Transmission Lines General Properties of TEM Transmission Lines. Impedance, Inductance, and Capacitance

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1 398. Transmission Lines Transmission Lines Fig... Two-conductor transmission line.. General Properties of TEM Transmission Lines We saw in Sec. 9.3 that TEM modes are described by Eqs ) and 9.3.4), the latter being equivalent to a two-dimensional electrostatic problem: H T = η ẑ E T T E T = T E T = TEM modes)..) The second of..) implies that E T can be expressed as the two-dimensional) gradient of a scalar electrostatic potential. Then, the third equation becomes Laplace s equation for the potential. Thus, the electric field can be obtained from: T ϕ = E T = T ϕ equivalent electrostatic problem)..) Because in electrostatic problems the electric field lines must start at positively charged conductors and end at negatively charged ones, a TEM mode can be supported only in multi-conductor guides, such as the coaxial cable or the two-wire line. Hollow conducting waveguides cannot support TEM modes. Fig... depicts the transverse cross-sectional area of a two-conductor transmission line. The cross-section shapes are arbitrary. The conductors are equipotentials of the electrostatic solution. Let ϕ a,ϕ b be the constant potentials on the two conductors. The voltage difference between the conductors will be V = ϕ a ϕ b. The electric field lines start perpendicularly on conductor a) and end perpendicularly on conductor b). The magnetic field lines, being perpendicular to the electric lines according to Eq...), are recognized to be the equipotential lines. As such, they close upon themselves surrounding the two conductors. In particular, on the conductor surfaces the magnetic field is tangential. According to Ampère s law, the line integrals of the magnetic field around each conductor will result into total currents I and I flowing on the conductors in the z-direction. These currents are equal and opposite. Impedance, Inductance, and Capacitance Because the fields are propagating along the z-direction with frequency ω and wavenumber β = ω/c, the z, t dependence of the voltage V and current I will be: Vz, t)= Ve jωt jβz Iz, t)= Ie jωt jβz..3) For backward-moving voltage and current waves, we must replace β by β. The ratio Vz, t)/iz, t)= V/I remains constant and independent of z. It is called the characteristic impedance of the line: Z = V line impedance)..4) I In addition to the impedance Z, a TEM line is characterized by its inductance per unit length L and its capacitance per unit length C. For lossless lines, the three quantities Z, L,C are related as follows: L = μ Z η, C = ɛ η Z inductance and capacitance per unit length)..5) where η = μ/ɛ is the characteristic impedance of the dielectric medium between the conductors. By multiplying and dividing L and C, we also obtain: L Z = C, c = = ɛμ L C..6) These expressions explain why μ and ɛ are sometimes given in units of henry/m and farad/m.

2 .. General Properties of TEM Transmission Lines Transmission Lines The velocity factor of the line is the ratio c/c = /n, where n = ɛ/ɛ = ɛ r is the refractive index of the dielectric, which is assumed to be non-magnetic. Because ω = βc, the guide wavelength will be λ = π/β = c/f = c /fn = λ /n, where λ is the free-space wavelength. For a finite length l of the transmission line, the quantity l/λ = nl/λ is referred to as the electrical length of the line and plays the same role as the optical length in thin-film layers. Eqs...5) and..6) are general results that are valid for any TEM line. They can be derived with the help of Fig.... Q = C V = ɛηi C = ɛη I V = ɛ η Z Next, we consider an E-field line between points A and B on the two conductors. The magnetic flux through the infinitesimal area dl dz will be dφ = B T dl dz = μ H T dl dz because the vector H T is perpendicular to the area. If we integrate from a) to b), we will obtain the total magnetic flux linking the two conductors per unit z-length: Fig... Surface charge and magnetic flux linkage. Φ = dφ b dz = μ H T dl a replacing H T = E T /η and using Eq...7), we find: b Φ = μ H T dl = μ b E T dl = μ a η a η V The magnetic flux is related to the inductance via Φ = L I. Therefore, we get: The voltage V is obtained by integrating E T dl along any path from a) to b). However, if that path is chosen to be an E-field line, then E T dl = E T dl, giving: b V = E T dl..7) a Similarly, the current I can be obtained by the integral of H T dl along any closed path around conductor a). If that path is chosen to be an H-field line, such as the periphery C a of the conductor, we will obtain: I = H T dl..8) C a The surface charge accumulated on an infinitesimal area dl dz of conductor a) is dq = ρ s dl dz, where ρ s is the surface charge density. Because the conductors are assumed to be perfect, the boundary conditions require that ρ s be equal to the normal component of the D-field, that is, ρ s = ɛ E T. Thus, dq = ɛ E T dl dz. If we integrate over the periphery C a of conductor a), we will obtain the total surface charge per unit z-length: Q = dq dz = ɛ E T dl C a But because of the relationship E T =η H T, which follows from the first of Eqs...), we have: Q = ɛ E T dl = ɛη C a H T dl = ɛηi C a..9) where we used Eq...8). Because Q is related to the capacitance per unit length and the voltage by Q = C V, we obtain Transmitted Power Φ = L I = μ η V L = μ η V I = μ Z η The relationships among Z, L,C can also be derived using energy considerations. The power transmitted along the line is obtained by integrating the z-component of the Poynting vector over the cross-section S of the line. For TEM modes we have P z = E T /η, therefore, P T = E T dx dy = T ϕ dx dy η S η S..) It can be shown in general that Eq...) can be rewritten as: P T = ReV I)= Z I = Z V..) We will verify this in the various examples below. It can be proved using the following Green s identity: T ϕ + ϕ T ϕ = T ϕ T ϕ) Writing E T = T ϕ and noting that Tϕ =, we obtain: E T = T ϕ E T ) Then, the two-dimensional Gauss theorem implies:

