is causing it. Therefore, if for instance the current waveform takes shape of an ideal square pulse (i.e. a pulse whose rising and falling edges are vertical), by definition (4.1) the induced voltage amplitude is infinite, Fig. 16.1. It is very important to be aware of existence of these high-voltage spikes in RF circuits simply because the voltage amplitude may take values much higher than the corresponding power supply value, which is against the conventional wisdom assumed in the design of low frequency amplifiers where maximum amplitudes of the internal voltages are limited by the power supply. Consequently, the other circuit components may be damaged if they are not rated for high enough operational voltages. Naturally, in real circuits it is impossible to achieve infinitely short rise/fall pulse edges, which is to say that amplitude of the induced voltage can not really take the infinite value. However, with high quality inductors (as quantified by their respective Q factor) and fast pulses the generated voltage amplitude may become high enough to cause the damage. R. Sobot, Wireless Communication Electronics by Example, 143 DOI: 10.1007/978-3-319-02871-2_16, © Springer International Publishing Switzerland 2014 144 16 Electronic Devices: Solutions Fig. 16.1 Problem 4.1: time domain current diagram. Current waveform (blue dotted line), and voltage waveform (red solid line) 4.2. Given the initial conditions, at the time t = 0 s when the capacitor is discharged, i.e. vC (t = 0) = 0 V, and the pulse source changes its output from 0 V to v(max), the total source voltage is distributed across the resistor, i.e. v R (t = 0) = v(max). Over the first half of its period T a square function shape holds its high level, while over the second half period it holds its low level, Fig. 16.2. In accordance to Kirchhoff’s voltage law, for 0 √ t √ T/2 we write expression for the voltage inside the RC loop as v(max) = v R (t) + vC (t) T/2 1 = i(t) R + i(x) d x C 0 ∴ di(t) + i(t) 0 = RC dt (16.1) (16.2) where (16.2) is derivative of (16.1). Solution of a first–order differential equation (16.2) is well known as t v(max) t i(t) = I0 exp − exp − = RC R RC Fig. 16.2 Problem 4.2: time domain current diagram (16.3) 16 Electronic Devices: Solutions 145 where, the initial current at t = 0 is I0 = v(max)/R , and time constant of an RC circuits is defined as τ = RC. We note that the product of two electrical variables, i.e. R and C, is measured in units of time, which is arguably the most important variable in physics and engineering. After substituting (16.3) into (16.1) we easily write expression for voltage across the capacitor as t vC (t) = v(max) 1 − exp − τ t = 10V 1 − exp − 1ms (16.4) Within the first half period time interval, plot of (16.4) shows exponential growth of voltage across the capacitor and at the same time exponential reduction of the current flowing inside the circuit. Once the capacitor is fully charged, i.e. vC = v(max) then voltage across the resistor drops to zero, thus no current flow. It is important to note that, in accordance to (16.4), it takes infinite amount of time for the voltage vC to become truly equal to v(max). Thus, saying that the capacitor is “fully charged” is, strictly speaking, not correct. Indeed, when we say “fully charged” we mean charged “close enough” to the maximum possible level. This is another example of differences between perfect mathematics and the engineering approximations. It is convenient to express progression of time in units of τ = RC so that, for example, t = 5 τ expression (16.4) becomes 5ø = 10V × 0.993 = 9.93V vC (t) = 10V 1 − exp − ø (16.5) which leads into convenient engineering rule of thumb that after five timing constants a capacitor is considered fully charged (regardless of the absolute time value). In other words, any other event in the circuit (for example, falling edge of the pulse) should wait at least 5RC units long until the maximum capacitor voltage is achieved. An example of important practical implication of this conclusion is that pulse width of clocks used in our computers is limited to 5τ to 6τ , which sets the minimum clock period to T → 10τ to 12τ . Similarly, if the capacitor is already fully charged to vC = v(max) and the source voltage becomes zero, e.g. the second half period of a pulse function, then the capacitor releases its charge through the resistor in accordance to Kirchhoff’s current law, 146 16 Electronic Devices: Solutions Source pulse τ=5.0 RC τ=0.2 RC τ=1.0 RC 1 amplitude 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 time [RC] Fig. 16.3 Problem 4.2: influence of the timing constant τ relative to the pulse period T = 12τ on the charge discharge capacitor function C dvC (t) vC (t) + =0 dt R ∴ t vC (t) = v(max) exp − RC (16.6) which is well known formula for exponential decay. Therefore, full charge–discharge cycle of a capacitor is combination of (16.4) and (16.6), for the first half period and the second half period respectively. Figure 16.3 illustrates importance of RC timing constant relative to a clock pulse (red dotted line). If the timing constant is too long, e.g. τ = 5RC, then the capacitor voltage does not have enough time to charge up to the “high” voltage level. Instead, when the falling edge of the pulse comes, the capacitor must start the discharging process, which results in waveform whose shape is closer to a triangular instead of square waveform. On the other hand, if the timing constant is short, e.g. τ = 0.2RC, then the capacitor voltage (black dash–dot line) is very close to the ideal pulse shape. The blue solid line shows the capacitor voltage when τ = RC. After 6RC time units the capacitor is practically fully charged. 4.3. In analog signal processing circuits, one of the most common operations is to derive one voltage from another voltage source. Voltage division, or if you prefer voltage multiplication with a factor smaller than one, is used to set voltage reference point that is needed, for example, to set biasing condition of a transistor. Simplest voltage reference is derived from the power supply by the means of two resistors in series, Fig. 16.4 However, for high precision stable voltage reference are designed using sophisticated feedback circuits. For example, a stable and precise voltage reference standard commonly used in modern electronic systems is provided by a circuit known as a bandgap voltage reference. 16 Electronic Devices: Solutions 147 Fig. 16.4 Problem 4.3: voltage reference based on a resistive divider In this example we learn how to design the simplest voltage reference, Fig. 16.4, by calculating the appropriate resistor values. Naturally, since there are two unknown variables, R1 and R2 , there must be two equations so that the system is solvable. Otherwise, we would be forced to arbitrary pick value of one resistor and then to calculate the other. Based on the given data, the two constrains are derived as follows. The reference voltage Vref is derived as Vref = I R2 R2 = VD D R2 R1 + R2 ∴ R1 + R2 = 3.0 R2 (16.7) after substitution of Vref = 1 V and VD D = 3.0 V. Second constraint is derived from the requirement that the equivalent Thévenin resistance of this reference is Rth √ 100 ω. Output resistance, as seen by looking into Vref node is obviously parallel resistance R1 ||R2 , thus in the worst case Rth = R1 R2 R1 + R2 ∴ R1 R2 = 100 (R1 + R2 ) (16.8) It is straightforward to solve the two equations system (16.7) and (16.8) and calculate R1 = 300 ω, and R2 = 150 ω. We keep in mind that this simple voltage reference is directly proportional to the power supply, while by industrial standards the power supply voltage alone is allowed to vary ±10 %. Therefore, this voltage reference is not suitable for high precision applications, however it is almost exclusively used to setup biasing conditions for transistors. If the power supply voltage is increased by ±10 % to VD D = 3.3 V, then the reference voltage is found to be 148 16 Electronic Devices: Solutions Fig. 16.5 Problem 4.5: voltage reference based on a diode Vref = R2 VD D = 1.1V R1 + R2 while for VD D = 2.7 V we find that Vref = 0.9 V. In other words, for a 20 % variation in the power supply voltage, the reference voltage also varies by 20 %. 4.4. The three ideal linearized elements (R,L,C) are the cornerstone of the traditional circuit theory, while the memristance M is joining the group as the fourth element. However, these elements are only idealized mathematical models of the real material behaviour. We already know that, under various external conditions, even a single piece of metallic wire behaves as a complicated R LC network. Thus, we use the ideal linearized element models only within a certain limited range of operation. Nevertheless, as we will find later in this book, it is actually the property of nonlinearity of real materials that enables us to perform operations such as frequency shifting, which is essential for operation of a radio. In this example we take a look at basic active nonlinear element known as a diode. Voltage–current characteristics of this two terminal device obeys “exponential law”, hence a diode mathematical model illustrates its “exponential nature” as in (4.2), while the approximation is valid if (VD ∞ VT ). For the given data straightforward calculation results in: 1. for VD = 600 mV (i.e. VD /VT = 23.120), I D = 0.64684449361 A and I D ≤ 0.64684441671 A, therefore the approximation error is λ = 11.89 × 10−6 %; and 2. for VD = 50 mV (i.e. VD /VT = 1.927), I D = 290.40168723 nA and I D ≤ 213.50168723 nA, therefore the approximation error is λ = 26.5 % which illustrates validity of using the approximative model for strongly biased (VD ∞ VT ) diodes. 4.5. In the previous problem 4.3 we used a simple voltage divider to design a voltage reference. We found that main disadvantage of using that topology is that voltage variations of the power supply directly propagate to the reference voltage node Fig. 16.5. 16 Electronic Devices: Solutions 149 In this example we demonstrate a bit better voltage reference topology based on a diode. The operational principle is very simple, we already know that the diode current I D depends upon the diode voltage VD . However, another way of stating the same is to say that if current I D is forced through the diode, then the voltage VD across diode must obey the same (4.2) equation. If we rewrite (4.2) to solve VD , we have I D ≤ I S exp VD n VT ∴ VD ≤ n VT ln ID IS Where, for I D = 1 A we find that VD = 616 mV. If the current I D changes ±10 %, i.e. from I D = 1.1 A to I D = 0.9 A, then the diode voltage is within the range of VD = 620 mV to VD = 612 mV. In other words, for 20 % variation of the I D current value, the diode voltage VD changes only 1.3 %. This effect is due to “levelling–off” shape of the ln(x) function, which changes very little for the large argument values. 4.6. In principle, a BJT transistor is often modelled as two “back to back‘” diodes (not exactly correct, but very useful model), where the forward biased base–emitter diode is used to control the collector current. Similarly to the simple diode model, we use the exponential model to relate collector current IC and the corresponding base–emitter voltage VB E as VB E VB E − 1 ≤ I S exp IC = I S exp VT VT ∴ VB E ≤ VT ln IC IS where the approximative expression is valid under the condition that VB E ∞ VT . For the given data, I S = 5 × 10−15 A, IC = 1 mA, and VT = 25 mV we easily find that VB E = 650.540 mV. By definition, gm gain is gm = IC VT (16.9) therefore, at this biasing point, gm gain is 40 mS. In order to develop better sense of how the base–emitter voltage controls the collector current, we for example recalculate the following values VB E , V IC , mA 0.50 0.55 0.60 0.65 0.70 0.75 0.80 gm , mS 0.002 0.080 0.017 0.680 0.132 5.280 0.978 39.120 7.231 289.240 53.432 2,137.200 394.814 15,792.560 150 16 Electronic Devices: Solutions which illustrate the importance of tight control of base–emitter voltage VB E , i.e. correct biasing point setting, because the collector current IC is exponentially related and even a small change of VB E causes large variations in the collector current. However, this relationship also illustrates amplification effect of a transistor, in this case small change in the input voltage causes large change in the output current, thus in this mode the transistor is said to behave as a gm amplifier. 4.7. When considering resistive elements, first question to ask is if the element changes its resistance relative to the frequency of signal at its terminals. Based on this criteria, naturally there are two distinct groups, resistances are either frequency independent or not. Beauty of the frequency dependent resistances (commonly refereed to as impedances) is that networks containing such elements simultaneously provide different gain for single tones at different frequencies. When we couple that behaviour with the Fourier’s frequency spectrum principle, we achieve very power mechanism to arbitrary shape frequency spectrum of the given multi–tone signal, and therefore to modify its time–domain shape. This mechanism is fundamental principle used in signal processing techniques. In order to develop better sense of how the resistive network affects the frequency spectrum of a given signal, let us assume that a multi–tone signal s(t) is defined as the sum of single tone signals, all of them with the same amplitude normalized to one, s(t) = 1 + sin(2π t) + sin(2π × 1 × 103 t) + sin(2π × 1 × 104 t) + sin(2π × 1 × 106 t) + sin(2π × 1 × 108 t) (16.10) Each of the single–tone components will be affected differently after the signal s(t) is passed through the resistive networks. By following rules for serial/parallel equivalent resistances, and by knowing expression for impedance Z C of capacitive elements, we write the four respective equivalent impedances as: Ra = R + 10R = 11R ≤ 10R Rb = R + Z C = R − Rc = R||Z C = j ωC ∴ R2 R 2 ω2 C 2 Rd = R||10R ≤ R ∴ +1 ∴ Ra ≥= f (ω) |Rb | = ∴ R2 + 1 2 ω C2 ∴ Rb = f (ω) Rc = f (ω) Rd ≥= f (ω) For the given numerical values, in the following table we summarize the frequency dependance of these four equivalent impedances which illustrates how impedance of these four networks changes with the frequency. In the case of resistive networks only, i.e. Ra and Rd , one is series combination and one is parallel combination of two resistors. In the case of the series combination, the equivalent resistance is greater than the greater of the two, i.e. Ra > max(10R, R). However, since one is ten 16 Electronic Devices: Solutions 151 f, Hz Ra , ω Rb , ω Rc , ω Rd , ω DC 1 1k 10k 1M 100M ∗ 10k ∗ 10k 1G 10k 1M 10k 100k 10k 1.414k 10k 1k 10k 1k 1k 1k 1k 1k 707 10 0 1k 1k 1k 1k 1k 1k 1k or more times larger than the other, it is reasonable to approximate the equivalent resistance with the larger one, i.e. Ra = 10 kω + 1 kω = 11 kω ≤ 10 kω. In the case of parallel connection of two resistors, however, the equivalent resistance is smaller than the smaller of the two, i.e. Rd < min(10R, R). Again, if one resistance is ten or more times greater than the other, and approximative error of 10 % or less is acceptable, then it is reasonable to write Rd = 10R||R ≤ R. The situation becomes much more interesting for Rb and Rc equivalent impedances – they change with the frequency. We already know that impedance of a given capacitance changes from Z C = ∗ to Z C = 0 as the frequency changes form DC to ∗. Thus, in the case of series RC connection, impedance the sum R + Z C ) changes from ∗ to R, while in the case of parallel connection R||Z C , the equivalent impedance changes from R to zero. Considering that we already know Ohm’s law which states that voltage across a resistor is calculated as V = I R, it should be trivial to conclude that if a multi–tone signal is applied across a frequency dependent impedance, then each harmonic of the frequency spectrum is affected by different gain factor. Consequently, in accordance with Fourier’s theorem, amplitudes of all harmonics is changed relative to the original frequency spectrum, and therefore the signal also changes its shape in the time domain. That change of the waveform shape may be intentional and desired, which is what signal processing is all about. Or, it may be unwanted and beyond our control, for example caused by parasitic frequency dependent components, which is what signal distortion is. 4.8. Being able to mentally approximate gain boundaries of a voltage divider is one of the essential engineering skills. In this example we estimate the voltage gains by inspection of the circuit topologies and by knowing at least some relationship between the components. By now we should be able to just write down voltage gain expression of a general voltage divider similar to (Z 1 , Z 2 ) as Vout,B = i Z 1 ,Z 2 Z 2 = V A,B Z2 Z1 + Z2 ∴ Vout,B Z2 = V A,B Z1 + Z2 which is easily evaluated, first assuming Z 1 = 10 Z 2 and no frequency dependance (i.e. R1 = 10 R2 ) as 152 16 Electronic Devices: Solutions AV = Vout,B Z2 Z2 = = = 0.0909 ≤ −20dB V A,B Z1 + Z2 11 Z 2 in other words, the 10 : 1 impedance ratio results in about ten times attenuation (i.e. around −20dB). In the extreme case of Z 1 ∞ Z 2 we conclude that the voltage gain A V ∓ 0. Similarly, if Z 2 = 10 Z 1 we write AV = Vout,B Z2 10 Z 1 = = = 0.909 ≤ 1 V A,B Z1 + Z2 11 Z 1 which is to say that the (1 : 10) ratio causes almost no voltage signal loss. In the extreme case of Z 2 ∞ Z 1 the voltage gain A V ∓ 1. In the case when the voltage divider includes frequency dependent components, e.g. a capacitor, we first evaluate the voltage gain at DC and then at f ∓ ∗. At DC, a capacitor is an open connection (i.e. Z C ∓ ∗) and can be removed from the parallel combination. Under that condition, it is straightforward to conclude that Vout,B = i R R = V A,B R R+R ∴ AV = Vout,B = 0.5 = −6dB V A,B in other words, two equal resistors making voltage divider must split the voltage in half, which is equivalent of saying that the voltage gain equals A V = −6dB. Similarly, the other extreme is when f ∓ ∗, which leads into Z C ∓ 0. The parallel combination R||Z C , therefore, results in the zero equivalent impedance. Direct consequence is that the output voltage Vout,B develops across a zero resistance, therefore the output voltage must be zero. In other words, network in Fig. 4.3 (right) shapes the frequency spectrum by setting A V = 0 for high frequencies, while the maximum gain A V = 0.5 is for DC and low frequencies. Design of RLC networks used to shape frequency spectrum of continuous signals is engineering discipline known as analog filter design. If the same shaping function of the frequency spectrum is implemented using digital logic and discrete signals, then we say that we did digital filter design. 4.9. We now continue our discussion about mentally evaluating circuits and determining useful boundaries. In this example we learn to recognize voltage divider inside a typical amplifier section. We already know that one of the main applications of a voltage divider is to provide voltage levels required to setup biasing conditions of a transistor. In addition, for the purpose of setting up the biasing conditions, we keep in mind the back–to–back diode model of a BJT transistor. To set a BJT transistor in the forward active mode, we know that base–emitter diode must be forward biased, while at the same time the base–collector diode must be reverse biased. 1. Two resistors R1 , R2 in Fig. 4.4 (left) create a simple voltage divider used to scale down the VCC voltage level. So, what voltage is required at the node B? Resistor R2 is in parallel with the base–emitter diode, thus voltage across R2 resistor equals to VB E . In the case of ideal base–emitter diode, any forward bias 16 Electronic Devices: Solutions 153 voltage VB E > 0 turns the diode on. Therefore, assuming any current (in this case I R2 = 1 mA) being forced through R2 , any R2 > 0 generates the required non–zero positive V R2 = VB E > 0 voltage. Setting up the base–emitter diode is only the first step in setting up the BJT transistor biasing point. Second step is to keep the base–collector diode reverse biased. That is achieved by setting potential at the collector VC to the same or higher level relative to the base potential, i.e. VC → VB > 0. In this approximative analysis the base current is assumed zero. 2. If the base–emitter diode is realistic, that is, if its turn on voltage is greater than zero, procedure for setting up the biasing conditions is still as same as in the ideal case. In order to provide VB E = 1 V voltage across R2 must be set to V R2 → 1 V. If the current I R2 = 1 mA then R2 = 1 V/1 mA = 1 kω. Again, in order to keep base–collector diode reverse biased, potential at collector must be VC → VB > 1 V. More precise numerical analysis is always left to the circuit simulators, however by doing our “back of the envelope” analysis we are capable to quickly and intuitively predict the circuit operation before committing the time for simulations. 4.10. Following up on the results of 4.9, the problem is now reduced to design of a network to provide the required biasing voltage at the gate node. Circuit is Fig. 4.4 (right) contains emitter resistor R E whose role is separate the emitter potential from the ground level. Current through this resistor forces the emitter node potential to VE = i R E = 1 V. From this point on, there are two given scenarios. 1. If base–emitter diode of transistor Q 1 is ideal, then base potential must be at VB > VE = 1 V. For example, we can set VB = 1.1 V. Given the power supply voltage VCC = 10 V we write VB = ∴ VB = VCC R2 VCC R2 = VCC R1 + R2 R1 + R2 1 R1 R2 +1 ∴ VCC 10V R1 −1 = −1= R2 VB 1.1V ∴ R1 ≤8 R2 where, the any choice of the actual resistor values is acceptable as long is R1 : R2 = 8 : 1. In practice, we put the additional constrain to limit the total power consumption. Infinite number of (R1 , R2 ) values satisfy the proportion requirement, thus we are free to take practical resistor values that allow minimal current through (R1 + R2 ). Of course, it is not practical to take too large resistor values and make the current too small, because we keep in mind the noise floor as well as practical realization of large resistor values in integrated circuit (IC) technologies. 2. If base–emitter diode is not ideal, i.e. VB E > 0, then for the given data the base potential must be set to VB > VE + Vth (B E) = 1 V + 1 V = 2 V. For example, we can set VB = 2.1 V, which leads into 154 16 Electronic Devices: Solutions R1 VCC 10V −1 = −1= R2 VB 2.1V ∴ R1 ≤ 3.8 R2 This example illustrates one of the first design steps of setting up voltage divider that serves the purpose of providing stable DC voltage level for the base terminal. 4.11. Estimate of a resistance by “looking into a node” is very useful technique that is not difficult to master, as long as we can visualize serial/parallel combination of impedances. In this example, three typical cases commonly found in almost all amplifying stages, are demonstrated. Naturally, this technique is an approximation, however very good one while more precise results can be achieved using numerical simulators. Although it is usually not specifically drawn in circuit schematic diagrams, the power supply source is assumed to be an ideal voltage source, thus its internal impedance is zero. Therefore, for the purposes of finding an equivalent resistance between a certain node and the ground, the positive power supply rail is shorted to the ground line. That being the case, voltage divider R1 , R2 in Fig. 4.5a becomes parallel connection R1 ||R2 from the perspective of Z out node, i.e. Z out = R1 R2 R1 + R2 (16.11) However, if an active component is involved, Fig. 4.5b, then while looking into base we must take into account influence of the emitter resistor R E . It is shown in the textbook that due to feedback effect, resistance R E is projected to and perceived by the base node as if it were multiplied by (β + 1) factor, where β is gain of Q 1 . In this example it was indicated that i B ≤ 0, which implies that the internal base resistance r B itself is r B ∓ ∗, therefore Z in = r B ||(β + 1) R E ≤ (β + 1) R E (16.12) which is the upper bound, because for finite r B the equivalent parallel resistance Z in is lower. In order to estimate Z out resistance looking into the emitter node Fig. 4.5c we start with the voltage divider R1 , R2 . Resistance at the base node consists of three resistance in parallel R1 ||R2 ||r B , which is approximated with R1 ||R2 because the internal base resistance r B ∓ ∗ (i B ≤ 0). Whatever resistance is associated with the base node, it is perceived from the emitter side as if it were divided by the (β + 1) factor, i.e. in this case (R1 ||R2 )/(β + 1). This mental transformation of the base side resistance to the emitter side is illustrated in Fig 16.6. When looking into emitter, we should see the emitter resistor R E to the ground. In addition, the internal BJT emitter resistance is re = VT IC (16.13) 16 Electronic Devices: Solutions 155 Fig. 16.6 Problem 4.11: schematic of a BJT output impedance network and for IC = 1mA at room temperature (i.e. VT ≤ 25 mV) it follows that re ≤ 25 ω. Obviously, resistance value of parallel R1 ||R2 divided by relatively large value of β is also expected to be small. Typically, emitter resistor R E is much larger than the others, Fig 16.6 (left). Therefore, depending upon the biasing collector current, it is expected that resistance looking into BJT emitter is rather small (definitely smaller than re ), that is the emitter node behaves as a voltage source. 4.12. Value of the BJT thermal voltage VT is very important, because it directly influences small signal emitter resistance re , which in return controls voltage gain of the BJT device. Design engineers must be aware of the temperature dependance of VT and either try to reduce the circuit temperature sensitivity, or try use this sensitivity to design, for example, temperature sensing circuits. It is straightforward to calculate thermal voltage by using its definition VT = kT q (16.14) which for the given temperature results in the following three values VT (−55 ⇒ C, 25 ⇒ C, 125 ⇒ C) = (18.8, 25.7, 34.3 mV) (16.15) This simple example demonstrates that within the given temperature range, the thermal voltage changed almost 100 %, which is very significant and must be taken into account during the design process. 4.13. In this example we practice to design a voltage divider that involves a resistor and a nonlinear element, i.e. a diode. As we already learned this combinations of elements is very often used to provide voltage reference to the rest of the circuitry. However, due to existence of the nonlinear device, the main problem is that in order to set diode voltage, we need to set diode current, which is set by the diode voltage. Therefore, the solution is found by iterations. In this case, we start the iterative process by assuming an ideal diode, i.e. VD = 0, which enables us to calculate the initial current through resistor as I R = VCC /R = 9.000 mA. The diode and resistor currents must be equal, therefore, diode current is alsoI D = 9.000 mA. But then, this current forces the diode voltage to be 156 16 Electronic Devices: Solutions VD = VT ln ID IS = kT 9.000 mA ln = 336.030 mV q 18.8 nA which is definitely not zero. Taking this diode voltage as the starting point, in the second iteration we find resistor current as IR = 9 V − 336.030 mV VCC − VD = = 8.664 mA R 1 kω which is of course different than the initially assumed 9.000 mA, and thus we must start another iteration by using I D = 8.664 mA as the diode current. By repeating the last two steps again, the next iteration gives VD = 335.052 mV which leads into I R = 8.665 mA. The following iteration gives VD = 335.055 mV and again I R = 8.665 mA. Therefore, being satisfied with three decimal places, we converge to VD = 335.055 mV. In this simple example we demonstrated the same process that numerical simulators repeat millions of times per second when used to resolve operating points of complex electronic circuits. 4.14. In this example we demonstrate temperature dependance of a BJT transistor, that is as same as foe any other pn junction. Here we already know that VB E ≤ VT ln IC IS where the temperature dependance is included only through the VT = Assuming I S = 100 fA and by direct implementation of (16.16) we find VB E (−55 ⇒ C, 25 ⇒ C, 125 ⇒ C) = (432.8, 591.6, 790 mV) (16.16) kT/q term. (16.17) while for I S = 200 fA the base emitter voltage VB E takes the following values VB E (−55 ⇒ C, 25 ⇒ C, 125 ⇒ C) = (419.8, 573.8, 766.2 mV) (16.18) which illustrates importance of the environment temperature where the electronic equipment is intended to operate. Additional important point to note is that (16.17) and (16.18) imply that the base–emitter voltage VB E increases with the temperature, i.e. that it has positive temperature coefficient. However, that conclusion is not correct because base–emitter voltage actually has negative temperature coefficient, i.e. its value reduces with the increase of temperature. The reason for this wrong conclusion in our example is that we completely ignored temperature dependance of the leakage current (i.e. the reverse saturation diode current) I S = f (T ). This temperature dependance is much stronger than the VT = f (T ) dependance, thus the overall VB E = f (T ) temperature coefficient is negative. Detailed temperature sensitivity analysis of a BJT transistor 16 Electronic Devices: Solutions 157 is beyond scope of this course, thus for now we keep in mind only the information about the negative temperature coefficient. 