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Transmission Lines Complex Numbers, Phasors and Circuits Complex numbers are defined by points or vectors in the complex plane, and can be represented in Cartesian coordinates z a jb j 1 or in polar (exponential) form z A exp( j) A cos jA sin a A cos real part b A sin imaginary part where 2 A a b © Amanogawa, 2000 – Digital Maestro Series 2 tan 1 b a 1 Transmission Lines Im z b A a Note : Re z A exp( j) A exp( j j 2 n) © Amanogawa, 2000 – Digital Maestro Series 2 Transmission Lines Every complex number has a complex conjugate z* a jb * a jb so that z z* ( a jb) ( a jb) a2 b2 z 2 A2 In polar form we have z* A exp( j) * A exp( j) A exp j 2 j A cos jA sin © Amanogawa, 2000 – Digital Maestro Series 3 Transmission Lines The polar form is more useful in some cases. For instance, when raising a complex number to a power, the Cartesian form z n ( a jb) ( a jb) ( a jb) is cumbersome, and impractical for noninteger exponents. polar form, instead, the result is immediate In n z n A exp( j) An exp jn In the case of roots, one should remember to consider + 2k as argument of the exponential, with k = integer, otherwise possible roots are skipped: n z n A exp j j 2 k n A exp j j 2 k n n The results corresponding to angles up to 2 are solutions of the root operation. © Amanogawa, 2000 – Digital Maestro Series 4 Transmission Lines In electromagnetic problems it is often convenient to keep in mind the following simple identities j exp j 2 j exp j 2 It is also useful to remember the following expressions for trigonometric functions exp( jz) exp( jz) exp( jz) exp( jz) cos z ; sin z 2 2j resulting from Euler’s identity exp( jz) cos( z) j sin( z) © Amanogawa, 2000 – Digital Maestro Series 5 Transmission Lines Complex representation is very useful for time-harmonic functions of the form A cos t Re A exp j t j Re A exp j exp j t Re A exp j t The complex quantity A A exp j contains all the information about amplitude and phase of the signal and is called the phasor of A cos t If it is known that the signal is time-harmonic with frequency , the phasor completely characterizes its behavior. © Amanogawa, 2000 – Digital Maestro Series 6 Transmission Lines Often, a time-harmonic signal may be of the form: A sin t and we have the following complex representation A sin t Re jA cos t j sin t Re jA exp j t j Re A exp j / 2 exp j exp j t Re A exp j / 2 exp j t Re A exp j t with phasor A A exp j / 2 This result is not surprising, since cos( t / 2) sin( t ) © Amanogawa, 2000 – Digital Maestro Series 7 Transmission Lines Time differentiation can be greatly simplified by the use of phasors. Consider for instance the signal V ( t ) V0 cos t with phasor V V0 exp j The time derivative can be expressed as V ( t) V0 sin t t Re jV0 exp j exp j t jV0 exp j j V © Amanogawa, 2000 – Digital Maestro Series is the phasor of V ( t) t 8 Transmission Lines With phasors, time-differential equations for time harmonic signals can be transformed into algebraic equations. Consider the simple circuit below, realized with lumped elements R L i (t) v (t) C This circuit is described by the integro-differential equation d it 1 t v( t ) L Ri i( t ) dt dt C © Amanogawa, 2000 – Digital Maestro Series 9 Transmission Lines Upon time-differentiation we can eliminate the integral as d2 i t d v( t ) di 1 L R i( t ) dt dt C dt 2 If we assume a time-harmonic excitation, we know that voltage and current should have the form v( t ) V0 cos( t V ) phasor V V0 exp( jV ) i( t ) I0 cos( t I ) phasor I I0 exp( j I ) If V0 and V are given, I0 and I are the unknowns of the problem. © Amanogawa, 2000 – Digital Maestro Series 10 Transmission Lines The differential equation can be rewritten using phasors L Re 2 I exp j t R Re j I exp j t 1 Re I exp j t Re j V exp j t C Finally, the transform phasor equation is obtained as 1 V R j L j IZI C where Z Impedance R Resistance 1 j L C Reactance © Amanogawa, 2000 – Digital Maestro Series 11 Transmission Lines The result for the phasor current is simply obtained as V V I0 exp j I I 1 Z R j L j C which readily yields the unknowns I0 and I . The time dependent current is then obtained from i( t ) Re I0 exp j I exp j t I0 cos t I © Amanogawa, 2000 – Digital Maestro Series 12 Transmission Lines The phasor formalism provides a convenient way to solve timeharmonic problems in steady state, without having to solve directly a differential equation. The key to the success of phasors is that with the exponential representation one can immediately separate frequency and phase information. Direct solution of the timedependent differential equation is only necessary for transients. Integro-differential equations Transform Algebraic equations based on phasors I=? i(t)=? Direct Solution ( Transients ) i(t) © Amanogawa, 2000 – Digital Maestro Series Solution AntiTransform I 13 Transmission Lines The phasor representation of the circuit example above has introduced the concept of impedance. Note that the resistance is not explicitly a function of frequency. The reactance components are instead linear functions of frequency: Inductive component proportional to Capacitive component inversely proportional to Because of this frequency dependence, for specified values of L and C , one can always find a frequency at which the magnitudes of the inductive and capacitive terms are equal r L 1 r C r 1 LC This is a resonance condition. The reactance cancels out and the impedance becomes purely resistive. © Amanogawa, 2000 – Digital Maestro Series 14 Transmission Lines The peak value of the current phasor is maximum at resonance I0 | I0 | V0 1 2 R L C 2 IM r © Amanogawa, 2000 – Digital Maestro Series 15 Transmission Lines Consider now the circuit below where an inductor and a capacitor are in parallel I L R C V The input impedance of the circuit is 1 j C Zin R j L © Amanogawa, 2000 – Digital Maestro Series 1 R j L 1 2 LC 16 Transmission Lines When 0 Zin R 1 LC Zin Zin R At the resonance condition r 1 LC the part of the circuit containing the reactance components behaves like an open circuit, and no current can flow. The voltage at the terminals of the parallel circuit is the same as the input voltage V. © Amanogawa, 2000 – Digital Maestro Series 17 Transmission Lines Power in Circuits Consider the input impedance of a transmission line circuit, with an applied voltage v(t) inducing an input current i(t). i(t) v(t) Zin For sinusoidal excitation, we can write v ( t ) V0 cos( t ) i ( t ) I0 cos( t ) /2 , /2 where is the phase difference between voltage and current. Note that = 0 only when the input impedance is real (purely resistive). © Amanogawa, 2000 – Digital Maestro Series 18 Transmission Lines The time-dependent input power is given by P( t ) v ( t ) i ( t ) V0 I0 cos( t ) cos( t ) V0 I0 cos() cos(2 t ) 2 The power has two (Fourier) components: (A) an average value V0 I0 cos() 2 (B) an oscillatory component with frequency 2f V0 I0 cos(2 t ) 2 © Amanogawa, 2000 – Digital Maestro Series 19 Transmission Lines The power flow changes periodically in time with an oscillation like (B) about the average value (A). Note that only when = 0 we have cos() = 1, implying that for a resistive impedance the power is always positive (flowing from generator to load). When voltage and current are out of phase, the average value of the power has lower magnitude than the peak value of the oscillatory component. Therefore, during portions of the period of oscillation the power can be negative (flowing from load to generator). This means that when the power flow is positive, the reactive component of the input impedance stores energy, which is reflected back to the generator side when the power flow becomes negative. For an oscillatory excitation, we are interested in finding the behavior of the power during one full period, because from this we can easily obtain the average behavior in time. From the point of view of power consumption, we are also interested in knowing the power dissipated by the resistive component of the impedance. © Amanogawa, 2000 – Digital Maestro Series 20 Transmission Lines We can also rewrite the time-dependent current as i ( t ) I0 cos( t ) I0 cos cos t I0 sin sin t where we have used the trigonometry formula cos A B cos A cos B sin A sin B This result yields an equivalent expression for the power P( t ) V0 cos( t ) I0 cos( t ) cos() V0 cos( t ) I0 sin( t ) sin() V0 I0 1 2 sin() sin(2 t ) V0 I0 cos() cos ( t ) 1 2 Active (Real) Power © Amanogawa, 2000 – Digital Maestro Series Reactive Power 21 Transmission Lines The active power corresponds to the power dissipated by the resistive component of the impedance, and it is always positive. The reactive power corresponds to the power stored and then reflected by the reactive component of the impedance. It oscillates from positive to negative during the period. Until now we have discussed properties of instantaneous power. Since we are considering time-harmonic periodic signals, it is very convenient to consider the time-average power 1 T P( t ) P( t ) dt T 0 where T = 1 / f is the period of the oscillation. We can use either the Fourier or the active/reactive power formulation to determine the time-average power. © Amanogawa, 2000 – Digital Maestro Series 22 Transmission Lines Fourier representation 1 T V0 I0 P( t ) cos() dt T 0 2 1 T V0 I0 cos(2 t ) dt 0 T 2 V0 I0 cos() 0 2 As one should expect, the time-average power flow is simply given by the Fourier component corresponding to the average of the original signal. © Amanogawa, 2000 – Digital Maestro Series 23 Transmission Lines Active/Reactive power representation 1 T P( t ) V0 I0 cos() cos2 ( t ) dt T 0 1 T V0 I0 sin() sin(2 t ) dt 0 T 2 V0 I0 cos() 0 2 This result tells us that the time-average power flow is the average of the active power. The reactive power has zero time-average, since power is stored and completely reflected by the reactive component of the input impedance during the period of oscillation. © Amanogawa, 2000 – Digital Maestro Series 24 Transmission Lines The maximum of the reactive power is V I V I max {Preac } = max{ 0 0 sin (φ ) sin ( 2ω t )} = 0 0 sin ( φ ) 2 2 Since the time-average of the reactive power is zero, we often use the maximum value above as an indication of the reactive power. The sign of the phase φ tells us about the imaginary part of the impedance (reactance) φ > 0 The reactance is inductive Current is lagging with respect to voltage Voltage is leading with respect to current φ < 0 The reactance is capacitive Voltage is lagging with respect to current Current is leading with respect to voltage © Amanogawa, 2000 – Digital Maestro Series 25 Transmission Lines If the net reactance is inductive V Z I R I j L I Im j L I V Current lags >0 I © Amanogawa, 2000 – Digital Maestro Series RI Re 26 Transmission Lines If the net reactance is capacitive Im 1 VZI RI j I C I RI Re Voltage lags <0 V - j I / C © Amanogawa, 2000 – Digital Maestro Series 27 Transmission Lines In many engineering situations, we use the root-mean-square (r.m.s.) values of quantities. For a given signal v( t ) V0 cos( t ) the r.m.s. value is defined as 1 T 2 1 T 2 2 Vrms V t dt V cos cos t d t 0 0 T 0 T 0 1 2 2 1 V0 cos d V0 2 0 2 This result is valid for sinusoidal signals. Each signal shape corresponds to a specific coefficient ( peak factor = V0 / Vrms ) that allows one to convert directly from peak to r.m.s. values. © Amanogawa, 2000 – Digital Maestro Series 28 Transmission Lines The peak factor for sinusoidal signals is V0 Vrms 2 1.4142 For a symmetric triangular signal the peak factor is V0 3 1.732 Vrms V0 t For a symmetric square signal the peak factor is simply V0 1 Vrms © Amanogawa, 2000 – Digital Maestro Series V0 t 29 Transmission Lines For a non-sinusoidal periodic signal, we can determine the r.m.s. value by using a very important theorem of vector spaces. If we decompose the non-sinusoidal signal into its Fourier components V ( t ) Vav V1 ( t ) V2 ( t ) V3 ( t ) k Vk t then 2 V t rms k Vk ( t)rms 2 2 V t k k rms So, the r.m.s. value of the signal is computed as 2 Vrms Vav V1 rms V2 rms V3 rms 2 © Amanogawa, 2000 – Digital Maestro Series 2 2 30 Transmission Lines So far, we have used peak values for the amplitude of voltage and