3 .. General Properties of TEM Transmission Lines 4 4. Transmission Lines P T = E T dx dy = η S η = η = η C a ϕ E T ˆn)dl C a ϕ E T ˆn)dl+ η T ϕ E T )dx dy S ϕ E T ˆn)dl η C b C b ϕ E T ˆn)dl where ˆn are the outward normals to the conductors the quantity ˆn is the normal outward from the region S.) Because the conductors are equipotential surfaces, we have ϕ = ϕ a on conductor a) and ϕ = ϕ b on conductor b). Using Eq...9) and noting that E T ˆn =± E T on conductors a) and b), we obtain: P T = η ϕ a E T dl C a η ϕ b E T dl = Q C b η ϕ a ɛ Q η ϕ b ɛ = ϕ a ϕ b ) Q ɛη = ɛηi V ɛη = V I = Z I The distribution of electromagnetic energy along the line is described by the timeaveraged electric and magnetic energy densities per unit length, which are given by: W e = 4 ɛ E T dx dy, W m = S 4 μ H T dx dy S Using Eq...), we may rewrite: W e = ɛηp T = c P T, W m = μ η P T = c P T Thus, W e = W m and the total energy density is W = W e + W m = P T /c, which implies that the energy velocity will be v en = P T /W = c. We may also express the energy densities in terms of the capacitance and inductance of the line: Power Losses, Resistance, and Conductance W e = 4 C V, W m = 4 L I..) Transmission line losses can be handled in the manner discussed in Sec. 9.. The field patterns and characteristic impedance are determined assuming the conductors are perfectly conducting. Then, the losses due to the ohmic heating of the dielectric and the conductors can be calculated by Eqs. 9..5) and 9..9). These losses can be quantified by two more characteristic parameters of the line, the resistance and conductance per unit length, R and G. The attenuation coefficients due to conductor and dielectric losses are then expressible in terms R, G and Z by: α c = R Z, α d = G Z..3) They can be derived in general terms as follows. The induced surface currents on the conductor walls are J s = ˆn H T = ˆn ẑ E T )/η, where ˆn is the outward normal to the wall. Using the BAC-CAB rule, we find J s = ẑˆn E T )/η. But, ˆn is parallel to E T on the surface of conductor a), and anti parallel on b). Therefore, ˆn E T =± E T. It follows that J s =±ẑ E T /η =±ẑ H T, pointing in the +z direction on a) and z direction on b). Inserting these expressions into Eq. 9..8), we find for the conductor power loss per unit z-length: P loss = dp loss dz = R s H T dl + C a R s H T dl..4) C b Because H T is related to the total current I via Eq...8), we may define the resistance per unit length R through the relationship: P loss = R I conductor ohmic losses)..5) Using Eq...), we find for the attenuation coefficient: α c = P loss P T = R I = R Z Z I..6) If the dielectric between the conductors is slightly conducting with conductivity σ d or loss tangent tan δ = σ d /ɛω, then there will be some current flow between the two conductors. The induced shunt current per unit z-length is related to the conductance by I d = G V. The shunt current density within the dielectric is J d = σ d E T. The total shunt current flowing out of conductor a) towards conductor b) is obtained by integrating J d around the periphery of conductor a): I d C = J d ˆn dl = σ d E T dl a C a Using Eq...9), we find: I d = σ d Q ɛ = G V G = σ d ɛ C = σ d η Z It follows that the dielectric loss constant 9..5) will be: α d = σ dη = G Z Alternatively, the power loss per unit length due to the shunt current will be P d = ReI d V )/ = G V /, and therefore, α d can be computed from: α d = P d P T = G V Z V = G Z

4 .. Parallel Plate Lines Transmission Lines It is common practice to express the dielectric losses and shunt conductance in terms of the loss tangent tan δ and the wavenumber β = ω/c = ωɛη: The inductance and capacitance per unit length are obtained from Eq...5): L = μ h w, C = ɛ w h..3) α d = σ dη = ωɛη tan δ = β tan δ and G = σ d ɛ C = ωc tan δ..7) Next, we discuss four examples: the parallel plate line, the microstrip line, the coaxial cable, and the two-wire line. In each case, we discuss the nature of the electrostatic problem and determine the characteristic impedance Z and the attenuation coefficients α c and α d.. Parallel Plate Lines The parallel plate line shown in Fig... consists of two parallel conducting plates of width w separated by height h by a dielectric material ɛ. Examples of such lines are microstrip lines used in microwave integrated circuits. For arbitrary values of w and h, the fringing effects at the ends of the plates cannot be ignored. In fact, fringing requires the fields to have longitudinal components, and therefore TEM modes are not strictly-speaking supported. The surface current on the top conductor is J s = ˆn H = ŷ) H = ẑ H x. On the bottom conductor, it will be J s = ẑh x. Therefore, the power loss per unit z-length is obtained from Eq. 9..8): P loss = R s H x w = w R si Comparing with Eq...5), we identify the resistance per unit length R = R s /w. Then, the attenuation constant due to conductor losses will be:.3 Microstrip Lines α c = P loss P T = R Z = R s wz = R s hη..4) Practical microstrip lines, shown in Fig..3., have width-to-height ratios w/h that are not necessarily much greater than unity, and can vary over the interval. < w/h <. Typical heights h are of the order of millimeters. Fig... Parallel plate transmission line. However, assuming the width is much larger than the height, w h, we may ignore the fringing effects and assume that the fields have no dependence on the x-coordinate. The electrostatic problem is equivalent to that of a parallel plate capacitor. Thus, the electric field will have only a y component and will be constant between the plates. Similarly, the magnetic field will have only an x component. It follows from Eqs...7) and..8) that: V = E y h, I = H x w Therefore, the characteristic impedance of the line will be: Z = V I = E yh H x w = η h..) w where we used E y = ηh x. The transmitted power is obtained from Eq...): P T = η E y wh)= V η h wh = w η h V = Z V = ZI..) Fig..3. A microstrip transmission line. Fringing effects cannot be ignored completely and the simple assumptions about the fields of the parallel plate line are not valid. For example, assuming a propagating wave in the z-direction with z, t dependence of e jωt jβz with a common β in the dielectric and air, the longitudinal-transverse decomposition 9..5) gives: T E z ẑ jβ ẑ E T = jωμh T ẑ T E z + jβe T )= jωμh T In particular, we have for the x-component: y E z + jβe y = jωμh x The boundary conditions require that the components H x and D y = ɛe y be continuous across the dielectric-air interface at y = h). This gives the interface conditions: y Ez air + jβeair y ɛ E air y = ɛe diel y = yez diel + jβey diel

