Lecturer Note Sub: MWE Subject code: PCEC 4402 Sem: 8 th Prepared by: Mr. M. R. Jena Lecturer-10 Components in a signal flow graph Nodes: each port, i, of a microwave network has two nodes, a, and b. Node ai has a wave entering port i, while node bi has a wave reflected from node i. Branches: a path connecting a-node and b-node. Signal flows from node a to node b. Examples A one-port network and its flow graph Discontinuities in Transmission Lines
Discontinuities in transmission lines cannot be avoided ipractical implementation. Mechanical or electrical transitions from one medium to another (e.g., a junction between two waveguides, or a coax-tomicrostrip transition). Sometimes, discontinuities are deliberately introduced to perform a certain function. Usually, a transmission line discontinuity can be represented as an equivalent circuit at some point on the transmission line. Analysis of discontinuities requires complicated field solution, which is usually performed by sophisticated CAD tools. Signal Flow Graphs- Signal flow graph is an additional technique to analyse microwave networks in terms of reflected and transmitted. Three different forms of it are given below with nodes
The decomposition rules are also given below
Lecturer-11 At microwave frequencies, other factors must be considered. To accurately calculate the behavior of that same 50-ohm resistor, you need to consider its length, width, and thickness of metal (due to the skin effect), and its proximity to the ground plane. This is when we must consider it as a distributed element. At low frequencies, the metal that connects components together is treated as an ideal connection, with no loss, no characteristic impedance, and no transmission phase angle. When interconnects become an appreciable fraction of the signal wavelength, these interconnections themselves must be treated as distributed elements or transmission lines. An extreme example of the need to consider the distributed properties of transmission lines is when we are dealing with a quarter-wavelength. At this electrical length (90 degrees), an open circuit is transformed to a short circuit, and a short-circuit is transformed to an open circuit! Think about this: a short-circuited 90 degree "stub" hanging in shunt off of a transmission line will be invisible to signals propagating down the the transmission line, while an open circuited 90 degree stub shunting a transmission line will cause a short circuit and the propagating signal will get hosed! A whole lot of microwave engineering exploits this concept, so you'd better understand it. One "classic" distributed element is the quarter-wave transformer. The quarterwave transformer is used to shift the impedance of a circuit by the following simple formula: Z 2 = sqrt(z 0 Z L ) where Z 2 is the characteristic impedance of the transformer, Z L is the load impedance, and Z 0 is the characteristic impedance of the system you are trying to maintain. Do you detect a pattern? Most of the equations on this page use the square-root function... perhaps they put that button on your Casio calculator for a reason! The transmission line has a single purpose for both the transmitter and the antenna.
This purpose is to transfer the energy output of the transmitter to the antenna with the leaspossible power loss. How well this is done depends on the special physical and electrical characteristics (impedance and resistance) of the transmission line. The electrical characteristics of a two-wire transmission line depend primarily on the construction of the line. The two-wire line acts like a long capacitor. The change of its capacitive reactance is noticeable as the frequency applied to it is changed. Since the long conductors have a magnetic field about them when electrical energy is being passed through them, they also exhibit the properties of inductance. The values of inductance and capacitance presented depend on the various physical factors that we discussed earlier. For example, the type of line used, the dielectric in the line, and the length of the line must be considered. The effects of the inductive and capacitive reactance of the line depend on the frequency applied. Since no dielectric is perfect, electrons manage to move from one conductor to the other through the dielectric. Each type of two-wire transmission line also has a conductance value. This conductance value represents the value of the current flow that may be expected through the insulation, If the line is uniform (all values equal at each unit length), then one small section of the line may represent several feet. This illustration of a two-wire transmission line will be used throughout the discussion of transmission lines; but, keep in mind that the principles presented apply to all transmission lines. We will explain the theories using LUMPED CONSTANTS and DISTRIBUTED CONSTANTS to further simplify these principles. Single Stub Tuning- A single open-circuited (or short-circuited) transmission line is connected either in parallel or series with the feed line at a certain distance from the load.