current. In terms of r.m.s. values, the time-average power for a sinusoidal signal is V0 I0 cos() Vrms Irms cos() P( t ) 2 2 Finally, we can relate the time-average power to the phasors of voltage and current. Since v ( t ) V0 cos( t ) Re V0 exp( j t ) i ( t ) I0 cos( t ) Re I0 exp( j) exp( j t ) we have phasors V V0 © Amanogawa, 2000 – Digital Maestro Series I I0 exp( j) 31 Transmission Lines The time-average power in terms of phasors is given by 1 1 * P( t ) Re V I Re V0 I0 exp( j) 2 2 V0 I0 cos() 2 Note that one must always use the complex conjugate of the phasor current to obtain the time-average power. It is important to remember this when voltage and current are expressed as functions of each other. Only when the impedance is purely resistive, I = I* = I0 since = 0. Also, note that the time-average power is always a real positive quantity and that it is not the phasor of the time-dependent power. It is a common mistake to think so. © Amanogawa, 2000 – Digital Maestro Series 32 Transmission Lines Now, we consider power flow including explicitly the generator, to understand in which conditions maximum power transfer to a load can take place. ZR Vin VG ZG ZR 1 Iin VG ZG ZR 1 * Pin Re Vin Iin 2 © Amanogawa, 2000 – Digital Maestro Series Iin ZG VG ZR Vin Generator Load 33 Transmission Lines As a first case, we examine resistive impedances ZG RG ZR RR Voltage and current are in phase at the input. The time-average power dissipated by the load is RR 1 1 * VG P( t ) VG RG RR RG RR 2 RR 1 2 VG 2 ( RG RR )2 © Amanogawa, 2000 – Digital Maestro Series 34 Transmission Lines To find the load resistance that maximizes power transfer to the load for a given generator we impose d P( t ) 0 dRR from which we obtain d RR 0 dRR ( RG RR )2 ( RG RR )2 2 RR ( RG RR ) ( RG RR ) 4 ( RG RR ) 2 RR 0 0 RR RG We conclude that for maximum power transfer the load resistance must be identical to the generator resistance. © Amanogawa, 2000 – Digital Maestro Series 35 Transmission Lines Let’s consider now complex impedances ZR RR jX R ZG RG jXG For maximum power transfer, generator and load impedances must be complex conjugate of each other * ZR ZG RR RG X R XG This can be easily understood by considering that, to maximize the active power supplied to the load, voltage and current of the generator should remain in phase. If the reactances of generator and load are opposite and cancel each other along the path of the current, the generator will only see a resistance. Voltage and current will be in phase with maximum power delivered to the load. © Amanogawa, 2000 – Digital Maestro Series 36 Transmission Lines The total time-average power supplied by the generator in conditions of maximum power transfer is 1 1 2 1 1 2 1 * Ptot Re VG Iin VG VG RR 2 2 2 RR 4 The time-average power supplied to the load is * ZR 1 1 1 * * VG Pin Re Vin Iin Re VG 2 2 ZG ZR ZG ZR ! RR jX R 1 2 1 2 1 VG Re VG RR 2 4 RR ! 8 © Amanogawa, 2000 – Digital Maestro Series 37 Transmission Lines The power dissipated by the internal generator impedance is 1 PG Re 2 1 2 VG 4 * (VG Vin ) Iin 1 1 2 1 1 2 1 VG VG RR RR RR 8 8 We conclude that, in conditions of maximum power transfer, only half of the total active power supplied by the generator is actually used by the load. The generator impedance dissipates the remaining half of the available active power. This may seem a disappointing result, but it is the best one can do for a real generator with a given internal impedance! © Amanogawa, 2000 – Digital Maestro Series 38 Transmission Lines Transmission Line Equations A typical engineering problem involves the transmission of a signal from a generator to a load. A transmission line is the part of the circuit that provides the direct link between generator and load. Transmission lines can be realized in a number of ways. Common examples are the parallel-wire line and the coaxial cable. For simplicity, we use in most diagrams the parallel-wire line to represent circuit connections, but the theory applies to all types of transmission lines. Load Generator ZG VG © Amanogawa, 2000 – Digital Maestro Series Transmission line ZR 39 Transmission Lines Examples of transmission lines d d D d Two-wire line D Coaxial cable W t D Microstrip © Amanogawa, 2000 – Digital Maestro Series 40 Transmission Lines If you are only familiar with low frequency circuits, you are used to treat all lines connecting the various circuit elements as perfect wires, with no voltage drop and no impedance associated to them (lumped impedance circuits). This is a reasonable procedure as long as the length of the wires is much smaller than the wavelength of the signal. At any given time, the measured voltage and current are the same for each location on the same wire. Load Generator ZG VG ZR VR VG ZG ZR © Amanogawa, 2000 – Digital Maestro Series VR VR VR ZR L << 41 Transmission Lines Let’s look at some examples. The electricity supplied to households consists of high power sinusoidal signals, with frequency of 60Hz or 50Hz, depending on the country. Assuming that the insulator between wires is air ( 0), the wavelength for 60Hz is: c 2.999 108 5.0 106 m 5, 000 km 60 f which is the about the distance between S. Francisco and Boston! Let’s compare to a frequency in the microwave range, for instance 60 GHz. The wavelength is given by c 2.999 108 5.0 10 3 m 5.0 mm f 60 109 which is comparable to the size of a microprocessor chip. Which conclusions do you draw? © Amanogawa, 2000 – Digital Maestro Series 42 Transmission Lines For sufficiently high frequencies the wavelength is comparable with the length of conductors in a transmission line. The signal propagates as a wave of voltage and current along the line, because it cannot change instantaneously at all locations. Therefore, we cannot neglect the impedance properties of the wires (distributed impedance circuits). V (z) V e j z V e j z Generator Load ZG VG V( 0 ) V( z ) V( L ) ZR L © Amanogawa, 2000 – Digital Maestro Series 43 Transmission Lines Note that the equivalent circuit of a generator consists of an ideal alternating voltage generator in series with its actual internal impedance. When the generator is open ( ZR ) we have: Iin 0 and Vin VG If the generator is connected to a load ZR VG Iin ZG ZR VG ZR Vin ZG ZR If the load is a short ( ZR VG Iin ZG and © Amanogawa, 2000 – Digital Maestro Series Iin ZG VG Vin ZR 0) Vin 0 Generator Load 44 Transmission Lines The simplest circuit problem that we can study consists of a voltage generator connected to a load through a uniform transmission line. In general, the impedance seen by the generator is not the same as the impedance of the load, because of the presence of the transmission line, except for some very particular cases. Zin ZR Zin Transmission line ZR only if Ln 2 [ n integer ] L Our goal is to determine the equivalent impedance seen by the generator, that is, the input impedance of the line terminated by the load. Once that is known, standard circuit theory can be used. © Amanogawa, 2000 – Digital Maestro Series 45 Transmission Lines Load Generator ZG Transmission line VG ZR Equivalent Load Generator ZG VG © Amanogawa, 2000 – Digital Maestro Series Zin 46 Transmission Lines A uniform transmission line is a “distributed circuit” that we can describe as a cascade of identical cells with infinitesimal length. The conductors used to realize the line possess a certain series inductance and resistance. In addition, there is a shunt capacitance between the conductors, and even a shunt conductance if the medium insulating the wires is not perfect. We use the concept of shunt conductance, rather than resistance, because it is more convenient for adding the parallel elements of the shunt. We can represent the uniform transmission line with the distributed circuit below (general lossy line) L dz R dz C dz dz © Amanogawa, 2000 – Digital Maestro Series L dz G dz R dz C dz G dz dz 47 Transmission Lines The impedance parameters L, R, C, and G represent: L = series inductance per unit length R = series resistance per unit length C = shunt capacitance per unit length G = shunt conductance per unit length. Each cell of the distributed circuit will have impedance elements with values: Ldz, Rdz, Cdz, and Gdz, where dz is the infinitesimal length of the cells. If we can determine the differential behavior of an elementary cell of the distributed circuit, in terms of voltage and current, we can find a global differential equation that describes the entire transmission line. We can do so, because we assume the line to be uniform along its length. So, all we need to do is to study how voltage and current vary in a single elementary cell of the distributed circuit. © Amanogawa, 2000 – Digital Maestro Series 48 Transmission Lines Loss-less Transmission Line In many cases, it is possible to neglect resistive effects in the line. In this approximation there is no Joule effect loss because only reactive elements are present. The equivalent circuit for the elementary cell of a loss-less transmission line is shown in the figure below. L dz I (z)+dI I (z) V (z) V (z)+dV C dz dz © Amanogawa, 2000 – Digital Maestro Series 49 Transmission Lines The series inductance determines the variation of the voltage from input to output of the cell, according to the sub-circuit below L dz I (z) V (z) V (z)+dV dz The corresponding circuit equation is (V dV ) V j L dz I which gives a first order differential equation for the voltage dV j L I dz © Amanogawa, 2000 – Digital Maestro Series 50 Transmission Lines The current flowing through the shunt capacitance determines the variation of the current from input to output of the cell. I (z) dI I (z)+dI C dz V (z)+dV The circuit equation for the sub-circuit above is dI j Cdz(V dV ) j CVdz j C dV dz The second term (including dV dz) tends to zero very rapidly for infinitesimal length dz and can be ignored, giving a first order differential equation for the current dI j C V dz © Amanogawa, 2000 – Digital Maestro Series 51 Transmission Lines We have obtained a system of two coupled first order differential equations that describe the behavior of voltage and current on the uniform loss-less transmission line. The equations must be solved simultaneously. dV j L I dz dI j C V dz These are often called “telegraphers’ equations” of the loss-less transmission line. © Amanogawa, 2000 – Digital Maestro Series 52 Transmission Lines One can easily obtain a set of uncoupled equations by differentiating with respect to the space coordinate. The first order differential terms are eliminated by using the corresponding telegraphers’ equation dI j C V dz d 2V dI j L j L j CV 2 LC V dz dz 2 d2 I dV j C j C j L I 2 LC I dz dz 2 dV j L I dz These are often called “telephonists’ equations”. © Amanogawa, 2000 – Digital Maestro Series 53 Transmission Lines We have now two uncoupled second order differential equations for voltage and current, which give an equivalent description of the loss-less transmission line. Mathematically, these are wave equations and can be solved independently. The general solution for the voltage equation is V (z) V e j z V e j z where the wave propagation constant is LC Note that the complex exponential terms including have unitary magnitude and purely “imaginary” argument, therefore they only affect the “phase” of the wave in space. © Amanogawa, 2000 – Digital Maestro Series 54 Transmission Lines We have the following useful relations: 2 2 f vp vp r r 0 0 r r c Here, v p f is the wavelength of the dielectric medium surrounding the conductors of the transmission line and 1 1 vp 0 r 0 r is the phase velocity of an electromagnetic wave in the dielectric. As you can see, the propagation constant can be written in many different, equivalent ways. © Amanogawa, 2000 – Digital Maestro Series 55 Transmission Lines The current distribution on transmission line can be readily obtained by differentiation of the result for the voltage dV j z j z j V e j V e j L I dz which gives 1 C j z j z j z j z I (z) V e V e V e V e L Z0 The real quantity Z0 L C is the “characteristic impedance” of the loss-less transmission line. © Amanogawa, 2000 – Digital Maestro Series 56 Transmission Lines Lossy Transmission Line The solution for a uniform lossy