5 .3. Microstrip Lines Transmission Lines Combining the two conditions, we obtain: y E diel z E air z ) ɛ ɛ = jβ ɛ E air y = jβ ɛ ɛ ɛ E diel y.3.) Because E y is non-zero on either side of the interface, it follows that the left-hand side of Eq..3.) cannot be zero and the wave cannot be assumed to be strictly TEM. However, E y is small in both the air and the dielectric in the fringing regions to the left and right of the upper conductor). This gives rise to the so-called quasi-tem approximation in which the fields are assumed to be approximately TEM and the effect of the deviation from TEM is taken into account by empirical formulas for the line impedance and velocity factor. In particular, the air-dielectric interface is replaced by an effective dielectric, filling uniformly the entire space, and in which there would be a TEM propagating mode. If we denote by ɛ eff the relative permittivity of the effective dielectric, the wavelength and velocity factor of the line will be given in terms of their free-space values λ,c : λ = λ ɛeff, c = c ɛeff.3.) There exist many empirical formulas for the characteristic impedance of the line and the effective dielectric constant. Hammerstad and Jensen s are some of the most accurate ones 886,89]: ɛ eff = ɛ r + + ɛ r + ) ab, u = w u h.3.3) where ɛ r = ɛ/ɛ is the relative permittivity of the dielectric and the quantities a, b are defined by: ] a = + u 4 49 ln + u/5) + ) ] u 3 u ln + 8. ).3.4) ɛr.9.53 b =.564 ɛ r + 3 The accuracy of these formulas is better than.% for u< and.3% for u<. Similarly, the characteristic impedance is given by the empirical formula: η Z = π ln fu) ɛ eff u + + 4u.3.5) where η = μ /ɛ and the function fu) is defined by: ) ] fu)= 6 + π 6)exp.3.6) u The accuracy is better than.% for. u and ɛ r < 8. In the limit of large ratio w/h, or, u, Eqs..3.3) and.3.5) tend to those of the parallel plate line of the previous section: ɛ eff ɛ r, Z η ɛr h w = η h w Some typical substrate dielectric materials used in microstrip lines are alumina, a ceramic form of Al O 4 with e r = 9.8, and RT-Duroid, a teflon composite material with ɛ r =.. Practical values of the width-to-height ratio are in the range. u and practical values of characteristic impedances are between ohm. Fig..3. shows the dependence of Z and ɛ eff on u for the two cases of ɛ r =. and ɛ r = 9.8. Z ohm) Characteristic Impedance ε r =. ε r = w/h Fig..3. ε eff Effective Permittivity ε r =. ε r = w/h Characteristic impedance and effective permittivity of microstrip line. The synthesis of a microstrip line requires that we determine the ratio w/h that will achieve a given characteristic impedance Z. The inverse of Eq..3.5) solving for u in terms of Z is not practical. Direct synthesis empirical equations exist 887,89], but are not as accurate as.3.5). Given a desired Z, the ratio u = w/h is calculated as follows. If u, and, if u>, u = ɛ r πɛ r where A, B are given by: u = lnb ) ɛ r A = π ɛ r + ) Z + ɛ r η ɛ r + B = π η ɛ r Z 8 e A e A.3.7) ] + ] B lnb ) π.3 +. ) ɛ r.3.8).3.9) The accuracy of these formulas is about %. The method can be improved iteratively by a process of refinement to achieve essentially the same accuracy as Eq..3.5). Starting with u computed from Eqs..3.7) and.3.8), a value of Z is computed through Eq..3.5). If that Z is more than, say,.% off from the desired value of the line

6 .3. Microstrip Lines Transmission Lines impedance, then u is slightly changed, and so on, until the desired level of accuracy is reached 89]. Because Z is monotonically decreasing with u, if Z is less than the desired value, then u is decreased by a small percentage, else, u is increased by the same percentage. The three MATLAB functions mstripa, mstrips, and mstripr implement the analysis, synthesis, and refinement procedures. They have usage: eff,z] = mstripaer,u); % analysis equations.3.3) and.3.5) u = mstripser,z); % synthesis equations.3.7) and.3.8) u,n] = mstriprer,z,per); % refinement The function mstripa accepts also a vector of several u s, returning the corresponding vector of values of ɛ eff and Z. Inmstripr, the output N is the number of iterations required for convergence, and per is the desired percentage error, which defaults to.% if this parameter is omitted. Example.3.: Given ɛ r =. and u = w/h =, 4, 6, the effective permittivities and impedances are computed from the MATLAB call: u = ; 4; 6]; eff, Z] = mstripaer,u); The resulting output vectors are: u = 4 ɛ eff =.9, Z = ohm Example.3.: To compare the outputs of mstrips and mstripr, we design a microstrip line with ɛ r =. and characteristic impedance Z = 5 ohm. We find: u = mstrips., 5)= ɛ eff,z]= mstripa.,u)=.88, 5.534] u = mstripr., 5)= 3.89 ɛ eff,z]= mstripa.,u)=.883, ] The first solution has an error of.7% from the desired 5 ohm impedance, and the second, a.% error. As another example, if Z = Ω, the function mstrips results in u =.8949, Z = Ω, and a.5% error, whereas mstripr gives u =.8939, Z = Ω, and a.% error. Typical values of the loss tangent are of the order of. for alumina and duroid substrates. The conductor losses are approximately computed from Eq...4):.4 Coaxial Lines α c = R s wz.3.) The coaxial cable, depicted in Fig..4., is the most widely used TEM transmission line. It consists of two concentric conductors of inner and outer radii of a and b, with the space between them filled with a dielectric ɛ, such as polyethylene or teflon. The equivalent electrostatic problem can be solved conveniently in cylindrical coordinates ρ, φ. The potential ϕρ, φ) satisfies Laplace s equation: T ϕ = ρ ϕ ) + ϕ ρ ρ ρ ρ φ = Because of the cylindrical symmetry, the potential does not depend on the azimuthal angle φ. Therefore, ρ ϕ ) ρ ρ ρ = ρ ϕ = B ϕρ)= A + B ln ρ ρ where A, B are constants of integration. Assuming the outer conductor is grounded, ϕρ)= atρ = b, and the inner conductor is held at voltage V, ϕa)= V, the constants A, B are determined to be B = V lnb/a) and A = B ln b, resulting in the potential: V ϕρ)= lnb/ρ).4.) lnb/a) It follows that the electric field will have only a radial component, E ρ = ρ ϕ, and the magnetic field only an azimuthal component H φ = E ρ /η: V E ρ = lnb/a) ρ, H V φ = η lnb/a) ρ Integrating H φ around the inner conductor we obtain the current:.4.) In using microstrip lines several other effects must be considered, such as finite strip thickness, frequency dispersion, dielectric and conductor losses, radiation, and surface waves. Guidelines for such effects can be found in ]. The dielectric losses are obtained from Eq...7) by multiplying it by an effective dielectric filling factor q: α d = q ω c tan δ = f c πq ɛ eff tan δ = λ πq ɛ eff tan δ, q= ɛ eff ɛ r.3.) Fig..4. Coaxial transmission line.