Single Sub Tuning can match any load impedance to line, but suffer from disadvantage of requiring a variable length between the load and stub. Since lumped elements are not required, single stub is convenient and easy to fabricate in microstrip form. Two adjustable parameters and susceptance or reactance provided by stub. Although for microstrip lines open circuit is easy to Since fabricate since a via-hole is enough, for coax or waveguide short circuit is preferred since open circuit line may be large for radiation. If the impedance has the form of at the distance. Then stub reactance can be chosen as, resulting for matching condition. For a given susceptance or reactance the difference in lengths of open and short-circuited stub is λ/4. The above method, using one stub, is a common matching technique. The general procedure is: 1. Find the point at which r = 1. Cut it there. 2. Find the value of jx at that point. 3. Find the appropriate stub length to cancel out jx and connect the stud in series at that point. Obviously, the matching stub can also be connected in shunt (parallel), but the stub length may be different since we are now adding up admittance! On the Smith chart, we should get the admittance of the required stub by rotating the impedance by 180 o.
Lecturer-12 ELECTROMAGNETIC FIELDS The distributed constants of resistance, inductance, and capacitance are basic properties common to all transmission lines and exist whether or not any current flow exists. As soon as current flow and voltage exist in a transmission line, another property becomes quite evident. This is the presence of an electromagnetic field, or lines of force, about the wires of the transmission line. The lines of force themselves are not visible; however, understanding the force that an electron experiences while in the field of these lines is very important to your understanding of energy transmission. There are two kinds of fields; one is associated with voltage and the other with current. The field associated with voltage is called the ELECTRIC (E) FIELD. It exerts a force on any electric charge placed in it. The field associated with current is called a MAGNETIC (H) FIELD, because it tends to exert a force on any magnetic pole placed in it. Figure 3-6 illustrates the way in which the E fields and H fields tend to orient themselves between conductors of a typical two-wire transmission line. The illustration shows a cross section of the transmission lines. The E field is represented by solid lines and the H field by dotted lines. The arrows indicate the direction of the lines of force. Both fields normally exist together and are spoken of collectively as the electromagnetic field. Inductance of a Transmission Line When current flows through a wire, magnetic lines of force are set up around the wire. As the current increases and decreases in amplitude, the field around the wire expands and collapses accordingly.
The energy produced by the magnetic lines of force collapsing back into the wire tends to keep the current flowing in the same direction. This represents a certain amount of inductance, which is expressed in microhenrys per unit length. Figure 3-2 illustrates the inductance and magnetic fields of a transmission line. Capacitance of a Transmission Line Capacitance also exists between the transmission line wires, as illustrated in figure 3-3. Notice that the two parallel wires act as plates of a capacitor and that the air between them acts as a dielectric. The capacitance between the wires is usually expressed in picofarads per unit length. This electric field between the wires is similar to the field that exists between the two plates of a capacitor. Double Stub Tuning- The double stub tuner can not match all load impedances, but shortest matching section.load may be arbitrary distance from the first stub. Distance between the stubs should be generally chosen as λ/4 or 3λ/4 to reduce the frequency sensitivity. Why we use double-stub tuning? The single-stub tuner has one limitation: it must be placed at the proper distance from the load, which is a variable that is difficult to adjust in practice. The double-stub tuner uses two tuning stubs, partially removes the requirement for variable distance from the load, and is widely used in laboratory practice as a single frequency matching device. Smith Chart Solution for finding the matching parameters: d, l1, and l2
Fig 12.1: Smith chart Step 1: Draw the normalized load impedance yl in the Smith Chart. Step 2: Taking into account the effect of the transmission line d between two stubs --- rotated 1+jb circle Step 3: Design l1 so that yl moves to y1 (or y1 for solution 2).