transmission line can be obtained with a very similar procedure, using the equivalent circuit for the elementary cell shown in the figure below. L dz R dz I (z)+dI I (z) V (z) C dz G dz V (z)+dV dz © Amanogawa, 2000 – Digital Maestro Series 57 Transmission Lines The series impedance determines the variation of the voltage from input to output of the cell, according to the sub-circuit L dz V (z) R dz I (z) V (z)+dV dz The corresponding circuit equation is (V dV ) V ( j Ldz Rdz) I from which we obtain a first order differential equation for the voltage dV ( j L R) I dz © Amanogawa, 2000 – Digital Maestro Series 58 Transmission Lines The current flowing through the shunt admittance determines the input-output variation of the current, according to the sub-circuit I (z) I (z)+dI dI C dz G dz V (z)+dV The corresponding circuit equation is dI ( j Cdz Gdz)(V dV ) ( j C G)Vdz ( j C G)dV dz The second term (including dV dz) can be ignored, giving a first order differential equation for the current dI ( j C G)V dz © Amanogawa, 2000 – Digital Maestro Series 59 Transmission Lines We have again a system of coupled first order differential equations that describe the behavior of voltage and current on the lossy transmission line dV ( j L R) I dz dI ( j C G)V dz These are the “telegraphers’ equations” for the lossy transmission line case. © Amanogawa, 2000 – Digital Maestro Series 60 Transmission Lines One can easily obtain a set of uncoupled equations differentiating with respect to the coordinate z as done earlier by dI ( j C G)V dz d 2V dI ( j L R) ( j L R)( j C G)V dz dz2 d2 I dV ( j C G ) ( j C G)( j L R) I dz dz2 dV ( j L R) I dz These are the “telephonists’ equations” for the lossy line. © Amanogawa, 2000 – Digital Maestro Series 61 Transmission Lines The telephonists’ equations for the lossy transmission line are uncoupled second order differential equations and are again wave equations. The general solution for the voltage equation is V (z) V e z V e z V e z e j z V e z e j z where the wave propagation constant is now the complex quantity ( j L R)( j C G) j The real part of the propagation constant describes the attenuation of the signal due to resistive losses. The imaginary part describes the propagation properties of the signal waves as in loss-less lines. The exponential terms including are “real”, therefore, they only affect the “magnitude” of the voltage phasor. The exponential terms including have unitary magnitude and purely “imaginary” argument, affecting only the “phase” of the waves in space. © Amanogawa, 2000 – Digital Maestro Series 62 Transmission Lines The current distribution on a lossy transmission line can be readily obtained by differentiation of the result for the voltage dV ( j L R) I V e z V e z dz which gives ( j C G) I (z) V ez V ez ( j L R) 1 V ez V ez Z0 with the “characteristic impedance” of the lossy transmission line ( j L R) Z0 ( j C G) © Amanogawa, 2000 – Digital Maestro Series Note: the characteristic impedance is now complex ! 63 Transmission Lines For both loss-less and lossy transmission lines the characteristic impedance does not depend on the line length but only on the metal of the conductors, the dielectric material surrounding the conductors and the geometry of the line crosssection, which determine L, R, C, and G. One must be careful not to interpret the characteristic impedance as some lumped impedance that can replace the transmission line in an equivalent circuit. This is a very common mistake! Z0 © Amanogawa, 2000 – Digital Maestro Series ZR Z0 ZR 64 Transmission Lines We have obtained the following solutions for the steady-state voltage and current phasors in a transmission line: Loss-less line V (z) V e jz V e j z 1 I (z) V e jz V e j z Z0 Lossy line V (z) V ez V ez 1 I (z) V ez V ez Z0 Since V (z) and I (z) are the solutions of second order differential (wave) equations, we must determine two unknowns, V+ and V , which represent the amplitudes of steady-state voltage waves, travelling in the positive and in the negative direction, respectively. Therefore, we need two boundary conditions to determine these unknowns, by considering the effect of the load and of the generator connected to the transmission line. © Amanogawa, 2000 – Digital Maestro Series 65 Transmission Lines Before we consider the boundary conditions, it is very convenient to shift the reference of the space coordinate so that the zero reference is at the location of the load instead of the generator. Since the analysis of the transmission line normally starts from the load itself, this will simplify considerably the problem later. ZR New Space Coordinate z d 0 We will also change the positive direction of the space coordinate, so that it increases when moving from load to generator along the transmission line. © Amanogawa, 2000 – Digital Maestro Series 66 Transmission Lines We adopt a new coordinate d = z, with zero reference at the load location. The new equations for voltage and current along the lossy transmission line are Loss-less line jd jd V (d) V e Lossy line d d V e 1 I (d) V e jd V e jd Z0 V (d) V e V e 1 I (d) V ed V e d Z0 At the load (d = 0) we have, for both cases, V (0) V V 1 I (0) V V Z0 © Amanogawa, 2000 – Digital Maestro Series 67 Transmission Lines For a given load impedance ZR , the load boundary condition is V (0) ZR I (0) Therefore, we have ZR V V V V Z0 from which we obtain the voltage load reflection coefficient V ZR Z0 R Z Z V R 0 © Amanogawa, 2000 – Digital Maestro Series 68 Transmission Lines We can introduce this result into the transmission line equations as Loss-less line Lossy line V e jd 2 j d I (d) e 1 R Z0 V (d) V e jd 1 R e2 j d V e d 2 d I (d) e 1 R Z0 V (d) V ed 1 R e2 d At each line location we define a Generalized Reflection Coefficient (d) R e2 j d (d) R e2 d and the line equations become V (d) V e jd 1 (d) V (d) V ed 1 (d) V e jd I (d) 1 (d) Z0 V ed I (d) 1 (d) Z0 © Amanogawa, 2000 – Digital Maestro Series 69 Transmission Lines We define the line impedance as 1 (d ) V (d ) Z0 Z( d ) 1 (d ) I (d ) A simple circuit diagram can illustrate the significance of line impedance and generalized reflection coefficient: Req = (d) d © Amanogawa, 2000 – Digital Maestro Series ZR Zeq=Z(d) 0 70 Transmission Lines If you imagine to cut the line at location d, the input impedance of the portion of line terminated by the load is the same as the line impedance at that location “before the cut”. The behavior of the line on the left of location d is the same if an equivalent impedance with value Z(d) replaces the cut out portion. The reflection coefficient of the new load is equal to (d) Req ( d ) ZReq Z0 ZReq Z0 If the total length of the line is L, the input impedance is obtained from the formula for the line impedance as 1 ( L) Vin V ( L) Z0 Zin 1 ( L) Iin I ( L) The input impedance is the equivalent impedance representing the entire line terminated by the load. © Amanogawa, 2000 – Digital Maestro Series 71 Transmission Lines An important practical case is the low-loss transmission line, where the reactive elements still dominate but R and G cannot be neglected as in a loss-less line. We have the following conditions: L R C G so that ( j L R)( j C G) R G 1 j L j C 1 j L j C R G RG j LC 1 j L j C 2 LC The last term under the square root can be neglected, because it is the product of two very small quantities. © Amanogawa, 2000 – Digital Maestro Series 72 Transmission Lines What remains of the square root can be expanded into a truncated Taylor series 1 R G j LC 1 2 j L j C 1 C L R G j LC 2 L C so that 1 C L R G 2 L C © Amanogawa, 2000 – Digital Maestro Series LC 73 Transmission Lines The characteristic impedance of the low-loss line is a real quantity for all practical purposes and it is approximately the same as in a corresponding loss-less line R j L L Z0 G j C C and the phase velocity associated to the wave propagation is 1 vp LC BUT NOTE: In the case of the low-loss line, the equations for voltage and current retain the same form obtained for general lossy lines. © Amanogawa, 2000 – Digital Maestro Series 74 Transmission Lines Again, we obtain the loss-less transmission line if we assume R0 G0 This is often acceptable in relatively short transmission lines, where the overall attenuation is small. As shown earlier, the characteristic impedance in a loss-less line is exactly real L Z0 C while the propagation constant has no attenuation term ( j L)( j C ) j LC j The loss-less line does not dissipate power, because = 0. © Amanogawa, 2000 – Digital Maestro Series 75 Transmission Lines For all cases, the line impedance was defined as 1 (d) V (d) Z0 Z(d) 1 (d) I (d) By including the appropriate generalized reflection coefficient, we can derive alternative expressions of the line impedance: A) Loss-less line 1 Re2 jd ZR jZ0 tan( d) Z0 Z(d) Z0 2 jd jZR tan( d) Z0 1 Re B) Lossy line (including low-loss) 1 R e2 d ZR Z0 tanh( d) Z0 Z(d) Z0 2 d ZR tanh( d) Z0 1 Re © Amanogawa, 2000 – Digital Maestro Series 76 Transmission Lines Let’s now consider power flow in a transmission line, limiting the discussion to the time-average power, which accounts for the active power dissipated by the resistive elements in the circuit. The time-average power at any transmission line location is 1 P(d , t ) Re V (d) I * (d) 2 This quantity indicates the time-average power that flows through the line cross-section at location d. In other words, this is the power that, given a certain input, is able to reach location d and then flows into the remaining portion of the line beyond this point. It is a common mistake to think that the quantity above is the power dissipated at location d ! © Amanogawa, 2000 – Digital Maestro Series 77 Transmission Lines The generator, the input impedance, the input voltage and the input current determine the power injected at the transmission line input. Iin Zin Vin VG ZG Zin ZG VG Vin Generator © Amanogawa, 2000 – Digital Maestro Series Zin Line 1 Iin VG ZG Zin 1 * Pin Re Vin Iin 2 78 Transmission Lines The time-average power reaching the load of the transmission line is given by 1 P(d=0 , t ) Re V (0) I * (0) 2 * 1 1 Re V 1 R V 1 R 2 Z0* This represents the power dissipated by the load. The time-average power absorbed by the line is simply the difference between the input power and the power absorbed by the load P line Pin P(d 0 , t ) Remember that the internal impedance of the generator dissipates part of the total power generated. © Amanogawa, 2000 – Digital Maestro Series 79 Transmission Lines The time-average power injected into the input of the transmission line is maximized when the input impedance of the transmission line and the internal generator impedance are complex conjugate of each other. Zin Load Generator ZG VG ZG Z*in © Amanogawa, 2000 – Digital Maestro Series Transmission line ZR for maximum power transfer 80 Transmission Lines In a loss-less transmission line no power is absorbed by the line, so the input time-average power is the same as the time-average power absorbed by the load. The characteristic impedance of the loss-less line is real and we can express the power flow as 1 P(d , t ) Re V (d) I* (d) 2 1 Re V e j d 1 Re j 2 d 2 1 * j d j 2 d * (V ) e 1 Re Z0 1 1 2 2 2 V V R 2 Z0 2 Z0 Incident wave © Amanogawa, 2000 – Digital Maestro Series Reflected wave 81 Transmission Lines In the case of low-loss lines, the characteristic impedance is again real, and the time-average power flow along the line is given by 1 P(d , t ) Re V (d) I * (d) 2 1 Re V e d e j d 1 Re2 d 2 1 * d j d 2 d * (V ) e e 1 Re Z0 1 1 2 2 2 d 2 2 d V e V e R 2 Z0 2 Z0 Incident wave © Amanogawa, 2000 – Digital Maestro Series Reflected wave 82 Transmission Lines Note that in a lossy line the reference for the amplitude of the incident voltage wave is at the load and that the amplitude grows exponentially moving towards the input. The amplitude of the incident wave behaves in the following way V e L V e d V input inside the line load The reflected voltage wave has maximum amplitude at the load, and it decays exponentially moving back towards the generator. The amplitude of the reflected wave behaves in the following way V R e L V R e d V R input inside the line load © Amanogawa, 2000 – Digital Maestro Series 83 Transmission Lines For a general lossy line the characteristic impedance is complex, and the time-average power is 1 P(d , t ) Re V (d) I * (d) 2 1 Re V e d e j d 1 (d) 2 * * * d j d Y0 (V ) e e 1 (d) G0 2 2 d G0 2 2 d 2 R V e V e 2 2 B0 V © Amanogawa, 2000 – Digital Maestro Series 2 e2 d Im((d)) 84 Transmission Lines We have introduced for convenience the characteristic admittance of the line Y0 1 G0 jB0 Z0 since a complex characteristic impedance would appear at denominator in the expression for the power. Note that for a low-loss transmission line the characteristic impedance is approximately real and B0 0 The previous result for the low-loss line can be readily recovered from the time-average power for the general lossy line. © Amanogawa, 2000 – Digital Maestro Series 85 Transmission Lines To completely specify the transmission line problem, we still have to determine the value of V+ from the input boundary condition. The load boundary condition imposes the shape of the interference pattern of voltage and current along the line. The input boundary condition, linked to the generator, imposes the scaling for the interference patterns. We have Zin Vin V ( L) VG ZG Zin or with 1 (L) Zin Z0 1 (L) ZR jZ0 tan( L) Zin Z0 jZR tan( L) Z0 loss - less line ZR Z0 tanh( L) Zin Z0 ZR tanh( L) Z0 lossy line © Amanogawa, 2000 – Digital Maestro Series 86 Transmission Lines For a loss-less transmission line: V (L) V e j L 1 (L) V e j L (1 Re j 2 L ) ! 