4.15. Following up on the results of 4.14, we now derive biasing voltage at the gate, while considering temperature dependance of VB E and presence of R E resistor. For large β values, it is reasonable to assume that I E ≤ IC , which simplifies calculation of potential at the emitter node, VE = I E R E = 100 mV. At 25 ⇒ C the base potential must be at least VB E = 591.6 mV higher, which sets VB → 691.6 mV. 4.16. This kind of estimate is done simply by inspection of the schematic diagram. The emitter resistor is “projected” to the base node as if being magnified by factor (β + 1), thus, Z in = (β + 1) R E , which for the given data results in Z in = 10 kω and Z in = ∗. In this case, the base current is negligible, i.e. rπ ∓ ∗, thus only the multiplied value of emitter resistor is seen from the input side. We note that very high input impedance (i.e. negligible input current) is property usually associated with CMOS type transistor. However, in the first approximation of BJT, it is often useful to ignore base current and estimate biasing parameters without the use of numerical solvers. 4.17. This problem is an example of under–constrained design specification set, which happens relatively often in engineering. In order to divide 3.3 V that is available at the positive power supply rail down to VB → 691.6 mV that is required by the base terminal, there is an infinite number of possible R1 , R2 values that would achieve that. Here are several possible reasons that could be used to add more constrains to the design specifications. By building up on the results in Problem 4.16 we develop the following arguments. The input signal amplitude is important factor. For the sake of argument, let us say that in a given application the input signal amplitude is vin = 5 mVpp . That means VB = 691.6 mV voltage must be increased by 2.5 mV in order to accommodate for the full swing of the input signal. In other words, the input common–mode voltage must be set to VB = 694.1 mV. Therefore, in this setup the base voltage fluctuates as VB = 694.1 ± 2.5 mV, which guarantees that even the minimum input side voltage is high enough to keep the transistor VB E diode turned on. We can now proceed and determine the required ratio (R1 /R2 ), which is found by setting up the proportion R1 ≤ 3.754 R2 (16.19) At this point, there are at least two possible additional design goals the could help us to set constrains leading into unique solution for (R1 , R2 ) pair of resistors. One possible design goal would be to achieve certain input resistance Z in value, which, by inspection we simply write as VB VCC = R1 + R2 R2 ∴ R1 VCC 3.3V −1 = −1= R2 VB 694.1mV Z in = R1 ||R2 ||(β + 1) R E ∴ (16.20) 158 16 Electronic Devices: Solutions Another possible design goal would be to reduce the overall power consumption of the circuit. Reference current VCC /(R1 + R2 ) is for all practical purposes wasteful, because the goal is to set VB , which can be achieved by infinite number of resistor values, as long as the ratio (16.19) is satisfied. Larger (R1 , R2 ) resistor values allow lower current, which reduces power consumption. We already discussed practical limits of large resistors. In this example we illustrated some of the design compromises that must be made while designing electronic circuits, which makes analog circuit design as much art as it is engineering. Indeed, elegant design solutions are always admired by colleagues and users. 4.18. It is straightforward to find from the given data for VB E , T , and I S that IC = 1 mA. By definition, change of the output current i C due to change of the input voltage v B E at the operating point IC is refereed to as gm , which in the case of BJT transistor takes form VB E di C d I exp − 1 = gm ∼ S dv B E iC =IC dv B E VT i =I C C VB E IS i C IC = exp = = (16.21) V V V V T T i C =IC T i C =IC T which is also illustrated in Fig. 16.7. For the given data and (16.21) gm = 1 mA IC q IC = = 38.922 mS = VT kT 25.692 mV (16.22) Internal emitter diode resistance re is found by definition of Ohm’s law as vB E α v B E α 1 = = ≤ rE ∼ iE i C iC =IC gm gm which, for the given data gives r E = 25.692 ω. Fig. 16.7 Problem 4.18: illustration for gm definition (16.23) 16 Electronic Devices: Solutions 159 It is now easy to tabulate BJT parameter variations for various biasing currents, and various temperatures. Good rule of thumb is that at room temperature and IC = 1 mA, results in r E ≤ 25 ω. By noticing the inverse relation between r E and IC , one can easily conclude that if the collector current doubles, the emitter resistance drops by half, and so on. Chapter 17 Electrical Resonance: Solutions In the circuit theory an inductor and capacitor are two of the four fundamental electronic elements, resistor and memristor being the other two. In addition, the filter theory is very well developed, which enable us to cary out analytical work with high level of accuracy. In this chapter we use complex analysis to derive all important parameters of an LC resonator: the resonant frequency f 0 , bandwidth B, Q factor, and dynamic resistance R D . Solutions: 5.1. By inspection of the two RLC networks in Fig. 5.1, it is straightforward to write the two expressions. 1 jω C 1 Yab (ω ) = G + j jω C − jωL Z ab (ω ) = R + j jωL − 5.2. Another way of saying that the impedance Z ab (ω0 ) is real is to say that its imaginary part is √(Z ab (ω0 )) = 0, i.e. √(Z ab (ω0 )) = 0 → ω0 L − 1 ω0 C =0 → ω0 L = 1 ω0 C → ω0 C = 1 ω0 L ∴ ω0 = ∞ 1 ∴ f0 = 1 ∞ LC 2λ LC (and, for parallel RLC network) 1 =0 ω0 C − √(Yab (ω0 )) = 0 → ω0 L ∴ R. Sobot, Wireless Communication Electronics by Example, 161 DOI: 10.1007/978-3-319-02871-2_17, © Springer International Publishing Switzerland 2014 162 17 Electrical Resonance: Solutions ω0 = ∞ 1 LC ∴ f0 = 2λ 1 ∞ LC (17.1) In other words, for ideal case of lossless inductor and capacitor, both series and parallel LC resonator have the same resonant frequency ω0 . In other words, the “ideal case” means wire resistance of the inductor is R = 0, i.e. Q factor is Q = ≤. 5.3. By inspection of the two RLC networks in Fig. 5.2, it is straightforward to write the two expressions. Series configuration: j Z ab = Z L + Z C = jωL − ω tC 1 at resonance (i.e. ω t = ω0 ), |Z L | = |Z C | → ω t L = ω tC ∴ Z ab = jω0 L − jω0 L = 0 Parallel configuration: 1 1 1 j = + = jω C − Z ab ZC ZL ωtL at resonance (i.e. ω = ω0 ), |Z L | = |Z C | ∴ 1 1 1 = − =0 Z ab jω0 L jω0 L ∴ Z ab = ≤ 5.4. Model of an ideal LC resonator assumes no thermal losses whatsoever. In reality, inductor is made of a wire with a finite resistance R, while the capacitor uses realistic dielectric that allows for a small but not zero leakage current. Thus, there is a minor thermal energy loss in the resonator, and the important consequence is that the loss must be compensated for, otherwise the resonance is not sustainable. Quantitative measure of the internal thermal losses is expressed through the Q factor. 1 + jω C R + jωL R − jωL = 2 + jω C R + (ω L)2 R ωtL = 2 + j ω tC − R + (ω L)2 R 2 + (ω t L)2 Y (ω t) = at resonance (i.e. ω t = ω p0 ), |Z L | = |Z C | → √(Y ) = 0 (17.2) 17 Electrical Resonance: Solutions 163 ω p0 C = R2 ω p0 L + (ω p0 L)2 R 2 + (ω p0 L)2 = ∴ L C ω p0 = R2 1 − 2 LC L (17.3) In other words, the resonant frequency ω p0 of a parallel LC network that includes realistic inductance has the additional term (R/L)2 due to the finite wire resistance, which slightly reduces the resonant frequency relative to the case of ideal LC resonator. When R ≥ 0 Eq. (17.3) reduces to (17.1) for ideal LC resonator, i.e. ω p0 ≥ ω0 . 5.5. The imaginary part of √(Y ) determined the resonant frequency, while the real part ∗(Y ), from (17.2), determines the dynamic resistance R D , i.e. real resistance of the LC resonator at the resonant frequency, as, R + (ω p0 L)2 R = 1 R2 + LC − ∗(Y (ω p0 )) = = R2 R2 L2 2 L2 R L + C − R2 RC = L ∴ L 1 = RD = ∗(Y (ω p0 )) RC R2 (17.4) For ideal case, i.e. R = 0, the dynamic resistance of LC resonator becomes R D = ≤. It should be noted that from the perspective of the resonant current, which circulates inside the RLC loop, the three elements are in series. Hence, reducing the resistance associated with the inductive branch is desirable in order to increase the dynamic impedance being perceived by the network external to the RLC resonator. 5.6. By definition, the quality factor Q of a RLC circuit is the ratio between the energy stored in the imaginary parts (L or C) and the energy dissipated by the real part, i.e. R, of the network during a complete cycle, Q ∓ 2λ maximum energy stored energy dissipated per cycle (17.5) 164 17 Electrical Resonance: Solutions 1. In ideal case, energy stored in the magnetic field of the inductor is eventually converted without losses into energy of the electrostatic filed of the capacitor. At the resonant frequency, the maximum energy stored in the network keeps bouncing back and forth between the inductor and capacitor without losses and, therefore, is calculated either at the moment when the capacitor is fully discharged (and therefore the inductor holds the full amount of energy W L ), or when the capacitor is fully charged (and therefore temporarily holds the full amount of the energy WC ). That is to say, WL = T 0 v(t)i(t) dt = T i(t)L 0 Ip di(t) 1 2 dt = L i di = L I p2 = L Irms (17.6) dt 2 0 or, similarly Vp T T dv(t) 1 2 dt = C v(t)i(t) dt = v(t)C v dv = C V p2 = C Vrms WC = dt 2 0 0 0 (17.7) ∞ ∞ where, I p = 2Imax is the peak current through the inductor, and V p = 2Vmax is the peak voltage across the capacitor. The energy dissipated in the resistor W R during one full resonant cycle T0 = 1 2λ = f0 ω0 (17.8) is simply, by definition, power times the time, i.e. 2 W R = PR × T0 = R Irms × T0 = 2λ 2 R Irms ω0 (17.9) which means that (17.5) becomes (using either W L or WC ) for series RLC Q s = 2λ WL = 2λ WR 2 L Irms 2λ ω0 R 2 Irms = ω0 L R (17.10) 2. At resonance, the resonant frequency ω0 , inductance L, and capacitance C are connected as in (17.1), therefore the three equivalent formulations of Q s are ω0 = ∞ 1 LC ∴ 1 1 ω0 L = = Qs = R ω0 RC R L C (17.11) 17 Electrical Resonance: Solutions 165 which are the answers (b) and (c) for series RLC network. Expressions (17.10) and (17.11) for the quality factor Q are very important and are used to quantify a number of specifications in radio design. It is important to note that for ideal inductor, i.e. R = 0 the Q factor becomes Q = ≤. It is desirable to increase the Q factor for many reasons that will be seen throughout this book. In addition, it should be noted that in series configuration Q factor is inversely proportional to the resistance R. For parallel RLC configuration Fig. 5.1 (right), however, it is desirable to have the R as high as possible in order to reduce the power dissipation (i.e. to reduce the current through the R branch of the RLC network), which is to say that the three equivalent quality factor Q p formulations for a parallel RLC network are C R Qp = = ω0 RC = R ω0 L L (17.12) To elaborate the point, it is also useful to find out how much is the difference between resonant frequencies of series and parallel resonant RLC networks. In problem 5.4 we already derived expression (17.3) for resonant frequency ω p0 of a parallel RLC network, which can be reformulated as ω p0 = 1 R2 − 2 = LC L 2 2 R R 1 2 − ωs0 = ωs0 1 − 2 2 = ωs0 1 − 2 L2 Q ωs0 L s ∴ ω p0 ⇒ ωs0 for (Q s > 10) (17.13) where, we appropriately introduced series resonant frequency ωs0 through series Q s factor. It is now obvious that for ideal or high Q networks (i.e. Q > 10) there is very small error in calculating resonating frequencies ωs0 and ω p0 of the series and parallel circuits, hence they can be used interchangeably, as long as the high Q condition applies. Finally, expression for dynamic resistance (17.4) can also be reformulated in terms of the Q factor, after using (17.11), as: RD = L Q = ω0 L Q = = Q2 R RC ω0 C (17.14) which is, again, resistance of a realistic RLC tank in resonance as perceived by the external network. Important distinction to make is that the resistance R is physical entity and in series RLC network needs to be as small as possible, while in parallel configuration it needs to be as large as possible. However, at resonance, this small resistance is perceived by the external network as if being magnified by the Q 2 factor. In ideal case, i.e. when serial the resistance R = 0, the Q factor goes to infinity. Hence, the last expression (17.14) is only mathematical approximation. 166 17 Electrical Resonance: Solutions 5.7. Often, it is useful to transform a series RLC network into its equivalent parallel configuration and vice versa. This transformation must be done only at a single frequency, which does not affect the Q factor of the networks. Serial and parallel Q factors are Xs Rs Rp Qp = Xp Qs = (17.15) (17.16) so that, assuming Q s = Q p = Q at the given frequency (17.17) Z s = Rs + j X s = Rs + j Q s Rs = Rs (1 + j Q s ) 1 − jQ 1 1 Q 1 1 = = −j = Yp = Zs Rs (1 + j Q) Rs (1 + j Q) 1 − j Q Rs (1 + Q 2 ) Rs (1 + Q 2 ) (17.18) Q 1 1 1 −jX −j (17.19) = = s 2 Rs (1 + Q 2 ) R X p p (1 + Q ) Q ∴ R p = Rs (1 + Q 2 ) 1 X p = Xs 1 + 2 Q (17.20) (17.21) after replacing (17.15) and (17.16) into (17.17) to (17.19). Again, for large Q, i.e. Q > 10, R p ⇒ Q 2 Rs (17.22) X p ⇒ Xs (17.23) 5.8. Let us consider series RLC network from the perspective of voltage source with resistance R driving impedance Z = j (ω L − 1/ω C ), Fig. 17.1. Maximum power transfer, therefore, happens when the source is matched to the load, i.e. R = |Z |. Otherwise, at DC the capacitor becomes open while inductor becomes short connection; and at the other side of the frequency spectrum, at very high frequencies, the Fig. 17.1 Solution 5.8: schematic diagram of a series LC network 17 Electrical Resonance: Solutions 167 capacitor becomes short while inductor becomes open connection. In both extreme cases there is no power transfer because the loop current must drop to zero. Hence, condition for maximum power transfer R = |Z |, leads into Vout = Vin Vin Vin Vin |Z | = R= =∞ |R + Z | |R ± j R| |1 ± j| 2 (17.24) which happens at two frequency points; let us label them (for the time being) as ωU and ω L (for “upper” and “lower” frequency, respectively), so that R = |Z | is written as 1 R = ωU L − (17.25) ωU C 1 −R = ω L L − ωL C (17.26) which, after using (17.1) to replace C, leads into R = ωU L − 1 ωU −R = ω L L − 1 ω02 L 1 ωL 1 ω02 L ∴ R = ωU L − ∴ −R = ω L L − ω02 L ωU ω02 L ωL ∴ R ω0 L = ωU ω0 ∴ − ωR0 L = ωL ω0 − ω0 ωU − ω0 ωL (17.27) and Q = ω0 L/R, which is to say 1 ωU ω0 = − Q ω0 ωU ωL 1 ω0 − = − Q ω0 ωL (17.28) (17.29) First, by adding (17.28) and (17.29) it follows that ωU ωL ω0 ω0 + = + ω0 ω0 ωU ωL ∴ ω02 = ωU ω L (17.30) and now, using (17.28) and (17.30) we write ω2 − ω02 ωU 1 ω0 = − = U Q ω0 ωU ωU ω0 ∴ ωU ω0 ωU ω0 ω0 ω0 Q= 2 = 2 = = 2 ω − ω ∆ω ωU − ω0 ωU − ωU ω L U L (17.31) The last expression is very important, because the two frequencies ωU and ω L are used to define the resonator’s bandwidth, as the two of them are at −3 dB points 168 17 Electrical Resonance: Solutions 0 Fig. 17.2 Solution 5.8: frequency domain plot to illustrate the BP bandwidth definition relative to the maximum amplitude of the resonator (which is at ω0 ). Also, (17.31) shows that narrow band is achieved by using high Q components, or in other words, high Q is needed to achieve narrow band around the resonant frequency. As a side note, in series RLC configuration high Q also means very low resistance R and high inductance L, which is to say that it is good for matching with a low impedance source, such as an antenna for example, whose impedance is usually in order of 50 π. Otherwise, if the source impedance is very high then the parallel RLC configuration must be used, where high Q means high resistance and very low inductance. 5.9. By definition, Q factors of inductive and capacitive branches in the LC network (after substitution ESR = r2 ) at resonant frequency ω0 are: XL ω0 L = = tan φ1 ∴ φ1 = arctan Q 1 r1 r1 XC 1 Q2 ∓ = = tan φ2 ∴ φ2 = arctan Q 2 r2 ω0 C r2 Q1 ∓ (17.32) (17.33) where φ1 and φ2 are respective phase angles in inductor and capacitor due to the thermal losses (resistances r1,2 denote the internal resistances of the coil and the effective series resistance of the capacitor respectively). We also define, after including (17.11), ω0 L Z 1 = r1 + jω0 L = + jω0 L Q1 ∴ ω0 L 2 1 2 |Z 1 | = + (ω0 L) = ω0 L 1 + 2 (17.34) Q1 Q1 17 Electrical Resonance: Solutions 169 as well as, Z 2 = r2 + ∴ 1 1 1 = + jω0 C Q 2 ω0 C jω0 C |Z 2 | = 1 1 1 + = 2 2 (Q 2 ω0 C) (ω0 C) ω0 C 1+ 1 Q 22 (17.35) From (17.32) and (17.33) in addition to straightforward application of trigonometric identities,1, 2 we write sin φ1 = sin φ2 = Q1 1 ; cos φ1 = 1 + Q 21 1 + Q 21 Q2 1 ; cos φ2 = 1 + Q 22 1 + Q 22 (17.36) (17.37) Derivation for the resonant frequency ω0 If an AC voltage Vin = V cos φ1 is applied to the resonator, Fig. 17.3, the total current needed to compensate for the thermal losses i = i 1 + i 2 is split between the two branches. The inductive branch current i 1 has two components: one that is in phase with the source voltage Vin , i.e. (V cos φ1 )/Z 1 , and one that is lagging by 90∼ , (V sin φ1 )/Z 1 . At the same time, the capacitive branch current i 2 also has two components: one that is in phase with the source voltage Vin , i.e. (V cos φ2 )/Z 2 , and one that is leading the source voltage Vin by 90∼ , i.e. (V sin φ2 )/Z 2 . At resonance, the two quadrature current components must be opposite and equal (so that the vector sum is zero), which leads into the following expressions (after using results (17.34) to (17.37)) Fig. 17.3 Notification in the realistic parallel LC network for solution of the problem 5.9 1 2 ∞ cos[arctan x] = 1/ 1 + x 2 . ∞ sin[arctan x] = x/ 1 + x 2 . 170 17 Electrical Resonance: Solutions 1 1 = (V sin φ2 ) Z1 Z2 ∴ Q1 1 ω0 C Q2 = 1 1 1 + Q 21 ω0 L 1 + Q 2 1 + Q 22 1 + Q 2 (V sin φ1 ) 1 Q1 2 1 + Q1 2 Q2 = 1 2 1 + Q2 1 + Q2 1 + 2 ω0 LC (17.38) 1 Q 22 1 where both the left and right side of (17.38) contain algebraic term that can be simplified as follows: x (1 + x 2 )(1 + x12 ) x2 x = 2 x+ x + x1 = 1 x x2 (1 + x 2 )(1 + = 1 ) x2 = x2 x2 + 2 + 1 1+ 1 x2 = (17.39) 1 x2 Using (17.39) it is straightforward to rewrite (17.38) as: 1 1+ 1 Q 21 = 1 1+ 1 Q 22 ω02 LC ∴ 1 1 + ω0 = ∞ 1+ LC 1 Q 22 1 Q 21 ⇒∞ 1 LC ; (Q 1,2 1) (17.40) which is the solution for the resonant frequency of an LC resonator with non-ideal inductor and non-ideal capacitor. Naturally, for very good L and C components the thermal losses are negligible, in other words Q 1,2 1, hence (17.40) can be approximated with the expression for the resonant frequency ω0 that was already derived in problem 5.2. for the ideal case of LC resonator. However, note that the assumption of high Q is not always valid, a very dramatic example is the case of on-chip inductors manufactured in standard CMOS process that are used in modern wireless devices—their Q is in order of five. Consequently, additional design and technological techniques must be employed to improve performance of integrated LC resonators. Derivation for the dynamic resistance R D At resonance, the sum of complex quadrature components of the two branch currents is zero, which leaves only the two in-phase current components. Similarly to the 17 Electrical Resonance: Solutions 171 previous derivation, we write, i=V ⎧ cos φ2 cos φ1 + Z1 Z2 ⎨ 1 1 = V ⎪ 1 + Q 2 ω0 L 1 + 1 ∴ =V 1 Q 21 + Q1 Q 2 ω0 C + 2 ω0 L(1 + Q 1 ) 1 + Q 22 1 1 + Q 22 ω0 C 1+ 1 Q 22 ⎩ (17.41) It is now convenient to introduce substitution for the ω0 C term in (17.41) by rewriting (17.40) as follows, ω02 LC = 1 + 12 Q2 1 + 12 Q1 ∴ Q 22 ω0 C 1 + Q 22 = Q 21 (1 + Q 21 )ω0 L ∴ ω0 C = 1 + Q 22 Q 21 Q 22 (1 + Q 21 )ω0 L (17.42) and after the substitution (17.41) becomes, i=V Q1 Q2 + ω0 L(1 + Q 21 ) 1 + Q 22 1 + Q 22 Q 21 Q 22 (1 + Q 21 )ω0 L =V Q1 ω0 L(1 + Q 21 ) 1+ Q1 Q2 (17.43) which now leads straight into the expression for dynamic resistance R D as RD ∓ Q 1 + Q11 V = ω0 L Q1 i 1+ Q 2 (17.44) for the case of non-ideal inductor and non-ideal capacitor. As is, (17.44) shows dependance of dynamic resistance versus Q factors of L and C components. In case of very good (but still not perfect) inductors, i.e. Q 1 1 or in other words (1/Q 1 ) ⇒ 0, equation (17.44) is than written as very close approximation, R D = ω0 L Q1 1+ Q1 Q2 = ω0 L Q1 Q2 Q1 + Q2 (17.45) Modern capacitors, however, are made using very good dielectrics, which is to say that Q 2 is not only large but could be approximated as Q 2 ≥ ≤, in other words Q 2 Q 1 , in other words Q 1 /Q 2 ⇒ 0. Therefore, in case of lossless capacitor (17.45) is further approximated as R D = ω0 L Q 1 (17.46) 172 17 Electrical Resonance: Solutions which is the case most commonly used in practice because, in comparison with capacitors, inductors are much harder components to build. Finally, in the extreme approximation that is good only for fast “back of the envelope” hand analysis even the inductor is assumed to be perfectly lossless, i.e. Q 1 ≥ ≤, which means that (17.46) becomes simply RD ≥ ≤ (17.47) which is what already was concluded earlier in (17.14). All four expressions (17.44) to (17.47), in addition to (17.31), for dynamic resistance R D are useful, as long as the applied assumptions are kept in mind. 5.10. In a series RL model, Q factor of an inductor is simply the ratio of its complex impedance ω0 L and its real resistance R. High quality, i.e. high Q factor, inductors are made of very low resistance wire. However, in parallel RL model, high Q values is achieved if the parallel resistance is very high. For the given data, first we note that because Q 10 series and parallel Q values are approximately same, i.e. Qs ⇒ Q p = Q ω0 L Rs Rp Qp = ω0 L Qs = 2λ f L = 628 mπ Q ∴ Rs = ∴ R p = 2λ f Q L = 25 kπ Known inductance value L resonates with only one capacitance value C at the given resonant frequency f 0 , f0 = 2λ 1 ∞ ∴ LC C= 1 ⇒ 126 pF (2λ f 0 )2 L Once Q factor and the resonant frequency f 0 are known, expression for bandwidth B is already found in 5.8 (17.31) as Q= f0 B ∴ B = 50 kHz With the given resonant frequency f 0 , the required bandwidth B = 500 kHz is achieved if Q factor of the LC resonator is Q 1 = 20. By working backwards, we first conclude that the total parallel resistance must be R P = 2λ f Q 1 L = 2.5 kπ which is to say that the problem is reduced to finding what resistance R in parallel with R p = 25 kπ gives the total resistance of R P = 2.5 kπ. From rules for the equivalent parallel resistance we find 17 Electrical Resonance: Solutions 173 Fig. 17.4 Problem 5.11: Low pass RC filter schematic (left), and high pass RC filter (right) 1 1 1 = + RP Rp R ∴ R = 2.778 kπ In this example we have exercised numerical relationships among an LC resonator’s parameters, and also we explored differences between serial and parallel LC resonators. We realize that series and parallel resonators have different impedances, which is important data point to know for circuit design because we want to maximize power transfer with the following stages inside radio circuits. 5.11. When LP filter is followed by a HP filter and their respective poles overlap, the outcome is effectively a bandpass filter whose bandwidth is the overlapped frequency region. In order to illustrate the point, in this example we consider the following simplified analysis of lowpass (LP) and highpass (HP) RC filters, Fig. 17.4. Transfer function HL P ( jω ) of an RC LP filter is derived following the same procedure as for the voltage divider. By taking the output voltage vout across the capacitor we have HL P ( jω ) = vout ZC 1 = = vin R + ZC jω RC + 1 ∴ |HL P ( jω )| = 1 1 + (ω RC)2 (17.48) where, we already know that the −3 dB point is defined as the frequency value 1 |H ( jω −3 dB )| = ∞ = −3 dB 2 (17.49) which is relative to its maximum voltage value that is normalized to one. Therefore, from (17.48) and (17.49) it follows that pole of LP RC filter is ω −3 dB L P = 1 RC (17.50) 174 17 Electrical Resonance: Solutions Similarly, transfer function H HP ( jω ) of an RC LP filter is derived as H HP ( jω ) = vout R jω RC = = vin R + ZC jω RC + 1 ∴ |H HP ( jω )| = ω RC (17.51) 1 + (ω RC)2 which leads us into expression for HP RC filter pole as ω −3 dB HP = 1 RC (17.52) We realize that expressions (17.48) and (17.51) are scalable and can be used to set the poles to any frequency. In general, we normalize all filter transfer function expressions to 1 Hz so that the designers can scale them to any other frequency. Thus, normalized versions of LP and HP filter transfer functions are found by setting pole locations at ω −3 dB = 1, that is RC = 1, as 1 HL P (ω ) = ∞ 1+ω2 ω H HP (ω ) = ∞ 1+ω2 (17.53) (17.54) where both maximum amplitudes are one, which is easy to verify by setting ω = 0 for LP, and ω ≥ ≤ for HP filter. From the circuit theory, we know that if two transfer functions are chained one after another, the resulting function is found as their product, thus we write |HL P−HP ( jω )| = |HL P ( jω )| |H HP ( jω )| (17.55) However, from (17.50), (17.52) and (17.55) we conclude that maximum amplitude ∞ ∞ of |HL P−HP ( jω )| must equal (1/ 2) (1/ 2) = 1/2, which means that we need to multiply (17.55) by two so that its amplitude is also normalized to one, Fig. 17.5. Series RLC network transfer function is derived, for example, with reference to schematic in Fig. 17.6, as vout = vin R= R + X L + XC vin R + jω L − j ∴ vout = |H R LC (ω )| = vin R 1 2 R2 + ω L − ωC 1 ωC R (17.56) 17 Electrical Resonance: Solutions 175 At the same time, we already know that Q factor for series RLC network is Q= 1 ω0 L = R ω0 RC (17.57) where, ω0 is the resonant frequency. Simple substitution of (17.57) into (17.56) leads into 1 (17.58) |H R LC (ω )| = ω0 2 2 ω − Q 1+ ω0 ω Fig. 17.5 Problem 5.11: normalized transfer functions of LP, HP, R LC, and 2 × (LP × HP) amplitudes normalized v out In order to normalize resonant frequency of (17.58) we set ω0 = 1 and made plot, Fig. 17.5 for Q = 10, which is very modest value of Q factor for RLC resonators (typically, RLC resonators have Q factors in the range of few hundred). With a little bit of algebra, it could be shown that RLC resonator whose Q = 0.5 degenerates in RC LPHP bandpass filter. Alternatively, by re-plotting Fig. 17.5 with linear frequency axis and by experimenting with the Q factor, it is easy to reach the same conclusion. Thus, it should be obvious why we typically do not use LPHP RC filter when a narrowband BP filter is needed. Although a bit simplified, this example illustrates a line of thought in the design process of a BP filter. Parallel RLC resonator is solved by following the same reasoning as in this example, while keeping in mind the duality of voltage/current series/parallel passive circuits. While the filter specifications are very clear, the choice and design of required components is constrained by practical issues related to the achievable component values and material properties. That discussion, however, is beyond the scope of this book. RLC, Q=10 LP HP LPHP 0dB -3dB 0.1 1 normalized frequency Fig. 17.6 Problem 5.11: series RLC circuit network 10 176 17 Electrical Resonance: Solutions 5.12. If a radio receiver were built by using only fixed discrete components, as it was at the very beginning when there were only a few radio transmitters in the whole world, the whole wireless transmission concept would have been completely unpractical. And, consequently, we would not have our cellphones, WiFi, radio, TV, satellites all transmitting at the same time, all the time. Indeed, very dark prospect by today’s standards. In order to enable the receiving equipment to isolate the wanted carrier frequency (out of the very noisy and loud EM environment, where everyone is talking at the same time and all the time) and to receive the message, we use the concept of BP filter. Effectively, a BP filter serves as a very narrow “door” into the receiver and allows only one carrier frequency at ω0 to pass in, while all the other signals at different frequencies are suppressed by the BP transfer function, i.e. they are rejected at the entrance. Instead of building one receiver equipment for each possible frequency, which is practically the impossible task, we developed concept of tuneable components. For technological reasons, first tuneable component that was developed for the use in RF equipment was a rotating metal plate capacitor, Fig. 17.7. Centre frequency ω0 of an ideal LC resonator is set by the two component values ω0 = ∞ 1 LC (17.59) where for a given L = 2.533 nH it follows that f 0 (C = 1.3 nF) ⇒ 87.7 MHz, while f 0 (C = 857.345 pF) ⇒ 108 MHz. Therefore, this tuneable capacitor is suitable for the use in FM radio receivers designed for North American region. 