Lecturer-13 WAVEGUIDE THEORY The two-wire transmission line used in conventional circuits is inefficient for transferring electromagnetic energy at microwave frequencies. At these frequencies, energy escapes by radiation because the fields are not confined in all directions, as illustrated in figure. Coaxial lines are more efficient than two-wire lines for transferring electromagnetic energy because the fields are completely confined by the conductors, as illustrated in figure. Waveguides are the most efficient way to transfer electromagnetic energy. WAVEGUIDES are essentially coaxial lines without center conductors. They are constructed from conductive material and may be rectangular, circular, or elliptical in shape, as shown in figure. Waveguides are used to transfer electromagnetic power efficiently from one point in space to another. Some common guiding structures are shown in the figure below like co-axial line, twowire line, microstrip line, rectangular waveguide, di-electric waveguide. These include the typical coaxial cable, the two-wire and mictrostrip transmission lines, hollow conducting waveguides, and optical fibers. In practice, the choice of structure is dictated by: (a) the desired operating frequency band, (b) the amount of power to be transferred, and (c) the amount of transmission losses that can be tolerated. Coaxial cables are widely used to connect RF components. Their operation is practical for frequencies below 3 GHz. Above that the losses are too excessive. For example, the attenuation might be 3 db per 100 m at 100 MHz, but 10 db/100 m at 1 GHz, and 50 db/100 m at 10 GHz. Their power rating is typically of the order of one kilowatt at 100 MHz, but only 200 W at 2 GHz, being limited primarily because of the
heating of the coaxial conductors and of the dielectric between the conductors (dielectric voltage breakdown is usually a secondary factor.) However, special short-length coaxial cables do exist that operate in the 40 GHz range. Another issue is the single-mode operation of the line. At higher frequencies, in order to prevent higher modes from being launched, the diameters of the coaxial conductors must be reduced, diminishing the amount of power that can be transmitted. Two-wire lines are not used at microwave frequencies because they are not shielded and can radiate. One typical use is for connecting indoor antennas to TV sets. Microstrip lines are used widely in microwave integrated circuits. Rectangular waveguides are used routinely to transfer large amounts of microwave power at frequencies greater than 3 GHz. For example at 5 GHz, the transmitted power might be one megawatt and the attenuation only 4 db/100 m. Optical fibers operate at optical and infrared frequencies, allowing a very wide bandwidth. Their losses are very low, typically, 0.2 db/km. The transmitted power is of the order of milliwatts. Longitudinal-Transverse Decompositions In a waveguiding system, we are looking for solutions of Maxwell s equations that are propagating along the guiding direction (the z direction) and are confined in the near vicinity of the guiding structure. Thus, the electric and magnetic fields are assumed to have the form: E(x, y, z, t)= E(x, y) e jωt jβz H(x, y, z, t)= H(x, y) e jωt jβz
where β is the propagation wavenumber along the guide direction. The corresponding wavelength, called the guide wavelength, is denoted by λg = 2π/β. The precise relationship betweenωand β depends on the type of waveguiding structure and the particular propagating mode. Because the fields are confined in the transverse directions (the x, y directions,) they cannot be uniform (except in very simple structures) and will have a non-trivial dependence on the transverse coordinates x and y. Next, we derive the equations for the phasor amplitudes E(x, y) and H(x, y). Because of the preferential role played by the guiding direction z, it proves convenient to decompose Maxwell s equations into components that are longitudinal, that is, along the z-direction, and components that are transverse, along the x, y directions. Thus, we decompose: E(x, y)= ˆx Ex(x, y)+ˆy Ey(x, y)+ˆz Ez(x, y) ET(x, y)+ˆz Ez(x, y) [Transverse] [longitudinal] Figure 13.1: Fields confined in two directions only. Operating Bandwidth All waveguiding systems are operated in a frequency range that ensures that only the lowest mode can propagate.
If several modes can propagate simultaneously, one has no control over which modes will actually be carrying the transmitted signal. This may cause undue amounts of dispersion, distortion, and erratic operation. A mode with cutoff frequency ωc will propagate only if its frequency is ω ωc, or λ < λc. Ifω < ωc, the wave will attenuate exponentially along the guide direction. This follows from the ω,β relationship (9.1.10): ω 2 = ω c 2 + β 2 c 2 =» β 2 = (ω 2 ω c 2 )/c 2 If ω ωc, the wavenumber β is real-valued and the wave will propagate. But if ω < ωc, β becomes imaginary, say, β = jα, and the wave will attenuate in the zdirection, with a penetration depth δ = 1/α: e jβz = e αz If the frequency ω is greater than the cutoff frequencies of several modes, then all of these modes can propagate. Conversely, if ω is less than all cutoff frequencies, then none of the modes can propagate. If we arrange the cutoff frequencies in increasing order, ω c1 < ω c2 < ω c3 <, then, to ensure single-mode operation, the frequency must be restricted to the interval ω c1 <ω<ω c2, so that only the lowest mode will propagate. This interval defines the operating bandwidth of the guide. These remarks apply to all waveguiding systems, not just hollow conducting waveguides. For example, in coaxial cables the lowest mode is the TEM mode having no cutoff frequency, ω c1 = 0. However, TE and TM modes with non-zero cutoff frequencies do exist and place an upper limit on the usable bandwidth of the TEM mode. Similarly, in optical fibers, the lowest mode has no cutoff, and the single-mode bandwidth is determined by the next cutoff frequency. In rectangular waveguides, the smallest cutoff frequencies are f 10 = c/2a, f20 = c/a = 2f 10, and f 01 = c/2b. Because we assumed that b a, it follows that always f 10 f 01. If b a/2, then 1/a 1/2b and therefore, f 20 f 01, so that the two lowest cutoff frequencies are f 10 and f 20.