1 Zin V VG ZG Zin e j L (1 Re j 2 L ) For a lossy transmission line: V (L) V e L 1 (L) V e L 1 Re2 d ! 1 Zin V VG ZG Zin e L (1 Re2 L ) © Amanogawa, 2000 – Digital Maestro Series 87 Transmission Lines In order to have good control on the behavior of a high frequency circuit, it is very important to realize transmission lines as uniform as possible along their length, so that the impedance behavior of the line does not vary and can be easily characterized. A change in transmission line properties, wanted or unwanted, entails a change in the characteristic impedance, which causes a reflection. Example: Z01 Z02 ZR 1 Z01 © Amanogawa, 2000 – Digital Maestro Series Zin Zin Z01 1 Zin Z01 88 Transmission Lines Special Cases ZR 0 (SHORT CIRCUIT) ZR = 0 Z0 The load boundary condition due to the short circuit is V (0) = 0 V (d 0) V e j 0 (1 R e j 2 0 ) V (1 R ) 0 © Amanogawa, 2000 – Digital Maestro Series R 1 89 Transmission Lines Since R V V V V We can write the line voltage phasor as V (d) V e j d V e j d V e j d V e j d V ( e j d e j d ) 2 jV sin( d) © Amanogawa, 2000 – Digital Maestro Series 90 Transmission Lines For the line current phasor we have 1 (V e j d V e j d ) I (d) Z0 1 (V e j d V e j d ) Z0 V j d (e e j d ) Z0 2V cos( d) Z0 The line impedance is given by 2 jV sin( d) V (d) jZ0 tan( d) Z(d) I (d) 2V cos( d) / Z0 © Amanogawa, 2000 – Digital Maestro Series 91 Transmission Lines The time-dependent values of voltage and current are obtained as V (d, t ) Re[V (d) e j t ] Re[2 j | V | e j sin( d) e j t ] 2| V |sin( d) Re[ j e j( t ) ] 2| V |sin( d) Re[ j cos(t ) sin(t )] 2| V |sin( d) sin(t ) I (d, t ) Re[ I (d) e j t ] Re[2 | V | e j cos( d) e j t ] / Z0 2| V |cos( d) Re[ e j ( t ) ] / Z0 2| V |cos( d) Re[ (cos(t ) j sin(t )] / Z0 | V | cos( d) cos(t ) 2 Z0 © Amanogawa, 2000 – Digital Maestro Series 92 Transmission Lines The time-dependent power is given by P(d, t ) V (d, t ) I (d, t ) | V |2 sin( d) cos( d) sin(t ) cos(t ) 4 Z0 | V |2 sin(2 d) sin (2t 2) Z0 and the corresponding time-average power is 1 T P(d, t ) P(d, t ) dt T 0 | V |2 1 T sin(2 d) sin (2t 2) 0 Z0 T 0 © Amanogawa, 2000 – Digital Maestro Series 93 Transmission Lines ZR (OPEN CIRCUIT) ZR Z0 The load boundary condition due to the open circuit is I (0) = 0 V j 0 I (d 0) e (1 R e j 2 0 ) Z0 V (1 R ) 0 Z0 © Amanogawa, 2000 – Digital Maestro Series R 1 94 Transmission Lines Since R V V V V We can write the line current phasor as 1 (V e j d V e j d ) I (d) Z0 1 (V e j d V e j d ) Z0 2 jV V j d j d (e ) sin( d) e Z0 Z0 © Amanogawa, 2000 – Digital Maestro Series 95 Transmission Lines For the line voltage phasor we have V (d) (V e j d V e j d ) (V e j d V e j d ) V ( e j d e j d ) 2V cos( d) The line impedance is given by 2V cos( d) V (d) Z0 j Z(d) tan( d) I (d) 2 jV sin( d) / Z0 © Amanogawa, 2000 – Digital Maestro Series 96 Transmission Lines The time-dependent values of voltage and current are obtained as V (d, t ) Re[V (d) e j t ] Re[2 | V | e j cos( d) e j t ] 2| V |cos( d) Re[ e j( t ) ] 2| V |cos( d) Re[ (cos(t ) j sin(t )] 2| V |cos( d) cos( t ) I (d, t ) Re[ I (d) e j t ] Re[2 j | V | e j sin( d) e j t ] / Z0 2| V |sin( d) Re[ j e j ( t ) ] / Z0 2| V |sin( d) Re[ j cos( t ) sin(t )] / Z0 | V | sin( d) sin(t ) 2 Z0 © Amanogawa, 2000 – Digital Maestro Series 97 Transmission Lines The time-dependent power is given by P(d, t ) V (d, t ) I (d, t ) | V |2 cos( d) sin( d) cos(t ) sin(t ) 4 Z0 | V |2 sin(2 d) sin (2t 2) Z0 and the corresponding time-average power is 1 T P(d, t ) P(d, t ) dt T 0 | V |2 1 T sin(2 d) sin (2t 2) 0 Z0 T 0 © Amanogawa, 2000 – Digital Maestro Series 98 Transmission Lines ZR Z0 (MATCHED LOAD) Z0 ZR = Z0 The reflection coefficient for a matched load is ZR Z0 Z0 Z0 R 0 ZR Z0 Z0 Z0 no reflection! The line voltage and line current phasors are V (d) V e j d (1 R e2 j d ) V e j d V j d V (1 R e2 j d ) I( d ) e e j d Z0 Z0 © Amanogawa, 2000 – Digital Maestro Series 99 Transmission Lines The line impedance is independent of position and equal to the characteristic impedance of the line V (d) V e j d Z0 Z(d) I (d) V j d e Z0 The time-dependent voltage and current are V (d, t ) Re[| V | e j e j d e j t ] | V | Re[e j( t d ) ] | V | cos(t d ) I (d, t ) Re[| V | e j e j d e j t ] / Z0 | V | | | V j ( t d ) ] cos(t d ) Re[e Z0 Z0 © Amanogawa, 2000 – Digital Maestro Series 100 Transmission Lines The time-dependent power is | | V cos(t d ) P(d, t ) | V | cos(t d ) Z0 | V |2 cos2 (t d ) Z0 and the time average power absorbed by the load is 1 t | V |2 cos2 (t d ) dt P(d) T 0 Z0 | V |2 2 Z0 © Amanogawa, 2000 – Digital Maestro Series 101 Transmission Lines ZR jX (PURE REACTANCE) ZR = j X Z0 The reflection coefficient for a purely reactive load is ZR Z0 jX Z0 R ZR Z0 jX Z0 ( jX Z0 )( jX Z0 ) ( jX Z0 )( jX Z0 ) © Amanogawa, 2000 – Digital Maestro Series X 2 Z02 Z02 X 2 2j XZ0 Z02 X 2 102 Transmission Lines In polar form R R exp( j) where R 2 2 X Z0 4 X 2 Z02 2 2 Z02 X 2 Z02 X 2 2 2 2 2 Z0 X 1 2 Z02 X 2 1 2 XZ0 tan X 2 Z2 0 The reflection coefficient has unitary magnitude, as in the case of short and open circuit load, with zero time average power absorbed by the load. Both voltage and current are finite at the load, and the time-dependent power oscillates between positive and negative values. This means that the load periodically stores and returns powers to the line without dissipation. © Amanogawa, 2000 – Digital Maestro Series 103 Transmission Lines Reactive impedances can be realized with transmission lines terminated by a short or by an open circuit. The input impedance of a loss-less transmission line of length L terminated by a short circuit is purely imaginary 2 f 2 L Zin j Z0 tan L j Z0 tan L j Z0 tan vp At a specified frequency f, any reactance value can be obtained by changing the length of the line from 0 to /2. Since the tangent function is periodic, the behavior of the impedance will repeat identically for each line increment of length /2. Note that a similar periodic behavior is obtained when the length of the transmission line is fixed and the frequency of operation is changed. © Amanogawa, 2000 – Digital Maestro Series 104 Transmission Lines Shorted transmission line – Fixed frequency L L 0 0 L L 4 4 L 4 2 L 2 3 L 2 4 3 L 4 3 L 4 Z in 0 s h o rt c irc u it Im Z in 0 in d u c ta n c e Z in o p e n c irc u it c a p a c ita n c e Im Z in 0 Z in 0 s h o rt c irc u it Im Z in 0 in d u c ta n c e Z in o p e n c irc u it Im Z in 0 c a p a c ita n c e © Amanogawa, 2000 – Digital Maestro Series 105 Z(L)/Zo = j tan( L) Transmission Lines Impedance of a short circuited transmission line (fixed frequency, variable length) 40 30 20 inductive Normalized Input Impedance 10 inductive inductive 0 -10 capacitive cap. -20 -30 -40 0 100 200 300 400 500 L [deg] Line Length L © Amanogawa, 2000 – Digital Maestro Series 106 Z(L)/Zo = j tan( L) Transmission Lines Impedance of a short circuited transmission line (fixed length, variable frequency) 40 30 20 inductive Normalized Input Impedance 10 inductive inductive 0 -10 capacitive cap. -20 -30 -40 0 100 vp / (4L) 200 vp / (2L) 300 3vp / (4L) 400 vp / L 500 5vp / (4L) [deg] f Frequency of operation © Amanogawa, 2000 – Digital Maestro Series 107 Transmission Lines For a transmission line of length L terminated by an open circuit, the input impedance is again purely imaginary Z0 j Zin j tan L Z0 Z0 j 2 2 f tan L tan L v p We can use the open circuited line to realize any reactance, but starting from a capacitive value when the line length is very short. The reactance varies linearly with frequency in the case of lumped elements since X L (inductance) or 1 (capacitance) X C This is not the case when the reactance is realized with a section of transmission line. © Amanogawa, 2000 – Digital Maestro Series 108 Transmission Lines Open transmission line – Fixed frequency L L 0 0 L L 4 4 L 4 2 L 2 3 L 2 4 3 L 4 3 L 4 Z in o p e n c irc u it Im Z in 0 c a p a c ita n c e Z in 0 s h o rt c irc u it Im Z in 0 in d u c ta n c e Z in o p e n c irc u it Im Z in 0 c a p a c ita n c e Z in 0 s h o rt c irc u it Im Z in 0 in d u c ta n c e © Amanogawa, 2000 – Digital Maestro Series 109 Normalized Input Impedance Z(L)/ Zo = - j cotan( L) Transmission Lines Impedance of an open circuited transmission line (fixed frequency, variable length) 40 30 20 inductive inductive inductive 10 0 -10 capacitive capacitive capacitive -20 -30 -40 0 100 200 300 400 500 [deg] L Line Length L © Amanogawa, 2000 – Digital Maestro Series 110 Normalized Input Impedance Z(L)/ Zo = - j cotan( L) Transmission Lines Impedance of an open circuited transmission line (fixed length, variable frequency) 40 30 20 inductive inductive inductive 10 0 -10 capacitive capacitive capacitive -20 -30 -40 0 100 vp / (4L) 200 vp / (2L) 300 3vp / (4L) 400 vp / L 500 5vp / (4L) [deg] f Frequency of operation © Amanogawa, 2000 – Digital Maestro Series 111 Transmission Lines It is possible to realize resonant circuits by using transmission lines as reactive elements. For instance, consider the circuit below realized with lines having the same characteristic impedance: I L1 short circuit Z0 L2 V Zin1 Zin1 j Z0 tan L 1 © Amanogawa, 2000 – Digital Maestro Series Z0 short circuit Zin2 Zin2 j Z0 tan L 2 112 Transmission Lines The circuit is resonant if L1 and L2 are chosen such that an inductance and a capacitance are realized. A resonance condition is established when the total input impedance of the parallel circuit is infinite (or, equivalently, when the input admittance of the parallel circuit is zero) 1 1 0 j Z0 tan r L1 j Z0 tan r L 2 or r r tan L1 tan L 2 vp vp with 2 r r r vp Since the tangent is a periodic function, there is a multiplicity of possible resonant angular frequencies r that satisfy the condition above. The solutions can be found by using a numerical procedure. © Amanogawa, 2000 – Digital Maestro Series 113 Transmission Lines Transient and Steady-State on a Transmission Line We need to give now a physical interpretation of the mathematical results obtained for transmission lines. First of all, note that we are considering a steady-state regime where the wave propagation along the transmission line is perfectly periodic in time. This means that all the transient phenomena have already decayed. To give a feeling of what the steady-state regime is, consider a transmission line that is connected to the generator by closing a switch at a reference time t = 0. For simplicity we assume that all impedances, including the line characteristic impedance, are real. Switch Generator Load RG t=0 VG © Amanogawa, 2000 – Digital Maestro Series Transmission line RR 114 Transmission Lines After the switch is closed, the voltage at the input of the transmission line will vary nearly instantaneously from the open voltage of the generator VG to a value V+ , with a current I+ beginning to flow into the line. A transient takes place in the transmission line, as charges in the conductors move, transporting the current towards the load. Until the front first reaches the end of the transmission line, the load voltage remains zero. Initially, the input impedance of the transmission line appears to be the same as the line characteristic impedance, because the current cannot yet sense the value of the load impedance. Therefore, a voltage front V+ propagates with the current front I+ , where Z0 V I Z0 VG RG Z0 V V0 I Z0 RG Z0 © Amanogawa, 2000 – Digital Maestro Series 115 Transmission Lines If the load does not match exactly the characteristic impedance of the line, the voltage V+ and the current I+ cannot be established across the load RR , when the front reaches the end of the line, because V I RR Therefore, voltage and current adjust themselves at the load by – reflecting back a wave front with voltage V and current I such that V V ( I I ) RR © Amanogawa, 2000 – Digital Maestro Series 116 Transmission Lines Since also the reflected front will see an impedance Z0 we have V Z0 I ; V V Z0 I RR Z0 V RR Z0 The quantity V RR Z0 R R Z V R 0 is the load reflection coefficient. © Amanogawa, 2000 – Digital Maestro Series 117 Transmission Lines The wave reflected by the load propagates in the negative direction and interferes with the oscillating values of voltage and current found along the transmission line, which continue to be injected by the generator. When the reflected wave reaches the input of the transmission line, it terminates on the generator impedance RG and if this does not match the line characteristic impedance, reflection back into the line again occurs, generating now a forward wave with associated voltage RG Z0 V2 V RG Z0 Remember that the ideal voltage source of the generator will behave simply as a short for the reflected wave attempting to exit the line from the input. © Amanogawa, 2000 – Digital Maestro Series 118 Transmission Lines The front reflected by the generator side will again reach the load, and waves of ever decreasing amplitude will keep bouncing back and forth along the line until the process associated to that initial wave front dies out. Every subsequent wave front injected over time by the oscillating generator undergoes an identical phenomenon of multiple reflections. If we assume a sinusoidal generator, voltage and current injected into the line repeat periodically, according to the period of the oscillation. Therefore, successive reflections at the ends of the line obey the same reflection coefficients, but involving in time different values of amplitude and phase. If the generator continues to supply the line with a stable oscillation, after a sufficient time the combined interference of forward and backward waves becomes stable, and one can identify two well-defined incoming and reflected steady-state waves, arising from the superposition of the infinite transient components travelling on the line. © Amanogawa, 2000 – Digital Maestro Series 119 Transmission Lines Note that the wave fronts travel with a phase velocity equal to the speed of light in the medium surrounding the wires. Also, note that the length of the line will affect the interference pattern of the wave superposition, so that lines of different length will result in different voltage and current distributions along the line. When we study the steady-state voltages and currents in a transmission line, we only need to know the phasors that represent the stable steady-state oscillations at each line location. The phasors provide a snapshot of how values of voltage and current relate to each other in space, at a reference time, in terms of amplitude and phase. The actual time oscillation can be easily recovered, because in steady-state we know that voltage and current are perfectly periodic at each line location, according to the period of the generator. If the generator provides more than one frequency of oscillation, at steady-state the behavior of each frequency in the spectrum can be studied independently and the total result is obtained by superposition. © Amanogawa, 2000 – Digital Maestro Series 120 Transmission Lines Standing Wave Patterns In practical applications it is very convenient to plot the magnitude of phasor voltage and phasor current along the transmission line. These are the standing wave patterns: V (d) V 1 (d) Loss - less line V 1 (d) I (d) Z0 V (d) V e d 1 d Lossy line V e d 1 d I (d) Z0 © Amanogawa, 2000 -–Digital Maestro Series 121 Transmission Lines The standing wave patterns provide the top envelopes that bound the time-oscillations of voltage and current along the line. In other words, the standing wave patterns provide the maximum values that voltage and current can ever establish at each location of the transmission line for given load and generator. The standing wave pattern gives a clear representation of wave interference in a transmission line. The patterns present a succession of maxima and minima which repeat in space with a period of length /2, due to constructive or destructive interference between forward and reflected waves. The patterns for a loss-less line are periodic in space, repeating exactly with a /2 period. Again, note that although we talk about maxima and minima of the standing wave pattern we are always examining a maximum of voltage or current that can be achieved at a transmission line location during any period of oscillation. © Amanogawa, 2000 -–Digital Maestro Series 122 Transmission Lines We limit now our discussion to the loss-less transmission line case where the generalized reflection coefficient varies as (d) R exp j 2 d R exp j exp j 2 d Note that the magnitude of an exponential with imaginary argument is always unity, so exp j exp j 2 d 1 and that in a loss-less line it is always true that, at any line location, (d) R When d increases, moving from load to generator, the generalized reflection coefficient on the complex plane moves clockwise on a circle with radius |R| and is identified by the angle - 2 d . © Amanogawa, 2000 -–Digital Maestro Series 123 Transmission Lines The voltage standing wave pattern has a maximum at locations where the generalized reflection coefficient is real and positive (d) R exp j exp j 2 d 1 2 d 2 n At these locations we have 1 (d) 1 R Vmax V (d max ) V 1 R The phase angle - 2 d changes by an amount 2, when moving from one maximum to the next. This corresponds to a distance between successive maxima of /2. © Amanogawa, 2000 -–Digital Maestro Series 124 Transmission Lines The voltage standing wave pattern has a minimum at locations where the generalized reflection coefficient is real and negative (d) R exp j exp j 2 d 1 2 d 2 n 1 At these locations we have 1 (d) 1 R Vmin V (d min ) V 1 R Also when moving from one minimum to the next, the phase angle - 2 d changes by an amount 2. This again corresponds to a distance between successive minima of /2. © Amanogawa, 2000 -–Digital Maestro Series 125 Transmission Lines The voltage standing wave pattern provides immediate information on the transmission line circuit If the load is matched to the transmission line ( ZR = Z0 ) the voltage standing wave pattern is flat, with value | V+ |. If the load is real and ZR > Z0 , the voltage standing wave pattern starts with a maximum at the load. If the load is real and ZR < Z0 , the voltage standing wave pattern starts with a minimum at the load. If the load is complex and Im(ZR ) > 0 (inductive reactance), the voltage standing wave pattern initially increases when moving from load to generator and reaches a maximum first. If the load is complex and Im(ZR ) < 0 (capacitive reactance), the voltage standing wave pattern initially decreases when moving from load to generator and reaches a minimum first. © Amanogawa, 2000 -–Digital Maestro Series 126 Transmission Lines Since in all possible cases d 1 the voltage standing wave pattern V (d) V 1 (d) cannot exceed the value 2 | V+ | in a loss-less transmission line. If the load is a short circuit, an open circuit, or a pure reactance, there is total reflection with d 1 since the load cannot consume any power. The voltage standing wave pattern in these cases is characterized by Vmax 2 V © Amanogawa, 2000 -–Digital Maestro Series and Vmin 0 . 127 Transmission Lines The quantity 1 + (d) is in general a complex number, that can be constructed as a vector on the complex plane. The number 1 is represented as 1 + j0 on the complex plane, and it is just a vector with coordinates (1,0) positioned on the Real axis. The reflection coefficient (d) is a complex number such that |(d)| 1. Im 1+ 1 © Amanogawa, 2000 -–Digital Maestro Series Re 128 Transmission Lines We can use a geometric construction to visualize the behavior of the voltage standing wave pattern V (d) V 1 (d) simply by looking at a vector plot of |(1 + (d))| . |V+| is just a scaling factor, fixed by the generator. For convenience, we place the reference of the complex plane representing the reflection coefficient in correspondence of the tip of the vector (1, 0). Example: Load with inductive reactance Im( ) 1+R R 1 © Amanogawa, 2000 -–Digital Maestro Series Re ( ) 129 Transmission Lines Im( ) Maximum of voltage standing wave pattern R 1+(d) (d) 1 2 dmax Re ( ) d 2 d max 0 Im( ) Minimum of voltage standing wave pattern 1+(d) 1 d 2 d min © Amanogawa, 2000 -–Digital Maestro Series (d) R Re ( ) 2 dmin 130 Transmission Lines The voltage standing wave ratio (VSWR) is an indicator of load matching which is widely used in engineering applications Vmax 1 R VSWR Vmin 1 R When the load is perfectly matched to the transmission line R 0 VSWR 1 When the load is a short circuit, an open circuit or a pure reactance R 1 VSWR We have the following useful relation VSWR 1 R VSWR 1 © Amanogawa, 2000 -–Digital Maestro Series 131 Transmission Lines Maxima and minima of the voltage standing wave pattern. Load with inductive reactance Im ZR 0 The load reflection coefficient is in this part of the domain ZR Z0 Im R Im 0 ZR Z0 Im() Re() 1 The first maximum of the voltage standing wave pattern is closest to the load, at location d 2 d max 0 © Amanogawa, 2000 -–Digital Maestro Series d max 4 132 Transmission Lines Load with capacitive reactance Im ZR 0 ZR Z0 Im R Im 0 ZR Z0 The load reflection coefficient is in this part of the domain Im() 1 Re() The first minimum of the voltage standing wave pattern is closest to the load, at location d 2 d min © Amanogawa, 2000 -–Digital Maestro Series d min 4 133 Transmission Lines A measurement of the voltage standing wave pattern provides the locations of the first voltage maximum and of the first voltage minimum with respect to the load. The ratio of the voltage magnitude at these points gives directly the voltage standing wave ratio (VSWR). This information is sufficient to determine the load impedance ZR , if the characteristic impedance of the transmission line Z0 is known. STEP 1: The VSWR provides the magnitude of the load reflection coefficient VSWR 1 R VSWR 1 © Amanogawa, 2000 -–Digital Maestro Series 134 Transmission Lines STEP 2: The distance from the load of the first maximum or minimum gives the phase of the load reflection coefficient. Vmax |V| For an inductive reactance, a voltage maximum is closest to the load and 4 d max 2 d max Vmin dmax 0 |V| Vmax For a capacitive reactance, a voltage minimum is closest to the load and 4 2 d min d min Vmin dmin 0 © Amanogawa, 2000 -–Digital Maestro Series 135 Transmission Lines STEP 3: The load impedance is obtained by inverting the expression for the reflection coefficient ZR Z0 R R exp j ZR Z0 © Amanogawa, 2000 -–Digital Maestro Series 1 R exp j ZR Z0 1 R exp j 136 Transmission Lines Smith Chart The Smith chart is one of the most useful graphical tools for high frequency circuit applications. The chart provides a clever way to visualize complex functions and it continues to endure popularity decades after its original conception. From a mathematical point of view, the Smith chart is simply a representation of all possible complex impedances with respect to coordinates defined by the reflection coefficient. Im(Γ ) 1 Re(Γ ) © Amanogawa, 2000 - Digital Maestro Series The domain of definition of the reflection coefficient is a circle of radius 1 in the complex plane. This is also the domain of the Smith chart. 