Complex Numbers, Phasors and Circuits

Transcription

1 Complex Numbers, Phasors and Circuits Transmission Lines Complex numbers are defined by points or vectors in the complex plane, and can be represented in Cartesian coordinates or in polar (exponential) form z= a+ jb j = 1 ( ) ( ) ( ) ( ) z= Aexp( jφ ) = Acos φ + jasin φ a= Acos φ b= Asin φ real part imaginary part where 1 b A= a + b φ= tan a Amanogawa, 6 Digital Maestro Series 1

2 Im b z A φ a Re Note : z = Aexp( jφ ) = Aexp( jφ± j nπ) Amanogawa, 6 Digital Maestro Series

3 Every complex number has a complex conjugate so that In polar form we have ( ) z* = a+ jb * = a jb z z* = ( a+ jb) ( a jb) = a + b = z = A ( ) Aexp( j j ) Acos( ) jasin ( ) z* = Aexp( jφ ) * = Aexp( jφ) = π φ = φ φ Amanogawa, 6 Digital Maestro Series 3

4 The polar form is more useful in some cases. For instance, when raising a complex number to a power, the Cartesian form n z = ( a+ jb) ( a+ jb) ( a+ jb) is cumbersome, and impractical for non integer exponents. In polar form, instead, the result is immediate n n n [ exp( )] exp( ) z = A jφ = A jnφ In the case of roots, one should remember to consider φ + kπ as argument of the exponential, with k = integer, otherwise possible roots are skipped: n n n φ kπ z = Aexp( jφ+ jkπ ) = Aexp j + j n n The results corresponding to angles up to π are solutions of the root operation. Amanogawa, 6 Digital Maestro Series 4

5 In electromagnetic problems it is often convenient to keep in mind the following simple identities π π j = exp j j = exp j It is also useful to remember the following expressions for trigonometric functions exp( ) exp( ) exp( ) exp( ) cos ( z) = jz + jz ; sin ( z) = jz jz j resulting from Euler s identity exp( ± jz) = cos( z) ± j sin( z) Amanogawa, 6 Digital Maestro Series 5

6 Complex representation is very useful for time-harmonic functions of the form The complex quantity ( ω +φ ) = [ ( ω + φ) ] = Re Aexp( jφ) exp( jωt) = Re Aexp( jωt) Acos t Re Aexp j t j [ ] [ ] A = Aexp( jφ ) contains all the information about amplitude and phase of the signal and is called the phasor of Acos( ω t+φ ) If it is known that the signal is time-harmonic with frequency ω, the phasor completely characterizes its behavior. Amanogawa, 6 Digital Maestro Series 6

7 Often, a time-harmonic signal may be of the form: Asin( ω t+φ ) and we have the following complex representation Transmission Lines ( ω +φ ) = ( ω +φ ) + ( ω +φ) = Re[ jaexp( jω t + jφ) ] = Re Aexp ( jπ/ ) exp( jφ) exp( jωt) = Re Aexp ( j( φ π/ ) ) exp( jωt) = Re Aexp( jωt) ( ) Asin t Re ja cos t jsin t [ ] [ ] ( ) with phasor A= Aexp j( φ π / ) This result is not surprising, since cos( ω t+φ π / ) = sin( ω t+φ ) Amanogawa, 6 Digital Maestro Series 7

8 Time differentiation can be greatly simplified by the use of phasors. Consider for instance the signal ( ) with phasor ( ) Vt () = Vcos ω t+φ V= Vexp jφ The time derivative can be expressed as Vt () t ( t ) = ωv sin ω +φ ( ) ( ) { j V j j t } = Re ω exp φ exp ω jωv exp jφ = jωv ( ) is the phasor of Vt () t Amanogawa, 6 Digital Maestro Series 8

9 With phasors, time-differential equations for time harmonic signals can be transformed into algebraic equations. Consider the simple circuit below, realized with lumped elements L R v (t) i (t) C This circuit is described by the integro-differential equation ( ) 1 di t t vt () = L + Ri+ it () dt dt C Amanogawa, 6 Digital Maestro Series 9

10 Upon time-differentiation we can eliminate the integral as () dv() t d i t di 1 = L + R + i () t dt dt dt C Transmission Lines If we assume a time-harmonic excitation, we know that voltage and current should have the form vt () = V cos( ω t+α ) phaor s V= V exp( jα ) V it () = I cos( ω t+α ) phasr o I= I exp( jα ) I V I If V and α V are given, I and α I are the unknowns of the problem. Amanogawa, 6 Digital Maestro Series 1

11 The differential equation can be rewritten using phasors { ω ( ω )} + ω ( ω ) { } LRe Iexp j t RRe j Iexp j t 1 + Re exp = Re exp C { I ( jωt) } jωv ( jωt) { } Finally, the transform phasor equation is obtained as 1 V = R+ jωl j I = I ωc where 1 = R + j ωl ωc Impedance Resistance Reactance Amanogawa, 6 Digital Maestro Series 11

12 The result for the phasor current is simply obtained as V V I = = = I exp( jαi ) 1 R+ jωl j ωc which readily yields the unknowns I and α I. The time dependent current is then obtained from ( I ) ( ) ( t ) { } it () = Re I exp jα exp jωt = I cos ω +α I Amanogawa, 6 Digital Maestro Series 1