On the other hand, if a/2 b a, then f 01 f 20 and the two smallest frequencies are f 10 and f 01 (except when b = a, in which case f 01 = f 10 and the smallest frequencies are f10 and f20.) The two cases b a/2 and b a/2 are depicted in Fig. It is evident from this figure that in order to achieve the widest possible usable bandwidth for the TE10 mode, the guide dimensions must satisfy b a/2 so that the bandwidth is the interval [fc, 2fc], where fc = f 10 = c/2a. In terms of the wavelength λ = c/f, the operating bandwidth becomes: 0.5 a/λ 1, or, a λ 2a. We will see later that the total amount of transmitted power in this mode is proportional to the cross-sectional area of the guide, ab. Thus, if in addition to having the widest bandwidth, we also require to have the maximum power transmitted, the dimension b must be chosen to be as large as possible, that is, b = a/2. Most practical guides follow these side proportions. If there is a canonical guide, it will have b = a/2 and be operated at a frequency that lies in the middle of the operating band [fc, 2fc], that is, f = 1.5fc = 0.75c/a WAVEGUIDE MODES OF OPERATION- The waveguide analyzed in the previous paragraphs yields an electric field configuration known as the half-sine electric distribution. This configuration, called a MODE OF OPERATION, is shown in figure. Recall that the strength of the field is indicated by the spacing of the lines; that is, the closer the lines, the stronger the field. The regions of maximum voltage in this field move continuously down the waveguide in a sine-wave pattern. To meet boundary conditions. the field must always be zero at the b walls.
Figure 13.2: Half-sine E field distribution. The half-sine field is only one of many field configurations, or modes, that can exist in a rectangular waveguide. A full-sine field can also exist in a rectangular waveguide because, as shown in figure, the field is zero at the b walls. Figure 13.3: Full-sine E field distribution. The magnetic field in a rectangular waveguide is in the form of closed loops parallel to the surface of the conductors. The strength of the magnetic field is proportional to the electric field. Figure illustrates the magnetic field pattern associated with a half-sine electric field distribution. The magnitude of the magnetic field varies in a sine-wave patterndown the center of the waveguide in time phase with the electric field. TIME PHASE means that the peak H lines and peak E lines occur at the same instant in time, although not necessarily at the same point along the length of the waveguide. The dominant mode is the most efficient mode. Waveguides are normally designed so that only the dominant mode will be used. To operate in the dominant mode, a waveguide must have an a (wide) dimension of at least one half-wavelength of the frequency to be propagated. The a dimension of the
waveguide must be kept near the minimum allowable value to ensure that only the dominant mode will exist. In practice, this dimension is usually 0.7 wavelength. Figure 13.4: Magnetic field caused by a half-sine E field. Of the possible modes of operation available for a given waveguide, the dominant mode has the lowest cutoff frequency. The high-frequency limit of a rectangular waveguide is a frequency at which its a dimension becomes large enough to allow operation in a mode higher than that for which the waveguide has been designed. Circular waveguides are used in specific areas of radar and communications systems, such as rotating joints used at the mechanical point where the antennas rotate. Figure illustrates the dominant mode of a circular waveguide. The cutoff wavelength of a circular guide is 1.71 times the diameter of the waveguide. Since the a dimension of a rectangular waveguide is approximately one halfwavelength at the cutoff frequency, the diameter of an equivalent circular waveguide must be 2/1.71, or approximately 1.17 times the a dimension of a rectangular waveguide. Figure 13.5: Dominant mode in a circular waveguide.