5.13. We are now ready for drill design problems, in this example we design LC resonator using only components found on the shelf. Typically, the inductor must be designed specifically for the intended application, thus by using L = 2.533 nH we need to connect in parallel C = 100 pF. This example is a short exercise in using the available resources to design required circuit. To create the equivalent C = 100 pF we consider series/parallel combinations of available capacitors, (a) C = C1 + C2 + C3 = 10 nF + 40 nF + 50 nF = 100 pF, i.e we connect the four capacitors in parallel. Fig. 17.7 Problem 5.12: tuneable capacitor using rotating metal plates with air gap in between each pair of capacitive plates 17 Electrical Resonance: Solutions 177 (b) In this case we find that serial combination of the four capacitors gives C = 100 nF as 1 1 1 1 1 1 1 1 1 = + + + (17.60) + + + = C C1 C2 C3 C4 200 nF 400 nF 0.5 µF 2 µF (c) Combination of capacitor C1 in parallel with series connection of C2 and C3 gives the right value of C = 100 nF as C = C1 + C2 ||C3 = 70 nF + 60 nF × 60 nF 60 nF + 60 nF (17.61) and we have one capacitor to spare. It is common practice to “hack” into prototype boards and make unusual component combinations until the circuit being developed is completely debugged and ready for volume production. 5.14. This example is a simple drill design problem to demonstrate sensitivity of Q factor relative to the quality of wire and the resonant frequency. In this example we design LC resonator using only components found on the shelf. With the given inductor of L = 2.0 nH and the given wire resistance, it is straightforward to calculate Q factor as 2λ × 10 × 106 Hz × 2 nH ωL = = 40 r λ × 10−3 π ωL 2λ × 100 × 106 Hz × 2 nH Q= = = 400 r λ × 10−3 π Q= which illustrates dependance of Q factor, and therefore the bandwidth, versus frequency. This relation is linear through the inductor impedance Z L = jω L, while Q factor is inverse function of the wire resistance, which is also correlated to the inductance value, and all together affecting the bandwidth. That is why the inductors are designed specifically for the given frequency and intended signal bandwidth. 5.15. Given high enough frequency even a single piece of wire starts behaving as a complicated RLC network. Existence of both inductive and capacitive elements implies unavoidable condition for resonance. To our surprise, given a stimuli at the right frequency a single piece of metal becomes LC resonator. This phenomena is refereed to as self-resonance. Due to this effect, at very high frequencies it is not possible to used discrete components, instead designers exploit electrical properties of various metallic shapes to design passive networks, albeit for very narrow band of operation. In this example, we explore operation of a realistic inductive device, Fig. 17.8, and effects of approaching the self-resonating mode. First, for the given data let us establish the self-resonant frequency f L0 of this coil as 1 f L0 = = 71.2 MHz 2λ 1 µH × 5 pF 178 17 Electrical Resonance: Solutions Fig. 17.8 Problem 5.15: RLC model af a realistic inductor and, its Q factor is 2λ × 25 MHz × 1 µH = 31.4 5π Q= With the self resonant frequency known, under normal circumstances of operation, we assume that the operational frequency is at least one decade (i.e. ten times) below the self resonant frequency. Otherwise, we must use effective values for the realistic inductance, which are derived from model in Fig. 17.8 as 1 1 RL ∼ + jω C L = + j ω CL − YL = R L + jω L (ω L)2 ωL 2 RL 1 − ω LC L = = ∗(Y L ) + j √(Y L ) −j 2 (ω L) ωL (17.62) where, effective resistance Reff and inductance L eff of this inductor are found from the real and imaginary parts of (17.62) respectively, as Reff = (ω L)2 RL and L eff = L L = 2 1 − ω LC L 1 − (ω/ω0L )2 (17.63) which we use to calculate L L eff = 1− f f L0 2 = 1− 1 µH 25 MHz 71.2 MHz 2 = 1.14 µH At the circuit resonance frequency ω0 the dynamic resistance R D of the LC tank in Fig. 17.8 is (17.64) R D = Q ω0 L In the same time, dynamic resistance R D of the equivalent circuit is described using effective values (17.65) R D = Q eff ω0 L eff 17 Electrical Resonance: Solutions 179 Due to equivalence of these two circuits it follows that Q ω0 L = Q eff ω0 L eff ∴ 2 ω 2 Q eff = Q 1 − ω LC L = Q 1 − ω0 (17.66) (17.67) where Q = ω0 L/R as defined in (17.4). We now find the numerical value Q eff = Q 1 − f f L0 2 = 31.4 1 − 25 MHz 71.2 MHz 2 = 27.5 This result shows that effective Q factor Q eff of realistic LC tank decreases with increased frequency and the difference relative to the ideal resonator model. 5.16. In this short drill problem we explore behaviour of a resonant RLC circuit that is counterintuitive relative to behaviour of passive circuits studied in linear electronics. At f = 10 kHz we have, X L = 2λ × 10 kHz × 3 mH = 188.5π 1 = 159.2π XC = 2λ × 10 kHz × 100 nF Z = R 2 + (X L − X C )2 = 302 + (188.5 − 159.2)2 π = 41.9π We note that at 10 kHz the serial RLC circuit looks more inductive. At f = 5 kHz we have, X L = 2λ × 5 kHz × 3 mH = 94.2π 1 = 318.3π XC = 2λ × 5 kHz × 100 nF Z = R 2 + (X L − X C )2 = 302 + (94.2 − 318.3)2 π = 226.1π We note that at 5 kHz the serial RLC circuit looks more capacitive. Obviously, the results are on two opposite slopes of RLC frequency transfer curve, thus the resonant frequency must be somewhere in between, at f 0 = 9188.8 Hz to be more precise. 5.17. Arguably the most important transfer function in communication electronics, an LC resonator’s band-pass (BP) transfer function, contains all relevant information about the resonator and its operation, thus practicing visual inspection of the BP plot and extracting the information is very important. Wireless communication is based on the idea that a single tone high-frequency signal is used as a carrier of the wanted information, i.e. speech, music, video. The problem is that a single tone by itself does not carry much information, aside from 180 17 Electrical Resonance: Solutions revealing the very existence of the signal source. Thus, the single tone communication systems are limited to on/off coding mechanism, Morse code for instance. On the other side the human speech consists of a number of single-tone signals organized in a very specific manner so that our hearing system and brain interpret them as an articulate and meaningful message. In the case of the human hearing system, typically a healthy person can distinguish tones from 20 Hz to 20 kHz, i.e. the human hearing system covers frequency bandwidth of approximately B = 20 kHz . The human speech, however, is synthesized by using frequencies in the range of approximately 300 Hz–3.4 kHz, which explains why a phone was relatively easy to invent. It becomes more difficult to transmit classic music that stimulates the full hearing range, or digital video signal that requires about 6 MHz wide transmission channel. We already learned about role of a noise in signal transmission mechanism and concluded that in order to increase the communication distance we must keep the noise level at the minimum. Consequently, we must design bandwidth of the resonator to allow passage of only desired frequencies, and in addition the bandwidth must be centred around the carrier frequency f 0 , as in Fig. 5.5. By inspection of the given data and frequency plot, we write B ∓ f 2 − f 1 = (460 − 450) kHz = 10 kHz (17.68) which is more than needed for a human speech transmission, but not enough for Hi–Fi reproduction of classical music. By definition, Q factor is ratio of the centre frequency f 0 and bandwidth B Q∓ f0 455 kHz = = 45.5 B 10 kHz (17.69) This value is reasonable and relatively easy achieved by today’s technology. In contrast, for the same bandwidth specification, cell phone communication system that works at 2.4 GHz would require a resonator with Q∓ 2.4 GHz f0 = = 240,000 B 10 kHz (17.70) which is very difficult (but not impossible) to manufacture, and it is good to note that in this case it is not possible to use the traditional LC resonator due to its wire resistance. For this reason, LC resonators can achieve Q factors in the range of up to a hundred, while quartz oscillators are manufactured with Q factors in the range of ten thousand to a million. If a C = 100 pF capacitor is available then, in order to set the resonant frequency at 455 kHz, f0 = 2λ 1 ∞ LC ∴ L= 1 = 1.223 µH (2λ f 0 )2 C (17.71) 17 Electrical Resonance: Solutions 181 inductor must be used in this design. Obviously, because the B > 0 it means that this inductor is realistic, thus its internal resistance is found from the Q factor as ωL R ωL 2λ × 455 kHz × 1.223 µH = = 76.877 mπ Q 45.5 (17.72) Based on these specifications, designer now can choose type of metallic wire, the number of turns, and other geometrical parameters of this inductor. Q= ∴ R= 5.18. In the textbook we find that series and parallel RLC network are different in their effective impedances presented to the rest of the circuit. For the power matching purposes, one configuration is chosen over the other. In this drill example we practice to recalculate serial to parallel resonator parameters. The two networks must have the same Q factor, which is found by definition as QS = XS 1 1 = = =2 RS 2λ C S R S 2λ × 7.95 pF × 10 π (17.73) then it is straight forward to use the conversion formulas derived in the textbook and to we write R p = Rs (1 + Q 2 ) = 10 π (1 + 22 ) = 50 π 1 1 1 1 + 2 = 25.024 π X p = Xs 1 + 2 = Q 2λ × 1 GHz × 7.95 pF 2 ∴ C P = 6.36 pF Chapter 18 Matching Networks: Solutions Although there is number of different ways to design matching networks of various levels of complexity, in this book we study Q-matching design methodology based on a single-stage passive matching network topology that uses only two components, an inductor and a capacitor. This most basic matching network topology is also known as L-network, where one of the two components is connected either in series and the other one in parallel with the source and load respectively, or vice versa. Aside from its simplicity, this methodology does not require sophisticated numerical simulators, instead hand calculations are sufficient. More elaborated two-stage variant of the topology based on the use of three components is commonly refereed to as either ω or T -network, depending how the three components are oriented. Solutions: 6.1. Filter theory and matching network design procedures are very well developed and they are one of a few examples where the design flow is actually very well defined and straightforward to implement. In this book we consider only L-shape passive LC matching networks, and there are therefore only four basic cases of the network topology to learn. The source impedance Rs is either greater or smaller than the load impedance R L and, at the same time, connection between the source and the load terminals must be either DC or AC. In this example, we consider the case of load resistance being larger then the source resistance, R L > Rs with the additional requirement to maintain DC connection between the source and load impedances. Therefore, to reach “the middle ground” impedance of the source side R S resistance needs to be increased (i.e. an impedance X S must be added in series to Rs ) while, at the same time, load side resistance R L must be reduced (i.e. an impedance X p must be added in parallel with R L ). In accordance to notation in Fig. 18.1 and Q-matching technique (where Z s = Z ∗p ) we design the LC matching network by executing the following steps: R. Sobot, Wireless Communication Electronics by Example, 183 DOI: 10.1007/978-3-319-02871-2_18, © Springer International Publishing Switzerland 2014 184 18 Matching Networks: Solutions Fig. 18.1 Problem 6.1: Typical case of non matched source and load resistances, Rs < R L 1. Calculation of Q factor: RL −1= Rs Qs = Q p = 50 λ −1=3 5λ 2. Calculation series impedance X s that needs to be added to the source side: Qs = Xs Rs ∴ X s = Q s Rs = 3 × 5 λ = 15 λ 3. Calculation parallel impedance X p that needs to be added to the load side: Qp = RL Xp ∴ Xp = 50 λ RL = = 16.667 λ Q 3 4. The above calculation of two complex impedances X s and X p still does not say which component (i.e. either L or C) should be used to implement each of them. At this point of design procedure, there are two possible solutions and both are equally valid if there are no additional constrains. In this example, the given condition is that there must be DC connection between the source and load, which means that X s component must be able to conduct DC, in other words an inductor must be used. Once X s is set as an inductor, X p must be set to the opposite, i.e. it must be a capacitor. With this additional constrain, the solution for matching network becomes unique and the actual component values at 10 MHz are calculated as X s = 15 λ ∴ X p = 16.667 λ Xs = 238.732 nH 2π × 10 MHz 1 Cp = = 954.9 pF 2π × 10 MHz × X p Ls = ∴ 18 Matching Networks: Solutions 185 and the final schematic of the solution is shown in Fig. 18.2. In the following examples, we illustrate some other possible cases of relationships between the source and load impedances. 6.2. Matching of two real resistance should be solved first when working on design of matching networks, i.e. all complex impedances should be added later as shown in the following examples. If the case of source resistance being greater than the load resistance, Fig. 6.1 (right), the design procedure is still as same as in problem 6.1, with the only difference that greater resistance is used as a reference to calculate the Q factor, In order to reach “the middle ground” this time impedance of the source side R S needs to be reduced (i.e. an impedance X p must be added in parallel to Rs ) while, at the same time, load side resistance R L must be increased (i.e. an impedance X p must be added in series with R L ). According to notation in Fig. 18.3 and Q-matching technique we design the LC matching network by executing the following steps: 1. Calculation of Q factor: Qs = Q p = Rs −1= RL 50 λ −1=3 5λ Fig. 18.2 Solution 6.1: Matching network for the case of Rs = 5λ < R L = 50λ at f = 10MHz Fig. 18.3 Problem 6.2: Typical case of non matched source and load resistances, Rs > R L 186 18 Matching Networks: Solutions 2. Calculation parallel impedance X p that needs to be added to the source side: Qp = Rs Xp ∴ Xp = 50 λ Rs = = 16.667λ Q 3 3. Calculation series impedance X s that needs to be added to the load side: Qs = Xs RL ∴ X s = Q s R L = 3 × 5 λ = 15 λ 4. The above calculation of two complex impedances X s and X p is valid for two possible solutions, depending whether AC or DC connection should be maintained through the matching network. In this example, the given condition is that there must be AC connection between the source and load, which means that X s component must be able to conduct AC, in other words a capacitor must be used. Once X s is set as capacitor, X p must be set to the opposite, i.e. it must be an inductor. With this additional constrain, the solution for matching network becomes unique and the actual component values at 10 MHz are calculated as X p = 16.667 λ X s = 15 λ ∴ Xp = 265.258 nH 2π × 10 MHz 1 Cs = = 1.061 nF ≈ 1 nF 2π × 10 MHz × X s ∴ Lp = and the final schematic of the solution is shown in Fig. 18.4. In the following examples, we illustrate some other possible cases of relationships between the source and load impedances. 6.3. In this short drill example we practice conversion from series to parallel matching network configuration by implementing formulas already derived in the textbook. Series inductor impedance at 10 MHz is calculated as X s = π L S = 2π × 10 MHz × 238.732 nH = 15 λ which is followed by calculation of serial Q as Fig. 18.4 Solution 6.2: Matching network for the case of Rs = 50 λ > R L = 5 λ at f = 10 MHz (18.1) 18 Matching Networks: Solutions 187 Qs = Xs =3 Rs (18.2) Therefore, serial to parallel conversion formulas give: R p = Rs (1 + Q 2 ) = 50 λ 1 X p = X s (1 + 2 ) = 16.667 λ Q ∴ L p = 265.258 nH 6.4. One of the consequences of having non-equal source and load impedances is that part of the incoming signal energy is reflected back at the interface point and never reaches the load. This situation is very similar to a sun beam entering water in a lake where part of the incoming light energy is reflected at the air–water surface while part of the incoming light energy is absorbed in the water. Depending upon how large is the mismatch between the source and load impedances, the incoming energy is reflected to the various degrees, form none to all, and therefore the received signal is weekended accordingly. In this example we illustrate quantitative measure of signal reflection at the interface of source and load impedances. By definition mismatch coefficient φ is calculated as: φ = Zs − Z p Zs + Z p One way to compare Z s and Z p impedances is to transform (X p , R L ) parallel network into its equivalent serial network (which is not rally necessary, the existing parallel configuration gives the same result), and then to confirm that they are actually matched at the given frequency. With reference to Figs. 18.1 and 18.2, by looking into the source side impedance to the left it is straightforward to write by inspection Z s = Rs + j X s = (5 + j 15) λ (18.3) By looking into the load side impedance to the right, and after converting the parallel R L , C impedances into their equivalent series impedances (Rs (R p ), X s (X p )) by using the transformation formulas as R p = Rs (1 + Q 2 ) ∴ 1 ) Q2 ∴ X p = X s (1 + Rp 50 λ = = 5λ (1 + Q 2 ) (1 + 32 ) Xp 16.667 λ = = 15 λ X s (X p ) = 1 (1 + Q 2 ) (1 + 312 ) Rs (R p ) = 188 18 Matching Networks: Solutions we write expression for the equivalent series network at the load side as X p = (5 − j15) λ (18.4) In other words, impedances on both sides are equal |Z s | = |Z p | = 15.811 λ. Therefore, at matching frequency of 10 MHz, from (18.3) and (18.4),we have, φ = Zs − Z p (15.811 − 15.811)λ = 0 = ∞ dB = Zs + Z p (15.811 + 15.811)λ which, again by definition, gives the mismatch loss ML as ML = 1 = 1 = 0 dB 1−φ2 In other words, mismatch loss ML = 0dB obtained in this case indicates perfect loss– less matching that is valid only if inductor L s = 238.732 nH and capacitor C p = 954.9 pF components are used at 10 MHz. At all other frequencies, the calculated series/parallel impedances map into different component values. 6.5. As same as any other passive network that contains RLC components, matching network is a narrow–band circuit whose bandwidth is determined by the RLC component values. In the first approximation we assume that the frequency response curve of the matching network is symmetrical, i.e. equivalent to an ideal LC resonator, which in reality is not the case. In order to find the network’s impedance, first step is to transform the series part of the network (i.e. at the source side) into its equivalent parallel network, see example 6.3, which gives component values of R p = 50 λ and L p = 265.258 nH. After this series to parallel transformation, the equivalent matching network looks as in Fig. 18.5 (right). A quick check confirms that the equivalent parallel LC resonator indeed resonates at Fig. 18.5 Solution 6.3: Conversion of the series R–L subnetwork portion of the matching network into its equivalent parallel R–L subnetwork, for purposes of calculating the overall bandwidth 18 Matching Networks: Solutions f0 = 2π 1 √ LC = 2π 189 1 265.258 nH × 954.9 pF ≈ 10 MHz (18.5) which means that the LC resonator’s dynamic impedance R D is infinite, hence the network resistance is determined only with the two real resistors Rs and R L in parallel. In other words, the equivalent resonator’s resistance R = 50 λ||50 λ = 25 λ at the resonant frequency. Resonator bandwidth can now be estimated by using either inductor or capacitor impedance (they are equal at the resonant frequency) versus the equivalent parallel resistor value to calculate Q factor, and therefore the bandwidth. At f 0 = 10 MHz, Q= 25 R ≈ 1.5 = XC 16.667 ∴ BW3dB = 10 MHz f0 = = 6.667 MHz Q 1.5 Once again, in this approximated estimate it was assumed that the matching network is symmetrical, which is not the case. In reality, in this case the matching network has BW3dB a bit wider that estimated in this example. For more precise estimate, however, we must use a numerical simulator. 6.6. Once the components of the matching network are fixed, effectiveness of the network is tied to the particular frequency that was used in the design process. As demonstrated in example 6.4, in the ideal case mismatch losses are reduced to ML = 0 dB. In reality, that is impossible goal to reach because of finite bandwidth required by realistic signals (as oppose to a single tone signal) and also because of the component value variations, thus specifications of realistic systems include allowed mismatch losses levels for the given communication standard. For the sake of argument, in this example we examine behaviour of matching network in Fig. 18.2 if a signal whose frequency bandwidth B is in 8 to 12 MHz range, which is centred around the carrier frequency of f 0 = 10 MHz. On the lower side of the frequency spectrum at f = 8 MHz the impedance values are: X s = π L S = 2π × 8 MHz × 238.732 nH = 12 λ ∴ Q= Xs 12 λ = 2.4 = Rs 5λ As earlier, it is more practical to convert the parallel RC connection on the load side into its serial connection and derive Xp = 1 1 = = 20.833 λ πC 2π × 954.931 pF and then convert parallel RC network on the load side into its equivalent series network 190 18 Matching Networks: Solutions 50 λ RL = = 7.396 λ 2 (1 + Q ) (1 + 2.42 ) Xp 20.833 λ = = 17.751 λ X s (X p ) = 1 (1 + Q 2 ) (1 + 2.41 2 ) Rs (R L ) = That is, the left side and right side looking impedances are Z s = (5 + j 12) λ ∴ |Z s | = 13 λ Z p = (7.936 − j 17.751) λ ∴ |Z p | = 19.231 λ which clearly illustrates the mismatch between the source and the load side of the matching network. Consequentially, there is a portion of signal energy being reflected, which at 8 MHz is quantified as φ = Z p − Zs (19.231 − 13)λ = −0.193 = −14.275 dB = Zs + Z p (19.231 + 13)λ And, therefore mismatch loss is ML = 1 = 1.039 = 0.331 dB 1−φ2 Repetition of the above procedure at 12 MHz gives the following results ∴ |Z s | = 18.682 λ Z s = (5 + j 18) λ Z p = (3.582 − j 12.894) λ ∴ |Z p | = 13.382 λ which at 12 MHz translates into φ = −0.165 = −15.636 dB and ML = 1.028 = 0.241 dB Moving away from the matching frequency results in the matching network becoming less effective, a portion of the incoming signal power is reflected back and dissipated while creating destructive interference with the incoming signal wave. Direct consequence is that amplitude of the signal is lowered, thus SNR reduced. 6.7. Realistic sources and loads are never purely resistive except only as an approximation at very low frequencies. Very elegant technique for designing matching networks is to include any pre-existing parasitic either inductance or capacitance into the calculations and “absorb” them either partially or completely into the required LC matching components that are calculated for the case of pure resistive source driving pure resistive load. This technique is based on the fact that dynamic resistance R D of a pure LC resonator at resonant frequency is resistive and ideally infinite, thus it does not change value of the total parallel resistance of the matching network. 18 Matching Networks: Solutions 191 In this example, the source is not ideal and it contains the inductive component. In the first step of the matching network design procedure, the parasitic inductance is ignored and network in Fig. 6.2 (left) is assumed to be as shown in Fig. 6.1 (left), (because in this problem we use the same values for source and load resistances). In problem 6.1 we already found values of the matching network for the case of a 5 λ resistive source driving a 50 λ resistive load as L s = 238.732 nH and C p = 954.9 pF. However, the source already contains 138.732 nH parasitic inductance. In this case we are lucky because of the DC connection requirement, thus in order to achieve L s = 238.732 nH all that is needed is to add additional X s = 100 nH inductance in series, Fig. 18.6, and therefore completely absorb the parasitic component into the required inductive value. 6.8. In this example we demonstrate two of the possible ways to treat the parasitic components, such as C L in Fig. 6.2 (left). Both methods are based on the idea that adding an inductor in parallel with capacitor creates LC resonator whose dynamic resistance R D is infinite at the resonant frequency, and therefore it does not contribute to the total equivalent parallel resistance. The two possible ways to “resonate out” the parasitic capacitance are: 1. to resonate out the full value of the parasitic component; and, 2. to resonate out only the “excess” part of the parasitic value. Both approaches are equally valid and the final decision is made by considering additional criteria for instance, the number of used components or other practical aspects of the design. Let us take a look at these two possible solutions and illustrate the point. 1. In the first approach, we take the whole value of the parasitic capacitance and use it to create LC resonator. By “resonating out” the whole value of the C L parasitic capacitor, the given network is reduced to the problem of matching two resistances as in problem 6.1. Hence, an LC resonator at f 0 = 10 MHz connected in parallel with the full value of the parasitic requires inductor whose value is calculated as f0 = 2π 1 √ L CL Fig. 18.6 Solution 6.7: Inductive source driving resistive load ∴ L= 1 = 253.303 nH (2 π 10 MHz)2 1 nF 192 18 Matching Networks: Solutions Therefore, at 10 MHz the capacitor will “disappear” because the newly build LC resonator is transparent at this particular frequency (i.e. 10 MHz tone is not attenuated at all). After addition of the inductor, the current problem is reduced to matching 5–50 λ resistor, hence we already know the solution, L s = 238.732 nH and C p = 954.9 pF, Fig. 18.7. 2. The second possibility is, after already solving the case of the 5–50 λ matching case, to resonate out only the excess portion of C L and used only C0 = 1 nF − 954.9 pF = 45.1 pF and inductor L 0 = 5.616 nH to create ideal “virtual” L 0 C0 resonator at 10 MHz. Thus, we temporarily imagine that the C L = 1 nF capacitor consists of two capacitors in parallel, i.e. C L = 954.9 pF + 45.1 pF, where at 10 MHz the 45.1 pF portion of the load capacitor becomes part of the L 0 C0 resonator alongside with L 0 , Fig. 18.8. The resulting circuit is more elegant engineering solution because it uses only two inductive components, as oppose the solution in Fig. 18.7 that needs two inductive plus one capacitive component. Note that, for the convenience, in the last several examples we keep reusing the 5–50 λ matching case, it is understood however that R S and R L resistances can take any values. Still, regardless of the component values, the single–stage matching network design flow presented here stays the same. What is more, now it should not be difficult to also work out other cases including R S > R L , AC or Fig. 18.7 Solution 1 for 6.8: Inductive source driving resistive load Fig. 18.8 Solution 2 for 6.8: Imaginary splitting of the C L capacitor for the purpose of partial resonating out 18 Matching Networks: Solutions 193 DC connection, and with various combinations of (L, C) parasitic components on each side. 6.9. As simple and elegant as they are, single stage L–shaped matching networks have limited number of degrees of freedom. Once the components of a single stage matching network are calculated based on Q matching requirements, bandwidth of the matching network is also set and can not be changed anymore. In order to add more degrees of freedom, two or more stages network is required. In the following examples we explore some of simpler cases of two stage matching network design by using a technique that enables us to reduce the problem of two-stage to the problem of one-stage network design repeated twice. This design technique is based on concept of the intermediate (i.e. “ghost”) resistance R I N T that serves as imaginary load resistance for the first single stage and imaginary source resistance for the second single stage of the two stage matching network. However, there are two possibilities, shown below, depending how R I N T is related to Rs , R L resistances, Fig. 18.9. When one of the design goals is to decrease bandwidth of the matching network then ghost resistance R I N T should be set outside of the resistance range set by Rs and R L values, i.e. either R I N T < min(R S , R L ) or R I N T > max(R S , R L ). Both possibilities are equally valid, and it is convenient to arbitrarily set R I N T value such that the resulting Q factor is a round number. Then simply proceed with the design of two single stage LC matching networks, the first one matching Rs to R I N T , and the second one matching R I N T to R L . Fig. 18.9 Problem 6.9: Two stage matching network design concept, Z out = R I N T = Z in 194 18 Matching Networks: Solutions 1. Note that the following two calculated Q values are not equal. However, bandwidth is limited by the higher Q, which should be used in the design. For the given data let us arbitrary set R I N T to a value greater than R L , i.e. R I N T > max(5 λ, 50 λ) = 250 λ, Fig. 18.9 (bottom), which produces round numbers for Q factors Q1 = RI N T − 1 = 7.0 Rs Q2 = and RI N T − 1 = 2.0 RL Obviously, Q 2 = 7 is the one limiting the total bandwidth. 