13 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 i ( t ) =? Transform Algebraic equations based on phasors I =? Direct Solution ( Transients ) Solution i ( t ) Anti- Transform I Amanogawa, 6 Digital Maestro Series 13

14 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 ω Capacitive component inversely proportional to ω 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 1 1 ω r L = ω r = ω C LC r This is a resonance condition. The reactance cancels out and the impedance becomes purely resistive. Amanogawa, 6 Digital Maestro Series 14

15 The peak value of the current phasor is maximum at resonance I I = R V 1 + ωl ωc I M ω r ω Amanogawa, 6 Digital Maestro Series 15

16 Consider now the circuit below where an inductor and a capacitor are in parallel I R L V C The input impedance of the circuit is 1 1 jω L in = R+ + jω C R j L = + ω 1 ω LC Amanogawa, 6 Digital Maestro Series 16

17 When ω = = in in R 1 ω = in LC ω = R At the resonance condition ω = 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, 6 Digital Maestro Series 17

18 Power in Circuits Transmission Lines 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) in For sinusoidal excitation, we can write v( t) = V cos( ω t+φ) i() t = I cos( ωt) [ /, /] φ π π where V and I are peak values and φ is the phase difference between voltage and current. Note that φ = only when the input impedance is real (purely resistive). Amanogawa, 6 Digital Maestro Series 18

19 The time-dependent input power is given by Pt () = v() t i() t = V I cos( ω t+φ)cos( ωt) V I = φ + ω +φ [ cos( ) cos( t )] The power has two (Fourier) components: (A) an average value V I cos( φ ) (B) an oscillatory component with frequency f V I cos( ω t +φ ) Amanogawa, 6 Digital Maestro Series 19

20 The power flow changes periodically in time with an oscillation like (B) about the average value (A). Note that only when φ = we have cos(φ) = 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, 6 Digital Maestro Series

21 cos A + B = cos Acos B sin Asin B one can write Using ( ) in phase with current in quadrature with current Transmission Lines v( t) = Vcos( ω t+φ ) = Vcosφcosωt Vsin φsin ωt This gives an alternative expression for power: Pt ( ) = V cos( ωt) I cos( ωt) cos( φ) V cos( ωt) I sin( ωt) sin( φ) V I = V I cos( φ) cos ( ωt) sin( φ) sin( ωt) Real Power Reactive Power V I V I V I = cos( φ ) + cos( φ) cos( ωt) sin( φ)sin( ωt) Real Power Reactive Power Amanogawa, 6 Digital Maestro Series 1

22 The real power corresponds to the power dissipated by the resistive component of the impedance, and it is always positive. The reactive power corresponds to 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 Pt () = Pt () dt T where T = 1 / f is the period of the oscillation. To determine the time-average power, we can use either the Fourier or the real/reactive power formulation. Amanogawa, 6 Digital Maestro Series

23 Fourier representation 1 T V I 1 T V () cos( ) I Pt dt cos( t ) dt T T = φ + ω φ V I = cos( φ) = 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, 6 Digital Maestro Series 3

24 Real/Reactive power representation 1 T T P( t) = ( V I cos( φ ) dt + V I cos( φ)cos( ωt) dt ) T = 1 T V I sin( φ)sin( ω t ) dt T V I cos( φ) = This result tells us that the time-average power flow is the average of the real 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, 6 Digital Maestro Series 4

25 The maximum of the reactive power is VI VI max{ P } max{ sin sin } reac = φ ω = sin φ ( ) ( t) ( ) 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 or reactance: φ > The reactance is inductive Current is lagging with respect to voltage Voltage is leading with respect to current φ < The reactance is capacitive Voltage is lagging with respect to current Current is leading with respect to voltage Amanogawa, 6 Digital Maestro Series 5

26 If the total reactance is inductive V = I = R I + jω LI Im jωl I V Current lags φ > I φ R I Re Amanogawa, 6 Digital Maestro Series 6

27 If the total reactance is capacitive 1 V I R I j I = = ωc Im I φ R I Re Voltage lags φ < V - j I /ω C Amanogawa, 6 Digital Maestro Series 7

28 In many situations, we may use the root-mean-square (r.m.s.) values of quantities, instead of the peak value. For a given signal the r.m.s. value is defined as vt () = V cos( ω t) π cos ( ) 1 T 1 π Vrms = V cos ( ω t) dt = V cos ( ωt) dωt T ωt 1 1 = V ϑ dϑ = V π This result is valid for sinusoidal signals. Any given signal shape V V ) that corresponds to a specific coefficient ( peak factor = / rms allows one to convert directly from peak value to r.m.s. value. Amanogawa, 6 Digital Maestro Series 8

29 The peak factor for sinusoidal signals is V = V rms For a symmetric triangular signal the peak factor is V = V rms V t For a symmetric square signal the peak factor is simply V V 1 rms = V t Amanogawa, 6 Digital Maestro Series 9