Lecturer-14 MODE NUMBERING SYSTEMS So far, only the most basic types of E and H field arrangements have been shown. More complicated arrangements are often necessary to make possible coupling, isolation, or other types of operation. The field arrangements of the various modes of operation are divided into two categories: TRANSVERSE ELECTRIC (TE) and TRANSVERSE MAGNETIC (TM). In the transverse electric (TE) mode, the entire electric field is in the transverse plane, which isperpendicular to the waveguide, (direction of energy travel). Part of the magnetic field is parallel to the length axis. In the transverse magnetic (TM) mode, the entire magnetic field is in the transverse plane and has no portion parallel to the length axis. Since there are several TE and TM modes, subscripts are used to complete the description of the field pattern. In rectangular waveguides, the first subscript indicates the number of half-wave patterns in the a dimension, and the second subscript indicates the number of half-wave patterns in the b dimension. The dominant mode for rectangular waveguides is shown in figure 3-38. It is designated as the TE mode because the E fields are perpendicular to the a walls. The first subscript is 1, since there is only one half-wave pattern across the a dimension. Figure 14.1: Dominant mode in a rectangular waveguide
There are no E-field patterns across the b dimension, so the second subscript is 0. The complete mode description of the dominant mode in rectangular waveguides is TE1,0. Subsequent description of waveguide operation in this text will assume the dominant (TE1,0) mode unless otherwise noted. A similar system is used to identify the modes of circular waveguides. The general classification of TE and TM is true for both circular and rectangular waveguides. In circular waveguides the subscripts have a different meaning. The first subscript indicates the number of fill-wave patterns around the circumference of the waveguide. The second subscript indicates the number of half-wave patterns across the diameter. In the circular waveguide in figure, the E field is perpendicular to the length of the waveguide with no E lines parallel to the direction of propagation. Thus, it must be classified as operating in the TE mode. If you follow the E line pattern in a counterclockwise direction starting at the top, the E lines go from zero, through maximum positive (tail of arrows), back to zero, through maximum negative (head of arrows), and then back to zero again. This is one full wave, so the first subscript is 1. Along the diameter, the E lines go from zero through maximum and back to zero, making a half-wave variation. The second subscript, therefore, is also 1. TE1,1 is the complete mode description of the dominant mode in circular waveguides. Several modes are possible in both circular and rectangular waveguides. Figure illustrates several different modes that can be used to verify the mode numbering system. Figure 14.2: Counting wavelengths in a circular waveguide.
Figure 14.3: Various modes of operation for rectangular and circular waveguides. L. Rayleigh, 1897 TE and TM modes propagation in hollow waveguides with rectangular or circular cross sections. Experiments in 1936: (1) G. C. Southworth at AT&T: Rectangular waveguide. 1 (2) M. L. Barrow at MIT : Circular waveguide. General Solutions for TEM, TE, and TM Waves
General two-conductor transmission line Closed waveguide as a transmission line Assume that all fields have a time dependence of e j ω t and propagation factor e jω z. EM fields in a waveguide or transmission line are decomposed into longitudinal and transverse components.