2. First stage: Matching 5–250 λ results in X s1 = Q 1 Rs = 35 λ and X p1 = RI N T = 35.714 λ Q1 3. Second stage: Matching 250–50 λ results in X s2 = Q 2 R L = 100 λ and X p2 = RI N T = 125 λ Q2 4. General solution, i.e. with no specified types of components is in Fig. 18.10. Each complex value is either “+” or “−” depending whether the component is an inductor or capacitor respectively. Numerical AC simulation would confirm decrease in the bandwidth. In addition, if possible then the two parallel components should be merged into one to reduce the number of components, thus this general solution would become a T–network. 6.10. When one of the design goals is to increase bandwidth√of the matching network then the intermediate resistance value is set to R I N T = Rs R L , i.e. geometrical mean of the two termination resistances. Then, again the two-stage design problem is reduced to design of two single-stage LC matching networks, the first one designed to match source resistance Rs to R I N T , and the second one to match R I N T to the load R L . Once again, resistance R I N T is not a real resistor component added in the Fig. 18.10 Solution 6.9: General solution that does not yet specify exactly the type, i.e. L or C, of the complex components 18 Matching Networks: Solutions 195 Fig. 18.11 Problem 6.10: Two stage matching network design concept, Z out = R I N T = Z in network, it is only a fictitious number used to set impedance of the inner node in between the two matching network stages (Fig. 18.11). For the given data given in this example, the design flow is as follows. 1. Calculation of the intermediate resistance: √ R I N T = R S R L = 5 λ × 50 λ = 15.811 λ (18.6) 2. Calculation of Q factor at R I N T looking left into the first stage, and looking right into the second stage: Q= Q= RI N T −1= RS RL −1= RI N T 15.811 λ − 1 = 1.470 5λ 50 λ − 1 = 1.470 15.811 λ 3. First stage (LP) components: matching 5–15.81 λ results in X S1 = Q Rs = 7.352 λ ∴ L S1 = 117 nH RI N T = 10.753 λ ∴ C P1 = 1.480 nF X P1 = Q where, series inductor and parallel capacitor are used to create LP filter. 4. Second stage (HP): matching 15.811–50 λ results in X S2 = Q R I N T = 23.250 λ RL = 10.753 λ ∴ X P2 = Q ∴ C S2 = 684.533 pF L P1 = 117.113 nH AC simulation of this matching network would confirm the increase in the bandwidth. With different requirement in terms of use of LP–HP sections other valid solutions can also be derived (Fig. 18.12). 196 18 Matching Networks: Solutions Fig. 18.12 Solution 6.10: Two stage matching network schematic 6.11. In this classical design problem, the design of a single stage LC matching network is straightforward, i.e. 1. Q–calculation: Rin −1= RA Qs = Q p = 2 kλ − 1 = 6.25 50 λ 2. Source side impedance calculation: Qs = Xs Rs ∴ X s = Q s Rs = 6.25 × 50 λ = 312.5 λ 3. Load side impedance calculation: Qp = Rin Xp ∴ Xp = 2 kλ Rin = = 320 λ Q 6.25 4. Calculation of the complex impedances X s and X p still does not say which component (L or C) should be used to implement them, Fig. 18.13. Both possible solutions are equally valid. Possible component values are (at f 0 = 665 kHz) depending whether AC or DC connection through the matching network is required: Fig. 18.13 Solution 6.11: General matching network solution 18 Matching Networks: Solutions 197 X s = 312.5 λ, ∴ L s = 74.79 µH or Cs = 765.9 pF X p = 320.0 λ ∴ C p = 747.9 pF or L p = 76.56 µH 6.12. One of the first problems to solve in RF receiver design process is to calculate components of the antenna-to-RF amplifier input matching network. Typically, the antenna presents itself to the amplifier as a voltage source V A with the internal resistance of R A = 50 λ. On the input side of the amplifier, impedance of infinitely large capacitance C0 is obviously zero at all frequencies, while input impedance of a FET device is assumed infinite. Thus, the input side of the amplifier presents itself to the antenna through node 1ias parallel connection of the voltage divider resistors Rin = R1 ||R2 , which is also connected in parallel with capacitor C2 , Fig. 18.14. Therefore, the original problem is reduced to the design of single stage matching network for the case of a 50 λ source driving (R1 ||R2 ) load with parasitic C2 capacitance. Equivalent resistance Rin at the amplifier input side, assuming infinite gate impedance of FET, is simply the voltage divider’s parallel resistnace (R1 , R2 ), i.e. Rin = R1 ||R2 (18.7) From the given condition that V (1) = 1/4 VD D and the laws of voltage divider, it follows that VD D 1 R2 = V D D R1 + R2 4 ∴ 3R2 = R1 ∴ R2 = 5.2 kλ = 1733.333 λ 3 (18.8) which after substitution into (18.7) gives Rin = 1300 λ. Therefore, the matching problem is reduced to matching the R A = 50 λ antenna resistance to the C2 ||Rin load impedance. It is now easy to calculate the matching network component values by first solving the case of 50–1300 λ resistance matching. Fig. 18.14 Problem 6.12: Equivalent schematic diagram of RF amplifier’s input stage connected to antenna through matching network 198 18 Matching Networks: Solutions 1300 λ −1=5 50 λ X s = Q s Rs = 5 × 50 λ = 250 λ Rin 1300 λ = 260 λ = Xp = Qp 5 Qs = Q p = Q = (18.9) Duo the required DC connection between the input and output terminals of the matching network, it follows that X s must be inductive and X p capacitive impedance. Therefore, X s = π L s = 250 λ 1 Xp = = 260 λ π Cp ∴ ∴ L s = 3.979 µH C p = 61.213 pF Obviously, because there is already C2 = 11.213 pF capacitor in parallel with the load resistance Rin , thus, an additional C = 50 pF must be added in parallel to make up for the required C p = 61.213 pF, which is the final solution for this problem, Fig. 18.15. 6.13. From the given data, we conclude that real source resistance of R S = 5 λ must be matched to real load resistance R L = 50 λ. In addition, there is series parasitic inductor at the source side, X S = 10 λ, and parallel parasitic capacitance at the load side, X L = 30 λ. At 10 MHz the source inductance L S is calculated as XS = π LS ∴ LS = 10 λ XS = = 159.155 nH π 2π × 10 MHz while the load capacitance C L is calculated as XL = 1 π CL ∴ CL = 1 = 530.516 pF 2π × 10 MHz × 30 λ Fig. 18.15 Solution 6.12: Schematic of matching network 18 Matching Networks: Solutions 199 Fig. 18.16 Solution 6.13: Schematic of matching network From example 6.13 we already know that matching network for the 5–50 λ case consists of L s = 238.732 nH and C p = 954.9 pF. Therefore, both the input side inductance and the load side capacitance should be absorbed into the solution for the matching network, which means only the differences to L s = 238.732 nH and C p = 954.9 pF values should be added, i.e. L = L s − L S = (238.732 − 159.155) nH = 79.577 nH C = C p − C L = (954.9 − 530.516) pF = 424.383 pF In this example, both parasitic inductance and parasitic capacitance are absorbed into the matching network components, Fig. 18.16. Chapter 19 RF and IF Ampliﬁers: Solutions By no means complete and self sufficient, the following tutorials should give enough guidance to the reader, so that the very difficult subject of analogue circuit design and analysis becomes more understandable and intuitively acceptable. Of course, analogue circuit design work is based on conceptual thinking developed in applied mathematics, thus the reader is encouraged to review mathematical techniques learned in the previous schooling and to use this subject as a playground to apply all those abstract concepts in real life engineering problems. Solution: 7.1. In this example we practice to quickly estimate impedances associated with the internal nodes of amplifier circuit simply by inspection and by knowing a few basic techniques. Circuit network in Fig. 7.1b demonstrates technique for estimating resistance Z out as seen by looking into (R1 , R2 ) voltage divider. It is simply Z out = R1 ||R2 . We reached this conclusion because there are only three components present in the network, resistors R1 , R2 and voltage source VCC . The internal resistance of a voltage source is zero, thus from the perspective of circuit’s internal resistance nodes G N D and VCC are shorted through the voltage source. Which leave only two resistors whose respective terminals are shorted together in parallel. Circuit network in Fig. 7.1c demonstrates technique for estimating the input impedance of a BJT transistor whose current gain ω is very large (thus, base current is i √ 0). We keep in mind the “magnification effect” of emitter resistor R E when it is looked through the base terminal. It is seen as magnified by ω factor, i.e. Z in √ ω R E (19.1) which makes the base terminal look like very high resistance (base resistance r B is much smaller then ω R E ). We note that resistance ω R E is not real resistance, instead it is only perceived resistance at the base node, which is caused by a feedback effect in this circuit (as derived in textbook), R. Sobot, Wireless Communication Electronics by Example, 201 DOI: 10.1007/978-3-319-02871-2_19, © Springer International Publishing Switzerland 2014 202 19 RF and IF Amplifiers: Solutions Circuit network in Fig. 7.1d demonstrates technique for estimating the output impedance as seen at BJT emitter node. First, between the output node and ground there is R E resistance. In addition, there is projected resistance due to resistors (R1 , R2 ) at the input side. We already found that the equivalent resistance of a voltage divider is R1 ||R2 , which is then projected to the emitter node by being scaled down by ω factor. Subsequently, the output resistance is found as Z out = R E || [ω (R1 ||R2 )] (19.2) which makes the emitter node look like very low resistance for large values of ω. 7.2. In this short drill example we practice the magnifying emitter resistance concept by solving the following equations RB ω = ω RE Rout = Rin ∴ ∴ RB ∴ R B = 10 kλ 100 100 kλ = 100 R E ∴ R E = 1 kλ 100 λ = This approximative technique is very useful to do mentally and quickly acquire a feeling for the circuit in hands. 7.3. From results found in the previous problems, by inspection we write: 1. Assuming the required Vth (B E) = 0 V, because I R2 > 0 then any positive value of R2 > 0 results in VB = Vth (B E) → 0 V, which is sufficient to turn base–emitter diode on. With the first condition for turning on BJT transistor, the second condition becomes simply that VC > VB → 0 V. 2. Assuming the required Vth (B E) = 1 V, then because VB = Vth (B E) = 1 V, it follows that R2 → VB = 1 kλ I R2 (19.3) which means that potential at collector node must be held at VC → VB → 1 V. As a reminder, the condition that base–collector diode must be reverse biased is satisfied even if terminals of the base–collector diode are shorted, i.e. at the same potential, thus we include the equality sign in the solution. 7.4. In this example we practice to apply and combine the previous results within a typical case of CE amplifier. 1. assuming ideal base–emitter diode, and by knowing both current and resistance at the emitter node, it is straightforward to find the emitter node potential as VE = I R E R E = 1 V. That means potential at base node must be VB → VE = 1 V. The biasing design problem is now reduce to finding scaling factor that is required to derive 1 V from the VCC = 10 V voltage source. Obviously, given that BJT base does not draw any current from the voltage divider, we write 19 RF and IF Amplifiers: Solutions 203 VB = VCC VCC R2 = R 1 R1 + R2 R2 + 1 ∴ R1 VCC = −1∞9 R2 VB (19.4) where, because VB → 1 V, the resistor ratio must be equal or less then nine. The transistor is turned on as long as VC → VB → 1 V. 2. assuming realistic base–emitter diode, procedure for finding the resistor ration stays the same, except that the base potential is shifted up to VB → 2 V. Therefore, VCC R1 = −1∞4 R2 VB (19.5) which illustrates reasoning behind values calculated for the voltage divider. As always, VC → VB → 2 V. 7.5. In the following drill examples we use the “back of the envelope” approach to to estimate bounds of various circuits parameters in isolation from all other aspects of circuit behaviour. By focusing at one aspect of circuit behaviour at the time, we learn to recognize them within a content and mentally evaluate their bounds. By inspection of Fig. 7.2d we recognize that the input AC signal (capacitor C stops DC component of the input signal) is connected to the gate terminal, while the output voltage is taken at the collector terminal of BJT transistor, thus the circuit a classical common–emitter (CE) amplifier. After carefully going through CE amplifier related derivations in textbook, with the given set of numerical data, and assuming that the transistor is set in the active gain mode we conclude that it is possible to estimate upper bound of CE amplifier voltage gain as the ratio of collector and emitter resistance, RC 10 kλ = 100 V/V = 40 dB = (19.6) Av ∞ RE 100 λ where the equality sign is valid only in the ideal case of infinite base resistance, that is equal collector and emitter currents, i.e. infinite current gain of BJT transistor. Thus, voltage in all realistic cases must be lower than 40 dB. By visual inspection of (19.6) one could make the following argument. Most often our design goal is achieve as high gain of an amplifier stage as possible, then why not choose either RC ≤ ≥ or R E ≤ 0 ? In both cases the voltage gain A V ≤ ≥ becomes immensely high. In short, it is not realistic approach because (19.6) is an overly simplified approximation, nevertheless exceptionally useful, that helps us to evaluate the upper gain limit by simply looking at resistances attached to collector and emitter terminals of the transistor. 204 19 RF and IF Amplifiers: Solutions Fig. 19.1 Solution 7.5: Illustration of expression (19.6) To illustrate the point, let us take a closer look what happens with the amplifier’s gain if we really push for RC ≤ ≥ or R E ≤ 0, which causes circuit in 7.2d to morph into the one in Fig. 19.1 (right). By setting RC = ≥ collector terminal becomes effectively disconnected from the positive power supply line. In other words, the collector current becomes IC = 0, which inevitably forces the gain to become zero, i.e. the transistor ceases its primary function. Therefore, everything else being equal, the increase of RC value causes the collector current to reduce, thus the gain must drop. The second choice of reducing R E initially appears more promising, until we eventually find by experiment that even if R E = 0 the voltage gain is still limited to a value much lower than the infinite gain. Obviously, there are some other aspects of the circuit that are not accounted for in (19.6) that limit the gain. At this moment we state only the first two most important reasons. First, inside a linear amplifier the maximum signal amplitude is limited by the power supply voltage, and second, in order to more accurately estimate the gain there is an additional internal resistance inside the transistor itself that must be included in (19.6), as we see in the following examples. 7.6. With the given set of numerical data we calculate the following circuit parameters one after the other. 1. From the given temperature it follows that transistor thermal voltage VT is VT = kT = 25.693 mV q (19.7) 2. From the given leakage current I S , base emitter voltage VB E , and the calculated thermal voltage VT , collector current IC is calculated as IC = I S VB E exp VT − 1 = 987.728 mA (19.8) 3. At the same time, from the calculated thermal voltage VT and collector current IC , small signal emitter resistance re is calculated as 19 RF and IF Amplifiers: Solutions 205 re √ 1 VT = = 26 λ gm IC (19.9) 4. From the given emitter resistance R E , input signal frequency f , and the parallel capacitance C, the total impedance Z E in the emitter branch is Z E = R E ||Z C = 1 ωC R E + ω1C RE (19.10) where, at f = 10 MHz capacitor impedance is calculated as Z C (1 µF) = 15.915 λ, and Z C (≥) = 0 λ. By substituting these two impedance values into (19.10), it follows that Z E (1 µF) = 15.912 λ Z E (≥) = 0 λ (19.11) 5. With the calculated impedances Z E in the emitter branch alongside the calculated small emitter resistance re , it is now possible to calculate more accurate voltage gain of CE amplifier, again as the ratio of collector and emitter impedances, as AV = RC re + Z E ∴ 10 kλ = 238.595 V/V = 47.553 dB 26 λ + 15.912 λ 10 kλ = 384.615 V/V = 51.7 dB A V (≥) = 26 λ A V (1 µF) = The above calculations demonstrate very important technique used in amplifier circuit design to control the small signal gain while maintaining desired biasing point. Fixed resistances in collector and emitter branches are used to set BJT biasing point, which also sets voltage to current gain gm as the first derivative of I (C) = f (VB E ) function (19.8) at biasing point (IC , VB E ). One of the implications of (19.6) is that reduction of R E leads into higher gain values. But, change of R E value also changes the biasing point, which controls the gain. Thus, this is classical case of the “chicken and egg” problem. Fortunately, biasing point is set by DC current/voltage values, while a signal is by definition time variance of the biasing point, i.e. it is an AC quantity. After recognizing these fundamental relations, it is now straightforward to come up with the solution that satisfies both requirements, one to keep the biasing point stable and the other to enable high gain of the signal. The solution is to provide two signal paths in parallel, one for DC and one for AC component of the signal. We already have devices that can be used to implement this idea, a resistor for DC path and a capacitor for AC path, Fig. 19.2. (Strictly speaking, AC path is provided by Z E ||R E .) 206 19 RF and IF Amplifiers: Solutions Fig. 19.2 Solution 7.6: Illustration of two–paths, DC and AC, enabled by addition of C E , with base–emitter diode resistance re explicitly shown Fig. 19.3 Solution 7.7: Illustration of CE voltage gain dependance versus emitter resistance This example illustrates how formal decoupling of DC and AC signal components enables very powerful practical technique for circuit design, as illustrated by the effects due to presence of emitter capacitor C E , Fig. 19.2. 7.7. To continue discussion from the previous examples, let us take a look what happens when there is no external R E resistor in the circuit, instead both gain and biasing points are set by relying on the internal emitter resistance, Fig. 19.3. In this example, there is only small emitter resistance to set the gain, which is derived by following the following reasoning. IC = VCC − Vout RC ∴ re = VT VT RC = IC VCC − Vout ∴ AV = RC VCC − Vout = = (100, 200, 392) V/V re VT when Vout = (7.5, 5, 0.2) V, which illustrates how voltage gain keeps changing with the output signal. If the output voltage swing becomes too high, the biasing conditions for setting mode of transistor operation may start falling apart, which pushes 19 RF and IF Amplifiers: Solutions 207 transistor first into its linear region, i.e. transistor changes its behaviour from the current source into voltage controlled resistor, and then eventually the transistor turns off. Recognizing this dynamic behaviour of a transistor is important for determining correct range of operation of an amplifier. 7.8. By now we realize that amplifier design using active components is very fine balancing act of looking for the best compromise among large number of competing parameters. That is why it is very difficult to formalize the analogue circuit design flow, similarly to what has been achieved with the digital circuit design. Instead, analogue circuit design is as much art as it is engineering, which sometimes makes analogue designers look like magicians. Nevertheless, fundamental physical principles always apply, and it is only matter of learning to recognize as many or them as possible within the content, and then to find the optimal solution. This particular skill is what clearly separates brilliant analogue from the rest. In this example we again evaluate relationship between the controlling VB E voltage and the output collector current IC . Two distinct base–emitter voltages force two distinct collector currents, assuming VB E ∗ VT , the approximative formula for collector current is valid, thus for two distinct biasing voltages IC1 IC2 IC2 IC1 VB E = I S exp VT VB E + πVB E = I S exp VT ∴ BE exp VB E +πV VT πVB E = = exp VT exp VB E VT which yields the following two numerical results πVB E = 18 mV ∴ πVB E = 60 mV ∴ IC2 √ 5 = 6.254 dB IC1 IC2 √ 11 = 21 dB IC1 which illustrates the need for keeping very tight control over VB E voltage, due to the exponential realtionship the output current changes too easily by factor two or more, which is not negligible by any measure. 7.9. Let us now start looking into important frequency dependent phenomena hiding inside active circuits. For instance, another way of seeing a pn junction of a diode is that in reality it is a capacitance. There are two electrodes, anode and cathode, at different potential and there is an isolation layer in between, depletion region, thus all elements of a capacitor are present. These internal parasitic capacitances are 208 19 RF and IF Amplifiers: Solutions Fig. 19.4 Solution 7.9: Illustration of Miller capacitance small and unavoidable because they are fundamental to the diode function. In the case of a BJT transistor, there is parasitic capacitance in between each pair of the three electrodes that are, due to their size, usually noticeable at higher frequencies. However, under special circumstances the collector–gate capacitance CC B becomes visible at much lower frequencies. Specifically, if the following three conditions are met 1. Parasitic impedance provides the return path from the output terminal back to the input terminal; 2. There is relatively large voltage gain (A V ∗ 1) between the input and output terminals; and 3. The output signal is inverted. then, due to the feedback action, the bridging impedance Z C is perceived at the gate node as (19.12) C M = (A V + 1) CC E which, for large voltage gain becomes significant. Similarly to the magnification effect of emitter resistance, Miller capacitance is not a real capacitor, however, it is perceived as real large capacitance when looking into the input terminal, Fig. 19.4. For the given numerical data, first estimated upper bound of voltage gain is AV = CM RC = 99 RE ∴ = (99 + 1) 1 pF = 100 pF (19.13) which illustrates the point. Due to this large capacitance, for realistic signal sources with finite source resistance the amplifier input side becomes frequency limited due to the RC product and therefore unavoidable creation of low–pass (LP) filter. 7.10. In this example we combine several of the previously learned concepts and demonstrate how their combined effect creates yet another limitation of an amplifier, Fig. 19.5. 19 RF and IF Amplifiers: Solutions 209 Fig. 19.5 Solution 7.10: Illustration of input side LP filter due to Miller capacitance CM By inspection of Fig. 7.2b , we write in the following order 1. Input impedance: 1 1 1 1 = + + Rin R1 R2 ω RE ∴ Rin = 950 λ (19.14) 2. Voltage gain: AV = RC = 99 RE (19.15) 3. Miller capacitance: C M = (ω + 1) CC B = 100 pF φ (19.16) 4. Bandpass of LP filter crated by Rin ||C M is found by ω 3 dB = 1 Rin C M ∴ f 3 dB = 1 2φ 950 λ × 100 φ pF = 5.263 MHz The calculated LP bandwidth is clearly not infinite and, what is most important, definitely not suitable for amplification for RF signals that are routinely above 10 MHz. For this reason, it is common to use a two stage cascoded amplifier topology instead of the single stage CE emitter, where the second common–base (CB) stage is used mostly to cancel the first of three conditions required to by Miller effect, i.e. to break the direct feedback from the output back to the input node. 7.11. In this example we practice another typical case of Miller capacitance limiting performance of an RF amplifier. Here, there is existing inductive component connected to the base node. From the previous examples we already learned to recognize existence of Miller capacitance, which in this case becomes parallel to the existing inductor L. By definition, an LC resonator is crated whose centre frequency is calculated from the given data as 210 19 RF and IF Amplifiers: Solutions ω0 = ∓ 1 L CM ∴ f0 = 2φ 1 2.533 µH × 100 pF = 10 MHz (19.17) which gives us illustration how BP filter is created at the input side of an RF amplifier based on CE single stage. From the given data we conclude that this amplifier should be used for a 10 MHz signal, however unless we have data about the inductor’s resistance we can not estimate working bandwidth. 7.12. Concept of Miller capacitance is very important for circuit design, and in this drill example, technically, same question is asked as in the previous example 7.11 where we learned how collector–base capacitance CC B is reflected into base node in parallel with the existing inductance L so that LC resonator is formed with resonant LC frequency set as 1 (19.18) f0 = ∓ 2φ L C M where, the Miller capacitance, which is calculated as C M = A V C = 100 pF Therefore, we conclude that L = 1 µH inductance should be used if all no other capacitance is present in the circuit. 7.13. In this example we illustrate how frequency characteristics of an amplifier becomes complicated as we include more and more details at the same time. In doing hand analysis we trade the numerical accuracy for simplicity and the quick estimate of the circuit behaviour, whose numerical accuracy can be improved by using simulators. From the previous examples we already estimated the input side resistance of the amplifier as Rin = Rin ||R E ||ω R E = 950 λ, and that Miller capacitance is C M = 100 pF. With the input signal generator being AC coupled through capacitor C = 1 µF, the input side network is reduced to the network in Fig. 19.6. Considering that C ∗ C M , it is reasonable approximation to account only for C at low frequencies and to account only for C M at high frequencies. Fig. 19.6 Solution 7.13: Illustration of input side HP–LP filter due to the coupling capacitor C and Miller capacitance C M 19 RF and IF Amplifiers: Solutions 211 At low frequencies capacitor C and input resistance Rin form a high pass (HP) filter whose −3 dB point is approximately (Miller capacitance is neglected) f 3 dB (HP) √ 1 = 167.6 Hz 2φ Rin C (19.19) while, at high frequencies Miller capacitance C M and input resistance Rin form a low pass (LP) filter (impedance of capacitor C is very low) and its −3 dB point is approximately 1 = 1.675 MHz (19.20) f 3 dB (L P) √ 2φ Rin C M In both cases the frequencies are underestimated, for the HP filter the actual −3 dB point is only slightly higher (about 188 Hz), however, the error at larger for the LP −3 dB point because Z C is in parallel with Rin , which pushes the bandpass of LP filter above 1.675 MHz. It is an iterative procedure to calculate the exact numerical values, for example simulator finds f 3 dB (L P) √ 14 MHz. Nevertheless, this quick approximative analysis correctly revealed reasons for the trend of this amplifier that is band–limited between f 3 dB (HP) and f 3 dB (L P), i.e. it rejects both low frequencies from DC to about a couple of hundred hertz, and high frequencies above several megahertz. 