30 For a non-sinusoidal periodic signal, we can use a decomposition into orthogonal Fourier components to obtain the r.m.s. value: V() t = V + V () t + V () t + V () t = V t av [ V ] 1 3 () 1 T 1 T rms = V () t dt = [ ()] V k t dt = k T T 1 T 1 T = [ V ( )] [ ( ) ( )] k k t dt + V i j i t Vj t dt T T = orthogonal 1 T 1 T () [ ( ) ( ) k V k t dt + V i t V i j j T T k k t dt = ] = ( V k krms ) ( ) rms ( 1 ) ( k krms av rms rms V = V = V + V + V ) + The final result holds for any decomposition into orthogonal functions and it is known in mathemics as Parseval s identity. Amanogawa, 6 Digital Maestro Series 3

31 In terms of r.m.s. values, the time-average power for a sinusoidal signal is then V () I Pt = cos( φ ) = Vrms Irms cos( φ ) Finally, we can relate the time-average power to the phasors of voltage and current. Since v() t = V cos( ω t) = Re V exp( jωt) { } { } i( t) = I cos( ωt φ ) = Re I exp( jφ)exp( jωt) we have phasors V = V I = I exp( jφ ) Amanogawa, 6 Digital Maestro Series 31

32 The time-average power in terms of phasors is given by 1 * 1 Pt ( ) = Re{ VI } = Re{ V Iexp( jφ) } V I = cos( φ) 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* = I since φ =. 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, 6 Digital Maestro Series 3

33 Now we consider power flow including explicitly the generator, to understand in which conditions maximum power transfer to a load can take place. V V R in = G G + R G I in 1 I = V + in G G R V G V in R 1 Re{ * in in in } P = V I Generator Load Amanogawa, 6 Digital Maestro Series 33

34 As a first case, we examine resistive impedances G = RG R = RR Voltage and current are in phase at the input. The time-average power dissipated by the load is 1 R 1 Pt () = VG V R R R R = R * G G + R G + R 1 R V R G ( R + R ) G R Amanogawa, 6 Digital Maestro Series 34

35 To find the load resistance that maximizes power transfer to the load for a given generator we impose d P() t = dr R from which we obtain d R = dr R R R R ( G + R) G + R R G + R 4 ( RG + RR) ( R R ) R ( R R ) ( R + R ) R = R = R G R R R G We conclude that for maximum power transfer the load resistance must be identical to the generator resistance. = Amanogawa, 6 Digital Maestro Series 35

36 Let s consider now complex impedances R = RR + jx R G = RG + jxg For maximum power transfer, generator and load impedances must be complex conjugate of each other: * R = G R = G R R X R = X G jx G jx G V G R G R G 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, 6 Digital Maestro Series 36

37 The total time-average power supplied by the generator in conditions of maximum power transfer is P = V I = V = V R * tot Re{ G in } G G RR 4 R The time-average power supplied to the load is 1 * 1 * 1 Re{ } Re R Pin = Vin Iin = VG VG G + R G + R 1 R + jx 1 1 = VG = 8 R Re R R VG 4RR R * Amanogawa, 6 Digital Maestro Series 37

38 The power dissipated by the internal generator impedance is 1 Re { ( ) * G G in in } P = V V I = V V = V G G G RR RR RR 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, 6 Digital Maestro Series 38

39 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. Generator Load V G G Transmission line R Amanogawa, 6 Digital Maestro Series 39

40 Examples of transmission lines d d D d Two-wire line D Coaxial cable w t h Microstrip Amanogawa, 6 Digital Maestro Series 4

41 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. Generator Load G V G V R V R V R R V V R R = G G + R L << λ Amanogawa, 6 Digital Maestro Series 41

42 Let s look at some examples. The electricity supplied to households consists of high power sinusoidal signals, with frequency of 6Hz or 5Hz, depending on the country. Assuming that the insulator between wires is air (ε ε ), the wavelength for 6Hz is: 8 c λ= = 5. 1 m= 5, f 6 km which is the about the distance between S. Francisco and Boston! Let s compare to a frequency in the microwave range, for instance 6 GHz. The wavelength is given by 8 c λ= = 5. 1 m= 5. f mm which is comparable to the size of a microprocessor chip. Which conclusions do you draw? Amanogawa, 6 Digital Maestro Series 4

43 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). Generator + jβz jβz V(z) = V e + V e Load G V G V( ) V( z ) V( L ) R L Amanogawa, 6 Digital Maestro Series 43

44 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 ( R ) we have: I = = in and Vin VG If the generator is connected to a load I V in in = = G ( + ) G V G ( + ) G V R R R If the load is a short ( R = ) V I G in = and Vin = G R V G G Generator V in I in R Load Amanogawa, 6 Digital Maestro Series 44

45 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: in = R in only if λ L= n = [ n integer ] Transmission line L R Our first goal is to determine the equivalent impedance seen by the generator, that is, the input impedance of a line terminated by the load. Once that is known, standard circuit theory can be used. Amanogawa, 6 Digital Maestro Series 45

46 Generator G Load V G Transmission line R Generator G Equivalent Load V G in Amanogawa, 6 Digital Maestro Series 46

47 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 L dz R dz C dz G dz C dz G dz dz dz Amanogawa, 6 Digital Maestro Series 47