7 .4. Coaxial Lines Transmission Lines π π I = H φ ρdφ = V η lnb/a) ρ ρdφ = πv η lnb/a).4.3) It follows that the characteristic impedance of the line Z = V/I, and hence the inductance and capacitance per unit length, will be: Z = η π lnb/a), L = μ π lnb/a), C = πɛ lnb/a) Using Eq..4.3) into.4.), we may express the magnetic field in the form:.4.4) H φ = I.4.5) πρ This is also obtainable by the direct application of Ampère s law around the loop of radius ρ encircling the inner conductor, that is, I = πρ)h φ. The transmitted power can be expressed either in terms of the voltage V or in terms of the maximum value of the electric field inside the line, which occurs at ρ = a, that is, E a = V/ a lnb/a) ) : P T = Z V = π V η lnb/a) = η E a πa )lnb/a).4.6) Example.4.: A commercially available polyethylene-filled RG-58/U cable is quoted to have impedance of 5 Ω, velocity factor of 66 percent, inner conductor radius a =.46 mm AWG -gauge wire), and maximum operating RMS voltage of 4 volts. Determine the outer-conductor radius b, the capacitance and inductance per unit length C,L, the maximum power P T that can be transmitted, and the maximum electric field inside the cable. Solution: Polyethylene has a relative dielectric constant of ɛ r =.5, so that n = ɛ r =.5. The velocity factor is c/c = /n =.667. Given that η = η /n = /.5 = 5.5 Ω and c = c /n = /.5 = m/sec, we have: Z = η π lnb/a) b = aeπz/η =.46 e π5/5.5 =.483 mm Therefore, b/a = The capacitance and inductance per unit length are found from: C = ɛ η Z = cz = =.7 pf/m L = μ Z η = Z c = 5 =.5 μh/m The peak voltage is related to its RMS value by V = V rms. It follows that the maximum power transmitted is: P T = Z V = V rms Z = 4 = 39. kw 5 see, for example, the 93 Coax RG-58/U cable from The peak value of the electric field occurring at the inner conductor will be: V E a = a lnb/a) = Vrms 4 a lnb/a) = = 3.9 MV/m.46 3 ln.483/.46) This is to be compared with the dielectric breakdown of polyethylene of about MV/m. Example.4.: Most cables have a nominal impedance of either 5 or 75 Ω. The precise value depends on the manufacturer and the cable. For example, a 5-Ω cable might actually have an impedance of 5 Ω and a 75-Ω cable might actually be a 73-Ω cable. The table below lists some commonly used cables with their AWG-gauge number of the inner conductor, the inner conductor radius a in mm, and their nominal impedance. Their dielectric filling is polyethylene with ɛ r =.5 or n = ɛ r =.5. type AWG a Z RG-6/U RG-8/U.5 5 RG-/U RG-58/U.46 5 RG-59/U.3 75 RG-74/U RG-3/U The most commonly used cables are 5-Ω ones, such as the RG-58/U. Home cable-tv uses 75-Ω cables, such as the RG-59/U or RG-6/U. The thin ethernet computer network, known as base-, uses RG-58/U or RG-58A/U, which is similar to the RG-58/U but has a stranded inner copper core. Thick ethernet base-5) uses the thicker RG-8/U cable. Because a dipole antenna has an input impedance of about 73 Ω, the RG-, RG-6, and RG Ω cables can be used to feed the antenna. Next, we determine the attenuation coefficient due to conductor losses. The power loss per unit length is given by Eq...4). The magnetic fields at the surfaces of conductors a) and b) are obtained from Eq..4.5) by setting ρ = a and ρ = b: H a = I πa, H b = I πb Because these are independent of the azimuthal angle, the integrations around the peripheries dl = adφ or dl = bdφ will contribute a factor of πa) or πb). Thus, P loss = R s πa) Ha + πb) H b ] = R s I 4π a + ) b It follows that: R s I α c = P loss 4π = P T a + ) b Z I.4.7)

8 .4. Coaxial Lines 4 4. Transmission Lines Using Eq..4.4), we finally obtain: α c = R s a + ) b ).4.8) η b ln a The ohmic losses in the dielectric are described by Eq...7). The total attenuation constant will be the sum of the conductor and dielectric attenuations: ) db/ ft total conductor dielectric nominal RG 8/U and RG 3/U α = α c + α d = R s η a + b ) + ω tan δ attenuation).4.9) b c ln a The attenuation in db/m will be α db = α. This expression tends to somewhat underestimate the actual losses, but it is generally a good approximation. The α c term grows in frequency like f and the term α d, like f. The smaller the dimensions a, b, the larger the attenuation. The loss tangent tan δ of a typical polyethylene or teflon dielectric is of the order of.4.9 up to about 3 GHz. The ohmic losses and the resulting heating of the dielectric and conductors also limit the power rating of the line. For example, if the maximum supported voltage is 4 volts as in Example.4., the RMS value of the current for an RG-58/U line would be I rms = 4/5 = 8 amps, which would likely melt the conductors. Thus, the actual power rating is much smaller than that suggested by the maximum voltage rating. The typical power rating of an RG-58/U cable is typically kw, W, and 8 W at the frequencies of MHz, MHz, and GHz. Example.4.3: The table below lists the nominal attenuations in db per feet of the RG-8/U and RG-3/U cables. The data are from 353]. f MHz) α db/ft) Both are 5-ohm cables and their radii a are.5 mm and.95 mm for RG-8/U and RG- 3/U. In order to compare these ratings with Eq..4.9), we took a to be the average of these two values, that is, a =.3 mm. The required value of b to give a 5-ohm impedance is b = 3.6 mm. Fig..4. shows the attenuations calculated from Eq..4.9) and the nominal ones from the table. We assumed copper conductors with σ = S/m and polyethylene dielectric with n =.5, so that η = η /n = /.5 = 5.5 Ω and c = c /n = 8 m/sec. The loss tangent was taken to be tan δ =.7. The conductor and dielectric attenuations α c and α d become equal around.3 GHz, and α d dominates after that. It is evident that the useful operation of the cable is restricted to frequencies up to GHz. Beyond that, the attenuations are too excessive and the cable may be used only for short lengths. Fig..4. Optimum Coaxial Cables f GHz) Attenuation coefficient α versus frequency. Given a fixed outer-conductor radius b, one may ask three optimization questions: What is the optimum value of a, or equivalently, the ratio b/a that a) minimizes the electric field E a inside the guide for fixed voltage V), b) maximizes the power transfer P T for fixed E a field), and c) minimizes the conductor attenuation α c. The three quantities E a,p T,α c can be thought of as functions of the ratio x = b/a and take the following forms: E a = V b x ln x, P T = η E a πb ln x x, α c = R s x + ηb ln x.4.) Setting the derivatives of the three functions of x to zero, we obtain the three conditions: a) ln x =, b) ln x = /, and c) ln x = + /x, with solutions a) b/a = e =.783, b) b/a = e / =.6487 and c) b/a = Unfortunately, the three optimization problems have three different answers, and it is not possible to satisfy them simultaneously. The corresponding impedances Z for the three values of b/a are 6 Ω, 3 Ω, and 76.7 Ω for an air-filled line and 4 Ω, Ω, and 5 Ω for a polyethylene-filled line. The value of 5 Ω is considered to be a compromise between 3 and 76.7 Ω corresponding to maximum power and minimum attenuation. Actually, the minimum of α c is very broad and any neighboring value to b/a = 3.59 will result in an α c very near its minimum. Higher Modes The TEM propagation mode is the dominant one and has no cutoff frequency. However, TE and TM modes with higher cutoff frequencies also exist in coaxial lines 86], with the lowest being a TE mode with cutoff wavelength and frequency: λ c =.873 π a + b), f c = c λ c = c nλ c.4.)