7.14. A general amplifier circuit needs two input and two output terminals. In addition, the signal being amplified can take the form of either voltage or current, therefore there are four possible types of ideal amplifier functions: 1) voltage input – voltage output, referred to as voltage amplifier whose gain is A V = vout/vin ; 2) current input – current output, referred to as current amplifier whose gain is Ai = iout/iin ; 3) voltage in – current out, referred to as transconductance amplifier whose gain is G m = i out i in iout/vin ; and 4) current input – voltage output, referred to as transresistance amplifier whose gain is A R = vout/iin i in . We keep in mind that these four ideal mathematical amplifying functions are impossible to implement separately from each other. To various extents, all realistic amplifiers embody each of the four ideal amplifying functions at the same time. Therefore, classification of realistic amplifiers is based solely on which one of the four functions is dominant. Conclusion about the dominant role is reached by observing whether the input/output terminals are better suited for voltage or current signals, which brings us back to the study of voltage/current dividers and their properties. And, of course, voltage/current dividers are modelled by ideal voltage/current source whose main functionality is related to their terminal impedances; a voltage source presents low (ideally zero) impedance, while a current source presents high (ideally infinite) impedance at their respective output terminals. A transistor is very versatile device that can amplify both voltage and current signals, Furthermore, transistor is modelled as a three terminal active device, which means if transistor is to be used as an amplifier then one terminal must be shared between the input and output sides of the single stage amplifier network. Thus, there are three possible configurations that a 212 19 RF and IF Amplifiers: Solutions Fig. 19.7 Solution 7.14: Common base amplifier schematics single stage amplifier can have, each named after the sharing terminal, i.e. common emitter (CE), common base (CB), and common collector (CC). Each of the three configuration has its own set of characteristics in terms of input/output impedance and voltage/current gain. By combining single stage amplifiers it is possible to implement any of the four fundamental amplifying functions with high level of fidelity. In this example we practice how to quickly evaluate common–base (CB) amplifier. Assumption that ω = ≥ implies approximations I B = 0 and IC = I E , which simplifies the analysis very much. By inspection of Fig. 19.7, we note that the base terminal is connected to the small signal ground (details of biasing setup are not shown in Fig. 19.7), while both collector and emitter terminals are accessible by the external world. By identifying the grounded terminal it is straightforward to classify this amplifier as common base. From the CB amplifier analysis we known that it is suitable to serve as a current buffer, because the input terminal is taken at the emitter node (which is low impedance node, thus suitable to accept current signals) while the output terminals is taken at collector node (which is high impedance node, thus suitable for generating current). With the ω = ≥ approximation, which leads into IC = I E , i.e. the output current equals the input current, we easily conclude that the current gain of CB is Ai = 1. The other characteristics of CB amplifier are as follows. 1. By knowing potential at VCC rail, collector resistance RC , and collector current IC = I E = I = 1 mA it is straightforward to conclude that collector potential VC is below VCC as VC = VCC − RC IC = 5 V − 7.5 kλ × 0.5 mA = 1.25 V (19.21) 2. By convention capital letter G m is used to denote voltage to current gain of a multi–device circuit, while small letter gm is used to denote voltage to current gain of a single transistor. By definition, gm is change (i.e. derivative) of the output current due to change of input voltage at the biasing point (IC , VB E ), thus for small signal changes we write gm = IC 0.5 mA = 20 mS = VT 25 mV (19.22) 19 RF and IF Amplifiers: Solutions 213 3. Infinitely large capacitance C implies short connection for AC signal, and DC signals are blocked. Thus an input AC signal can be injected into emitter node without disturbing DC biasing point set by the current source. That leaves collector as the output terminal. By inspection, we recognize that for this BJT transistor VB E = −vi , which leads into the following signal analysis i o = gm vi = gm (−VB E ) (19.23) therefore, by taking output current at the terminal node i o = i C the small signal voltage gain is found as the ratio of voltage across the load resistor RC and the input voltage vi , i.e. vC = RC i C = −RC gm VB E = RC gm vi ∴ AV = vC = gm RC = 20 mS × 7.5 kλ = 150 V/V = 43.522 dB vi (19.24) which illustrates that CB amplifier is effectively used to force current through the resistive load RC , thus to provide high voltage gain, while at the same time the current gain is unity. These properties make CB amplifier very useful, especially in combination with the other single–stage amplifiers to design multi–stage amplifiers such as cascoded amplifier. Chapter 20 Sinusoidal Oscillators: Solutions The following tutorial exercises are based on simple idea that the study of oscillators can be done by splitting this new concept into three already existing concepts, i.e. concept of a signal amplifier, concept of resonant LC circuit, and concept of feedback systems. Solutions: 8.1. All oscillations are, by definition, caused by a feedback path from output of an amplifier back through the feedback network to the amplifier input, Fig. 20.1. If the total gain around the loop is one or more, even a small noise signal found at the input terminal of the amplifier is sufficient to start the oscillations by being amplified and then routed back to the input, where it is amplified more, and the cycle repeats until the output signal reaches maximum possible amplitude. Depending upon the actual loop gain, within a few cycles, the closed loop enters a steady state with stable periodic waveform present at the amplifier’s output terminal. In this example we review the general feedback loop equation. By following the input signal through the loop, first along the forward path through the amplifier, and then back through the feedback network, the loop equations are written by inspection as, Vout = A (Vin + V f b ) V f b = ω Vout ∴ Vout = A (Vin + ω Vout ) ∴ Vout A = Vin 1 − Aω (20.1) which implies that for Aω = 1 the loop gain becomes infinite, i.e. the loop is unstable and it starts to oscillate. R. Sobot, Wireless Communication Electronics by Example, 215 DOI: 10.1007/978-3-319-02871-2_20, © Springer International Publishing Switzerland 2014 216 20 Sinusoidal Oscillators: Solutions Fig. 20.1 Solution 8.1: a general closed loop block diagram Fig. 20.2 Solution 8.2: simplified schematic diagram of a phase oscillator There are number of criteria for establishing conditions for oscillations, none of them truly general and all applicable, for purposes of our discussion we will adopt Barkhausen stability criterion, which can be summarized as 1. The loop gain is equal unity, |ω A| = 1, where in reality |ω A| is taken just a little bit more than unity to compensate for the internal losses; 2. The phase shift around the loop must be an integer multiple of 2λ . 8.2. A phase shift oscillator produces sinusoidal waveform and its operational principle is very educational for purposes of studying oscillating closed loops. It consists of an inverting amplifier in the forward path, and RC feedback passive network. Common emitter amplifier is of inverting type, therefore, by inspection, the total loading resistance Ro and output voltage vout are R o = RC vout = −Ro i c = −Ro ω i b From the perspective of the feedback network CE amplifier behavies as a voltage source with Ro source resistance, while the feedback loop is maintained through the base branch with current i b . Therefore, the circuit equations are set in accordance with the equivalent circuit in Fig. 20.2. Voltage loop equations are: j i1 − R i2 −Ro ω i b = Ro + R − ωC j i2 − R ib 0 = −R i 1 + R + R − ωC j ib 0 = −R i 2 + R + R − ωC (20.2) (20.3) (20.4) 20 Sinusoidal Oscillators: Solutions 217 The system (20.2) to (20.4) can be solved in number of ways, one possible approach is to introduce substitution Z = R − j/ωC and then eliminate the three currents to arrive at (Ro + Z )(Z 2 + 2R Z ) − R(R 2 + R Z − ω R Ro ) = 0 (20.5) which, after simple algebra rearangements, results in the following polynomial Z 3 + (2R + Ro )Z 2 + (2R Ro − R 2 )Z + ω R 2 Ro − R 3 = 0 (20.6) A complex number equals zero if both its real and complex part are zero, which is to say that the real and complex parts of (20.6) are √: R3 − 3R 1 2 + (2R + R ) R − o (ωC)2 (ωC)2 + (2R Ro − R 2 )R + ω R 2 Ro − R 3 = 0 1 3R 2 2R 1 2 + − (2R Ro − R ) =0 →: j − − (2R + Ro ) ωC (ωC)3 ωC ωC (20.7) (20.8) Solving the imaginary part (20.8) for ω , after substituting Z , leads into formula for the resonant frequency ω 0 as ω0 = 1 Ro +6 RC 4 R (20.9) At the same time, the real part (20.7) is solved for ω as ω = 23 + 4 Ro R + 29 R Ro (20.10) Equation (20.10) is solvable for ω = f ( Ro/R ) as function of the resistor ratio x = Ro /R, Fig. 20.3. Minimum of this function is easily found by setting its first derivative to zero as ω(x) = 29 + 4x + 23 x ∴ ω ∞ (x) = − 29 +4 x2 ∴ x ≤ ±2.6926 218 20 Sinusoidal Oscillators: Solutions Fig. 20.3 Solution 8.2: plot of ω(x) and ω ∞ (x) functions There are two possible solutions for x, in this case the positive value is taken to calculate the resistors, hence Ro /R ≤ 2.6926 that, after being substituted in (20.10), produces ωmin ≤ 44.5 (20.11) It is very interesting to note that the ω value does not depend upon specific values of Ro and R, only on their ratio. From (20.11) and Ro = RC = 10 kπ, it follows that R = Ro /2.6926 = 3.714 kπ. As the last step, at f = 10 MHz from (20.9) it follows that C ≤ 1 pF. 8.3. Similarly to any other LC resonator circuit, assuming Q > 10, we can easily estimate the resonant frequency by following the energy flow inside LC loop. From the energy flow perspective, energy stored in the capacitor must be delivered to the series connection of the two inductors, and then returned back to the capacitor. Therefore, circulating resonant current i c in a tapped L, centre grounded network perceives the two inductors L 1 , L 2 , and C components in series, which means that L eff = L 1 + L 2 = 2 µH. It is then straightforward to calculate the resonant frequency as f0 = 2λ ≥ 1 = 10 MHz L eff C This example illustrates how resonant frequency of an oscillating closed loop circuit is still set by values of its internal LC components, which are fundamentally responsible for the oscillations in any case. 8.4. When using our strategy of analyzing the forward path (i.e. amplifier) and the feedback path (i.e. LC network) separately, we must keep in mind that output terminal 20 Sinusoidal Oscillators: Solutions 219 of the amplifier is loaded by input impedance of the feedback path, while at the same time, output node of the feedback network is loaded by the input impedance of the amplifier. By inspection, the feedback network input voltage (i.e. voltage generated by the amplifier) is distributed across L 2 inductor (note the central ground node in between the two inductors), while the output voltage of the feedback network is distributed across L 1 inductor. As we already stated in example 8.3, the same resonating current i c is circulating through both inductive components, thus it is straightforward to write vin = i c jω L 2 vout = −i c jω L 1 ∴ vout i c jω L 1 L1 0.5 µH = −0.333 ω= =− =− =− vin i c jω L 2 L2 1.5 µH (20.12) that is, amplitude of signal passing through the feedback network is multiplied by factor ω that is smaller than one, which must be later compensated for by the amplifier gain. 8.5. As it was shown in textbook, theoretical derivation of this relationship is rather involved, and very approximative. For the purposes of hand analyses, exercises of this kind are therefore reduced to direct implementation of the derived formulas. Using formula provided in textbook, and knowing that the R L resistor of the feedback network, Fig. 8.1 (middle), is in fact the input resistance of the amplifier, straightforward implementation of the textbook formula yields L 2 2 Qω 0 L 22 || L1 L1 + L2 1.5 µH 2 50 × 2λ × 10 MHz × (1.5 µH)2 = 10 kπ || 0.5 µH 2 µH = 90 kπ || 3.534 kπ = 3.4 kπ Reff = R L For purposes of practicsing, repeat problems 8.3 to 8.5 for the other three types of feedback network shown in the textbook. 8.6. Although, by now we already know how to estimate the resonant frequency of this circuit, let us take the opportunity and develop one possible methodology to solve this kind of circuits in a more general way. First, let us rearrange the circuit network so that it becomes more obvious how the network equations are going to be written. In Fig. 20.4 (left) two paths, p1 and p2 , from the collector node through the feedback network to ground are identified. By following the components on each of the two paths it is straightforward to redraw the equivalent circuit diagram to look as Fig. 20.4 (right). 220 20 Sinusoidal Oscillators: Solutions By doing so it becomes easy to generalize components in each branch of the circuit and to introduce the equivalent subnetwork that represents the amplifier itself (grey box that contains BJT and RC ), whose function is to be a voltage controlled current source, i.e. collector current i C = f (V1 ). Second, by inspection, equivalent resistance Ro at the collector node A is Ro = RC || rC || R D (20.13) where, RC is the given real resistor (looking up into the RC from the A node), rC is the output resistance of BJT at its biasing point (looking down into the collector), and R D (looking left into the LC resonator) is the dynamic resistance at resonant frequency. From the signal perspective the three resistances are in parallel, i.e. connected between the A node and signal ground. In addition, the condition Q ∗ ∓ implies that R D = f (Q 2 ) ∗ ∓, i.e. it has no influence on Ro and can be ignored in (20.13). Third, the BJT amplifier (the grey box in Fig. 20.4), is then replaced with its equivalent current source whose current is gm v1 and the output resistance is Ro , Fig. 20.5 (left). Note that the feedback loop is maintained through the controlling voltage v1 . Finally, in order to simplify the incoming analytical expressions, the last step in this network transformation is to substitute the real RLC components with general Z 1 , Z 2 , and Z 3 impedances, Fig. 20.5 (right), where Fig. 20.4 Solution 8.6: simplified schematic transformation of LC oscillator Fig. 20.5 Simplified schematic transformation of LC oscillator, problem 8.6 20 Sinusoidal Oscillators: Solutions 221 1 ω C1 1 Z 2 = Ro || − j ω C2 Z 3 = jω L Z1 = − j With this last substitution and transformation, it becomes straightforward to write the KCL current equation at the node A, after recognizing that the same current flows through Z 1 and Z 3 , as, v2 v2 − v1 + Z2 Z3 v1 v2 = Z1 Z1 + Z3 −gm v1 = (20.14) (20.15) which leads into, v2 v2 v1 + − Z2 Z3 Z3 Z1 v1 = v2 Z1 + Z3 −gm v1 = (20.16) (20.17) by substituting (20.17) into (20.16) it follows that 1 Z1 Z1 1 −gm − v2 = v2 + v2 Z1 + Z3 Z2 Z3 Z1 + Z3 ∴ Z1 + Z3 Z1 + Z3 Z1 Z1 + Z3 + − −gm Z 1 = Z2 Z3 Z3 Z1 + Z3 Z1 Z3 Z1 Z1 −gm Z 1 = + + +1− Z2 Z2 Z3 Z3 −gm Z 1 Z 2 = Z 1 + Z 2 + Z 3 (20.18) The result (20.18) is rather general in sense that the three impedances Z 1 , Z 2 , and Z 3 may be any other combination of RLC components, not necessarily the ones we started with in this problem. Indeed, CE amplifiers that use the other feedback networks studied in this course could be solved by applying the same methodology as in this example. In this particular example, it is easier to switch to admittances for the reactive components (of course, the final result must be the same). Following up with (20.18) we write 222 20 Sinusoidal Oscillators: Solutions −gm = ∴ Z1 + Z2 Z3 1 1 1 1 + = + + Z 3 = Y1 + Y2 + Y1 Y2 Z 3 Z1 Z2 Z1 Z2 Z1 Z2 Z1 Z2 1 1 + jωC2 + ( jωC1 ) + jωC2 jωL −gm = jωC1 + Ro Ro 1 ω 2 LC1 = + jω (C1 + C2 ) − ω 2 LC1 C2 − Ro Ro (20.19) Condition of resonance is that the imaginary part of (20.19) equals zero, which directly leads into expression for the resonant frequency as, C1 + C2 = ω 2 LC1 C2 ∴ ω 20 = 1 C2 L CC11+C 2 = 1 LCs (20.20) where, Cs is equivalent series capacitance of C1 and C2 . Expression (20.20) is what we already have known by inspection of the resonant loop in Fig. 8.2, and it is only reconfirmed by this derivation. For the given data, from (20.20) it follows that f 0 = 10 MHz. Under condition of oscillation (20.19) is left only with its real part, i.e. after substituting ω 0 from (20.20), we write −gm = ω 2 LC1 1 − 0 Ro Ro ∴ gm = 1 C1 Ro C 2 (20.21) Therefore, from (20.21) for the given data Ro = RC || rc = 5 kπ ∴ 1 × 1 = 200 µs gm = 5 kπ The case of finite Q = 50 is worked out by using Z 3 = r + jωL all along, where r = ω L/Q = 2.513 π. Since the expressions become a bit more complicated, it maybe be beneficial to use some of the analytical software tools, for example MAPLE. It is also recommended that the same problem is solved for other types of LC feedback networks by using the same methodology. 20 Sinusoidal Oscillators: Solutions 223 8.7. To conclude this chapter on oscillators, we revisit a modern version of tuneable oscillators used virtually in all modern wireless circuits, that is based on a voltage controlled capacitive element, a varicap. In the previous chapters we already met mechanical tuneable rotary capacitor that served its role for decades. Modern varicap element is nothing more than our old friend a semiconductor diode, disguised as a voltage controlled capacitor. From the fundamental operation of a pn junction we remember that the width of depletion region between p and n layers varies based on the diode biasing voltage VD . At the same time we know that p type layer, depletion region, and n type layer form a capacitor. From the definition of capacitance C we know that C =φ S d (20.22) where, φ is property of the dielectrics, S is area of the capacitor plates, and d is distance between the two plates forming the capacitor, which in the case of varicap diode equals the width of of the depletion region. Therefore, by changing diode biasing voltage VD , width d of the depletion region changes, which changes C, which means that varicap capacitance is function of diode voltage, C = f (VD ). Thus, we created voltage controlled capacitance that can be used as part of LC resonator to design voltage controlled LC resonator, which is to say tuneable radio receiver. In this example we practice to evaluate resonators whose capacitance C D is due to a biased varicap diode, CD = C0 |VD | 1+ 0.5 = 20 pF | − 0.7| 1+ 0.5 = 5.16 pF By inspection of Fig. 8.2, from the perspective of the resonating current, the three capacitances, C1 , C2 , and C D are perceived by the resonating loop as being in series, hence the total loop capacitance C at zero bias is, 1 1 1 1 = + + = 0.0567 1/pF C C1 C2 C0 ∴ C = 17.65 pF which means that the zero biasing frequency is f0 = 1 ≤ 3.789 MHz 2λ 17.65 pF × 100 µH If the biasing voltage is set to VD = −7V the total loop capacitance becomes 1 1 1 1 = + + C C1 C2 CD ∴ C = 4.998 pF 224 20 Sinusoidal Oscillators: Solutions which means that the zero biasing frequency is now found as f0 = 1 ≤ 7.126 MHz ≥ 2λ 4.998 pF × 100 µH Therefore, this oscillator is continuously tuneable from 3.789–7.126 MHz by the means of a varicap diode whose biasing voltage changes by 7 V. This tiny element that replaced the bulky rotary capacitor enabled virtually all modern miniature wireless circuits. Chapter 21 Frequency Shifting: Solutions In the high-school algebra we already learned that multiplication of two sinusoidal functions is done by directly applying trigonometry identities 1 [cos(x − y) + cos(x + y)] 2 1 sin x sin y = [cos(x − y) − cos(x + y)] 2 1 sin x cos y = [ sin (x + y) + sin (x − y)] 2 1 cos x sin y = [ sin (x + y) − sin (x − y)] 2 cos x cos y = which, in the context of radio communication theory, are fundamental for the concept of frequency shifting. Once the two general mathematical variables x and y take physical meaning of two frequencies, f 1 and f 2 , then it becomes clear that the operation of multiplication of two sinusoidal function produces the sum of two other sinusoidal functions, where one of the two new functions has high frequency argument ( f 1 + f 2 ) while the second new sinusoidal function has low frequency argument of ( f 1 − f 2 ). That is, if we start, for instance, with 10 and 11 MHz tones at the input of a multiplying circuit, then output signal’s frequency spectrum consists of only two new tones widely separated, one is found at 1 MHz and one at 21 MHz. Surprisingly, the two original tones, 10 and 11 MHz, are not anymore present in the output spectrum. This is extremely useful property of mathematical multiplication of two sinusoidal functions that can be exploited for practical applications. Obviously, by controlling one of the two input frequencies, we can “search and isolate” any other tone already existing in the space, thus we can “tune in” and receive message carried by that particular carrier. In summary, the problem of shifting a single tone waveform whose frequency is A, is solved by multiplying it with another single tone waveform whose frequency is B, so that the output waveform is a dual-tone waveform that is synthesized from two R. Sobot, Wireless Communication Electronics by Example, 225 DOI: 10.1007/978-3-319-02871-2_21, © Springer International Publishing Switzerland 2014 226 21 Frequency Shifting: Solutions single tone waveforms at frequencies ( A + B) and (A − B). All that it left to do is to design the multiplying circuit, and then filter out either the high or low frequency tone, and -voilà!- you just learned the most basic principle a radio. Solutions: 9.1. Direct implementation of the trigonometry identities leads into S(t) = S1 (t) × S2 (t) = 3V sin (2π × 1 MHz × t) × 4V sin (2π × 20 MHz × t) = 12V [cos(2π × 19 MHz × t) − cos(2π × 21 MHz × t)] (21.1) First point to observe is that the output waveform S(t) is not a single tone signal, instead there are two tones present, one at 19 MHz (the difference tone) and one at 21 MHz (the sum tone). Its main frequency is relatively high, here it is 20 MHz, however, its amplitude varies in time in accordance with the low-frequency signal, here 1 MHz. This conclusion becomes obvious once the product of the two signals S(t) is plot together with the low-frequency single-tone signal S1 (t), where we note that both positive and negative versions of the low-frequency tone S1 (t) are embedded into amplitude of the high-frequency tone as its envelope, Fig. 21.1. This particular example illustrates how an amplitude modulated (AM) waveforms are created mathematically. The high frequency AM signal is referred to as the carrier, while the envelope itself is the actual low frequency message signal. This example is trivial because the envelope is just another single tone signal, not for example modulated speech or music. If that were the case, the envelope shape would be much more complicated, nevertheless, it would still be exact replica of the transmitted message. Finally, we note that there are two identical envelopes, both carrying identical signal waveform, thus by recovering either of the two envelopes results in the same recovered message. 9.2. The trigonometric identity sin x × sin y = 1/2[cos(|x − y|) − cos(x + y)] shows that frequency spectrum of the product of two single tones contains another two tones (and not the original tones), one low–frequency tone with frequency calculated Fig. 21.1 Solution 9.1: waveform plots 20 S(t) S1(t) -S1(t) S(t), S1(t), S2(t) 15 10 5 0 -5 -10 -15 -20 0 0.5 1 time 1.5 2 21 Frequency Shifting: Solutions 227 as difference f L F = | f 1 − f 2 |, and one with high-frequency calculated as the sum f H F = f1 + f2 . If the carrier frequency tone is known, for example f L O = 10 MHz, in order to find carrier frequencies f C1 and f C2 , which after multiplication with the 10 MHz tone both produce a 1 kHz signal, we need to look at the right side of trigonometric identity to realize that there are actually two possible signals, one on each sides of the carrier frequency at f L O ± 1 kHz, nevertheless at the same distance of 1 kHz from f 0 , i.e. f L F = f 2 − f 0 = 10.001 MHz − 10.000 MHz = 1 kHz f L F = | f 1 − f 0 | = |9.999 MHz − 10.000 MHz| = 1 kHz Hence, multiplication of a 10 MHz tone with these two tones, i.e. 9.999 and 10.001 MHz, results in two overlapping 1 kHz LF tones in the output frequency spectrum, Fig. 21.2. At the same time, the summing tones in the output spectrum are not identical, one is at 20.001 MHz while the other is at 19.999 MHz, still, their difference is double of the LF 1 kHz tone. This example illustrates very important (and not desirable) side effect of multiplication of sinusoidal functions. For a given waveform frequency f L O there are two other HF signals that produce the same LF signals. However, only one of them carries the wanted signal, while the second therefore produces unwanted interference. Imagine how it would sound if at the same time a radio receivers plays a classical music concert and live commentary of a hockey game. Depending on which of the two programs was wanted by the person listening to the radio, one of the two carriers is perceived as unwanted and it is referred to as the ghost or image frequency. Existence of the ghost frequencies is unavoidable because it comes from the rules of mathematics not from the physics, and in practice it must be resolved by regulations and laws. That is why a radio transmitting equipment must have the licence to operate at a given frequency, so that the assigned frequency is not creating ghost image to the already existing radio station. Fig. 21.2 Solution 9.2: frequency domain plot of two image signals 228 21 Frequency Shifting: Solutions 9.3. Function of frequency multiplier, i.e. frequency shifter, is implemented by a circuit known as mixer. As oppose to a mixer used to combine musical soundtracks, which is implements by liner amplifiers, the radio mixer is implemented using nonlinear circuits. In the following examples we practice the frequency shifting principle. 1. Received signal and the local oscillator waveforms are mixed (i.e. multiplied) inside the receiver’s mixer, therefore an ideal output waveform contains only two tones that the sum and the difference of the two input frequencies, that is: (sum:)1435 kHz + 980 kHz = 2415 kHz (difference:) 1435 kHz − 980 kHz = 455 kHz 2. As oppose to a transmitter whose function is to upconvert the message frequency, a receiver is intended to downconvert frequency of the incoming carrier waveform, thus in receivers the lower frequency tone is used as IF, f I F = 455 kHz. By declaring standard numbers to the intermediate frequencies simplifies design of commercial radio equipment. 3. Working backwards from the mixer output, it is straightforward to find two frequencies presented at the mixer’s input terminals that can produce same f I F = 455 kHz for the given local oscillator f L O = 1435 kHz as: (sum:)1435 kHz + 455 kHz = 1890 kHz (difference:) 1435 kHz − 455 kHz = 980 kHz In other words, aside from the wanted station operating at 980 kHz, another station operating at 1890 kHz also produces the same IF frequency after its carrier waveform is multiplied with f L O by the mixer, therefore it would not be possible to separate the two transmitted programs, which means that the other frequency is the image of the wanted frequency. 4. Graphical representation, for case of f L O > f R F , is shown in Fig. 21.3. 9.4. Even though frequency spectrum space is regulated, the number of radio transmitting stations is huge, which is to say that it is not practical to assign neighbouring frequencies too far apart because the number of available frequencies would be very small. Instead, practical solution is to also control maximum power level of the image signals that falls close to the frequency band of the wanted signal. This is possible because the image signal, by definition, is not at resonant frequency of LC resonator in radio receiver’s front end electronics. That means, any signal that is not at the resonant frequency f 0 is attenuated to some extent depending how far its frequency is from f 0 , Fig. 21.4. In the textbook we already derived formula for relative amplitudes Ar of frequencies that fall inside bandwidth of LC resonator as: 21 Frequency Shifting: Solutions 229 1 Ar = 1+ Q2 f f0 − f0 f 2 Therefore, for the given data Q = 20, f / f 0 = 1.1/1 = 1.1 and f 0 / f = 1/1.1 = 0.909, we find relative amplitude of an image tone that is 10 % away from f 0 as Ar = 0.253 = −11.925 dB. For sake of comparison, if high quality resonator is used, for example one whose Q = 200, then the image tone amplitude would be attenuated to Ar = −31.640 dB, and if Q = 500 then the attenuation is Ar = −39.596 dB relative to the tone at resonant frequency. 