48 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, 6 Digital Maestro Series 48

49 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. I (z) L dz I (z)+di V (z) C dz V (z)+dv dz Amanogawa, 6 Digital Maestro Series 49

50 The series inductance determines the variation of the voltage from input to output of the cell, according to the sub-circuit below L dz V (z) I (z) V (z)+dv dz The corresponding circuit equation is ( V+ d V) V = jω Ldz I which gives a first order differential equation for the voltage dv dz = jω L I Amanogawa, 6 Digital Maestro Series 5

51 The current flowing through the shunt capacitance determines the variation of the current from input to output of the cell. I (z) C dz di I (z)+di V (z)+dv The circuit equation for the sub-circuit above is di = jω Cdz( V+ d V) = jωcvdz jω CdVdz The second term (including dv dz) tends to zero very rapidly in the limit of infinitesimal length dz leaving a first order differential equation for the current di dz = jω CV Amanogawa, 6 Digital Maestro Series 51

52 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 = jωl I dz di = jωcv dz These are often called telegraphers equations of the loss-less transmission line. Amanogawa, 6 Digital Maestro Series 5

53 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 d dz d dz V I di dz = jω CV di = jω L = jωl jω CV= ω LCV dz dv = jω C = jωc jω L I= ω LC I dz dv dz = jω L I These are often called telephonists equations. Amanogawa, 6 Digital Maestro Series 53

54 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 + jβz jβz V(z) = V e + V e where the wave propagation constant is β=ω LC Note that the complex exponential terms including β have unitary magnitude and purely imaginary argument, therefore they only affect the phase of the wave in space. Amanogawa, 6 Digital Maestro Series 54

55 We have the following useful relations: π π f ω β= = = λ v v v p p ω ε r µ = r =ω ε µ ε r µ r =ω εµ c f Here, λ= p is the wavelength of the dielectric medium surrounding the conductors of the transmission line and 1 1 v p = = ε ε µ µ εµ r is the phase velocity of an electromagnetic wave in the dielectric. As you can see, the propagation constant β can be written in many different, equivalent ways. r Amanogawa, 6 Digital Maestro Series 55

56 The current distribution on the transmission line can be readily obtained by differentiation of the result for the voltage dv dz + jβz jβz = jβ V e + jβ V e = jω L I which gives C 1 I(z) = V e V e = V e V e L The real quantity ( + jβz jβ z) ( + jβz jβz) = L C is the characteristic impedance of the loss-less transmission line. Amanogawa, 6 Digital Maestro Series 56

57 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. I (z) L dz R dz I (z)+di V (z) C dz G dz V (z)+dv dz Amanogawa, 6 Digital Maestro Series 57

58 The series impedance determines the variation of the voltage from input to output of the cell, according to the sub-circuit L dz R dz V (z) I (z) V (z)+dv dz The corresponding circuit equation is ( V+ d V) V = ( jω Ldz + Rdz) I from which we obtain a first order differential equation for the voltage dv dz = ( jω L+ R) I Amanogawa, 6 Digital Maestro Series 58

59 The current flowing through the shunt admittance determines the input-output variation of the current, according to the sub-circuit I (z) di I (z)+di C dz G dz V (z)+dv The corresponding circuit equation is d I = ( jω Cdz + Gdz)( V+ d V) = ( jω C+ G) Vdz ( jω C+ G)dVdz The second term (including dv dz) can be ignored, giving a first order differential equation for the current di dz = ( jω C+ G) V Amanogawa, 6 Digital Maestro Series 59

60 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 = ( jω L+ R) I dz di = ( jω C+ G) V dz These are the telegraphers equations for the lossy transmission line case. Amanogawa, 6 Digital Maestro Series 6

61 One can easily obtain a set of uncoupled equations by differentiating with respect to the coordinate z as done earlier d dz d dz V I di dz = ( jω C+ G) V di = ( jω L+ R) = ( jω L+ R)( jω C+ G) V dz dv = ( jω C+ G) = ( jω C+ G)( jω L+ R) I dz dv dz = ( jω L+ R) I These are the telephonists equations for the lossy line. Amanogawa, 6 Digital Maestro Series 61

62 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 + γz γ z + αz jβ z αz jβ z V(z) = V e + V e = V e e + V e e where the wave propagation constant is now the complex quantity γ= ( jω L+ R)( jω C+ G) =α+ jβ The real part α of the propagation constant γ describes the attenuation of the signal due to resistive losses. The imaginary part β describes the propagation properties of the signal waves as in loss-less lines. The exponential terms including α are real, therefore, they only affect the magnitude of the voltage phasor. The exponential terms including β have unitary magnitude and purely imaginary argument, affecting only the phase of the waves in space. Amanogawa, 6 Digital Maestro Series 6

63 The current distribution on a lossy transmission line can be readily obtained by differentiation of the result for the voltage which gives dv dz + γz γz = ( jω L+ R) I = γ V e +γ V e ( jω C+ G) + γz γz I(z) = ( V e V e ) ( jω L+ R) 1 ( + γz γz V e V e ) = with the characteristic impedance of the lossy transmission line = ( jω L+ R) ( jω C+ G) Note: the characteristic impedance is now complex! Amanogawa, 6 Digital Maestro Series 63