9 .5. Two-Wire Lines Transmission Lines This is usually approximated by λ c = πa + b). Thus, the operation of the TEM mode is restricted to frequencies that are less than f c. Example.4.4: For the RG-58/U line of Example.4., we have a =.46 mm and b =.548 mm, resulting in λ c =.873πa + b)/ = mm, which gives for the cutoff frequency f c = /.5749 = GHz, where we used c = c /n = GHz cm. For the RG-8/U and RG-3/U cables, we may use a =.3 mm and b = 3.6 as in Example.4.3, resulting in λ c = 3.6 mm, and cutoff frequency of f c = 4.68 GHz. The above cutoff frequencies are far above the useful operating range over which the attenuation of the line is acceptable..5 Two-Wire Lines The two-wire transmission line consists of two parallel cylindrical conductors of radius a separated by distance d from each other, as shown in Fig..5.. ρ = ρ ρb cos φ + b, ρ = ρ ρb cos φ + b.5.) Therefore, the potential function becomes: ) ϕρ, φ)= Q πɛ ln ρ = Q ρ πɛ ln ρ ρb cos φ + b ρ ρb cos φ + b.5.3) In order that the surface of the left conductor at ρ = a be an equipotential surface, that is, ϕa, φ)= V/, the ratio ρ /ρ must be a constant independent of φ. Thus, we require that for some constant k and all angles φ: ρ ρ which can be rewritten as: = ρ=a a ab cos φ + b a ab cos φ + b = k a ab cos φ + b = k a ab cos φ + b ) This will be satisfied for all φ provided we have: a + b = k a + b ), b = k b These may be solved for b,b in terms of k: b = ka, b = a k.5.4) Fig..5. Two-wire transmission line. The quantity k can be expressed in terms of a, d by noting that because of symmetry, the charge Q is located also at distance b from the center of the right conductor. Therefore, b + b = d. This gives the condition: We assume that the conductors are held at potentials ±V/ with charge per unit length ±Q. The electrostatic problem can be solved by the standard technique of replacing the finite-radius conductors by two thin line-charges ±Q. The locations b and b of the line-charges are determined by the requirement that the cylindrical surfaces of the original conductors be equipotential surfaces, the idea being that if these equipotential surfaces were to be replaced by the conductors, the field patterns will not be disturbed. The electrostatic problem of the two lines is solved by invoking superposition and adding the potentials due to the two lines, so that the potential at the field point P will be: ) ϕρ, φ)= Q πɛ ln ρ Q πɛ ln ρ = Q πɛ ln ρ.5.) ρ where the ρ,ρ are the distances from the line charges to P. From the triangles OP+Q ) and OP Q ), we may express these distances in terms of the polar coordinates ρ, φ of the point P: with solution for k: b + b = d ak + k )= d k + k = d a k = d ) d a +.5.5) a An alternative expression is obtained by setting k = e χ. Then, we have the condition: ) d b + b = d ae χ + e χ )= a cosh χ = d χ = acosh a Because χ = ln k, we obtain for the potential value of the left conductor: ϕa, φ)= Q Q ln k = πɛ πɛ χ = V This gives for the capacitance per unit length:.5.6)

10 .6. Distributed Circuit Model of a Transmission Line Transmission Lines C = Q V = πɛ χ = πɛ ).5.7) d acosh a The corresponding line impedance and inductance are obtained from C = ɛη/z and L = μz/η. We find: Z = η π χ = η ) d π acosh a L = μ π χ = μ π acosh d a ).5.8) Z = R + jωl Y = G + jωc.6.) This leads to a so-called distributed-parameter circuit, which means that every infinitesimal segment Δz of the line can be replaced by a series impedance Z Δz and a shunt admittance Y Δz, as shown in Fig..6.. The voltage and current at location z will be Vz), Iz) and at location z + Δz, Vz + Δz), Iz + Δz). In the common case when d a, we have approximately k d/a, and therefore, χ = ln k = lnd/a). Then, Z can be written approximately as: Z = η lnd/a).5.9) π To complete the electrostatic problem and determine the electric and magnetic fields of the TEM mode, we replace b = ak and b = a/k in Eq..5.3) and write it as: ϕρ, φ)= Q πɛ ln ρ akρ cos φ + a k k.5.) ρ k akρ cos φ + a Fig..6. Distributed parameter model of a transmission line. The electric and magnetic field components are obtained from: E ρ = ηh φ = ϕ ρ, Performing the differentiations, we find: E ρ = Q ρ ak cos φ πɛ ρ akρ cos φ + a k E φ = Q ak sin φ πɛ ρ ak cos φ + a k E φ = ηh ρ = ϕ ρ φ ] ρk ak cos φ ρ k akρ cos φ + a ak sin φ ρ k akρ cos φ + a.5.) ].5.) The resistance per unit length and corresponding attenuation constant due to conductor losses are calculated in Problem.3: R = R s πa d, α c = R d 4a Z = R s ηa d.5.3) acoshd/a) d 4a.6 Distributed Circuit Model of a Transmission Line We saw that a transmission line has associated with it the parameters L,C describing its lossless operation, and in addition, the parameters R,G which describe the losses. It is possible then to define a series impedance Z and a shunt admittance Y per unit length by combining R with L and G with C : The voltage across the branch a b is V ab = Vz + Δz) and the current through it, I ab = Y Δz)V ab = Y Δz Vz + Δz). Applying Kirchhoff s voltage and current laws, we obtain: Vz) = Z Δz) Iz)+V ab = Z Δz Iz)+Vz + Δz) Iz) = I ab + Iz + Δz)= Y Δz Vz + Δz)+Iz + Δz).6.) Using a Taylor series expansion, we may expand Iz + Δz) and Vz + Δz) to first order in Δz: Iz + Δz) = Iz)+I z)δz Vz + Δz) = Vz)+V z)δz and Y Δz Vz + Δz)= Y Δz Vz) Inserting these expressions in Eq..6.) and matching the zeroth- and first-order terms in the two sides, we obtain the equivalent differential equations: V z)= Z Iz)= R + jωl )Iz) I z)= Y Vz)= G + jωc )Vz).6.3) It is easily verified that the most general solution of this coupled system is expressible as a sum of a forward and a backward moving wave: Vz) = V + e jβcz + V e jβcz Iz) = Z c V+ e jβcz V e jβcz).6.4)