137 Transmission Lines The goal of the Smith chart is to identify all possible impedances on the domain of existence of the reflection coefficient. To do so, we start from the general definition of line impedance (which is equally applicable to the load impedance) 1 + Γ (d ) V (d ) = Z0 Z( d ) = 1 − Γ (d ) I (d ) This provides the complex function Z( d ) = f {Re ( Γ ) , Im ( Γ )} that we want to graph. It is obvious that the result would be applicable only to lines with exactly characteristic impedance Z0. In order to obtain universal curves, we introduce the concept of normalized impedance Z(d ) 1 + Γ (d ) z( d ) = = 1 − Γ (d ) Z0 © Amanogawa, 2000 - Digital Maestro Series 138 Transmission Lines The normalized impedance is represented on the Smith chart by using families of curves that identify the normalized resistance r (real part) and the normalized reactance x (imaginary part) z ( d ) = Re ( z ) + j Im ( z ) = r + jx Let’s represent the reflection coefficient in terms of its coordinates Γ ( d ) = Re (Γ ) + j Im ( Γ ) Now we can write 1 + Re ( Γ ) + j Im ( Γ ) r + jx = 1 − Re ( Γ ) − j Im ( Γ ) = 1 − Re 2 ( Γ ) − Im2 ( Γ ) + j 2 Im ( Γ ) © Amanogawa, 2000 - Digital Maestro Series 2 1 Re Im − Γ + ( )) (Γ ) ( 2 139 Transmission Lines The real part gives r= 1 − Re 2 Add a quantity equal to zero ( Γ ) − Im 2 ( Γ ) (1 − Re ( Γ ))2 + Im2 ( Γ ) ( 2 r ( Re ( Γ ) − 1 )2 + (Re r ( Re ( Γ ) − 1 ) + Re 2 (Γ ) − 1 2 =0 ) + r Im (Γ ) − 1 ) 2 ( Γ ) + Im ( Γ ) + 2 1 1+ r − 1 1+ r =0 1 2 + + (1 + r ) Im ( Γ ) = 1 + r 1+ r 1 2 2 1 r r 2 (1 + r ) Re ( Γ ) − 2 Re ( Γ ) + + (1 + r ) Im ( Γ ) = 2 1 + r (1 + r ) 1+ r ⇒ Re ( Γ ) − 1 + r r 2 © Amanogawa, 2000 - Digital Maestro Series + Im 2 (Γ ) = ( ) 1 2 Equation of a circle 1+ r 140 Transmission Lines The imaginary part gives x= x 2 Multiply by x and add a quantity equal to zero 2 Im ( Γ ) (1 − Re ( Γ ))2 + Im 2 ( Γ ) =0 (1 − Re ( Γ ))2 + Im 2 ( Γ ) − 2 x Im ( Γ ) + 1 − 1 = 0 (1 − Re ( Γ ))2 + Im 2 ( Γ ) − 2 Im ( Γ ) + 1 = 1 x 2 2 x x 2 1 1 2 2 (1 − Re ( Γ )) + Im ( Γ ) − Im ( Γ ) + 2 = 2 x x x ⇒ 2 1 = ( Re ( Γ ) − 1 ) + Im ( Γ ) − 2 x x 2 © Amanogawa, 2000 - Digital Maestro Series 1 Equation of a circle 141 Transmission Lines The result for the real part indicates that on the complex plane with coordinates (Re(Γ), Im(Γ)) all the possible impedances with a given normalized resistance r are found on a circle with { } 1 Radius = 1+ r r ,0 Center = 1+ r As the normalized resistance r varies from 0 to ∞ , we obtain a family of circles completely contained inside the domain of the reflection coefficient | Γ | ≤ 1 . Im(Γ ) r=1 r=5 r=0 Re(Γ ) r = 0.5 © Amanogawa, 2000 - Digital Maestro Series r →∞ 142 Transmission Lines The result for the imaginary part indicates that on the complex plane with coordinates (Re(Γ), Im(Γ)) all the possible impedances with a given normalized reactance x are found on a circle with { } 1 Center = 1 , x 1 Radius = x As the normalized reactance x varies from -∞ to ∞ , we obtain a family of arcs contained inside the domain of the reflection coefficient | Γ | ≤ 1 . Im(Γ ) x = 0.5 x=1 x →±∞ Re(Γ ) x=0 x = - 0.5 © Amanogawa, 2000 - Digital Maestro Series x = -1 143 Transmission Lines Basic Smith Chart techniques for loss-less transmission lines Given Z(d) ⇒ Find Γ(d) Given ΓR and ZR Find dmax and dmin (maximum and minimum locations for the voltage standing wave pattern) Find the Voltage Standing Wave Ratio (VSWR) Given Z(d) ⇒ Find Y(d) Given Γ(d) ⇒ Find Z(d) ⇒ Find Γ(d) and Z(d) Given Γ(d) and Z(d) ⇒ Find ΓR and ZR Given Y(d) ⇒ Find Z(d) © Amanogawa, 2000 - Digital Maestro Series 144 Transmission Lines Given Z(d) ⇒ Find Γ(d) 1. Normalize the impedance Z (d ) R X = + j = r+ j x z (d ) = Z0 Z0 Z0 2. 3. 4. Find the circle of constant normalized resistance r Find the arc of constant normalized reactance x The intersection of the two curves indicates the reflection coefficient in the complex plane. The chart provides directly the magnitude and the phase angle of Γ(d) Example: Find Γ(d), given Z ( d ) = 25 + j 100 Ω © Amanogawa, 2000 - Digital Maestro Series with Z0 = 50 Ω 145 Transmission Lines 1. Normalization z (d) = (25 + j 100)/50 = 0.5 + j 2.0 2. Find normalized resistance circle 3. Find normalized reactance arc 1 x = 2.0 05 2 3 r = 0.5 50.906 ° 0.2 0 0.2 0.5 1 2 5 4. This vector represents the reflection coefficient -0.2 Γ (d) = 0.52 + j0.64 -3 |Γ (d)| = 0.8246 ∠ Γ (d) = 0.8885 rad -0 5 = 50.906 ° -2 -1 1. 0.8246 © Amanogawa, 2000 - Digital Maestro Series 146 Transmission Lines Given Γ(d) ⇒ Find Z(d) 1. 2. 3. Determine the complex point representing the given reflection coefficient Γ(d) on the chart. Read the values of the normalized resistance r and of the normalized reactance x that correspond to the reflection coefficient point. The normalized impedance is z (d ) = r + j x and the actual impedance is Z(d) = Z0 z ( d ) = Z0 ( r + j x ) = Z0 r + j Z0 x © Amanogawa, 2000 - Digital Maestro Series 147 Transmission Lines Given ΓR and ZR ⇐⇒ Find Γ(d) and Z(d) NOTE: the magnitude of the reflection coefficient is constant along a loss-less transmission line terminated by a specified load, since Γ (d ) = Γ R exp ( − j 2β d ) = Γ R Therefore, on the complex plane, a circle with center at the origin and radius | ΓR | represents all possible reflection coefficients found along the transmission line. When the circle of constant magnitude of the reflection coefficient is drawn on the Smith chart, one can determine the values of the line impedance at any location. The graphical step-by-step procedure is: 1. Identify the load reflection coefficient ΓR and the normalized load impedance ZR on the Smith chart. © Amanogawa, 2000 - Digital Maestro Series 148 Transmission Lines 2. 3. Draw the circle of constant reflection coefficient amplitude |Γ(d)| =|ΓR|. Starting from the point representing the load, travel on the circle in the clockwise direction, by an angle 2π d θ= 2βd = 2 λ 4. The new location on the chart corresponds to location d on the transmission line. Here, the values of Γ(d) and Z(d) can be read from the chart as before. Example: Given ZR = 25 + j 100 Ω with Z0 = 50 Ω find Z( d ) and Γ( d ) © Amanogawa, 2000 - Digital Maestro Series for d = 0.18λ 149 Transmission Lines Circle with constant | Γ | 1 zR 05 ΓR 2 3 0.2 ∠ ΓR 0 0.2 0.5 -0.2 Γ(d) = 0.8246 ∠-78.7° = 0.161 – j 0.809 1 2 = 2 (2π/λ) 0.18 λ = 2.262 rad = 129.6° θ Γ (d) -3 z(d) -2= 0.236 – j1.192 Z(d) = z(d) × Z0 = 11.79 – j59.6 Ω -0 5 -1 © Amanogawa, 2000 - Digital Maestro Series 5 θ=2βd z(d) 150 Transmission Lines Given ΓR and ZR 1. 2. 3. ⇒ Find dmax and dmin Identify on the Smith chart the load reflection coefficient ΓR or the normalized load impedance ZR . Draw the circle of constant reflection coefficient amplitude |Γ(d)| =|ΓR|. The circle intersects the real axis of the reflection coefficient at two points which identify dmax (when Γ(d) = Real positive) and dmin (when Γ(d) = Real negative) A commercial Smith chart provides an outer graduation where the distances normalized to the wavelength can be read directly. The angles, between the vector ΓR and the real axis, also provide a way to compute dmax and dmin . Example: Find dmax and dmin for ZR = 25 + j 100 Ω ; ZR = 25 − j100Ω © Amanogawa, 2000 - Digital Maestro Series (Z0 = 50 Ω ) 151 ZR 25 j 100 Im(ZR) > 0 ( Z0 50 ) Transmission Lines 2β dmax = 50.9° dmax = 0.0707λ 1 ZR 05 ΓR 2 3 0.2 ∠ ΓR 0 0.2 0.5 1 2 5 -0.2 -3 -2 -0 5 -1 © Amanogawa, 2000 - Digital Maestro Series 2β dmin = 230.9° dmin = 0.3207λ 152 ZR 25 j 100 Im(ZR) < 0 ( Z0 50 ) Transmission Lines 2β dmax = 309.1° dmax = 0.4293 λ 1 05 2 3 0.2 0 0.2 0.5 1 2 5 ∠ ΓR -0.2 ΓR -3 -2 -0 5 2β dmin = 129.1° dmin = 0.1793 λ © Amanogawa, 2000 - Digital Maestro Series ZR -1 153 Transmission Lines Given ΓR and ZR ⇒ Find the Voltage Standing Wave Ratio (VSWR) The Voltage standing Wave Ratio or VSWR is defined as Vmax 1 + Γ R = VSWR = Vmin 1 − Γ R The normalized impedance at a maximum location of the standing wave pattern is given by 1 + Γ ( dmax ) 1 + Γ R z ( dmax ) = = = VSWR!!! 1 − Γ ( dmax ) 1 − Γ R This quantity is always real and ≥ 1. The VSWR is simply obtained on the Smith chart, by reading the value of the (real) normalized impedance, at the location dmax where Γ is real and positive. © Amanogawa, 2000 - Digital Maestro Series 154 Transmission Lines The graphical step-by-step procedure is: 1. 2. 3. 4. Identify the load reflection coefficient ΓR and the normalized load impedance ZR on the Smith chart. Draw the circle of constant reflection coefficient amplitude |Γ(d)| =|ΓR|. Find the intersection of this circle with the real positive axis for the reflection coefficient (corresponding to the transmission line location dmax). A circle of constant normalized resistance will also intersect this point. Read or interpolate the value of the normalized resistance to determine the VSWR. Example: Find the VSWR for ZR1 = 25 + j 100 Ω ; ZR2 = 25 − j100Ω © Amanogawa, 2000 - Digital Maestro Series (Z0 = 50 Ω ) 155 Transmission Lines Circle with constant | Γ | 1 zR1 05 2 3 ΓR1 0.2 0 0.2 0.5 1 2 5 ΓR2 -0.2 Circle of constant conductance r = 10.4 z(dmax )=10.4 -3 -2 -0 5 zR2 -1 © Amanogawa, 2000 - Digital Maestro Series For both loads VSWR = 10.4 156 Transmission Lines Given Z(d) ⇐⇒ Find Y(d) Note: The normalized impedance and admittance are defined as 1 + Γ (d ) z( d ) = 1 − Γ (d ) 1 − Γ (d ) y( d ) = 1 + Γ (d ) Since λ Γ d + = −Γ ( d ) 4 ⇒ λ 1 + Γ d + λ 4 1 − Γ (d ) z d + = = = y(d ) λ 1 + Γ (d ) 4 1− Γ d + 4 © Amanogawa, 2000 - Digital Maestro Series 157 Transmission Lines Keep in mind that the equality λ z d + = y( d ) 4 is only valid for normalized impedance and admittance. The actual values are given by λ λ Z d + = Z0 ⋅ z d + 4 4 y( d ) Y ( d ) = Y0 ⋅ y( d ) = Z0 where Y0=1 /Z0 is the characteristic admittance of the transmission © Amanogawa, 2000 - Digital Maestro Series 158 Transmission Lines line. The graphical step-by-step procedure is: 1. 2. 3. Identify the load reflection coefficient ΓR and the normalized load impedance ZR on the Smith chart. Draw the circle of constant reflection coefficient amplitude |Γ(d)| =|ΓR|. The normalized admittance is located at a point on the circle of constant |Γ| which is diametrically opposite to the normalized impedance. Example: Given ZR = 25 + j 100 Ω with Z0 = 50 Ω find YR . © Amanogawa, 2000 - Digital Maestro Series 159 Transmission Lines Circle with constant | Γ | 1 z(d) = 0.5 + j 2.0 Z(d) = 25 + j100 [ Ω ] 05 2 3 0.2 0 0.2 0.5 1 2 5 θ = 180° = 2β⋅λ/4 -0.2 -3 -2 -0 5 © Amanogawa, 2000 - Digital Maestro Series y(d) = 0.11765 – j 0.4706 Y(d) = 0.002353 – j 0.009412 [ S ] z(d+λ/4)-1= 0.11765 – j 0.4706 Z(d+λ/4) = 5.8824 – j 23.5294 [ Ω ] 160 Transmission Lines The Smith chart can be used for line admittances, by shifting the space reference to the admittance location. After that, one can move on the chart just reading the numerical values as representing admittances. Let’s review the impedance-admittance terminology: Impedance = Resistance + j Reactance Z = R + jX Admittance = Conductance + j Susceptance Y = G + jB On the impedance chart, the correct reflection coefficient is always represented by the vector corresponding to the normalized impedance. Charts specifically prepared for admittances are modified to give the correct reflection coefficient in correspondence of admittance. © Amanogawa, 2000 - Digital Maestro Series 161 Transmission Lines Smith Chart for Admittances y(d) = 0.11765 – j 0.4706 -1 -0 5 Negative (inductive) susceptance Γ -2 -3 -0.2 2 5 1 0.5 0.2 0 0.2 3 Positive (capacitive) susceptance 2 05 1 z(d) = 0.5 + 2.0 © Amanogawa, 2000 - Digital Maestro Series 162 Transmission Lines Since related impedance and admittance are on opposite sides of the same Smith chart, the imaginary parts always have different sign. Therefore, a positive (inductive) reactance corresponds to a negative (inductive) susceptance, while a negative (capacitive) reactance corresponds to a positive (capacitive) susceptance. Numerically, we have 1 z= r+ jx y= g+ jb= r+ jx r− jx r − jx y= = 2 ( r + j x )( r − j x ) r + x2 r x g= 2 b=− 2 ⇒ 2 r +x r + x2 © Amanogawa, 