64 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! R R Amanogawa, 6 Digital Maestro Series 64

65 We have obtained the following solutions for the steady-state voltage and current phasors in a transmission line: Loss-less line Lossy line + jβz jβz + γz V(z) = V e + V e 1 I(z) V e V e ( + jβz jβz = ) ( + γz γz I(z) = V e V e ) V(z) = V e + V e 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. 1 γz Amanogawa, 6 Digital Maestro Series 65

66 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. New Space Coordinate R d z 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, 6 Digital Maestro Series 66

67 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 Lossy line + jβd jβd + γd γd V(d) = V e + V e 1 I(d) V e V e ( + jβd jβd = ) ( + γd γd I(d) = V e V e ) At the load (d = ) we have, for both cases, V() = V + V 1 I() = V V + V(d) = V e + V e ( + ) 1 Amanogawa, 6 Digital Maestro Series 67

68 For a given load impedance R, the load boundary condition is Therefore, we have V() = I() R + R + ( ) V + V = V V from which we obtain the voltage load reflection coefficient Γ R V = = + V R R + Amanogawa, 6 Digital Maestro Series 68

69 We can introduce this result into the transmission line equations as Loss-less line Lossy line ( jβ Γ ) R + jβd d V(d) = V e 1+ e ( γd Γ ) R + γd V(d) = V e 1+ e + jβd V e I(d) = 1 ( jβd Γ ) Re + γd V e I(d) = 1 ( γd Γ ) Re At each line location we define a Generalized Reflection Coefficient Γ (d) = Γ j d R e β and the line equations become + jβd jβd ( Γ ) V(d) = V e 1 + (d) + V e I(d) = 1 (d) ( Γ ) Γ (d) = Γ + d R e γ γd γd ( Γ ) V(d) = V e 1 + (d) + V e I(d) = 1 (d) ( Γ ) Amanogawa, 6 Digital Maestro Series 69

70 We define the line impedance as ( d) ( ) ( ) ( ) ( ) = V d 1 d I d = + Γ 1 Γ d A simple circuit diagram can illustrate the significance of line impedance and generalized reflection coefficient: Γ Req = Γ(d) eq =(d) R d Amanogawa, 6 Digital Maestro Series 7

71 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 (d) replaces the cut out portion. The reflection coefficient of the new load is equal to Γ(d) Γ Req ( d ) = Γ = Req Req If the total length of the line is L, the input impedance is obtained from the formula for the line impedance as in ( ) ( ) + ( ) ( ) = Vin V L 1 L I = Γ I L = + 1 Γ L in The input impedance is the equivalent impedance representing the entire line terminated by the load. Amanogawa, 6 Digital Maestro Series 71

72 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: ω L>> R ω C>> G so that γ = ( jωl+ R)( jωc+ G) R G = jωl jωc 1+ 1 jωl + jωc R G RG jω LC 1+ + jωl jωc ω LC The last term under the square root can be neglected, because it is the product of two very small quantities. Amanogawa, 6 Digital Maestro Series 7

73 What remains of the square root can be expanded into a truncated Taylor series γ 1 R G jω LC jωl jωc 1 C L = R + G + jω LC L C so that α 1 C L = R + G LC L C β = ω Amanogawa, 6 Digital Maestro Series 73

74 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+ jωl L = G+ jωc C and the phase velocity associated to the wave propagation is v p ω = β 1 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, 6 Digital Maestro Series 74

75 Again, we obtain the loss-less transmission line if we assume Transmission Lines R = G = 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 C while the propagation constant has no attenuation term γ= ( jωl)( jω C) = jω LC = jβ The loss-less line does not dissipate power, because α =. Amanogawa, 6 Digital Maestro Series 75

76 For all cases, the line impedance was defined as (d) V (d) 1 +Γ(d) = = I(d) 1 Γ(d) By including the appropriate generalized reflection coefficient, we can derive alternative expressions of the line impedance: A) Loss-less line jβd 1+ΓRe jβd 1 ΓRe tan( d) (d) R + j β = = j tan( β d) + B) Lossy line (including low-loss) d γ 1+ΓRe d γ 1 ΓRe tanh( d) (d) R + γ = = tanh( γ d) + R R Amanogawa, 6 Digital Maestro Series 76

77 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) { * } 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, 6 Digital Maestro Series 77

78 The generator, the input impedance, the input voltage and the input current determine the power injected at the transmission line input. G I in V V in in = G G + in V G V in in 1 I = V + in G G in Generator Line 1 P = Re V I { * } in in in Amanogawa, 6 Digital Maestro Series 78

79 The time-average power reaching the load of the transmission line is given by the general expression 1 { * P(d=, t) = Re V() I () } = Re V 1+ R 1 * ( ) R ( Γ ) V ( Γ ) 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 = P P(d =, t ) line in 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. Remember that the internal impedance of the generator dissipates part of the total power generated. * Amanogawa, 6 Digital Maestro Series 79