11 .7. Wave Impedance and Reflection Response Transmission Lines where β c,z c are the complex wavenumber and complex impedance: β c = j R + jωl )G + jωc ) = ω ) L C j ) R ωl j G ωc Z Z c = Y = R + jωl G + jωc.6.5) The time-domain impulse response of such a line was given in Sec The real and imaginary parts of β c = β jα define the propagation and attenuation constants. In the case of a lossless line, R = G =, we obtain using Eq...6): β c = ω L C = ω μɛ = ω c = β, Z c = L C = Z.6.6) In practice, we always assume a lossless line and then take into account the losses by assuming that R and G are small quantities, which can be evaluated by the appropriate expressions that can be derived for each type of line, as we did for the parallel-plate, coaxial, and two-wire lines. The lossless solution.6.4) takes the form: Vz) = V + e jβz + V e jβz = V + z)+v z) Iz) = V+ e jβz V e jβz) = V+ z) V z) ).6.7) Z Z This solution is identical to that of uniform plane waves of Chap. 5, provided we make the identifications: Vz) Ez) Iz) Hz) Z η and V + z) E + z) V z) E z).7 Wave Impedance and Reflection Response All the concepts of Chap. 5 translate verbatim to the transmission line case. For example, we may define the wave impedance and reflection response at location z: Zz)= Vz) Iz) = Z V + z)+v z) V + z) V z), Γz)= V z) V + z).7.) To avoid ambiguity in notation, we will denote the characteristic impedance of the line by Z. It follows from Eq..7.) that Zz) and Γz) are related by: + Γz) Zz)= Z Γz), Γz)= Zz) Z.7.) Zz)+Z For a forward-moving wave, the conditions Γz)= and Zz)= Z are equivalent. The propagation equations of Zz) and Γz) between two points z,z along the line separated by distance l = z z are given by: Z + jz tan βl Z = Z Γ = Γ e jβl.7.3) Z + jz tan βl where we have the relationships between Z,Z and Γ,Γ : + Γ + Γ Z = Z, Z = Z.7.4) Γ Γ We may also express Z in terms of Γ : + Γ + Γ e jβl Z = Z = Z Γ Γ e jβl.7.5) The relationship between the voltage and current waves at points z and z is obtained by the propagation matrix: V I ] cos βl jz sin βl = jz sin βl cos βl ] V I ] propagation matrix).7.6) Similarly, we may relate the forward/backward voltages at the points z and z : ] ] ] V+ e jβl V+ = V e jβl propagation matrix).7.7) V It follows from Eq..6.7) that V ±,V ± are related to V,I and V,I by: V ± = V ± Z I ), V ± = V ± Z I ).7.8) Fig..7. depicts these various quantities. We note that the behavior of the line remains unchanged if the line is cut at the point z and the entire right portion of the line is replaced by an impedance equal to Z, as shown in the figure. This is so because in both cases, all the points z to the left of z see the same voltage-current relationship at z, that is, V = Z I. Sometimes, as in the case of designing stub tuners for matching a line to a load, it is more convenient to work with the wave admittances. Defining Y = /Z, Y = /Z, and Y = /Z, it is easily verified that the admittances satisfy exactly the same propagation relationship as the impedances: Y = Y Y + jy tan βl Y + jy tan βl.7.9) As in the case of dielectric slabs, the half- and quarter-wavelength separations are of special interest. For a half-wave distance, we have βl = π/ = π, which translates to l = λ/, where λ = π/β is the wavelength along the line. For a quarter-wave, we have βl = π/4 = π/ orl = λ/4. Setting βl = π or π/ in Eq..7.3), we obtain:

12 .8. Two-Port Equivalent Circuit Transmission Lines where the impedance elements are: Z = Z = jz cot βl Z = Z = jz sin βl.8.) The negative sign, I, conforms to the usual convention of having the currents coming into the two-port from either side. This impedance matrix can also be realized in a T-section configuration as shown in Fig..8.. Fig..7. Length segment on infinite line and equivalent terminated line. l = λ l = λ 4 Z = Z, Γ = Γ Z = Z Z, Γ = Γ.7.) The MATLAB functions zg.m and gz.m compute Γ from Z and conversely, by implementing Eq..7.). The functions gprop.m, zprop.m and vprop.m implement the propagation equations.7.3) and.7.6). The usage of these functions is: G = zgz,z); % Z to Γ Z = gzg,z); % Γ to Z G = gpropg,bl); % propagates Γ to Γ Z = zpropz,z,bl); % propagates Z to Z V,I] = vpropv,i,z,bl); % propagates V,I to V,I The parameter bl is βl. The propagation equations and these MATLAB functions also work for lossy lines. In this case, β must be replaced by the complex wavenumber β c = β jα. The propagation phase factors become now:.8 Two-Port Equivalent Circuit e ±jβl e ±jβcl = e ±αl e ±jβl.7.) Any length-l segment of a transmission line may be represented as a two-port equivalent circuit. Rearranging the terms in Eq..7.6), we may write it in impedance-matrix form: V V ] Z Z = Z Z ] I I ] impedance matrix).8.) Fig..8. Length-l segment of a transmission line and its equivalent T-section. Using Eq..8.) and some trigonometry, the impedances Z a,z b,z c of the T-section are found to be: Z a = Z Z = jz tanβl/) Z b = Z Z = jz tanβl/) Z c = Z = jz sin βl The MATLAB function tsection.m implements Eq..8.3). Its usage is: Za,Zc] = tsectionz,bl);.9 Terminated Transmission Lines.8.3) We can use the results of the previous section to analyze the behavior of a transmission line connected between a generator and a load. For example in a transmitting antenna system, the transmitter is the generator and the antenna, the load. In a receiving system, the antenna is the generator and the receiver, the load. Fig..9. shows a generator of voltage V G and internal impedance Z G connected to the load impedance Z L through a length d of a transmission line of characteristic