9.5. From KVL, for two equal resistors voltage potential at the node A is at half the sum of the two inputs, i.e. vD = vA = 1 1 (v1 + v2 ) = [V1 · cos(ω1 t) + V2 · cos(ω2 t)] 2 2 (21.2) Following the signal path after the node A, the diode voltage v D is converted into current. Sum of the input voltage signals is assumed to be small, (implying that v D < Vt so that higher order terms in (21.5) keep approaching zero), which allows the exponential term Fig. 21.3 Solution 9.3: frequency relationship of wanted waveform, its image, and the IF signal. (The sum of frequencies not shown.) Fig. 21.4 Solution 9.4: illustration of partial suppression of image frequency by LC resonator 230 21 Frequency Shifting: Solutions vD −1 i D = I S exp Vt (21.3) to be expanded into Taylor series around the diode’s biasing point. Series expansion for exponential function is well known as √ x2 x3 x4 xn e = =1+x + + + + ··· n! 2 6 24 x (21.4) n=0 hence, after substitution x = v D /Vt , equation (21.3) becomes 1 vD + 1+ Vt 2 I D = IS vD Vt 2 1 + 6 vD Vt 3 1 + 24 vD Vt 4 + ··· −1 (21.5) We can now examine each of the terms on the right side of (21.5) separately and find out about the signal’s total spectrum (note: the “1” and “-1” terms cancel). Obviously, the exact series include an infinite number of terms, hence, in the first approximation, since the assumption is that the signal is small, the third and higher orders terms are neglected (i.e. they are smaller and smaller numbers being divided by larger and larger numbers). After substituting (21.2) into (21.5), we have the following two terms: 1. The linear term: vD 1 = [V1 · cos(ω1 t) + V2 · cos(ω2 t)] = f (ω1 , ω2 ) Vt 2Vt (21.6) which is to say that the linear term of the series expansion contributes as same frequencies to the output spectrum as the original signal v D , i.e. ω1 and ω2 . We already have that at the input, hence this term is not much useful. 2. The square term: 1 2 vD Vt 2 2 1 1 · cos(ω t) + V · cos(ω t)] [V 1 1 2 2 2Vt2 2 1 2 V1 cos2 (ω1 t) + 2V1 V2 cos(ω1 t) cos(ω2 t) = 8Vt2 +V22 cos2 (ω2 t) 1 1 V12 (1 + cos(2ω1 t))+ = 2 2 8Vt V1 V2 (cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)+ 1 V22 (1 + cos(2ω2 t)) 2 = 21 Frequency Shifting: Solutions 231 = f [(ω1 − ω2 ), 2ω1 , 2ω2 , (ω1 + ω2 )] which is to say that aside from the wanted sum and difference tones ω1 and ω2 , there are the additional unwanted tones, i.e. 2ω1 , 2ω2 , that are not part of the ideal multiplication operation. The multiplying term, i.e. the square term, of interest in expression for diode current I Ds , after replacing the numerical data for this problem, is V1 V2 [cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)] 8Vt2 1V × 1V = 1μA [cos(2π × 1 kHz × t) + cos(2π × 20.001 MHz × t)] 8 × (25mV )2 = 200μA [cos(2π × 1 kHz × t) + cos(2π × 20.001 MHz × t)] I Ds = I S which is very often used because a diode is the simplest device that can serve as a mixer, and also works over wide range of frequencies. Moreover, the small signal condition can be now stated a bit more specifically as V1 V2 < Vt2 . In this particular case, the diode generates 1 kHz and 20.001 MHz signals that are widely separated and easily used in the subsequent circuits. The addition, unwanted tones are usually filtered out with an LC resonator. In conclusion, using a diode as a nonlinear element for the purpose of multiplying two single tone signals does produce the desired theoretical tones ((ω1 − ω2 ) and (ω1 + ω2 )), however it also produces tones that are not part of the ideal solution (i.e. ω1 , ω2 , 2ω1 , 2ω2 , ...). In addition, when higher order harmonics in (21.5) are not neglected, many more tones are created in the output frequency spectrum as the inter-multiplication terms. All tones that are not needed must be filtered out afterwards, implying that a diode is rather inefficient multiplying element. 9.6. Depending on the frequency range of interest, it is more common that the simple diode mixer is replaced with the active BJT based circuit. The two variants of BJT transistor mixers in Fig. 9.2 are very similar, therefore the following two equations are written by inspection : VB E (Q 1 ) = 1 (v1 + v2 ) 2 VB E (Q 2 ) = v1 − v2 (21.7) (21.8) Relationship between collector current IC of a BJT transistor versus the base-emitter voltage VB E is as same as for a diode, iC = I S vB E exp Vt −1 (21.9) 232 21 Frequency Shifting: Solutions which is to say that expression for the square term of interest in the output frequency spectrum of the circuit in Fig. 9.2 (left) is similar to the one for a diode, with the only addition of the β factor, ICs = β I S V1 V2 [cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)] 8Vt2 where the corresponding expression for the circuit in Fig. 9.2 (right) is only slightly different. It is important to note that due to the β factor a BJT mixer has much better efficiency relative to a simple diode mixer, and it is possible even to have a conversion gain. That means that it is possible to have the output tone (usually the low-frequency one, |ω1 − ω2 |) with higher power than the input signal power. On the other hand, BJT transistor mixer still has the same limitation as for the diode in terms of the input signal amplitude relative the the Vt voltage. Both circuits in Fig. 9.2 have LC resonators in the collector branch that can be tuned to either of the two tones of interest, i.e. either to |ω1 − ω2 | or ω1 + ω2 , and immediately filter out all other unwanted tones in the frequency spectrum. 9.7. Using similar procedure as in the previous problem with main difference that current-voltage characteristics between drain current I D and the gate-source voltage vG S of a JFET transistor obeys the following relationship I D = I DSS vG S 2 1− Vp (21.10) where, I DSS is the JFET saturation drain current, VG S is gate-source voltage, and V p is pinch-off voltage. In JFET case there is no exponential term, which makes the derivation a bit simpler. Therefore, expansion of (21.10) leads into I D = I DSS v2 vG S + G2S 1−2 Vp Vp (21.11) By focusing only on the non-linear terms in (21.11), the square term is I D → −I DSS → −I DSS 1 [V1 · cos(ω1 t) + V2 · cos(ω2 t)]2 4 V p2 V1 V2 [cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)] 2 V p2 (21.12) where, (21.12) focuses only at the cosine product term from the previous step. It is interesting to note that because there was only second order term in (21.11) without the higher order terms, there was no need to apply power series expansion as in the cases of a diode and BJT, which is to say that there is no strict limitation to amplitudes of V1 and V2 , as long as the JFET is not cut off or become forward 21 Frequency Shifting: Solutions 233 biased. Again, similar to the BJT circuits from the previous example, the LC resonator simultaneously filters out all other harmonics except the desired one. JFET transistors are commonly used in RF mixer applications due to tolerance to high signal levels and good conversion efficiency. 9.8. Miller effect drastically limits useful frequency range of single stage CE or CS amplifiers that are intended for RF applications. To overcome that limit, and remove the output to input node feedback path that is the cause of Miller effect, it is common to use casoded amplifier topology instead of the single stage amplifier. Effectively, casoced amplifier is a two stage ampliﬁer, where in a standard cascoded configuration, transistor M1 is set as the CS amplifier for v1 signal, while M2 serves as CG current buffer. Therefore, assuming v2 = const = VDC2 , we can write equations for M1 transistor in saturation (ignoring its nonlinear effects) as I D = k(VG S − Vth )2 = k(v1 − Vth )2 ∴ d ID = 2k(v1 − Vth ) = 2k[V1 sin (ω1 t) − Vth ] gm ∞ ≤ d VG S (21.13) where, k = (μn Cox W )/(2L) and Vth is MOS threshold voltage, and gm ∞ is the circuit’s overall gm under condition that the gate of M2 is at its small signal ground. Drain current I D passes through the current buffer M2 with no loss, (i.e. the two transistors have the same drain current), which is followed by LC load. Therefore, voltage across the LC load at resonance is found by its dynamic resistance R D multiplied by the drain current, which is as same as the output voltage Vout relative to VD D . In other words, Vout = I D R D = gm ∞ v1 R D = gm ∞ R D V1 sin (ω1 t) (21.14) That is to say, when gate of M2 is at the small signal ground, in respect to signal v1 the circuit works as a CS cascoded amplifier. However, when gate of M2 is used as the input terminal for another signal, for example in case of dual-gate MOS mixer signal v2 comes from local oscillator (LO), the common drain current is additionally controlled by v2 as well. Variation of drain current due to variation of v2 voltage is manifested as change of the circuit’s overall gm , as I D = k[(VDC2 + VG S2 ) − Vth ]2 ∴ d ID = 2k[(VDC2 + VG S2 ) − Vth ] gm ≤ d(VDC2 + VG S2 ) = 2kVDC2 + 2k(V2 sin (ω2 t) − Vth ) → gm ∞ + gmΔ sin (ω2 t) (21.15) 234 21 Frequency Shifting: Solutions where, gm ∞ is part of the circuit’s gm due to v1 (21.13), while gmΔ is variation of the circuit’s gm due to v2 whose common mode is (i.e. it is centred around) at VDC2 . It is important to note that VDC2 is not constant anymore and that this arrangement works because the two transistors are identical with the same drain current. After replacing gm ∞ in (21.14) with gm from (21.15) it follows Vout = [gm ∞ + gmΔ sin (ω2 t)] R D V1 sin (ω1 t) = gm ∞ R D V1 sin (ω1 t) + gmΔ R D V1 sin (ω2 t) sin (ω1 t) → gmΔ R D V1 [cos(|ω1 − ω2 |t) + cos((ω1 + ω2 )t)] (21.16) where (21.16) again focuses only on the cosine product term, the LC resonator is then tuned to either of the two desired tones and filters out all the other harmonics. In conclusion, dual-gate FET mixer is commonly used for design of RF mixers for purposes of multiplying the incoming RF signal with the local oscillator (LO). Setting appropriate LO frequency, the RF signal is then precisely shifted in frequency domain, i.e. either “down-converted” or “up-converted”. In addition, from (21.16) it becomes obvious that v2 signal amplitude should be as large as possible so that gmΔ term is maximized as well, which is another advantage of this circuit. Chapter 22 Modulation: Solutions Examples given in this chapter serve as an introduction in AM, FM, and phase modulation techniques and basic principles. It is recommended that other books are also consulted for more drill type problems. Solutions: 10.1. AM modulation is relatively simple and easy to implement, while some of its basic aspects are illustrated in this example. 1. Given 1.5 kHz audio signal is always positive because its common mode voltage is set to 3 V, while its amplitude is 1.5 V p , which means that its amplitude swing is between 1.5 V and 4.5 V. At the same time the 50 kHz carrier signal is set at zero common mode voltage, Fig. 22.1 (left). 2. AM modulation means that the amplitude of the carrier is multiplied by the audio signal waveform, thus it is straightforward to write S(t) = Aa (t) × Ac (t) = 6 [3 + 1.5 sin(2ω × 1500 × t)] × cos(2ω × 50, 000 × t) where maximum amplitude of modulated signal is 6 × (3 + 1.5) = 27, and therefore the minimum amplitude is −27, Fig. 22.1 (right). 3. Modulation factor m is also known as percentage modulation or modulation index, and it is defined the ratio of the baseband (i.e. audio) and the carrier maximum amplitudes. In this example, m= 1.5 = 0.25 = 25 % 6 Preferred AM transmission is when the modulation index is most of the time at 100 %, which implies the case when the signal and carrier amplitudes are equal. In that case of m = 1, AM modulation results in largest possible demodulated signal amplitude, which is important to maintain the overall signal to noise ratio (SNR) in the transmission system. R. Sobot, Wireless Communication Electronics by Example, 235 DOI: 10.1007/978-3-319-02871-2_22, © Springer International Publishing Switzerland 2014 236 22 Modulation: Solutions 10 Audio Carrier Amplitude, V Amplitude, V 5 0 -5 -10 0 20 40 60 time, µs 80 100 35 30 25 20 15 10 5 0 -5 -10 -15 -20 -25 -30 -35 AM 0 20 40 60 80 100 time, µs Fig. 22.1 Solution 10.1: illustration of a 1.5 kHz and 50 kHz waveforms (left), and AM modulated waveform (right) 4. By inspection of the given equations, the carrier frequency is f c = 50 kHz and the signal frequency is f s = 1.5 kHz. 5. After expanding trigonometry equations describing AM modulated waveform S AM (t) as S AM (t) = e(t) sin λ c t = (C + B sin λ b t) sin λ c t B = C 1 + sin λ b t sin λ c t C = C (1 + m sin λ b t) sin λ c t = sin λ c t + m sin λ b t sin λ c t m = sin λ c t + [cos |λ c − λ b | t − cos (λ c + λ b ) t] 2 where, index c indicates carrier, and index b indicates baseband audio waveform, C is the baseband common mode, B is the baseband maximum amplitude, and C is the carrier maximum amplitude. This analysis shows that spectrum of AM waveform contains three tones, 1. the carrier itself f c = 50 kHz; 2. the lower–sideband (LSB) tone at f c − f s = (50 − 1.5) kHz = 48.5 kHz; and 3. the upper–sideband frequency (USB) f c + f s = (50 + 1.5) kHz = 51.5 kHz. This result is very important, because it also shows that energy of AM signal is divided among the carrier itself (not much useful), and two other tones who carry the same message. This observation revels that AM modulation is not very power efficient, and it also gives hints about possible techniques for design of AM transmission systems with better power efficiency. 10.2. By inspection of Fig. 22.2 and the discussion in Problem 10.1 we observe that AM signal occupies two times signal frequency bandwidth, which is the distance 22 Modulation: Solutions 237 Fig. 22.2 Solution 10.1 frequency domain plot of two image signals between the upper–sideband (USB) and lower–sideband (LSB) frequencies, i.e. 10 kHz. Therefore only 10 stations can fit within a 100 kHz frequency space. 10.3. In this drill exercise we simply use formula developed in textbook that relates the total power to the carrier power, thus we write: m2 PT = Pc 1 + 2 0.852 1200 W = Pc 1 + 2 ∴ Pc = 881.5 W Sum of the carrier power and power in two sidebands PS B equals the total power, therefore, PS B = PT − Pc = 318.5 W One half of the total sideband power PS B is in the upper sideband (USB) and one half in the lower sideband (LSB), i.e. PU S B = PL S B = PS B = 159.25 W 2 which leads us to conclude that 72 % of the total energy of an AM modulated waveform is held in the carrier itself, Fig. 22.2, while each of the tones actually carrying the information uses only approximately 13 % of the total power. This result reveals that AM modulation, although very simple, is not very efficient in terms of energy used to actually transmit the message. 10.4. Change of modulation index for an AM waveform causes the change of power contained in the sidebands, while the carrier power is kept constant. By using the equation which relates total power and the carrier power, expression for power in each of the sidebands is 238 22 Modulation: Solutions 1. m = 0.7: PU S B = PL S B = m 2 Pc 0.72 × 1500 W = = 183.75 W 4 4 2. m = 0.5: PU S B = PL S B = 0.52 1500 W m 2 Pc = = 93.75 W 4 4 This example illustrates importance and the role of modulation index in AM waveforms, which should be compared with the role of modulation index in FM waveforms. 10.5. Frequency modulated waveform has spectrum that is significantly different than spectrum of AM waveform. The main difference is that the modulation index in FM waveform can take values outside from the (0, 1) range. In addition, due to the properties of Bessel functions, change of modulation index of FM waveform changes the internal distribution of the energy among the harmonics, however the total energy stays constant. 1. Using Carson’s rule we write: B F M = 2(m f + 1) f b = 2(1.5 + 1) 10 kHz = 50 kHz 2. Using Bessel’s function table (see the textbook) we write: PT = J02 + 2 (J12 + J22 + J32 + J42 + J52 ) Pc = 0.5122 + 2 (0.5582 + 0.2322 + 0.0612 + 0.0122 + 0.0022 ) = 1.000258 (22.1) where the PT → 1 for infinite number of terms, or, in other words, the total power is constant; it is just redistributed between the carrier and sidebands as function of the modulation index. 3. according to the Bessel’s function table, for m f = 1.5 the first sideband frequency signal has the highest amplitude, J1 = 0.558 relative to the amplitude of the unmodulated signal. We note that for certain values of the modulation index, the tone whose frequency corresponds to the carrier frequency can be completely suppressed, Fig. 22.3, however the total energy of the FM waveform is still the same. 10.6. In this case the intermediate frequency is difference between the carrier and the local oscillator frequencies, i.e. 1. for f L O > f c it is straightforward to write 22 Modulation: Solutions 239 Fig. 22.3 Solution 10.5: illustration of an FM waveform frequency spectrum for two modulation indexes 2. for f c > f L O f I F = f L O − fc ∴ f L O = 995 kHz f I F = fc − f L O ∴ f L O = 85 kHz This short example illustrates relationship between the local oscillator frequency and the carrier frequency, where regardless to which of the two frequency is higher, they are always multiplied in the mixer and an IF waveform is produced. Both of the possible solutions are used, and the choice which one to implement is left to the designer and possible practical limitations of the design at hand. 10.7. FM waveforms keep constant amplitude, however, depending on amplitude of the baseband signal that is used for the modulation, the carrier frequency keeps moving around the centre frequency, where the distance of the instantaneous frequency from the centre frequency (i.e. the carrier unmodulated frequency) is proportional to the baseband signal’s amplitude. 1. given frequency deviation of f = 50 kHz, it is another way to say that the carrier swing is 100 kHz, i.e. it is “deviating” on both sides of the carrier frequency). 2. the highest frequency is one deviation above the carrier, i.e. 107.65 MHz, and the lowest frequency is one deviation bellow the carrier, i.e. 107.66 MHz, 3. By definition, f 50 kHz = 7.143 = mf = fm 7 kHz which illustrates the role and range of FM modulation index, in comparison with AM modulation index. 10.8. We continue discussion on power of FM waveform by taking a look at powers of the individual harmonics. We note that for unmodulated FM signal, the total power equals to the carrier power, PT = Pc , (i.e. for m = 0). Also, the total power does not change for various modulation indexes, it only becomes redistributed. That means that the carrier power, which is initially 100 W, is reduced by its respective J0 coefficient, while the rest of the power will be “assigned” to the sideband signals. 240 22 Modulation: Solutions Fig. 22.4 Solution 10.1: equivalent schematic of reactance modulator used in parallel with LC resonator 1. Using equation for FM power and values of Bessel function (as tabulated in textbook) for m f = 2.0 it follows: PT = Pc J02 + 2 J12 + J22 + J32 + J42 + J52 + J62 and J0 = 0.224, J1 = 0.577, J2 = 0.353, J3 = 0.129, J4 = 0.034, J5 = 0.007, J6 = 0.001 In other words, P0 = 100 W × 0.2242 = 5.0176 W P1 = 100 W × 2 × 0.5772 = 66, 5858 W P2 = 100 W × 2 × 0.3532 = 24.9218 W P3 = 100 W × 2 × 0.1292 = 3.3282 W P4 = 100 W × 2 × 0.0342 = 0.2312 W P5 = 100 W × 2 × 0.0072 = 0.0098 W P6 = 100 W × 2 × 0.0012 = 0.0002 W which gives, again, the total power of 100 W. 2. Using Carson’s rule, estimated bandwidth is (for m f = 2), B F M = 2(m f + 1) f m = 6 × 1.0 kHz = 6 kHz This example illustrates practical use of Bessel function whose coefficients are used to scale the corresponding harmonics in FM waveform. 10.9. An elegant way of implementing a voltage controlled capacitance is by circuit known as reactance modulator, where its output capacitance is function of the control voltage Vctrl . In this example it is attached to the L T C T resonator to create a voltage controlled resonator, Fig. 22.4. For a MOS transistor implementation its equivalent capacitance at the f out node is given as Ceq = gm RC. Therefore, for the given data, resonant frequency of the 22 Modulation: Solutions 241 voltage controlled resonator is set by: f out = 1 2ω L T (C T + Ceq ) ∴ C T + Ceq = 103.4 nF ∴ Ceq = 20 nF 10.10. Another elegant method to create a voltage controlled resonator is to use varicap diode inside LC resonator. For phase modulator shown in Fig. 10.2 it was shown in textbook that phase deviation constant is calculated as K =− Q (1 + 2V0 )(1 + n) where, C = n Cd0 , and Cd0 is the varicap diode capacitance for the biasing voltage V0 . Straightforward calculation gives K = −0.2 rad/V. Chapter 23 Signal Demodulation: Solutions Design of peak detectors is done with help of numerical solver, however, preliminary calculations can be done using approximative model. In order to design peak detector whose operation becomes close to the circuit model based on ideal diode, in realistic peak detector it is common to use active diode circuit instead of a single diode. Problems: 11.1. Peak detector, or envelope detector, is a very versatile and useful circuit in analog signal processing. Its main function is to measure instantaneous time–varying amplitude of a signal waveform. In RF communication systems, information about the instantaneous amplitude is essential for analog algorithms used to extract the embedded message from both AM and FM waveforms. Functionality of a peak detector is based on an ideal diode model, i.e. voltage drop across the diode is assumed to be zero. Role of the diode is to serve as unidirectional current valve, which enables charges to flow into capacitor but not in the opposite direction. By itself, the capacitor serves as storage space for the incoming charge, where the instantaneous voltage difference across the capacitor indicates the amount stored charges, thus the voltage is proportional to the signal amplitude. As long as the amplitude of incoming waveform keeps increasing, the diode is turned on and the voltage across the capacitor can follow because the influx of charges is stored in the capacitor and, therefore the capacitor voltage increases too. However, once amplitude of the incoming waveform starts to drop relative to its previous value, then the diode is turned off because the capacitor voltage becomes higher than voltage amplitude of the incoming waveform, thus the diode becomes reverse biased. At this moment, unless there is another path to ground for the charges to flow, voltage across the capacitor would have stayed at the same level indefinitely, because the path back through the diode is closed. That is why resistor R is needed in parallel with the capacitor. It should now become apparent that with the right balance between RC time constant and period T of the incoming signal it is possible to adjust the overall dynamics of the circuit so that the output voltage becomes relatively close approximation of the signal’s envelope, Fig. 11.1 (right). R. Sobot, Wireless Communication Electronics by Example, 243 DOI: 10.1007/978-3-319-02871-2_23, © Springer International Publishing Switzerland 2014 244 23 Signal Demodulation: Solutions In practice, realistic diodes have the minimum turn on voltage in order of a few hundred mV, which is much higher than amplitude of weak RF signals arriving to the peak detector. As solution to this problem a circuit known as an “active diode”, which consists of operational amplifier and a realistic diode, was developed to approximate ideal diode behaviour by inserting very low voltage drop across its input and output terminals. In this example, we practice an approximate method to estimate component values of a peak detector. At the same time we keep in mind that methodology presented in the textbook is very approximate, and that the final component values are found with the help of numerical solvers. 1. It was shown in textbook that input impedance of the envelope detector should be approximated as Z in = R/2, therefore Z in = 1 kω. 2. Amplitude of unmodulated input signal (i.e. carrier amplitude vc ( pk)) equals to the average value of its envelope voltage venv . From the given data and inspection of Fig. 11.1, we write vi (avg) = 1.5 V + 0.5 V venv (max) + venv (min) = = 1.0 V = vc ( pk) 2 2 By definition, RMS power of the carrier is Pc = vc2 ( pk) = 0.5 mW 2Z in From the same data we can find the modulation index as m= 1.5 − 0.5 = 0.5 1.5 + 0.5 Therefore, 0.52 Pt = 1 + 0.5 mW = 562.5 µW 2 3. Output voltage of the envelope detector should be as same as the envelope of the input waveform, except for the shift due to diode voltage drop. From the given data (right side graph) by inspection we conclude that the carrier envelope is shifted by 0.2 V because of the diode voltage drop. Therefore, relative to the already calculated minimum and maximum values of the input signal, maximum value of the envelope is v0 (max) = 1.3V, minimum value of the envelope is v0 (min)0.3 V, and the average (DC) output is v0 (DC) = (1.3 V + 0.3 V)/2 = 0.8 V, i.e. also shifted by 200 mV relative to the peak detector’s input side level. 4. With known average (DC) current and output resistance (R), the output current must be I0 (DC) = 0.8 V /2 kω = 400 µA. 23 Signal Demodulation: Solutions 245 5. At the end of peak detector analyses, in textbook it was shown that good balance for choosing the capacitor value, with respect to the clipping conditions at 5 kHz (1/m a )2 − 1 C= = 7.7 nF 2λ R f m (max) Again, design of a peak detector circuits starts with very approximate hand analysis to find the initial solution, however correct solution requires iterative use of numerical solvers. 11.2. As we found in the previous chapters, AM carrier is bound by two envelopes (positive and negative) that are embodiments of the same message. It is in the envelop detector where it is decided which one of the two envelopes is going to be used. In Fig. 11.1 diode is forward biased if the input signal amplitude is greater than the capacitor voltage, thus the positive envelope is stored in the capacitor. In contrast, diode in Fig. 11.3 is forward biased when the input signal amplitude is lower than the capacitor voltage. That is to say that in this case negative envelope is stored in the capacitor. Another addition to note in Fig. 11.3 is presence of LP filter that consists of (R1 , C2 ) that removes the ragged edges (i.e. the high frequency components) and smoothens shape of the recovered envelope. 1. Output signal spectrum, at R L , is due to the nonlinear characteristics of the diode. Both the carrier and the signal are being processed by the diode and, therefore, they produce sidetones. With the stated assumption of the problem, the incoming IF signal has carrier at 665 kHz which was modulated by a 5 kHz signal. As a result, the incoming IF signal contains 665 kHz, as well as two sidetones at 660 and 670 kHz with the respective relative amplitudes, Fig. 23.1. These three tones are processed by the diode nonlinear characteristics, i.e. multiplied again, and Fig. 23.1 Solution 11.2: frequency spectrum of diode output signal 246 23 Signal Demodulation: Solutions produce the following tones: Sum frequencies: Difference frequencies: (660 + 665) kHz = 1325 kHz (660 + 670) kHz = 1330 kHz (665 + 670) kHz = 1335 kHz (670 − 665) kHz = 5 kHz (665 − 660) kHz = 5 kHz (670 − 660) kHz = 10 kHz Note that in the low frequency range the 5 kHz tone is easily isolated by a LP fillter. 2. Fast changing signal is the carrier, slow changing signal is the signal envelope. Note that due to the diode orientation the negative amplitude signal envelope is recovered. The last stage serves purpose of removing the DC offset in the signal envelope by using the blocking capacitor C, Fig. 23.2. 3. At 5 kHz: impedance of capacitors and diode resistances are as follows: 1 = 144.68 kω ≈ 145 kω 2λ × 5 kHz × 220 pF 1 Z C2 = = 1.4468 Mω ≈ 1.45 Mω 2λ × 5 kHz × 22 pF V 0.7V RD = = = 100 ω I 7 mA Z C1 = where diode resistance R D is found from the diode transfer characteristics graph, Fig. 11.2, and the transformer is modelled as an ideal voltage source element. Hence, the peak detector is modelled with the equivalent voltage divider, where first resistance consists of R D = 100 ω, and the second resistance R is R = Z C1 ||(R1 + (Z C2 ||R2 ||R L )) = 4.6 kω Fig. 23.2 Solution 11.2: envelope detector internal waveforms (23.1) 23 Signal Demodulation: Solutions 247 That means that the voltage amplitude at node 3 relative to the voltage amplitude of the input signal, is as same ratio as A= R V (3) = = 0.978 Vin RD + R (23.2) i.e. the 5 kHz signal is attenuated only approximately 2.2 % relative to its input side amplitude. 4. At 665 kHz: impedance of capacitors are as follows: 1 = 1087.86 ω ≈ 1.1 kω 2λ × 665 kHz × 220 pF 1 = 10.878 kω ≈ 11 kω = 2λ × 665 kHz × 22 pF Z C1 = Z C2 while the diode resistance is not function of frequency, i.e. diode resistance is still R D = 100 ω. The equivalent circuit network is as same as at 5 kHz, however this time R = 840 ω. Subsequently, A = 0.894, i.e. the 665 kHz carrier tone is attenuated approximately 10.6 % relative to its input side amplitude. 