80 It is instructive to develop further the general expression for the time-average power at the load, using =R +jx for the characteristic impedance, so that 1 R + jx R + jx = = = * * + R X Alternatively, one may simplify the analysis by introducing the line characteristic admittance 1 Y = = G + jb It may be more convenient to deal with the complex admittance at the numerator of the power expression, rather than the complex characteristic impedance at the denominator. Amanogawa, 6 Digital Maestro Series 8

81 (d=, ) 1 P t Re V 1 1 = +Γ V 1 Γ = R R * V = Re R + jx 1+ Re Γ + jim Γ 1 Re Γ + jim Γ V = Re R + + R R R R jx 1 Re ΓR Im jim + Γ R + Γ R V = Re R jx 1 jim + Γ + Γ = R R V = R R Γ X Im Γ R R * Amanogawa, 6 Digital Maestro Series 81

82 Equivalently, using the complex characteristic admittance: (d=, ) 1 P t Re V 1 Y* = + +Γ V+ 1 R Γ = R + V = Re G jb 1+ Re Γ + jim Γ 1 Re Γ + jim Γ + V = Re G jb R + + * Transmission Lines R R R R 1 Re Γ Im jim + Γ R + Γ R V = Re G jb 1 jim Γ + Γ = R R V = G G Γ + B Im Γ R R Amanogawa, 6 Digital Maestro Series 8

83 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. in Generator G Load V G Transmission line R G = * in for maximum power transfer Amanogawa, 6 Digital Maestro Series 83

84 The characteristic impedance of the loss-less line is real and we can express the power flow, anywhere on the line, as 1 * P(d, t) = Re{ V(d) I (d) } 1 Re + jβd 1 ( jβd Γ ) Re = V e + 1 ( V ) e 1 e ( j β Γ ) R + * jβd d + + V V ΓR 1 1 = Incident wave Reflected wave This result is valid for any location, including the input and the load, since the transmission line does not absorb any power. * Amanogawa, 6 Digital Maestro Series 84

85 In the case of low-loss lines, the characteristic impedance is again real, but the time-average power flow is position dependent because the line absorbs power. 1 { * P(d, t) = Re V(d) I (d) } 1 Re { + αd jβd( 1 γd = V e e + Γ ) Re 1 ( V ) e e 1 e ( Γ ) R + * αd jβd γd + αd + αd V e V e ΓR 1 1 = Incident wave Reflected wave * Amanogawa, 6 Digital Maestro Series 85

86 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 + α L + α d + V e V e 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 + α L + α d + ΓR ΓR ΓR V e V e V input inside the line load Amanogawa, 6 Digital Maestro Series 86

87 For a general lossy line the power flow is again position dependent. Since the characteristic impedance is complex, the result has an additional term involving the imaginary part of the characteristic admittance, B, as 1 P(d, t) = Re V(d) I (d) { * } 1 Re + αd jβd 1 (d) { = V e e + ( Γ ) * + * αd jβd * ( Γ ) } + αd G + αd V e V e ΓR Y ( V ) e e 1 (d) G = + B V + e αd Im( Γ(d)) Amanogawa, 6 Digital Maestro Series 87

88 For the general lossy line, keep in mind that * 1 R jx R jx = = = = = * + R + X Y G jb G R X = B = R + X R + X Recall that for a low-loss transmission line the characteristic impedance is approximately real, so that B and 1 G R. The previous result for the low-loss line can be readily recovered from the time-average power for the general lossy line. Amanogawa, 6 Digital Maestro Series 88

89 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 1 (L) (L) in + Γ Vin = V = VG with in = + 1 Γ(L) G in or j tan( β L) R in = j R tan( β L) tanh( γ L) R in = R tanh( γ L) + loss - less line lossy line Amanogawa, 6 Digital Maestro Series 89

90 For a loss-less transmission line: + j β L j L j L [ Γ ] + β ΓR β V(L) = V e 1 + (L) = V e (1 + e ) + 1 V = V e (1 e ) For a lossy transmission line: in G j L j L G + β β in + ΓR + γ L + γ L γ d [ Γ ] ( Γ ) V(L) = V e 1 + (L) = V e 1+ Re + 1 V = V γl e (1 e ) in G L G + γ in + ΓR Amanogawa, 6 Digital Maestro Series 9

91 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: 1 R Γ 1 1 in Γ in 1 1 = + in 1 Amanogawa, 6 Digital Maestro Series 91

92 Special Cases R (SHORT CIRCUIT) R = The load boundary condition due to the short circuit is V () = + jβ jβ R + (1 R) V(d = ) = V e (1 +Γ e ) = V +Γ = Γ = 1 R Amanogawa, 6 Digital Maestro Series 9

93 Since Γ = R V V + + V = V We can write the line voltage phasor as + jβd jβd V(d) = V e + V e + jβ d + jβd = V e V e + jβd jβd = V ( e e ) + = jv sin( βd) Amanogawa, 6 Digital Maestro Series 93

94 For the line current phasor we have 1 + jβd jβd I(d) = ( V e V e ) 1 ( + jβ d + jβd V e V e ) + V e jβd e jβd = + = ( + ) + The line impedance is given by V = cos( β d) + V(d) jv sin( βd) (d) = = = j tan( βd) I(d) + V cos( βd) / Amanogawa, 6 Digital Maestro Series 94