2000 - Digital Maestro Series 163 Transmission Lines Impedance Matching A number of techniques can be used to eliminate reflections when line characteristic impedance and load impedance are mismatched. Impedance matching techniques can be designed to be effective for a specific frequency of operation (narrow band techniques) or for a given frequency spectrum (broadband techniques). One method of impedance matching involves the insertion of an impedance transformer between line and load Z0 Impedance Transformer ZR In the following, we neglect effects of loss in the lines. ©Amanogawa, 2000 – Digital Maestro Series 164 Transmission Lines A simple narrow band impedance transformer consists of a transmission line section of length /4 ZA ZB Z01 Z02 Z01 ZR /4 dmax or dmin The impedance transformer is positioned so that it is connected to a real impedance ZA. This is always possible if a location of maximum or minimum voltage standing wave pattern is selected. ©Amanogawa, 2000 – Digital Maestro Series 165 Transmission Lines Consider a general load impedance with its corresponding load reflection coefficient ZR RR jX R ; ZR Z01 R R exp j ZR Z01 If the transformer is inserted at a location of voltage maximum dmax 1 d 1 R ZA Z01 Z01 1 d 1 R If it is inserted instead at a location of voltage minimum dmin 1 d 1 R ZA Z01 Z01 1 d 1 R ©Amanogawa, 2000 – Digital Maestro Series 166 Transmission Lines Consider now the input impedance of a line of length /4 Zin Z0 ZA L = /4 Since: 1 d 1 R ZA Z01 Z01 1 d 1 R we have Zin lim tan L ©Amanogawa, 2000 – Digital Maestro Series ZA jZ0 tan( L) Z0 jZA tan( L) Z0 Z02 ZA 167 Transmission Lines Note that if the load is real, the voltage standing wave pattern at the load is maximum when ZR > Z01 or minimum when ZR < Z01 . The transformer can be connected directly at the load location or at a distance from the load corresponding to a multiple of /4 . ZA=Real ZB Z01 Z02 Z01 ZR=Real d1 /4 n /4 ; n=0,1,2… ©Amanogawa, 2000 – Digital Maestro Series 168 Transmission Lines If the load impedance is real and the transformer is inserted at a distance from the load equal to an even multiple of /4 then ZA ZR ; d1 2 n n 4 2 but if the distance from the load is an odd multiple of /4 2 Z01 ZA ZR ©Amanogawa, 2000 – Digital Maestro Series ; d1 (2 n 1) n 4 2 4 169 Transmission Lines The input impedance of the impedance transformer after inclusion in the circuit is given by 2 Z02 ZB ZA For impedance matching we need 2 Z02 Z01 ZA Z02 Z01 ZA The characteristic impedance of the transformer is simply the geometric average between the characteristic impedance of the original line and the load seen by the transformer. Let’s now review some simple examples. ©Amanogawa, 2000 – Digital Maestro Series 170 Transmission Lines Real Load Impedance ZA ZB Z01 = 50 Z02 = ? RR = 100 /4 2 Z02 ZB Z01 Z02 Z01 RR 50 100 70.71 RR ©Amanogawa, 2000 – Digital Maestro Series 171 Transmission Lines Note that an identical result is obtained by switching Z01 and RR ZA ZB Z01 = 100 Z02 = ? RR = 50 /4 2 Z02 ZB Z01 Z02 Z01 RR 100 50 70.71 RR ©Amanogawa, 2000 – Digital Maestro Series 172 Transmission Lines Another real load case ZA ZB Z01 = 75 Z02 = ? RR = 300 /4 2 Z02 ZB Z01 Z02 Z01 RR 75 300 150 RR ©Amanogawa, 2000 – Digital Maestro Series 173 Transmission Lines Same impedances as before, but now the transformer is inserted at a distance /4 from the load (voltage minimum in this case) ZB Z01 = 75 2 752 Z01 ZA 18.75 RR 300 ZA Z02 /4 Z01 RR = 300 /4 2 Z02 ZB Z01 Z02 Z01 ZA 75 18.75 37.5 ZA ©Amanogawa, 2000 – Digital Maestro Series 174 Transmission Lines Complex Load Impedance – Transformer at voltage maximum ZA ZB Z01 = 50 Z02 Z01 /4 dmax ZR = 100 + j 100 100 j100 50 R 0.62 100 j100 50 1 R ZA Z0 213.28 1 R Z02 Z01 ZA 50 213.28 103.27 ©Amanogawa, 2000 – Digital Maestro Series 175 Transmission Lines Complex Load Impedance – Transformer at voltage minimum ZA ZB Z01 = 50 Z02 Z01 /4 dmin ZR = 100 + j 100 100 j100 50 R 0.62 100 j100 50 1 R ZA Z0 11.72 1 R Z02 Z01 ZA 50 11.72 24.21 ©Amanogawa, 2000 – Digital Maestro Series 176 Transmission Lines If it is not important to realize the impedance transformer with a quarter wavelength line, we can try to select a transmission line with appropriate length and characteristic impedance, such that the input impedance is the required real value ZA Z01 Z02 ZR = RR + jXR L RR jX R jZ02 tan( L) Z01 ZA Z02 Z02 j RR jX R tan( L) ©Amanogawa, 2000 – Digital Maestro Series 177 Transmission Lines After separation of real and imaginary parts we obtain the equations Z02 ( Z01 RR ) Z01 X R tan L tan L Z02 X R 2 Z01 RR Z02 with final solution 2 2 Z01 RR RR XR Z02 1 RR / Z01 tan L 1 RR / Z01 Z01 RR RR2 X R2 XR The transformer can be realized as long as the result for Z02 is real. Note that this is also a narrow band approach. ©Amanogawa, 2000 – Digital Maestro Series 178 Transmission Lines Single stub impedance matching Impedance matching can be achieved by inserting another transmission line (stub) as shown in the diagram below ZA = Z0 ZR Z0 Z0S dstub Lstub © Amanogawa, 2000 – Digital Maestro Series 179 Transmission Lines There are two design parameters for single stub matching: The location of the stub with reference to the load dstub The length of the stub line Lstub Any load impedance can be matched to the line by using single stub technique. The drawback of this approach is that if the load is changed, the location of insertion may have to be moved. The transmission line realizing the stub is normally terminated by a short or by an open circuit. In many cases it is also convenient to select the same characteristic impedance used for the main line, although this is not necessary. The choice of open or shorted stub may depend in practice on a number of factors. A short circuited stub is less prone to leakage of electromagnetic radiation and is somewhat easier to realize. On the other hand, an open circuited stub may be more practical for certain types of transmission lines, for example microstrips where one would have to drill the insulating substrate to short circuit the two conductors of the line. © Amanogawa, 2000 – Digital Maestro Series 180 Transmission Lines Since the circuit is based on insertion of a parallel stub, it is more convenient to work with admittances, rather than impedances. YA = Y0 YR = 1/ZR Y0 = 1/Z0 dstub Y0S Lstub © Amanogawa, 2000 – Digital Maestro Series 181 Transmission Lines For proper impedance match: 1 YA = Ystub + Y ( d stub ) = Y0 = Z0 Line admittance at location dstub before the stub is applied Input admittance of the stub line Ystub + Y (dstub) YR = 1/ZR Y0S Lstub dstub © Amanogawa, 2000 – Digital Maestro Series 182 Transmission Lines In order to complete the design, we have to find an appropriate location for the stub. Note that the input admittance of a stub is always imaginary (inductance if negative, or capacitance if positive) Ystub = jBstub A stub should be placed at a location where the line admittance has real part equal to Y0 Y (d stub ) = Y0 + jB (d stub) For matching, we need to have Bstub = − B ( d stub ) Depending on the length of the transmission line, there may be a number of possible locations where a stub can be inserted for impedance matching. It is very convenient to analyze the possible solutions on a Smith chart. © Amanogawa, 2000 – Digital Maestro Series 183 Transmission Lines First location suitable for stub insertion 1 y(dstub1) 05 2 3 θ2 θ1 0.2 0 0.2 0.5 zR 1 2 5 Constant |Γ(d)| circle -0.2 -3 yR Load location Unitary conductance circle -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series y(dstub2) Second location suitable for stub insertion 184 Transmission Lines The red arrow on the example indicates the load admittance. This provides on the “admittance chart” the physical reference for the load location on the transmission line. Notice that in this case the load admittance falls outside the unitary conductance circle. If one moves from load to generator on the line, the corresponding chart location moves from the reference point, in clockwise motion, according to an angle θ (indicated by the light green arc) 4π θ = 2β d = d λ The value of the admittance rides on the red circle which corresponds to constant magnitude of the line reflection coefficient, |Γ(d)|=|ΓR |, imposed by the load. Every circle of constant |Γ(d)| intersects the circle Re { y } = 1 (unitary normalized conductance), in correspondence of two points. Within the first revolution, the two intersections provide the locations closest to the load for possible stub insertion. © Amanogawa, 2000 – Digital Maestro Series 185 Transmission Lines The first solution corresponds to an admittance value with positive imaginary part, in the upper portion of the chart ( Y ( d stub1 ) = Y0 + j B d stub1 Line Admittance - Actual : y ( d stub1 ) = 1 + j b ( d stub1 ) Normalized : θ1 Stub Location : d stub1 = λ 4π ( ) − j b ( d stub1 ) − j B d stub1 Stub Admittance - Actual : Normalized : λ Stub Length : Lstub = tan −1 Z0 s B 2π λ Lstub = tan −1 Z0s B 2π ( © Amanogawa, 2000 – Digital Maestro Series ) 1 d stub1 ( short ) ( ) (d stub1 )) (open) 186 Transmission Lines The second solution corresponds to an admittance value with negative imaginary part, in the lower portion of the chart ( Y (d stub2 ) = Y0 − j B d stub2 Line Admittance - Actual : y (d stub2 ) = 1 − j b ( d stub2 ) Normalized : θ Stub Location : d stub2 = 2 λ 4π ( ) j b ( d stub2 ) j B d stub2 Stub Admittance - Actual : Normalized : λ 1 −1 Stub Length : Lstub = − tan Z0 s B d stub 2π 2 λ Lstub = tan −1 − Z0s B d stub2 2π ( © Amanogawa, 2000 – Digital Maestro Series ) ( ( ( short ) ) )) (open) 187 Transmission Lines If the normalized load admittance falls inside the unitary conductance circle (see next figure), the first possible stub location corresponds to a line admittance with negative imaginary part. The second possible location has line admittance with positive imaginary part. In this case, the formulae given above for first and second solution exchange place. If one moves further away from the load, other suitable locations for stub insertion are found by moving toward the generator, at distances multiple of half a wavelength from the original solutions. These locations correspond to the same points on the Smith chart. First set of locations Second set of locations © Amanogawa, 2000 – Digital Maestro Series λ = d stub1 + n 2 λ = d stub2 + n 2 188 Transmission Lines Second location suitable for stub insertion 1 y(dstub2) 05 2 3 0 yR θ2 0.2 0.2 0.5 Load location 1 θ1 2 5 Unitary conductance circle Constant |Γ(d)| circle -0.2 zR -3 -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series y(dstub1) First location suitable for stub insertion 189 Transmission Lines Single stub matching problems can be solved on the Smith chart graphically, using a compass and a ruler. This is a step-by-step summary of the procedure: (a) Find the normalized load impedance and determine the corresponding location on the chart. (b) Draw the circle of constant magnitude of the reflection coefficient |Γ| for the given load. (c) Determine the normalized load admittance on the chart. This is obtained by rotating 180° on the constant |Γ| circle, from the load impedance point. From now on, all values read on the chart are normalized admittances. © Amanogawa, 2000 – Digital Maestro Series 190 Transmission Lines (a) Obtain the normalized load 1 impedance zR=ZR /Z0 and find its location on the Smith chart 05 2 3 zR 0.2 0 0.2 0.5 -0.2 1 2 5 180° = λ /4 -3 yR (b) Draw the constant |Γ(d)| circle -2 -0 5 (c) Find the normalized load admittance knowing that yR = z(d=λ /4 ) From now on the represents admittances. chart © Amanogawa, 2000 – Digital Maestro Series -1 191 Transmission Lines (d) Move from load admittance toward generator by riding on the constant |Γ| circle, until the intersections with the unitary normalized conductance circle are found. These intersections correspond to possible locations for stub insertion. Commercial Smith charts provide graduations to determine the angles of rotation as well as the distances from the load in units of wavelength. (e) Read the line normalized admittance in correspondence of the stub insertion locations determined in (d). These values will always be of the form y (d stub ) = 1 + jb y (d stub ) = 1 − jb © Amanogawa, 2000 – Digital Maestro Series top half of chart bottom half of chart 192 Transmission Lines First Solution (d) Move from load toward generator and stop at a location where the real part of the normalized line admittance is 1. 