5. By choosing component values, designer has control over how much the carrier tone is attenuated relative to the envelope signal, as well as control over the internal timing constants. 11.3. In this drill example we practice to the use of dB and dBm power level units at the system level. 1. Input power is calculated by definition Pin = ∴ (8 µV)2 V2 = = 1.28 pW R 50 ω Pin ≡ 10 log 1.28 pW = −88.9 dBm = −118.9 dBW 1 mW where dBm is power relative to 1 mW reference, while dBW is relative to 1 W reference level. 2. One of the advantages of using dB unit is that the total system power is found by simple addition of either dB or dBm units along the system chain, i.e.: Pout = − 88.9 dBm + 8 dB + 3 dB + 24 dB + 26 dB + 26 dB − 2 dB + 34 dB = 30.1 dBm ∴ Pout ≡ 1W 248 23 Signal Demodulation: Solutions 11.4. Regardless of the modulation signal shape, AM operation of mixer produces modulated waveform, Fig. 23.4. By inspection, we observe that the carrier amplitude swings up to maximum possible value, thus the modulation index is m = 1, or m= 4−0 Vmax − Vmin =1 = Vmax + Vmin 4+0 11.5. Assuming constant antenna impedance, for AM waveform the total power PT of modulated waveform is related to the unmodulated carrier power PC and the modulation index m as m2 m2 2 2 ∴ I T R = IC R 1 + PT = PC 1 + 2 2 ∴ 2 I T m = 2 −1 IC2 thus, for the given data, 1.1A 2 m= 2 − 1 = 0.648 = 64.8 % 1A 11.6. Distribution of harmonic powers within an FM waveform is calculated by using Bessel function coefficients, where each coefficient is use to scale the respective harmonic tone. For the given example S F M (t) = 2000 sin 2λ × 108 + 2 × λ × 104 cos (λ × 104 t) t (23.3) it follows that: 1. By inspection of the given FM signal equation: fc = (2λ × 108 ) rad/s πc = = 100 MHz 2λ 2λ 2. Given FM peak voltage of 2000 V, by definition P= √ (2000 V/ 2)2 = 40 kW 50 ω 3. Modulation index of an FM waveform is found from definition of the waveform frequency π (t), which is described as 23 Signal Demodulation: Solutions 249 π (t) = π c + k B0 cos π b t (23.4) where, π c is the carrier frequency, k is frequency deviation constant, B0 is maximum amplitude of the modulating signal, and π b is frequency of the modulating signal. With this annotation, modulation index m f is defined as mf = k B0 πb (23.5) which, by inspection of the FM signal Eq. (23.3) with reference to (23.4) results in: mf = 2 × λ × 104 =2 λ × 104 4 4 3 3 2 2 amplitude amplitude Fig. 23.3 Solution 11.2: equivalent envelope detector circuit 1 0 -1 -2 -3 -4 1 0 -1 -2 -3 0 0.5 1 time 1.5 2 -4 0 0.5 1 time Fig. 23.4 Solution 11.4: envelope of a carrier modulated by triangular waveform 1.5 2 250 23 Signal Demodulation: Solutions 4. By definition, fb = (λ × 104 ) πb = = 5 kHz 2λ 2λ 5. By reading the table with Bessel functions, for m f = 2, significant bands exist to J4 , which means that on one side of the unmodulated carrier frequency 4 × f s = 4 × 5 kHz = 20 kHz bandwidth is needed. Thus, due to the symmetry of bandwidth, Fig. 23.3, relative to the carrier frequency whose scaling factor is J0 , B = 2 × 20 kHz = 40 kHz. 6. From table with Bessel functions, J1 is the largest sideband at 0.58 times the unmodulated carrier amplitude. The smallest sideband is J4 = 0.03. Hence √ (0.58 × 2000 V/ 2)2 = 13.5 kW (i.e. 27 kW for the two–sided band) 50 ω √ (0.03 × 2000 V/ 2)2 = 36 W (i.e. 72 W for the two–sided band) P4 = 50 ω P1 = In this example we reviewed basic definitions related to FM waveforms. Chapter 24 RF Receivers: Solutions System level analysis concepts presented in the following tutorials are taken one at the time, so that the reader can recognize them within the content. Although we use mostly AM radio receiver working at relatively low frequencies by today’s standard to demonstrate the principles, by now the reader should be comfortable with applying all relevant concepts and techniques, with the understanding that all the other wireless communication systems follow the same principles and similar design philosophy. Solutions: 12.1. The traditional heterodyne radio receiver architecture, Fig. 24.1, was invented by Edwin Armstrong in 1918 and has been used in virtually all radio equipment. Usually, if the radio receiver contains one VCO circuit, i.e. one IF frequency, it is referred to as heterodyne receiver. As the radio frequencies moved into higher ranges, due to problems related to image frequencies, it was necessary to introduce second IF frequency and performs down–converting operation in two steps. When two VCOs and two IF frequencies are used for down–conversion, the radio architecture is referred to as super heterodyne radio. The space around us is filled with RF EM waves v R F from various sources emitting all at the same time, and it is in a radio antenna that the EM waves are converted into the induced voltage signals. At the first radio stage, i.e. in the matching network and RF amplifier, the RF amplifier’s internal LC resonator allows only the desired waveform whose frequency is ω c to enter the system. The filtered v AM waveform is mixed (i.e. multiplied) with the local voltage controlled oscillator’s tone vV C O to produce down–converted waveform v AM− . It is role of IF amplifier to filter and amplify the lower-side band waveform v I F , which is then passed through envelope detector to extract the transmitted message v pk . The envelope waveform v pk filtered with low-pass filter to remove the ragged edges (i.e. HF components) and to deliver high fidelity version of the transmitted message v pk L P, which is then amplified with audio amplifier to deliver the recovered vb . 12.2. In this example we practice calculations related to tuneability of an AM receiver and variations in bandwidth due to the component values. R. Sobot, Wireless Communication Electronics by Example, 251 DOI: 10.1007/978-3-319-02871-2_24, © Springer International Publishing Switzerland 2014 252 24 RF Receivers: Solutions Fig. 24.1 Solution 12.1: heterodyne radio architecture and time domain plots of the internal waveforms 1. By definition, Q factor of an LC resonator is calculated as Q= 1050 kHz f0 = = 105 B 10 kHz Assuming constant Q factor at f 0 = 1600 kHz the achieved bandwidth is f max 1600 kHz = = 15.238 kHz Q 105 B f max = which requires capacitance C= 1 (2 π f max )2 L = 9.895 nF (24.1) 2. Similarly, at f min = 500 kHz for the given data we fand B f min = f min 500 kHz = = 4.762 kHz Q 105 which requires that the tuneable capacitor is set to C= 1 (2 π f min )2 L = 101.321 nF (24.2) In this example we illustrate that with the fixed inductance, the tuneablity comes with the price in increased overall bandwidth. Although at the centre of the tuning range the designed bandwidth is B = 10 kHz, at the low end of the tuning range a 24 RF Receivers: Solutions 253 signal whose bandwidth is B = 4.762 kHz can be processed. At the same time, at the high end of the tuning range, a signal whose bandwidth is up to 15.238 kHz can be processed. In summary, this receiver should be used only to process signals whose bandwidth is B √ 4.762 kHz, however if the channel spacing Δf between various AM radio stations is constant, then it must be set to Δf → B f max = 15.238 kHz to avoid the inter-channel interference. 12.3. Design of tuneable components is an important engineering issue because there are practical limits to the possible tuneable range that can be actually manufactured given the state of the art technology. In this example we take a look at design of tuneable dual capacitor that is intended for simultaneous use in the RF amplifier and LO oscillator sections. 1. Starting from the input RF signal frequency range it follows that, for a given inductor, frequency tuneability ratio R f R F and capacitor tuneability ratio Rc R F are f R F (max) 1600 kHz = = 3.2 f R F (min) 500 kHz ∴ 1 ∞ Cmax f R F (max) 2π LCmin = = 1 f R F (min) Cmin ∞ 2π LCmax ∴ C R F (max) f R F (max) 2 = = 10.24 = C R F (min) f R F (min) R f RF = Rc R F That is, in order to enable frequency tuning ratio of 3.2 times a tuneable capacitor whose tuning ratio is at least 10.24 is needed. 2. Frequency of a signal generated by the mixer, in this example the intermediate frequency IF, is calculated as absolute difference Δ between the local oscillator oscillator and RF frequencies, Δ = f I F = | f R F − f L O |. That is to say, there are two possible cases for relationships between the RF and LO frequencies that generate the same intermediate frequency. a. Case 1: f L O > f R F The graph in Fig. 24.2 shows how an RF signal is shifted down to IF frequency. Although in this example a range of frequencies is considered, for a band of RF frequencies it is sufficient to follow only what happens to the two frequencies that define the band edges, i.e. f min and f max (all frequencies in between are bounded by these two extremes). Hence, by simple inspection of the graph, if RF frequency shifts the LO frequency must follow at the same distance Δ, so that the same IF frequency is generated, i.e. 254 24 RF Receivers: Solutions Fig. 24.2 Problem 12.3: relationship among frequencies of the incoming RF signal f R F , local oscillator f L O , and the intermediate frequency f I F for the case of fLO > fRF f L O (min) = f R F (min) + f I F = 500 kHz + 465 kHz = 965 kHz f L O (max) = f R F (max) + f I F = 1600 kHz + 465 kHz = 2065 kHz Hence, the local oscillator’s frequency tuneability ratio R f L O and capacitor tuneabilility ratio RcL O are R f LO = f L O (max) 2065 kHz = = 2.1399 f L O (min) 965 kHz ∴ RcL O C L O (max) = = C L O (min) f L O (max) f L O (min) 2 = 4.5792 (24.3) b. Case 2: f L O < f R F Similarly to the Case 1, we calculate f L O (min) = f R F (min) − f I F = 500 kHz − 465 kHz = 35 kHz f L O (max) = f R F (max) − f I F = 1600 kHz − 465 kHz = 1135 kHz Hence, the local oscillator’s frequency tuneability ratio R f L O and capacitor tuneabilility ratio RcL O are R f LO = f L O (max) 1135 kHz = = 32.429 f L O (min) 35 kHz ∴ RcL O = C L O (max) = C L O (min) f L O (max) f L O (min) 2 = 1051.6 (24.4) 3. Results (24.3) and (24.4) give us a hint about reasonable choice for the LO frequency. Although both cases are mathematically equally valid, manufacturing tuneable capacitor whose capacitance tuneability range is about 4.5 times is not much difficult, however manufacturing a capacitor that is required to change its capacitance for more then 1,000 times is not easily done. At the same time, design of a VCO whose tuning range is just over 2 is much easier than design of a VCO 24 RF Receivers: Solutions 255 whose tuning range is more than 32 as required in the second case. Therefore, choosing Case 1: f L O > f R F , i.e. f L O = 965 − 2065 kHz is more practical engineering solution. 12.4. Continuing discussion from the previous examples, in this case intermediate frequency IF is generated as f I F = f L O − f R F = 1 MHz. However, the same IF is generated by another transmitter whose working frequency is fimage = f L O + f I F = 12 MHz, which must be declared forbidden frequency, i.e. the image frequency to the already existing transmitter’s frequency. 12.5. An ideal amplifier is intended to implement a simple functions of multiplying amplitude of the input signal and providing the multiplied version at its output terminal, i.e. ideal amplifying function is y=Gx (24.5) where, y is the output signal amplitude, x is the input signal amplitude, and G is the multiplication factor. In the ideal case, the multiplication factor can take any value for instance (G = −≤, ...−1, ...0, ...+1, ...+≤), where absolute value |G| represents the amplitude multiplication factor while the ± sign indicates the phase of output signal (i.e. negative gain implies inverted amplifier). In addition, if the multiplication factor |G| > 1 it is common to use term “gain”, if |G| < 1 it is common to use term “attenuation”, and if |G| = 1 it is common to use term “buffer” to indicate the outcome of the amplitude multiplication operation. We dully note that, as defined, the amplification function is linear. Realistic amplifiers, however, can not provide an arbitrary gain. Instead, only within a limited range of the input signal amplitudes an amplifier can be reasonably well approximated by the linear function. In order to design a system level function, which assumes linear amplification of the used amplifiers, is important to define boundaries of the linear operation for the realistic amplifier being used in the system. If the output signal level, as predicted by (24.5), becomes too high the amplifier starts to deliver only as high as it realistically can. In other words, realistic output becomes less then (24.5) calls for. Commonly accepted practice is to assume that an RF amplifier is linear as long as the difference between its realistic output and predicted linear output levels is less than 1 dB. The input level for which the difference becomes equal to 1 dB is appropriately named the 1 dB compression point. Alternate measure of amplifier nonlinearity is based on existence of higher harmonics, which always indicates non–linear signal processing. Extrapolated intercept point where powers of the first and third harmonics in the output signal spectrum become equal is referred to as the third order intercept point (IIP3). In this example we practice to read an amplifier specifications from measured relationship input–output signal levels. By inspection of the graph in Fig. 12.3 we can conclude the following: 1. Gain: in the linear part of the transfer characteristics for the input of −50 dBm the output power is −30 dBm, hence the gain is 20 dB. The same gain is expected 256 24 RF Receivers: Solutions Fig. 24.3 Problem 12.3: relationship among frequencies of the incoming RF signal f R F , local oscillator f L O , and the intermediate frequency f I F for the case of fLO < fRF for the input level of −20 dB because the predicted linear output level is 0 dB, i.e. 20 dB above the input level, which is by definition the amplifier gain. 2. 1 dB compression point: the linear part of the characteristics extends to approximately −20 dBm of the input power, when the output power becomes −1 dBm instead of the expected 0dBm. Therefore the 1 dBcompression point is at −20 dBm of the input power. 3. The third order intercept point IIP3: harmonics power of the third harmonic is extrapolated until its intersection with the extrapolated linear gain, and the crossing point is found at the output power of approximately +9.6 dBm, which is only extrapolated point, not the real measurement point. Keep in mind that the amplifier output never reaches that level of output power, it has already saturated close to the 1 dB compression point level. It should be apparent that 1 dB compression IIP3 points are closely related and can be used interchangeably to quantify the amplifier nonlinearity. 12.6. As already implied by the existence of image frequencies, creating RF transmission scheme that ic free of interference among the transmitting stations is rather complicated. We already learned that the image frequency entering the receiving mixer is shifted and it also falls into the desired IF band. In addition, in case when transmitted waveform contains (either intentionally or unintentionally) both the carrier and its second harmonic frequency it opens possibility that the second harmonic falls over the desired RF frequency band. In this example we explore one such possibility. Receiver’s RF front end serves as an entrance door that allows only tones whose frequencies are within the LC resonator’s bandwidth to enter the mixer stage. We also keep in mind that the local oscillator frequency f L O is not visible to the outside world, i.e. only frequencies that pass through the LC bandpass filter are multiplied by the oscillator’s frequency. We also note that, while frequency shifting merely translates bandwidth along the frequency axis without changing its width (because the shifting is result of addition/subtraction of two frequencies), plain frequency multiplication (i.e. multiplication by a constant) affects the bandwidth width as well. Depending upon relationship between f R F and f T x , (because f R F ≥ = f T x−I ) there are two possibilities we should consider. 1. Case f T x > f R F : If the local oscillator’s frequency f L O is lower than the interfering Tx signal, then both f T x−I and it’s second harmonic f T x−I I = 2 × f T x−I are 24 RF Receivers: Solutions 257 Fig. 24.4 Problem 2.6: relationship among frequencies of a receiver and the interfering transmitter for f T x > f R F case far away from the receiver’s input RF frequency band f R F , Fig. 24.4. Therefore, both frequencies are rejected by the input LC bandpass filter. 2. Case f T x < f R F : if the interfering signals are to reach the mixer their frequencies must be aligned with the LC bandpass filter’s frequency range f R F = 950 ± 200 kHz. The first harmonic is already prohibited from that range, thus we take a look what happens if the second harmonic is aligned with the f T x−I I = f R F band, Fig. 24.5. In that case, the first harmonic must be further down the frequency axis, i.e. f T x−I = 1/2 f T x−I = 475 ± 5 kHz. This frequency alignment results in the transmitter’s second harmonic f T x−I I entering the receiver along with the intended RF signal f R F and reaches the mixer’s input terminals. Output terminals of the mixer then deliver I F waveform that is modulated by both the intended RF signal as well as the unintended T x signal, thus the received message is ruined and can not be recovered. Therefore, in this case, we conclude that transmitting stations should not be allowed to operate in the f T x = 475 ± 5 kHz range. 12.7. A bundle of frequencies close to each other can be visualized as a multi–wire cable, where each wire is reserved for one broadcasting channel. In this example, each channel occupies 10 kHz wide frequency space (similar to wire diameter), thus within the medium wave AM band it is possible to have the total of n channels, i.e. n= (1610 − 540) kHz = 1070 10 kHz 258 24 RF Receivers: Solutions Fig. 24.5 Problem 12.6: relationship among frequencies of a receiver and the interfering transmitter for f T x < f R F case if the assumptions are that there is no need for “guard bands” (i.e. frequency spacing in between the neighbouring channels to account for manufacturing tolerances) and that all channels are available for communication (i.e. for the moment we ignore the image frequencies inside the AM band). In order to shift all channels down to 455 kHz the local VCO must be able to generate frequencies f L Omin = f I F + f R Fmin = 455 kHz + 540 kHz = 995 kHz f L Omax = f I F + f R Fmax = 455 kHz + 1610 kHz = 1965 kHz ∴ f L Omax = 1.975 f L Omin which is considered an easy tuning ratio to design. Let us now estimate bandpass filter’s Q factor that can provide B = 10 kHz for each channel. f R Fmax 1610 kHz = = 161 B 10 kHz f R Fmin 540 kHz = = = 54 B 10 kHz Q max = Q min Obviously, holding a constant B = 10 kHz over the whole AM band is not trivial requirement because Q factor is set by properties of RLC components in the RF resonator (see also the previous examples in this chapter). One way to deal with the problem would be to provide the fixed 10 kHz bandwidth filtering at the IF stage, 24 RF Receivers: Solutions 259 Fig. 24.6 Problem 12.7: illustration of using IF bandpass filter to set fixed bandwidth for each RF channel while the RF stage resonator is designed to provide minimal bandwidth that allows all individual channels to pass (i.e Q = 54), and then further trim the bandwidth at the IF frequency (which is constant), Fig. 24.6. If a bandpass filter is designed at IF frequency then 455 kHz fI F = = 45.5 Q= B 10 kHz which is fixed and easily achieved value. This example illustrates one advantage of using standard IF frequency for a given band, which simplifies the overall receiver design because it is easier to design tuneable VCO than tuneable fixed bandwidth bandpass filter. 12.8. When it is not practical to do the frequency downconversion in one step, it is necessary to use super–heterodyne receiver architecture where two IF frequencies are used. In this example we practice relationships among two LO frequencies and the internally generated frequencies. Following the previously reached conclusion that it is more practical to set the LO frequency higher than the second frequency at the mixer input terminals, in order to shift the carrier frequency f R F = 20 MHz down to f I F1 = 10.7 MHz obviously the first LO frequency must be f L O1 = f R F + f I F1 = 30.7 MHz.This relationship between f L O1 and f R F must also produce the sum frequency at the I F1 node, i.e. f R F + f L O1 = 50.7 MHz, Fig. 24.7. It is now more logical and practical to Fig. 24.7 Solution 12.8: super–heterodyne radio architecture 260 24 RF Receivers: Solutions choose frequency f L O2 = 11.155 MHz for the second V C O2 , which then produces f L O2 − 10.7 MHz = 455 kHz, instead of choosing f L O2 = 51.155 MHz to produce (51.155MHz − 50.7 MHz) = 455 kHz. Sums and differences between the pair (50.7, 10.7 MHz) and 11.155 MHz lead into the second IF frequencies taking values of (61.855, 39.545, 21.855 MHz, 455 kHz). We note that in this approach the wanted 455 kHz waveform is widely separated from the other three higher frequency waveforms, thus it is very easy to design IF amplifier whose LC bandwidth is centred at 455 kHz and narrow enough to attenuate the other three waveforms, Fig. 24.7. We recognize that the other available multiple choices would still result in mathematically correct solutions, however in the design process it is important to take into account practicality of the chosen component values, as well as relationship with the applicable industrial standards. 12.9. In this example we practice to use frequency multiplier, as oppose to frequency shifter (which requires a mixer). Plain frequency multiplication does not produce the sums and difference frequencies, and its only input terminal is to accept the waveform whose frequency needs to by multiplier by a constant. Therefore, it is straightforward to calculate: a) The output carrier frequency is f c = 27 × f 0 = 94.5 MHz, b) The output carrier deviation is Δ f c = 27×Δ f 0 = ±43.2 kHz, which is equivalnet to finding a difference between Δf cout = f cmax − f cmin = 94.5432 MHz − 94.4568 MHz = 86.4 kHz c) Frequency modulation is mf = Δf c 43.2 kHz × 100 % = × 100 % = 57.6 % B/2 75 kHz where we keep in mind that frequency modulation occupies half bandwidth for the positive deviation and half bandwidth for the negative frequency deviation. d) By using proportion Vmax 100 = 57.6 3.6 V pp ∴ Vmax = 6.25 V pp 12.10. Signal processing is, in reality, just implementation of system level mathematical functions. A hypothetical system in Fig. 12.5 takes a range of tones and applies mathematical operations to produce the output tone whose power is measured. In this example we practice to recognize difference between frequency shifting and frequency multiplication, and compare to linear addition of two functions. 1. The input FM signal occupies frequency range of 200 kHz ± 200 Hz, that is from 199.8 to 200.2 kHz range. Therefore, 24 RF Receivers: Solutions 261 64 × 200.2 kHz = 12.8128 MHz 64 × 199.8 kHz = 12.7872 MHz ∴ f 1 = 12.8 MHz ± 12.8 kHz 2. Similarly, 54 × 200 kHz = 10.8 MHz ∴ f 2 = 10.8 MHz ± 10.8 kHz 3. It is important to note that linear addition of two waveforms does not introduce new tones in the output waveform (frequency shifting is done by multiplication of two waveforms, which is non–linear operation). Thus, the output of the summing block contains only two frequency bands, f 1 ±Δ f 1 , and f 2 ±Δ f 2 , which is to say that LSB is centred around 10.8MHz with the bandwidth of B = Δf 2 = 21.6 kHz. Thus the LSB bandwidth occupies from 10.7892 to 10.8108 MHz and it requires Q= 10.8 MHz f2 = = 500 B 21.6 kHz 4. Waveform bandwidth B is defined at one half of the maximum power level at the centre frequency, which is by definition −3.010 dB, and in this example it is centred around LSB band at f 2 = 10.8 MHz. In other words, if the measured power at the edge of LSB waveform bandwidth is 1 mW, which is by definition equivalent to 0 dBm, than the maximum power is by definition P f 0 = 2 mW, which is equivalent to 3.010 dBm. The upper side band (USB) in this example occupies frequency space from f 1 (max) = 12.8128 MHz to f 1 (min) = 12.7872 MHz. In order to find attenuation of these two tones, we apply the attenuation formula for tones that are not centred at the resonant frequency f 0 and for the given Q, as f 1 (max) f2 12.8128 MHz 10.8 MHz − = − = 0.343 f2 f 1 (max) 10.8 MHz 12.8128 MHz f 1 (min) f2 12.7872 MHz 10.8 MHz y2 = − = − = 0.339 f2 f 1 (min) 10.8 MHz 12.7872 MHz y1 = which we use to calculate the attenuation relative to the maximum level as 262 24 RF Receivers: Solutions 1 = −22.349 dB 171.735 1 + (Q × y1 )2 1 1 = = −22.297 dB Ar (12.7872MHz) = 2 169.706 1 + (Q × y2 ) Ar (12.8128MHz) = 1 = These two equations gave us relative ratio of the maximum power (+3 dBm) and each of the tones, thus we write directly Ar (12.8128 MHz) = 3.010 dBm − 22.349 dB = −19.338 dBm Ar (12.7872 MHz) = 3.010 dBm − 22.297 dB = −19.287 dBm These results are summarized in Fig. 24.8. 12.11. By inspection of the equivalent block diagram of a three stage RF amplifier, Fig. 24.9, we write, 1. Power gain of the second stage A2 is found as difference between power levels at node 2iand node 1i(i.e. P2 − P1 ) and by following the signal power from the input node, thus we simply write Fig. 24.8 Solution 12.10: frequency spectrum of output signal (not to scale) Fig. 24.9 Solution 12.11: simplified block diagram a three stage RF amplifier 24 RF Receivers: Solutions Pin = −24 dBm and 263 A1 = 10 dB ∴ P1 = −24 dBm + 10 dB = −14 dBm P2 = 10 mW = 10 dBm ∴ A2 = P2 − P1 = 10 dBm − (−14 dBm) = 24 dB 2. Gain of the last stage in amplifier, is found by looking at difference between the output power and power level at the node 2ias 2 R = 1 W = 30 dBm ∴ Pout = i out L A3 = Pout − P2 = 30 dBm − 10 dBm = 20 dB which now gives us the total gain Atot of the three stage amplifier as Atot = A1 + A2 + A3 = 10 dB + 24 dB + 20 dB = 54 dB = 251, 188 V/V It is now possible to follow the input noise power as Pn (in) = k T, Beff = k T π B = 1.951 × 10−15 W = −117.0976 dBm 2 By knowing the input signal power Pin and the input noise power Pn (in) by definition we find the input side SNRin as SNRin = Pin = −24 dBm − (−117.0976 dBm) = 93.0976 dB Pn (in) (24.6) while signal to noise ratio at the output of the first stage is given as SNRo = 90.09655 dB. From these two numbers we find noise figure of the first stage by definition as NF1 = SNRo − SNRin = 3.010 dB Now we can summarize noise figure of the three stages and apply Friis’s formula for multistage amplifiers, as NF1 = 3.010 dB NF2 = 6.020 dB ∴ F1 = 2 ∴ ∴ F1 = 4 F1 = 8 NF3 = 9.030 dB ∴ F2 − 1 F3 − 1 3 7 F = F1 + + =2+ + = 2.303 A1 A1 A2 10 2511 which means that the total noise figure is NF = 10 log 2.303 = 3.622 dB. Therefore, the total output noise power is found as the sum of amplified input noise power plus the internally generated noise, i.e. 264 24 RF Receivers: Solutions Fig. 24.10 Solution 12.11: diagram of relationship among RF amplifier specifications and the total dynamic range Pn (out) = Pn (in) + A(tot) + NF(tot) = −59.4756 dBm ∴ Pn (out) = 1.128 nV vn (out) = Pn (out) R L ∗ 336 μV ∴ 3. We have now all information to establish dynamic range of this amplifier. It may be easier if the solution is presented graphically, Fig. 24.10. The last piece of information that is calculated is the total signal to noise ratio at the output node, SNRtot = Ps (out) = 30 dBm − (−54.4756 dBm) = 89.4756 dBm Pn (out) where we conclude that this amplifier needs quite a bit of improvement, because as it stands it can not process any useful signal, its DR = 0 dB. 12.12. Reception of wanted but weak signal from a faraway transmitter in the presence of unwanted but strong signal from a nearby transmitter, for example in a crowded bus when two persons standing next to each other use their cell phones at the same time, may be associated with a couple of quite unexpected side effects. In this example we explore some of the non–linear effects in an RF receiver. 