95 The time-dependent values of voltage and current are obtained as jω t + jθ jωt V(d, t) = Re[ V(d) e ] = Re[ j V e sin( βd) e ] + j( ω t+θ) = V sin( βd) Re[ j e ] + = V sin( βd) Re[ jcos( ω t+θ) sin( ω t+θ)] + = V sin( β d) sin( ω t+θ) I(d, t) = Re[ I(d) e ] = Re[ V e cos( βd) e ]/ + j( ω t+θ) = V cos( βd) Re[ e ]/ + = V cos( βd) Re[(cos( ω t+θ ) + jsin( ω t+θ)]/ + V = cos( β d) cos( ω t +θ) jω t + jθ jωt Amanogawa, 6 Digital Maestro Series 95

96 The time-dependent power is given by P(d, t) = V(d, t) I(d, t) + V = 4 sin( βd)cos( βd)sin( ω t+θ)cos( ω t+θ) + V = sin(βd) sin (ω t + θ) and the corresponding time-average power is 1 T < P(d, t) > = P(d, t) dt T + V 1 T = sin(βd) sin (ω t + θ ) = T Amanogawa, 6 Digital Maestro Series 96

97 R (OPEN CIRCUIT) R The load boundary condition due to the open circuit is I () = + V jβ jβ I(d = ) = e (1 ΓR e ) + V = (1 Γ R) = Γ = R 1 Amanogawa, 6 Digital Maestro Series 97

98 Since Γ = R V = V V V + + We can write the line current phasor as 1 + jβd jβd I(d) = ( V e V e ) 1 ( + jβ d V e + jβd V e ) + jβd jβd + jv = V = ( e e ) = sin( βd) Amanogawa, 6 Digital Maestro Series 98

99 For the line voltage phasor we have + jβd jβd V(d) = ( V e + V e ) + jβ d + jβd = ( V e + V e ) + jβd jβd = V ( e + e ) + = V cos( βd) The line impedance is given by (d) + V(d) V cos( βd) = = = j I(d) + jv sin( βd)/ tan( βd) Amanogawa, 6 Digital Maestro Series 99

100 The time-dependent values of voltage and current are obtained as jω t + jθ jωt V(d, t) = Re[ V(d) e ] = Re[ V e cos( βd) e ] + j( ω t+θ) = V cos( βd) Re[ e ] + = V cos( βd) Re[(cos( ω t+θ ) + jsin( ω t+θ)] + = V cos( β d) cos( ω t+θ) I(d, t) = Re[ I(d) e ] = Re[ j V e sin( βd) e ]/ + j( ω t+θ) = V sin( βd) Re[ je ]/ + = V sin( βd) Re[ jcos( ω t+θ) sin( ω t+θ)]/ + V = sin( β d) sin( ω t +θ) jω t + jθ jωt Amanogawa, 6 Digital Maestro Series 1

101 The time-dependent power is given by P(d,) t = V(d,) t I(d,) t = + V = 4 cos( βd)sin( βd)cos( ω t+θ)sin( ω t+θ) + V = sin(βd) sin (ω t + θ) and the corresponding time-average power is 1 T < P(d,) t > = P(d,) t dt T + V 1 T = sin(βd) sin (ω t + θ ) = T Amanogawa, 6 Digital Maestro Series 11

102 R = (MATCHED LOAD) R = The reflection coefficient for a matched load is R Γ R = = = + + R The line voltage and line current phasors are no reflection! + jβd jβ d + jβd R + + jβ d jβd V jβd R V(d) = V e (1 +Γ e ) = V e V Id ( ) = e (1 Γ e ) = e Amanogawa, 6 Digital Maestro Series 1

103 The line impedance is independent of position and equal to the characteristic impedance of the line (d) + jβ d = V(d) V e I(d) = + V = e jβ d The time-dependent voltage and current are + jθ jβd jωt V(d, t) = Re[ V e e e ] + j( ω t+β d +θ ) + = V Re[ e ] = V cos( ω t+β d +θ) + jθ jβd jωt + + j( ω+β t d +θ) V I(d, t) = Re[ V e e e ]/ V = Re[ e ] = cos( ω t+β d +θ) Amanogawa, 6 Digital Maestro Series 13

104 The time-dependent power is + V P(d, t) = V cos( ω t+β d +θ) cos( ω t+β d +θ) + V = cos ( ω t +β d +θ) + and the time average power absorbed by the load is 1 t V < P(d) > = cos ( ω t+β d +θ) dt T = + V + Amanogawa, 6 Digital Maestro Series 14

105 R = jx (PURE REACTANCE) R = j X The reflection coefficient for a purely reactive load is R jx Γ R = = = + jx+ R j ( jx )( jx ) X X = = + ( jx + )( jx ) + X + X Amanogawa, 6 Digital Maestro Series 15

106 In polar form where ( ) Γ R =ΓRexp( jθ ) ( ) 4X X X Γ R = + = = X X X ( ) ( ) ( ) 1 1 X θ= tan X 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 power and then returns it to the line without dissipation. Amanogawa, 6 Digital Maestro Series 16