05 First location suitable for stub insertion 1 dstub1=(θ1/4π)λ 2 3 zR 0.2 (e) Read here the value of the normalized line admittance y(dstub1) = 1+jb θ1 0 0.2 0.5 1 2 5 -0.2 -3 yR Load location -2 Unitary conductance circle -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series 193 Transmission Lines Second Solution (d) Move from load toward generator and stop at a location where the real part of the normalized line 05 admittance is 1. 1 2 3 θ2 0.2 0 0.2 0.5 zR 1 2 Unitary conductance circle 5 -0.2 -3 yR Load location -2 -0 5 (e) Read here the value of the normalized line admittance y(dstub2) = 1 - jb -1 Second location suitable for stub insertion dstub2=(θ2/4π)λ © Amanogawa, 2000 – Digital Maestro Series 194 Transmission Lines (f) Select the input normalized admittance of the stubs, by taking the opposite of the corresponding imaginary part of the line admittance line: y ( d stub ) = 1 + jb line: y ( d stub ) = 1 − jb → stub: ystub = − jb → stub: ystub = + jb (g) Use the chart to determine the length of the stub. The imaginary normalized admittance values are found on the circle of zero conductance on the chart. On a commercial Smith chart one can use a printed scale to read the stub length in terms of wavelength. We assume here that the stub line has characteristic impedance Z0 as the main line. If the stub has characteristic impedance Z0S must be renormalized as ≠ Z0 the values on the Smith chart Y0 Z0 s ± jb' = ± jb = ± jb Y0 s Z0 © Amanogawa, 2000 – Digital Maestro Series 195 Transmission Lines 1 05 2 3 Short circuit 0.2 0 0.2 -0.2 y=∞ 0.5 1 2 5 (g) Arc to determine the length of a short circuited stub with normalized input admittance - jb -3 (f) Normalized input admittance of stub -0 5 ystub = 0 - jb -2 -1 © Amanogawa, 2000 – Digital Maestro Series 196 Transmission Lines 1 05 2 3 (g) Arc to determine the length of an open circuited stub with normalized input admittance - jb 0.2 0 0.2 0.5 1 2 5 -0.2 y=0 -3 (f) Normalized input admittance of stub Open circuit -0 5 ystub = 0 - jb -2 -1 © Amanogawa, 2000 – Digital Maestro Series 197 Transmission Lines 1 (f) Normalized input admittance of stub ystub = 0 + jb 05 2 3 (g) Arc to determine the length of a short circuited stub with normalized input admittance + jb 0.2 0 0.2 0.5 1 2 y=∞ Short circuit 5 -0.2 -3 -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series 198 Transmission Lines 1 (f) Normalized input admittance of stub ystub = 0 + jb 05 2 3 0.2 0 (g) Arc to determine the length of an open circuited stub with normalized input admittance + jb 0.2 0.5 1 2 5 -0.2 y=0 -3 Open circuit -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series 199 Transmission Lines First Solution 1 05 2 3 0.2 0 matching condition 0.2 0.5 1 2 5 -0.2 -3 yR -2 -0 5 After the stub is inserted, the admittance at the stub location is moved to the center of the Smith chart, which corresponds to normalized admittance 1 and reflection coefficient 0 (exact matching condition). If you imagine to add gradually the negative imaginary admittance of the inserted stub, the total admittance would follow the yellow arrow, reaching the match point when the complete stub admittance is added. -1 © Amanogawa, 2000 – Digital Maestro Series 200 Transmission Lines First Solution If the stub does not have the proper normalized input admittance, the matching condition is not reached 1 05 2 Effect of a stub with negative susceptance of insufficient magnitude 3 Effect of a stub with positive susceptance 0.2 0 0.2 0.5 1 2 5 -0.2 -3 yR -2 -0 5 Effect of a stub with negative susceptance of excessive magnitude -1 © Amanogawa, 2000 – Digital Maestro Series 201 Transmission Lines Double stub impedance matching Impedance matching can be achieved by inserting two stubs at specified locations along transmission line as shown below YA = Y01 dstub2 Y01 = 1/Z01 dstub1 YR = 1/ZR Y0S2 Y0S1 Lstub2 © Amanogawa, 2000 – Digital Maestro Series Lstub1 202 Transmission Lines There are two design parameters for double stub matching: The length of the first stub line Lstub1 The length of the second stub line Lstub2 In the double stub configuration, the stubs are inserted at predetermined locations. In this way, if the load impedance is changed, one simply has to replace the stubs with another set of different length. The drawback of double stub tuning is that a certain range of load admittances cannot be matched once the stub locations are fixed. Three stubs are necessary to guarantee that match is always possible. © Amanogawa, 2000 – Digital Maestro Series 203 Transmission Lines The length of the first stub is selected so that the admittance at the location of the second stub (before the second stub is inserted) has real part equal to the characteristic admittance of the line Y’A = Y01 + jB dstub2 Y01 = 1/Z01 dstub1 YR = 1/ZR Y0S1 Lstub1 © Amanogawa, 2000 – Digital Maestro Series 204 Transmission Lines YA = Y01 + jB – jB = Y01 dstub2 Y01 = 1/Z01 dstub1 YR = 1/ZR Ystub = -jB Y0S1 Lstub1 Y0S2 Lstub2 The length of the second stub is selected to eliminate the imaginary part of the admittance at the location of insertion. © Amanogawa, 2000 – Digital Maestro Series 205 Transmission Lines 1 05 2 3 0.2 0 0.2 0.5 1 2 At the location where the second stub is inserted, the possible normalized admittances that can give matching are found on the circle of unitary conductance on the Smith chart. 5 -0.2 -3 The normalized admittance that we want at location -0 5 dstub2 -2 is on this circle -1 © Amanogawa, 2000 – Digital Maestro Series 206 Transmission Lines dstub Think of stub matching in a unified way. Single stub YR YA = Y01 The two approaches solve the same problem Y01 = 1/Z01 dstub2 Y0S2 Double stub YR Lstub2 Y0S1 Lstub1 © Amanogawa, 2000 – Digital Maestro Series 207 Transmission Lines If one moves from the location of the second stub back to the load, the circle of the allowed normalized admittances is mapped into another circle, obtained by pivoting the original circle about the center of the chart. At the location of the first stub, the allowed normalized admittances are found on an auxiliary circle which is obtained by rotating the unitary conductance circle counterclockwise, by an angle 4 4 aux d stub2 d stub1 d 21 © Amanogawa, 2000 – Digital Maestro Series 208 Transmission Lines This angle of rotation corresponds to a distance 1 d12 = dstub2 -dstub1 05 2 3 0.2 0 aux 0.2 0.5 1 2 5 Pivot here -0.2 The normalized admittance that we want at location dstub1 is on this auxiliary circle. -0 5 -3 -2 -1 © Amanogawa, 2000 – Digital Maestro Series 209 Transmission Lines 1 05 2 3 0.2 0 aux 0.2 0.5 1 2 5 This is-0.2 the auxiliary circle for distance between the stubs d21 = /8 + n /2. -3 -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series 210 Transmission Lines 1 05 2 3 0.2 0 aux 0.2 0.5 1 2 5 -0.2 -3 -2 -0 5 This is the auxiliary circle for distance between the stubs -1 d21 = /4 + n /2. © Amanogawa, 2000 – Digital Maestro Series 211 Transmission Lines 1 05 2 3 0.2 0 aux 0.2 0.5 1 2 5 -0.2 -3 -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series This is the auxiliary circle for distance between the stubs d21 = 3 /8 + n /2. 212 Transmission Lines 1 05 2 3 0.2 0 aux 0.2 0.5 1 2 5 -0.2 -3 This is the auxiliary circle for distance -0between the stubs 5 d21 = n /2. -2 NOTE: this is not a good -1 choice for double stub design! © Amanogawa, 2000 – Digital Maestro Series 213 Transmission Lines Given the load impedance, we need to follow these steps to complete the double stub design: (a) Find the normalized load impedance and determine the corresponding location on the chart. (b) Draw the circle of constant magnitude of the reflection coefficient || for the given load. (c) Determine the normalized load admittance on the chart. This is obtained by rotating -180 on the constant || circle, from the load impedance point. From now on, all values read on the chart are normalized admittances. (d) Find the normalized admittance at location dstub1 by moving clockwise on the constant || circle. © Amanogawa, 2000 – Digital Maestro Series 214 Transmission Lines (e) Draw the auxiliary circle (f) Add the first stub admittance so that the normalized admittance point on the Smith chart reaches the auxiliary circle (two possible solutions). The admittance point will move on the corresponding conductance circle, since the stub does not alter the real part of the admittance (g) Map the normalized admittance obtained on the auxiliary circle to the location of the second stub dstub2. The point must be on the unitary conductance circle (h) Add the second stub admittance so that the total parallel admittance equals the characteristic admittance of the line to achieve exact matching condition © Amanogawa, 2000 – Digital Maestro Series 215 Transmission Lines (d) Move to the first stub location (a) Obtain the normalized load 1 impedance zR=ZR /Z0 and find its location on the Smith chart 05 2 3 zR 0.2 0 0.2 0.5 -0.2 1 2 5 180 = /4 -3 yR (b) Draw the constant |(d)| circle -2 -0 5 (c) Find the normalized load admittance knowing that yR = z(d= /4 ) From now on the chart represents admittances. © Amanogawa, 2000 – Digital Maestro Series -1 216 Transmission Lines 1 05 2 3 (f) First solution: Add admittance of first stub to reach auxiliary circle 0.2 0 0.2 0.5 1 2 5 -0.2 -3 yR -2 -0 5 -1 (e) Draw the auxiliary circle © Amanogawa, 2000 – Digital Maestro Series (f) Second solution: Add admittance of first stub to reach auxiliary circle 217 Transmission Lines 1 (g) First solution: Map the normalized admittance from the auxiliary circle to the location of the 05 second stub dstub2. 2 3 0.2 0 (h) Add second stub admittance 0.2 0.5 1 2 5 -0.2 -3 -2 -0 5 First solution: Admittance at location dstub2 before insertion of second stub © Amanogawa, 2000 – Digital Maestro Series -1 218 Transmission Lines (g) Second solution: Map the normalized admittance from the auxiliary circle to the location of the second stub dstub2. 1 05 2 3 Second solution: Admittance at 0.2 Add second (h) stub admittance 0 0.2 0.5 location dstub2 before insertion of second stub 1 2 5 -0.2 -3 -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series 219 Transmission Lines As mentioned earlier, a double stub configuration with fixed stub location may not be able to match a certain range of load impedances. This is easily seen on the Smith chart. If the normalized admittance of the line, at the first stub location, falls inside a certain forbidden conductance circle tangent to the auxiliary circle (and always contained inside the unitary conductance circle), it is not possible to find a value for the first stub that can bring the normalized admittance to the auxiliary circle. Therefore, it is impossible to position the normalized admittance of the second stub location on the unitary conductance circle. When this condition occurs, the location of one of the stubs must be changed appropriately. Alternatively, a third stub could be added. Examples of forbidden regions follow. © Amanogawa, 2000 – Digital Maestro Series 220 Transmission Lines 1 05 2 Forbidden conductance circle. If the admittance at the first stub location falls inside this circle, match is not possible with the given two stub configuration. 3 0.2 0 aux 0.2 0.5 1 2 5 This is-0.2 the auxiliary circle for distance between the stubs d21 = /8 + n /2. -3 -2 -0 5 The normalized conductance circle for the normalized admittance does not intersect the auxiliary circle. -1 © Amanogawa, 2000 – Digital Maestro Series 221 Transmission Lines 1 05 2 3 aux 0.2 0 0.2 0.5 1 2 Forbidden 5 conductance circle -0.2 -3 -2 -0 5 This is the auxiliary circle for distance between the stubs -1 d21 = /4 + n /2. © Amanogawa, 2000 – Digital Maestro Series 222 Transmission Lines 1 Forbidden 2 conductance circle 05 aux 0.2 0 3 0.2 0.5 1 2 5 -0.2 -3 -2 -0 5 -1 © Amanogawa, 2000 – Digital Maestro Series This is the auxiliary circle for distance between the stubs d21 = 3 /8 + n /2. 223