13 .9. Terminated Transmission Lines 4 4. Transmission Lines VL I L ] = ] ] cos βd jz sin βd Vd sin βd cos βd I d jz.9.5) It is more convenient to express V d,i d in terms of the reflection coefficients Γ d and Γ G, the latter being defined by: Γ G = Z G Z + Γ G Z G = Z.9.6) Z G + Z Γ G It is easy to verify using Eqs..9.3) and.9.6) that: Z G + Z d = Z Γ G Γ d Γ G ) Γ d ), Z G + Z = Z Γ G Fig..9. Terminated line and equivalent circuit. impedance Z. We wish to determine the voltage and current at the load in terms of the generator voltage. We assume that the line is lossless and hence Z is real. The generator impedance is also assumed to be real but it does not have to be. The load impedance will have in general both a resistive and a reactive part, Z L = R L + jx L. At the load location, the voltage, current, and impedance are V L, I L, Z L and play the same role as the quantities V, I, Z of the previous section. They are related by V L = Z L I L. The reflection coefficient at the load will be: Γ L = Z L Z + Γ L Z L = Z.9.) Z L + Z Γ L The quantities Z L,Γ L can be propagated now by a distance d to the generator at the input to the line. The corresponding voltage, current, and impedance V d, I d, Z d play the role of V, I, Z of the previous section, and are related by V d = Z d I d. We have the propagation relationships: where Z d = Z Z L + jz tan βd Z + jz L tan βd Γ d = Γ L e jβd.9.) Γ d = Z d Z + Γ d + Γ L e jβd Z d = Z = Z Z d + Z Γ d Γ L e jβd.9.3) At the line input, the entire length-d line segment and load can be replaced by the impedance Z d, as shown in Fig..9.. We have now a simple voltage divider circuit. Thus, From these, it follows that: V d = V GZ + Γ d, I d = Z G + Z Γ G Γ d V G Z G + Z Γ d Γ G Γ d.9.7) where Γ d may be replaced by Γ d = Γ L e jβd. If the line and load are matched so that Z L = Z, then Γ L = and Γ d = and Z d = Z for any distance d. Eq..9.7) then reduces to: V d = V GZ V G, I d = matched load).9.8) Z G + Z Z G + Z In this case, there is only a forward-moving wave along the line. The voltage and current at the load will correspond to the propagation of these quantities to location l =, which introduces a propagation phase factor e jβd : V = V GZ e jβd V G, I = e jβd matched load).9.9) Z G + Z Z G + Z where V,I denote V L,I L when Z L = Z. It is convenient also to express V L directly in terms of V d and the reflection coefficients Γ d and Γ L. We note that: V L = V L+ + Γ L ), V L+ = V d+ e jβd, V d+ = V d + Γ d It follows that the voltage V L and current I L = V L /Z L are: V L = V d e jβd + Γ L, I L = I d e jβd Γ L.9.) + Γ d Γ d Expressing V L and also I L = V L /Z L directly in terms of V G, we have: V d = V G I d Z G = V GZ d V G, I d =.9.4) Z G + Z d Z G + Z d Once we have V d,i d in terms of V G, we can invert the propagation matrix.7.6) to obtain the voltage and current at the load: V L = V GZ + Γ L e jβd, I L = Z G + Z Γ G Γ d V G Z G + Z Γ L Γ G Γ d e jβd.9.) It should be emphasized that d refers to the fixed distance between the generator and the load. For any other distance, say l, from the load or, distance z = d l from

14 .. Power Transfer from Generator to Load Transmission Lines the generator,) the voltage and current can be expressed in terms of the load voltage and current as follows: V l = V L e jβl + Γ l + Γ L, I l = I L e jβl Γ l Γ L, Γ l = Γ L e jβl.9.). Power Transfer from Generator to Load The total power delivered by the generator is dissipated partly in its internal resistance and partly in the load. The power delivered to the load is equal for a lossless line) to the net power traveling to the right at any point along the line. Thus, we have: This follows from V G = V d + I d Z G, which implies P tot = P d + P G = P L + P G..) V G I d = V di d + Z G I d..) Eq...) is a consequence of..) and the definitions: P tot = ReV G I d)= Re V d + Z G I d ) I d ] Next, we calculate the reflection coefficients: Γ L = Z L Z Z L + Z = o, Γ G = Z G Z Z G + Z =.486 and Γ d = Γ L e jβd = o. It follows that: Z d = Z + Γ d Γ d = j9.83, V d = V GZ d Z G + Z d = j.36 = o The voltage across the load will be: V L = V d e jβd + Γ L + Γ d = j.65 = o V The current through the generator is: I d = V d Z d =.3 j. = o A It follows that the generated and dissipated powers will be: P tot = ReV G I d)=.678 W P G = ReZ GI d I d )= ReZ G) I d P d = ReV d I d)= ReV L I L)= P L..3) P G = ReZ G) I d =.838 W P L = P d = ReV d I d)=.488 W The last equality follows from Eq..9.5) or from V d± = V L± e ±jβd : We note that P tot = P G + P L. ReV d I d)= Vd+ V d ) = VL+ V L ) = Z Z ReV L I L) In the special case when the generator and the load are matched to the line, so that Z G = Z L = Z, then we find the standard result that half of the generated power is delivered to the load and half is lost in the internal impedance. Using Eq..9.8) with Z G = Z, we obtain V d = I d Z G = V G /, which gives: P tot = V G, P G = V G = 4Z 8Z P tot, P d = P L = V G = 8Z P tot..4) Example..: A load Z L = 5 + j Ω is connected to a generator V G = o volts with a -ft 3.48 m) cable of a 5-ohm transmission line. The generator s internal impedance is ohm, the operating frequency is MHz, and the velocity factor of the line, /3. Determine the voltage across the load, the total power delivered by the generator, the power dissipated in the generator s internal impedance and in the load. If the line is lossy, with a complex wavenumber β c = β jα, the power P L at the output of the line is less than the power P d at the input of the line. Writing V d± = V L± e ±αd e ±jβd, we find: P d = Z Vd+ V d ) = Z VL+ e αd V L e αd) P L = Z VL+ V L ) We note that P d >P L for all Γ L. In terms of the incident forward power at the load, P inc = V L+ /Z, we have: P d = P inc e αd Γ L e αd) = P inc e αd Γ d ) P L = P inc ΓL )..5) where Γ d = Γ L e αd. The total attenuation or loss of the line is P d /P L the inverse P L /P d is the total gain, which is less than one.) In decibels, the loss is: Solution: The propagation speed is c = c /3 = 8 m/sec. The line wavelength λ = c/f = m and the propagation wavenumber β = π/λ =.34 rads/m. The electrical length is d/λ = 3.48/ =.54 and the phase length βd = radians. ) ) Pd e αd Γ L e αd L = log = log P L Γ L total loss)..6)