107 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 π π f in = jtan( β L) = jtan L = jtan L λ v p For a specified frequency f, any reactance value (positive or negative!) can be obtained by changing the length of the line from to λ/. An inductance is realized for L < λ/4 (positive tangent) while a capacitance is realized for λ/4 < L < λ/ (negative tangent). When L = and L = λ/ the tangent is zero, and the input impedance corresponds to a short circuit. However, when L = λ/4 the tangent is infinite and the input impedance corresponds to an open circuit. Amarcord, 6 Digital Maestro Series 17

108 Since the tangent function is periodic, the same impedance behavior of the impedance will repeat identically for each additional line increment of length λ/. A similar periodic behavior is also obtained when the length of the line is fixed and the frequency of operation is changed. At zero frequency (infinite wavelength), the short circuited line behaves as a short circuit for any line length. When the frequency is increased, the wavelength shortens and one obtains an inductance for L < λ/4 and a capacitance for λ/4 < L < λ/, with an open circuit at L = λ/4 and a short circuit again at L = λ/. Note that the frequency behavior of lumped elements is very different. Consider an ideal inductor with inductance L assumed to be constant with frequency, for simplicity. At zero frequency the inductor also behaves as a short circuit, but the reactance varies monotonically and linearly with frequency as X = ω L (always an inductance) Amarcord, 6 Digital Maestro Series 18

109 Short circuited transmission line Fixed frequency L L = in = λ < L < 4 k p Im in > λ L = 4 in λ λ < L < 4 k p Im in < λ L = in = λ 3λ < L < 4 k p Im in > 3λ L = 4 in 3 λ < L < λ 4 k p Im in < short circuit inductance open circuit capacitance short circuit inductance open circuit capacitance Amarcord, 6 Digital Maestro Series 19

110 Normalized Input Impedance (L)/o = j tan( β L) inductive Impedance of a short circuited transmission line (fixed frequency, variable length) capacitive inductive inductive π/(β)= λ/4 π/β =λ/ 3π/(β)= 3λ/4 cap. π/β= λ 5π/(β)= 5λ/4 θ [deg] L Line Length L Amarcord, 6 Digital Maestro Series 11

111 Normalized Input Impedance (L)/o = j tan( β L) inductive Impedance of a short circuited transmission line (fixed length, variable frequency) capacitive inductive inductive θ [deg] v p / (4L) v p / (L) 3v p / (4L) v p / L 5v p / (4L) f cap. Frequency of operation Amarcord, 6 Digital Maestro Series 111

112 For a transmission line of length L terminated by an open circuit, the input impedance is again purely imaginary in = j = j = j tan( βl) π tan L π f tan L λ v p We can also use the open circuited line to realize any reactance, but starting from a capacitive value when the line length is very short. Note once again that the frequency behavior of a corresponding lumped element is different. Consider an ideal capacitor with capacitance C assumed to be constant with frequency. At zero frequency the capacitor behaves as an open circuit, but the reactance varies monotonically and linearly with frequency as X 1 = (always a capacitance) ω C Amarcord, 6 Digital Maestro Series 11

113 Open circuit transmission line Fixed frequency L L = in λ < L < 4 k p Im in < λ L = 4 in = λ λ < L < 4 k p Im in > λ L = in λ 3λ < L < 4 k p Im in < 3λ L = 4 in = 3 λ < L < λ 4 k p Im in > open circuit capacitance short circuit inductance open circuit capacitance short circuit inductance Amarcord, 6 Digital Maestro Series 113

114 Normalized Input Impedance (L)/ o = - j cotan( β L) Impedance of an open circuited transmission line (fixed frequency, variable length) capacitive inductive inductive inductive capacitive capacitive π/(β) = λ/4 π/β = λ/ 3π/(β) = 3λ/4 π/β = λ 5π/(β) = 5λ/4 θ [deg] L Line Length L Amarcord, 6 Digital Maestro Series 114

115 Normalized Input Impedance (L)/ o = - j cotan( β L) Impedance of an open circuited transmission line (fixed length, variable frequency) capacitive inductive inductive inductive capacitive capacitive θ [deg] v p / (4L) v p / (L) 3v p / (4L) v p / L 5v p / (4L) f Frequency of operation Amarcord, 6 Digital Maestro Series 115

116 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: L 1 I L short circuit V short circuit in1 in ( ) ( ) = j tan β L = j tan β L in1 1 in Amarcord, 6 Digital Maestro Series 116

117 The circuit is resonant if L 1 and L 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 j = tan L r ( β ) jtan( β L) r 1 or ω ω π ω tan L tan L v v λ v r 1 = r r with β r = = p p r p Since the tangent is a periodic function, there is a multiplicity of possible resonant angular frequencies ω r that satisfy the condition above. The values can be found by using a numerical procedure to solve the trascendental equation above. Amarcord, 6 Digital Maestro Series 117

118 Transient and Steady-State on a Transmission Line Transmission Lines 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 =. For simplicity we assume that all impedances, including the line characteristic impedance, are real. Generator R G Switch Load t = V G Transmission line R R Amanogawa, 6 Digital Maestro Series 118