1. Quantitative measure of an amplifier’s non–linearity is its 1dB compression point, which can be calculated directly from the non–linear transfer function as A1 dB = |a1 | 0.145 = |a3 | 0.145 2 = 1.042 0.267 ∴ A1 dB = 10 log 1.042 ∓ = 0.179 dB 2. Similarly, the third order intercept point IIP3 is also calculated directly from the non–linear transfer function 24 RF Receivers: Solutions 265 IIP3 = 4 |a1 | = 3 |a3 | 4 2 = 3.16 3 0.267 ∴ IIP3[dB] = 10 log 3.16 ∓ = 5 dB 3. Sensitivity of a receiver is calculated relative to the thermal noise floor. At the room temperature (290K), for the given data the receiver sensitivity is S = −174 dBm + NF + 10 log B + SNR = −174 dBm + 20 dB + 10 log 106 + 0 dB = −94 dBm 4. Dynamic range DR is calculated as difference between the sensitivity level (i.e. the noise floor) and either the IIP3[dB] or A1 dB [dB], hence 2 99 dB = 66 dB 3 2 DR = A1 dB − S = 0.179 dB − (−94 dBm) ∓ = 94 dB ∴ DReff = 94 dB ∗ 62.7 dB 3 DR = IIP3 − S = 5 dB − (−94 dBm) = 99 dB ∴ DReff = 5. The output signal level y(x) drops down to zero, and therefore the intended receiving signal is blocked, when y(x) = 0 ∴ A2 = ∴ 3 2 a 1 + a3 A 2 = 0 2 2a1 = 3(−a3 ) 2×2 = 2.234 V 3 × 0.267 Appendix A Physical Constants and Engineering Prefixes 1. Basic physical constants Physical constant Symbol Value Speed of light in vacuum Magnetic constant (vacuum permeability) Electric constant (vacuum permittivity) Characteristic impedance of vacuum Coulomb’s constant Elementary charge Bohr’s radius Boltzmann constant Plank’s constant Atomic mass unit Electron mass Proton mass Avogadro’s number c μ0 λ0 = 1/(μ0 c2 ) Z 0 = μ0 c ke = 1/4ω λ0 e, q a0 k h u me mp NA 2.99792458 × 108 m/s 4ω × 10−7 N/A2 8.854187817 × 10−12 F/m 376.730313461 8.987551787 × 109 Nm2 /C2 1.602176565 × 10−19 C 5.2917721092 × 10−11 m 1.3806488 × 10−23 J/K 6.62606957 × 1034 Js 1.660538921 × 10−27 kg 9.10938291 × 10−31 kg 1.672621777 × 10−27 kg 6.0221415 × 1023 1/mol 2. Basic engineering prefix system Tera Giga Mega Kilo Hecto Deca Deci Centi Milli Micro Nano Pico T G 1012 109 M 106 k 103 h 102 da 101 d c m µ 10−1 10−2 10−3 10−6 Femto Atto n p f a 10−9 10−12 10−15 10−18 R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 267 268 Appendix A: Physical Constants and Engineering Prefixes 3. Some useful physical quantities Name Symbol Quantity Unit Astronomical unit The Earth radius Mass of the Earth Mass of the Sun Gravitational acceleration Gravitational constant AU r mE mS g G 149.5978707 × 109 6.3568 × 106 5.9736 × 1024 1.9891 × 1030 9.806650 6.67384 × 10−11 m m kg kg ms−2 Nm2 kg−2 4. SI system of fundamental units Name Unit Quantity Symbol Metre Kilogram Second Ampere Kelvin Candela Mole m kg s A K cd mol Length Mass Time Electric current Thermodynamic temperature (−273.16√ C) Luminous intensity Amount of substance l m t I T Iv n 5. Derived units Name Unit Quantity Dimension Hertz Rad Newton Joule Watt Coulomb Volt Farad Ohm Siemens Weber Tesla Henry Hz rad N J W C V F S Wb T H Frequency Angle Force, weight Energy, work, heat Power Electric charge Voltage Electric capacitance Electric resistance Electric conductance Magnetic flux Magnetic field strength Inductance s−1 kg m s−2 N m = C V = W s = kg m2 s−2 V A = J s−1 = kg m2 s−3 As W A−1 = J C−1 = kg m2 s−3 A−1 C V−1 = kg−1 m−2 s4 A−2 V A−1 = kg m2 s−3 A−2 π −1 = kg−1 m−2 s3 A2 J A−1 = kg m2 s−2 A−1 V s m−2 = Wb m−2 N A−1 m−1 = kg s −2 A−1 V s A−1 = kg m2 s −2 A−2 Appendix B Maxwell’s Equations Complete set of Maxwell’s equation is listed here for the reference. 1. Gauss’s law for electric fields D · ds = q free, enc S ∇ · D = φ free integral form (B.1) differential form (B.2) 2. Gauss’s law for magnetic fields S B · ds = 0 integral form (B.3) ∇·B = 0 differential form (B.4) 3. Faraday’s law E · dl = − L ∇×E d dt =− B · ds integral form (B.5) S ∂B ∂t differential form (B.6) 4. Ampere–Maxwell law H · d l = I free, enc + L ∇×H d dt = J free + D · ds integral form (B.7) S ∂D ∂t differential form R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 (B.8) 269 Appendix C Second Order Differential Equation The three basic elements have voltages at their respective terminals as: v R = i R vL = L di dt vC = q C (C.1) and, if they are put together in a series circuit that includes a voltage source v(t), after applying KVL the circuit equation is v(t) = v L + v R + vC ∴ v(t) = L q di +i R+ dt C (C.2) However, we know that a current is derivative of charge in respect to time, hence we have the second order differential equation v(t) = L d 2q 1 dq + q +R dt 2 dt C ∴ v(t) = d 2q 1 R dq + q + 2 dt L dt LC (C.3) which is now solved, starting with its auxiliary quadratic equation 0 = x2 + 1 R x+ L LC R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 (C.4) 271 272 Appendix C: Second Order Differential Equation and its general solution with complex roots r1,2 ⎛ ⎞ 2 R 1⎝ R 4 ⎠ = − − ± 2 L L LC (C.5) Appendix D Complex Numbers A complex number is a neat way of presenting a point in (mathematical) space with two co-ordinates or, equivalently, it is a neat way to write two equations in the form of one. A general complex number is Z = a + jb, where a, b are real numbers referred to as real and imaginary parts, i.e. →(Z ) = a, and ∞(Z ) = b. Here is a reminder for basic operations with complex numbers, keep in mind that j 2 = −1. (a + jb) + (c + jd) = (a + c) + j (b + d) (D.1) (a + jb) − (c + jd) = (a − c) + j (b − d) (a + jb) (c + jd) = (ac − bd) + j (bc + ad) (a + jb) (c − jd) ac + bd (a + jb) bc − ad = = 2 +j 2 2 (c + jd) (c + jd) (c − jd) c +d c + d2 ≤ (a + jb) = (a − jb) |(a + jb)| = (a + jb)(a − jb) = (a 2 + b2 ) (D.2) (D.3) (D.4) (D.5) (D.6) It is much easier to visualize complex numbers and operations if we use vectors and trigonometry of right triangle, i.e. Pythagoras’ theorem. Imaginary part is always taking value at the y–axis and the real part is always on the x–axis. Therefore, alternative view of complex numbers is based on geometry, i.e. (a + jb) ≥ (|Z |, ε ) (D.7) where, of course, the absolute value of Z is just the length of the hypotenuse, and real and imaginary parts are just the two legs of the right–angled triangle, i.e. |Z | = ∗ Z Z≤ = (a 2 + b2 ), b ε = arctan a (D.8) where, ε is the phase angle, which, after using Euler’s formula becomes R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 273 274 Appendix D: Complex Numbers Fig. D.1 Illustration of complex numbers in [→(Z ), ∞(Z )] space, its equivalence to Pythagoras’ theorem and the vector arithmetics e j x ≥ cos x + j cos x (D.9) enables us to write really compact form of complex numbers Z = a + jb = |Z | e jε (D.10) which leads into another simple way of doing complex arithmetic, simply by using the absolute values and the arguments in combination with the algebra rules of exponential numbers, for example A e jε A B e jε B = AB e j (ε A +ε B ) (D.11) and we have the final link, A e jε ≥ A (cos ε + j sin ε ) (D.12) where, → A e jε = A cos ε ∞ A e jε = A sin ε (D.13) Appendix E Basic Trigonometric Identities sin(α + ω/2) = + cos α cos(α + ω/2) = − sin α (E.1) (E.2) sin(α + ω ) = − sin α cos(α + ω ) = − cos α (E.3) (E.4) sin(α ± ν) = sin α cos ν ± cos α sin ν cos(α ± ν) = cos α cos ν ∓ sin α sin ν (E.5) (E.6) sin2 α = 1/2 (1 − cos 2α) (E.7) cos α = 1/2 (1 + cos 2α) (E.8) sin α = 1/4 (3 sin α − sin 3α) (E.9) cos α = 1/4 (3 cos α + cos 3α) (E.10) 2 3 3 sin2 α cos2 α = 1/8 (1 − cos 4α) sin α cos α = 3 3 1/32 (3 sin 2α − sin 6α) (E.11) (E.12) cos α cos ν = 1/2 (cos(α − ν) + cos(α + ν)) sin α sin ν = 1/2 (cos(α − ν) − cos(α + ν)) (E.13) (E.14) sin α cos ν = 1/2 (sin(α + ν) + sin(α − ν)) cos α sin ν = 1/2 (sin(α + ν) − sin(α − ν)) (E.15) (E.16) R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 275 276 Appendix E: Basic Trigonometric Identities α∓ν α±ν cos 2 2 α−ν α+ν cos cos α + cos ν = 2 cos 2 2 α−ν α+ν sin cos α − cos ν = −2 sin 2 2 sin α ± sin ν = 2 sin (E.17) (E.18) (E.19) Appendix F Useful Algebra Equations 1. Binomial formula (x ± y)2 = x 2 ± 2x y + y 2 (F.1) (F.2) (x ± y) = x ± 3x y + 3x y ± y n(n − 1) n−2 2 n(n − 1)(n − 2) n−3 3 x x (x ± y)n = x n +nx n−1 + y + y · · · + yn 2! 3! (F.3) 3 3 2 2 3 where, n! = 1 · 2 · 3 · · · n and 0! ≥ 1. 2. Special cases x 2 − y 2 = (x − y)(x + y) (F.4) x 3 − y 3 = (x − y)(x 2 + x y + y 2 ) (F.5) x + y = (x + y)(x − x y + y ) 3 3 2 2 (F.6) x − y = (x − y )(x + y ) = (x − y)(x + y)(x + y ) 4 4 2 2 2 2 2 2 (F.7) 3. Useful Taylor series f (x) = ⇒ f (n) (a) (x − a)n (general expression around x = a point) n! n=0 ⇒ x2 x3 xn =1+x + + + ··· ex = n! 2! 3! (F.8) n=0 R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 277 278 Appendix F: Useful Algebra Equations sin x = cos x = tan x = ⇒ (−1)n 2n+1 x3 x5 =x− x + − ··· (2n + 1)! 3! 5! n=0 ⇒ n=0 ⇒ n=1 x4 (−1)n 2n x2 x =1− + − ··· (2n)! 2! 4! for all x for all x 2x 5 B2n (−4)n (1 − 4n ) 2n−1 x3 x + + ··· =x+ (2n)! 3 15 (F.9) (F.10) for |x| < ω 2 (F.11) Appendix G Bessel Polynomials 1. Bessel differential equation x2 d2 y dy +x + (x 2 − α 2 )y = 0 dx2 dx (G.1) 2. Relation with trigonometric functions cos(x sin α) =J0 (x) + 2 [J2 (x) cos 2α + J4 (x) cos 4α + · · · ] sin(x sin α) =2 [J1 (x) sin α + J3 (x) sin 3α + J5 (x) sin 5α + · · · ] cos(x cos α) =J0 (x) − 2 [J2 (x) cos 2α − J4 (x) cos 4α +J6 (x) cos 6 − J8 (x) cos 8α · · · ] (G.2) sin(x cos α) =2 [J1 (x) cos α − J3 (x) sin 3α + J5 (x) sin 5α + · · · ] (G.5) (G.3) (G.4) 3. Bessel series x2 x4 x6 J0 (x) = 1 − 2 + 2 2 − 2 2 2 + · · · 2 2 ·4 2 ·4 ·6 x4 x2 x + + · · · 1− 2 J1 (x) = 2 2 · 2 2 · 24 · 2 · 3 x4 x2 xn Jn (x) = n + + 1− 2 4 2 n! 2 · (n + 1) 2 · 2 · (n + 1) · (n + 2) (−1) p x 2 p + ··· p! 22 p (n + 1)(n + 2) · · · (n + p) R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 (G.6) (G.7) (G.8) 279 280 Appendix G: Bessel Polynomials 4. Bessel approximations For very large x Bessel function reduces to Jn (x) = nω ω 2 cos x − − ωx 2 4 (G.9) Glossary The following glossary of technical terms is provided for reference only. The reader is advised to further study the terms in appropriate books, for example a technical dictionary. Absolute zero is the theoretical temperature at which entropy would reach its minimum value. By international agreement, absolute zero is defined as 0 K on the Kelvin scale and as −273.15 √ C on the Celsius scale. Active device is an electronic component that has signal gain larger then one, for example a transistor. Compare to passive device. Active mode is a condition for a BJT transistor where emitter–base junction is forward biased, while the collector–base junction is reverse biased. Admittance reciprocal of impedance, is the measure of how easy AC current flows in a circuit, measured in Siemens [S]. Ampere [A] is the unit of electric current defined as the flow of one coulomb of charge per second. Ampère’s Law states that a current flowing into a wire generates a magnetic flux that encircles the wire following the “right hand rule”, where the thumb points the direction of the current flow and the other curled fingers show direction of the magnetic field. Study Maxwell’s equations for more details. Amplifier is a liner device that implements mathematical equation y = A x, where y is the amplified output signal, A is the gain coefficient, and x is the input signal. Analogue is the general class of devices and circuits meant to process a continuous signal. Compare with digital and sampled signals. Attenuation is gain lower then one. Attenuator is a device that reduces a gain, without introducing phase of frequency distortion. R. Sobot, Wireless Communication Electronics by Example, DOI: 10.1007/978-3-319-02871-2, © Springer International Publishing Switzerland 2014 281 282 Glossary Automatic Gain Control a closed loop feedback system designed to hold the overall gain as constant as possible. Average power is the power averaged over one period. Bandwidth is the difference between upper and lower frequencies at which the amplitude response is 3 dB bellow the maximum level. It is equivalent to half–power bandwidth. Base is the region of BJT transistor between the emitter and collector. Bel [B] is a dimensionless unit used to express ratio of two powers. More practical unit is decibel [dB]. Beta β is the current gain of a BJT transistor. It is the ratio of the change of collector current relative to change of the base current, ν = d IC /d I B . Bias is a steady current or voltage used to set operating conditions of the device. Breakdown voltage is the voltage at which the reverse current of a reverse biased pn junction suddenly rises. If the current is not limited the device is destroyed. Capacitance is the ratio between the electric charge two conductors and voltage between the two. Capacitor is a device made of two conductors separated by an insulating material for purposes of storing electric charge, i.e, energy. Celsius [◦ C] is a unit increment of temperature unit defined as 1/100 between the freezing (0 √ C) and boiling (100 √ C) point of water. Compare to Kelvin and Fahrenheit. Characteristic Curve a family of I–V plots shown for several parameter values. Characteristic impedance is the entry point impedance of an infinitely long transmission line. Charge a basic property of elementary particles of matter (electrons, protons, etc.) responsible for creating a force field. Circuit the interconnection of devices, both passive and active, for the purpose of synthesizing a mathematical function. Common mode average value of a sinusoidal waveform. Coulomb [C] the unit of electric charge defined as the charge transported through a unity area in one second by an electric current of one ampere. An electron has a charge of 1.602 × 10−19 C. Coulomb’s Law defines force between two electric charges in space. Common–base configuration a single BJT amplifier configuration where the base potential is fixed, emitter serves as the input and collector as the output terminal. Glossary 283 Also known as current buffer. Equivalent to common–gate configuration for MOS amplifiers. Common–collector combination a single BJT amplifier transistor configuration where the collector potential is fixed, the base serves as the input and emitter as the output terminal. Also known as voltage buffer or voltage follower. Equivalent to common–drain configuration for MOS amplifiers. Common–emitter a single BJT amplifier transistor configuration where the emitter potential is fixed, while the base serves as the input and collector as the output terminal. Also known as gm stage. Equivalent to common–source configuration for MOS amplifiers. Conductivity represents the ability of a matter to conduct electricity. Conductor a material that easily conducts electricity. Current transfer of electrical charge through a unity size area per unit of time. Current gain the ratio of current at output and current at the input terminals of a device or circuit. Current source a device capable of provided constant current value regardless of the voltage at its terminals. DC direct current, i.e. current that flows in one direction only. DC biasing a process of setting stable operating point of a device. DC load line a straight line across a family of I–V curves that shows movement of the operating point as the output voltage changes for a given load. DC analysis mathematical procedure to calculate stable operating point. Decibel [dB] is a dimensionless unit used to express ratio of two powers. Ten times smaller then bel [B]. Device a single discrete device, for instance resistor, transistor, capacitor, etc. Dielectric a material that is not good in conducting electricity, i.e. opposite of a conductor. Characterized by dielectric constant. Differential amplifier an amplifier that operates on differential signals. Differential signal a difference between two sinusoidal signals of same frequency, same amplitude, same common mode, and with phase difference of 180√ C. Digital is the general class of devices and circuits meant to process a sampled signal. Compare with analogue and continuous signals. Diode a nonlinear two terminal device which obeys exponential transfer function. Used as unidirectional switch. 284 Glossary Discrete device an individual electrical component that exhibits behaviour associated with resistor, transistor, capacitor, inductor, etc. Compare with distributed components. Dynamic range the difference of the maximum acceptable signal level and the minimum acceptable signal level. Electric field a field generated by an electric charge, detected by existence of the electric force within space surrounding the charge. Electrical noise any unwanted electrical signal. Electromagnetic (EM) wave a phenomena exhibited by flow of electromagnetic energy through the space. In special case of standing wave this definition may need more explanation. Electron a fundamental particle that carries negative charge. Electronics is the branch of science and technology which makes use of the controlled motion of electrons through different media and vacuum. Electrostatics is the branch of science that deals with the phenomena arising from stationary or slow–moving electric charges. Emitter a region of a BJT transistor from which charges are injected into the base. One of the three terminal points at BJT device. Energy a concept that can be loosely defined as the ability of a body to perform work. Equivalent circuit a simplified version of the original circuit that still performs the same function. Equivalent noise temperature the absolute temperature at which a perfect resistor would generate the same noise as its equivalent real component at the room temperature. Fall time the time during a pulse decreases from 90 to 10 % of its maximum value (sometimes defined between the 80 and 20 % points). Farad [F] the unit of capacitance of a capacitor. One Farad is very large, capacitance of the Earth’s ionosphere with respect to the ground is around 50 mF. Faraday’s Law the law of electromagnetic induction. See also Faraday’s cage. Faraday cage an enclosure that blocks out external static electric fields. Feedback the process of coupling output and input terminals through an external path. Negative feedback increases stability of an amplifier for the price of reduced gain, positive feedback boosts gain and is needed for creating oscillating circuits. Field a concept that describes a flow of energy through the space. Glossary 285 Field–Effect Transistor (FET) a transistor controlled by two perpendicular electrical fields used to change resistivity of semiconductor material underneath the gate terminal and force current between the source and drain terminals. Flicker noise or 1/f noise. A random noise in semiconductors whose power spectral density is, to the first approximation, inverse to frequency. Frequency the number of complete cycles per second. Frequency response a curves showing gain and phase change of a device as a function of frequency. Gain the ratio of signal values measured at output and input. Gauss’s Law is a law relating the distribution of electric charge to the resulting electric field. Ground an arbitrary potential reference point that all other potentials in a circuit are compared against. Difference between the ground potential and the node potential is expressed as voltage at that node. The ground node may or may not be the lowest potential in the circuit. Henry [H] the unit for self and mutual inductance. Hertz [Hz] the unit for frequency, equal to one cycle per second. Impedance resistance of a two terminal device at any frequency. Inductance is the property in an electrical circuit where a change in the electric current through that circuit induces an electromotive force (EMF) that opposes the change in current. Inductor is a passive electrical component that can store energy in a magnetic field created by the electric current passing through it. Input current, voltage, power, or other driving force applied to a circuit or device. Insertion loss the attenuation resulting from inserting a circuit between source and load. Insulator a material with very low conductivity. Intermediate Frequency (IF) is a frequency to which a carrier frequency is shifted as an intermediate step in transmission or reception. Intermodulation Products additional harmonics created due to non–linear device processing two or more single tone signals. Junction a joining of two semiconductor materials. Junction capacitance capacitance associated with pn junction region. Kelvin [K] the unit increment of temperature on the absolute temperature scale. Kirchhoff’s Current Law (KCL) is the law of conservation of charge. 286 Glossary Kirchhoff’s Voltage Law (KVL) is based on the conservation of “energy given/taken by potential field” (not including energy taken by dissipation). Large signal a signal with large enough amplitude to move the operating point of a device far away from its original biasing point. Hence, non–linear model of the device must be used. Law of conservation of energy is the fundamental law of nature. It states that energy can neither be created nor destroyed, it can only be transformed from one state to another. Large signal analysis a method used to describe behaviour of devices stimulated by large signals. Hence, the nonlinear devices in terms of the underlying nonlinear equations. Linear network A network in which the parameters of resistance, inductance, and capacitance are constant with respect to current or voltage, and in which the voltage or current of sources is independent of or directly proportional to other voltages and currents, or their derivatives, in the network. Load a device that absorbs energy and converts it into another form. Local Oscillator (LO) an oscillator used to generate single tone signal that is needed for upconversion and downconversion operations. Lossless a theoretical device that does not dissipate energy. Low Noise Amplifier (LNA) is an electronic amplifier used to amplify very weak signals captured by an antenna. Lumped element a self contained and localized element that offers one particular property, for instance, resistance over a range of frequencies. Magnetic field a field generated by magnetic energy, detected by existence of the magnetic force within space surrounding the magnet. Matching circuit a passive circuit designed to interface two networks for purpose of enabling maximum energy transfer between the two networks. Matching a concept of connecting two networks for purpose of enabling maximum energy transfer between them. Maxwell’s equations are a set of four partial differential equations that relate the electric and magnetic fields to their sources, charge density and current density. These equations can be combined to show that light is an electromagnetic wave. Individually, the equations are known as Gauss’s law, Gauss’s law for magnetism, Faraday’s law of induction, and Ampère’s law with Maxwell’s correction. These four equations, together with the Lorentz force law are the complete set of laws of classical electromagnetism. Metal Oxide Semiconductor Field Effect Transistor (MOSFET) Originally, a sandwich of aluminum–Silicone Dioxide–Silicon was used to manufacture FET transistors. Although, a metal is not used anymore for creating gates for FET transistors, the name has stuck. Glossary 287 Microwaves waves in the frequency range of 1–300 GHz, i.e. with a wavelength of three hundreds to one millimetre. Mixer a nonlinear three port device used for frequency shifting operation. Negative resistance a resistance of a device or circuit where an increase in the current entering a port results in a decreased voltage across the same port. Noise any unwanted signal that interferes with the wanted signal. Noise Figure (NF) is a measure of degradation of the signal–to–noise ratio (SNR), caused by components in a radio frequency (RF) signal chain. Nonlinear a system which does not satisfy the superposition principle, or whose output is not directly proportional to its input. Norton’s Theorem is the dual of Thévenin’s theorem, states that any collection of voltage sources, current sources, and resistors with two terminals is electrically equivalent to an ideal current source in parallel with a single resistor. NPN transistor a transistor with p–type base and n–type collector and emitter. Octave the interval between any two frequencies having a ration of 2:1 Ohm [] unit of resistance, as defined by Ohm’a law. Ohm’s Law states that the change of current through a conductor between two points is directly proportional to the change of voltage across the two points, and inversely proportional to the resistance between them. One–dB gain compression point the point at which the power gain at the output of a nonlinear device or circuits is reduced by 1 dB relative to its small signal linear model predicted value. Open loop gain the ratio of the output signal and the input signals of an amplifier with now feedback path present. Oscillator an electronic device the generates a single tone (or some other regular shape) signal at predetermined frequency. Output current, voltage, power, or driving force delivered at the output terminals. Pasive a component that does not have gain larger then one. Phase the angular property of a wave. Phase shifter a two port network which provides a controllable phase shift of the RF signals. Phasor a mathematical representation of a sine wave by a rotating vector. Power the rate at which work is performed. Quality factor (Q factor) is a dimensionless parameter that characterizes a resonator’s bandwidth relative to its centre frequency. 288 Glossary Radio frequency (RF) any frequency at which coherent electromagnetic radiation of energy is possible. Reactance is the opposition of a circuit element to a change of current, caused by the build–up of electric or magnetic fields in the element. Reactive element an inductor and capacitor. Reflected waves the waves reflected from a discontinuity in the medium they are traveling in. Resistance of an object is a measure of its opposition to the passage of a steady electric current. Resistor a lumped element designed to have a certain resistance. Resonant frequency the frequency at which a given system or circuit responds with maximum amplitude when driven by an external single tone. Root Mean Square (RMS) is the square root of the arithmetic mean (average) of the squares of the original values. Saturation a circuits condition whereby an increase of the input signal does not produce expected change at the output. Self–resonant frequency the frequency at which all real devices or circuits start to oscillate due to the internal parasitic inductances and capacitances. Signal an electrical quantity containing an information that is carried by voltage or current. Single ended circuit a circuit operating on single ended signals, as oppose to differential signals. Skin effect is the tendency of an alternating electric current (AC) to distribute itself within a conductor so that the current density near the surface of the conductor is greater than that at its core. That is, the electric current tends to flow at the “skin” of the conductor, at an average depth called the skin depth. Small signal a low amplitude signal the occupies very narrow region that is centred at the biasing point. Hence, linear model alway applies. Small signal amplifier an amplifier that operates only in the linear region. Space is the boundless, three–dimensional extent in which objects and events occur and have relative position and direction. Stability ability of a circuit to stay away from the self–resonating frequency. Standing wave is a wave that remains in a constant position. It can arise in a stationary medium as a result of interference between two waves traveling in opposite directions. For waves of equal amplitude traveling in opposing directions, there is on average no net propagation of energy. Glossary 289 Standing Wave Ratio (SWR) the ratio of the maximum to the minimum value of current or voltage in a standing wave. Thévenin’s theorem is the dual of Norton’s theorem, states that any combination of voltage sources, current sources and resistors with two terminals is electrically equivalent to a single voltage source and a single series resistor. Third order intercept point (IP3) is a measure for weakly nonlinear systems and devices, for example receivers, linear amplifiers and mixers. Time a concept used to order the sequence of events. Transmission line any system of conductors capable of efficiently conducting electromagnetic energy. Tuned circuit a circuit consisting of inductance and capacitance that can be adjusted for resonance at the desired frequency. Tuning process of adjusting resonant frequency of the tuned circuit. Varactor a two terminal pn junction used as a voltage controlled capacitor. Volt [V] a unit for potential difference. Voltage controlled oscillator (VCO) an oscillator whose output frequency is controlled by a voltage. Voltage divider is a simple linear circuit that produces an output voltage that is a fraction of its input voltage. Voltage follower amplifier also known as voltage buffer amplifier provides electrical impedance transformation from one circuit to another. Voltage source a device capable of provided constant voltage value regardless of the current at its terminals. Wave a disturbance that progresses from point in space to another. Wavefront a surface having constant phase. Wavelength space distance between two consecutive points having the same phase. Wave propagation journey of the wave through space. White noise random signal that consists of all possible frequencies from zero to infinity. Work the advancement in space of a point under application of force. 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