15 .. Open- and Short-Circuited Transmission Lines Transmission Lines If the load is matched to the line, Z L = Z, so that Γ L =, the loss is referred to as the matched-line loss and is due only to the transmission losses along the line: L M = log e αd ) = 8.686αd matched-line loss)..7) Denoting the matched-line loss in absolute units by a = LM/ = e αd, we may write Eq...6) in the equivalent form: ) a Γ L L = log a Γ L ) total loss)..8) The additional loss due to the mismatched load is the difference: ) ) ΓL e 4αd Γd L L M = log = log Γ L Γ L..9) Example..: A 5 ft long RG-58 coax is connected to a load Z L = 5 + 5j ohm. At the operating frequency of MHz, the cable is rated to have. db/ ft of matched-line loss. Determine the total loss of the line and the excess loss due to the mismatched load. Solution: The matched-line loss of the 5 ft cable is L M = 5./ =.8 db or in absolute units, a =.8/ =.5. The reflection coefficient has magnitude computed with the help of the MATLAB function zg: Γ L =abszg5 + 5j, 5)=.6 It follows that the total loss will be: ) ) a Γ L.5.6 L = log = log a Γ L ) = 3. db.5.6 ) The excess loss due to the mismatched load is 3..8 =.3 db. At the line input, we have Γ d = Γ L e αd = Γ L /a =.6/.5 =.4. Therefore, from the point of view of the input the line appears to be more matched.. Open- and Short-Circuited Transmission Lines Open- and short-circuited transmission lines are widely used to construct resonant circuits as well as matching stubs. They correspond to the special cases for the load impedance: Z L = for an open-circuited line and Z L = for a short-circuited one. Fig... shows these two cases. Knowing the open-circuit voltage and the short-circuit current at the end terminals a, b, allows us also to replace the entire left segment of the line, including the generator, with a Thévenin-equivalent circuit. Connected to a load impedance Z L, the equivalent circuit will produce the same load voltage and current V L,I L as the original line and generator. Setting Z L = and Z L = in Eq..9.), we obtain the following expressions for the wave impedance Z l at distance l from the open- or short-circuited termination: Fig... Open- and short-circuited line and Thévenin-equivalent circuit. Z l = jz cot βl Z l = jz tan βl The corresponding admittances Y l = /Z l will be: Y l = jy tan βl Y l = jy cot βl open-circuited) short-circuited) open-circuited) short-circuited)..)..) To determine the Thévenin-equivalent circuit that replaces everything to the left of the terminals a, b, we must find the open-circuit voltage V th, the short-circuit current I sc, and the Thévenin impedance Z th. The impedance Z th can be determined either by Z th = V th /I sc, or by disconnecting the generator and finding the equivalent impedance looking to the left of the terminals a, b. It is obtained by propagating the generator impedance Z G by a distance d: Z G + jz tan βd Z th = Z Z + jz G tan βd = Z + Γ th, Γ th = Γ G e jβd..3) Γ th The open-circuit voltage can be determined from Eq..9.) by setting Z L =, which implies that Γ L =, Γ d = e jβd, and Γ G Γ d = Γ G e jβd = Γ th. The shortcircuit current is also obtained from.9.) by setting Z L =, which gives Γ L =, Γ d = e jβd, and Γ G Γ d = Γ G e jβd = Γ th. Then, we find: V th = V GZ e jβd, I sc = Z G + Z Γ th V G Z G + Z e jβd + Γ th..4) It follows that V th /I sc = Z th, as given by Eq...3). A more convenient way of writing Eq...4) is by noting the relationships:

16 .. Open- and Short-Circuited Transmission Lines Transmission Lines Then, Eq...4) becomes: Γ th = Z Z th + Z, + Γ th = Z th Z th + Z Z th + Z Z th + Z V th = V, I sc = I..5) Z Z th where V,I are the load voltage and currents in the matched case, given by Eq..9.9). The intuitive meaning of these expressions can be understood by writing them as: Z Z th V = V th, I = I sc..6) Z th + Z Z th + Z These are recognized to be the ordinary voltage and current dividers obtained by connecting the Thévenin and Norton equivalent circuits to the matched load impedance Z, as shown in Fig.... The Thévenin equivalent circuit to the left of the terminals a, b is easily determined by shorting the generator and finding the Thévenin impedance and then finding the open-circuit voltage. We have: Z th = Z b + Z cz a + Z G ) V G Z c, V th =..7) Z c + Z a + Z G Z c + Z a + Z G where Z a,z b,z c for a length-d segment are given by Eq..8.3): ) βd Z a = Z b = jz tan, Z c = jz sin βd It is straightforward to verify that the expressions in Eq...7) are equivalent to those in Eq...3) and..4). Example..: For the generator, line, and load of Example.., determine the Thévenin equivalent circuit. Using this circuit determine the load voltage. Solution: We work with the T-section approach. The following MATLAB call gives Z a and Z c, with Z = 5 and βd = : Z a,z c ]= tsection5, )= 66.89j, 33.83j] Then, Eq...7) gives with Z b = Z a : Z th = Z b + Z cz a + Z G ) =.39 + j6.36 Ω Z c + Z a + Z G V G Z c V th = =.8 + j.6 = o V Z c + Z a + Z G Alternatively, Z th can be computed by propagating Z G = by a distance d: Fig... Thévenin and Norton equivalent circuits connected to a matched load. Z th = zprop, 5, )=.39 + j6.36 Ω The quantities V,I are the same as those obtained by connecting the actual line to the matched load, as was done in Eq..9.9). An alternative way of determining the quantities V th and Z th is by replacing the length-d transmission line segment by its T-section equivalent circuit, as shown in Fig...3. The load voltage is found from the Thévenin circuit: V L = V thz L Z L + Z th = j.65 = o V which agrees with that found in Example.... Standing Wave Ratio The line voltage at a distance l from the load is given by Eq..9.), which can be written as follows in terms of the forward wave V L+ = V L / + Γ L ): Fig...3 T-section and Thévenin equivalent circuits. The magnitude of V l will be: V l = V L+ e jβl + Γ l )..)