Lecturer-15 The Waveguide Model One method of characterizing transmission lines describes a transmission as a guide for electromagnetic waves. With this method, the source sends the electromagnetic signal, which consists of a coupled voltage wave and current wave, to the load. The voltage wave corresponds to the electric field and the current wave corresponds to the magnetic field of the wave. The transmission line, which effectively acts as a transmission medium, guides the signal along the way. The signal travels this medium at the speed of light within that medium. You can calculate the speed of light, v, in a transmission line from the permittivity (ε) and the permeability (µ) of the dielectric between the conductors (refer back to Maxwell EQs). = 1/ For example, if Teflon separates the pair of wires that makes up the transmission line, the wave travels at the speed of light in Teflon, which is approximately 70% of the speed of light in a vacuum. (vteflon 0.7c). As the signal travels along the transmission line, the voltage wave defines the voltage at each point, and the current wave defines the current at each point. (The waveguide model of the transmission line were the signal is represented as a voltage wave and a current wave that travel at a velocity equal to the speed of
light of the dielectric between the wires. The voltage wave is equal to the current wave multiplied by Z0, the line impedance.) As a signal travels down a transmission line the ratio of voltage to current is constant. This ratio is the characteristic impedance of the transmission line (Z0). This impedance is defined by the geometry of the line and the permittivity and permeability of the dielectric. For example the characteristic impedance of a coaxial transmission line is 0 60 / log [ / ] where d is the diameter of the inner conductor, D is the inside diameter of the outer conductor, and εr is the relative dielectric constant of the material. Now we have two models for a transmission line: a circuit comprising infinitesimal inductances and capacitances with parameters L and C, a wavequide for signals with parameters v and Z0. The following equations relate the parameters of the two lossless models. 0= / and = 1/ Although these models are interchangeable, the waveguide model is usually more useful for transmission line analysis. Power transmitted by a waveguide With the field solutions at hand, one can determine the amount of power transmitte along the guide, as well as the transmission losses. The total power carried by the fields along the guide direction is obtained by integrating the z-component of the Poynting vector over the cross-sectional area of the guide:
PT =ʃ s Pz ds, where Pz = 1/2 Re(E H * ) Z^ It is easily verified that only the transverse components of the fields contribute to the power flow, that is, Pz can be written in the form: Pz = ½ Re(E T H T * ) Z^ For waveguides with conducting walls, the transmission losses are due primarily to ohmic losses in (a) the conductors and (b) the dielectric medium filling the space between the conductors and in which the fields propagate. In dielectric waveguides, the losses are due to absorption and scattering by imperfections. The transmission losses can be quantified by replacing the propagation wavenumber β by its complex-valued version βc = β jα, where α is the attenuation constant. The z-dependence of all the field components is replaced by: e jβz e jβcz = e (α+jβ)z = e αz e jβz The quantity α is the sum of the attenuation constants arising from the various loss mechanisms. For example, if αd and αc are the attenuations due to the ohmic losses in the dielectric and in the conducting walls, then α = αd + αc The ohmic losses in the dielectric can be characterized either by its loss tangent, say tan δ, or by its conductivity σd the two being related by σd = ω ε tan δ. More generally, the effective dielectric constant of the medium may have a negative imaginary part ε I that includes both conductive and polarization losses, ε (ω)= ε j ε I, with ε I = ε tan δ. Then, the corresponding complex-valued wavenumber βc is obtained by the replacement:
β = (ω 2 µε k c 2 ) βc = (ω 2 µε (ω) k c 2 ) For weakly lossy dielectrics (ε I << ε), we may make the approximation: Consider TE mode. i.e. Ez = 0. The equations for ~E? = (Ex ;Ey ) and ~B? = (Bx ;By ) become Copper Losses One type of copper loss is I2R LOSS. In rf lines the resistance of the conductors is never equal to zero. Whenever current flows through one of these conductors, some energy is dissipated in the form of heat. This heat loss is a POWER LOSS. With copper braid, which has a resistance higher than solid tubing, this power loss is higher. Another type of copper loss is due to SKIN EFFECT. When dc flows through a conductor, the movement of electrons through the conductor s cross section is uniform, The situation is somewhat different when ac is applied. The expanding and collapsing fields about each electron encircle other electrons. This phenomenon, called SELF INDUCTION, retards the movement of the encircled electrons. The flux density at the center is so great that electron movement at this point is reduced. As frequency is increased, the opposition to the flow of current in the center of the wire increases. Current in the center of the wire becomes smaller and most of the electron flow is on the wire surface. When the frequency applied is 100 megahertz or higher, the electron movement in the center is so small that the center of the wire could be removed without any noticeable effect on current.
You should be able to see that the effective crosssectional area decreases as the frequency increases. Since resistance is inversely proportional to the cross-sectional area, the resistance will increase as the frequency is increased. Also, since power loss increases as resistance increases, power losses increase with an increase in frequency because of skin effect. Copper losses can be minimized and conductivity increased in an rf line by plating the line with silver. Since silver is a better conductor than copper, most of the current will flow through the silver layer. The tubing then serves primarily as a mechanical support. Dielectric Losses DIELECTRIC LOSSES result from the heating effect on the dielectric material between the conductors. Power from the source is used in heating the dielectric. The heat produced is dissipated into the surrounding medium. When there is no potential difference between two conductors, the atoms in the dielectric material between them are normal and the orbits of the electrons are circular. When there is a potential difference between two conductors, the orbits of the electrons change. The excessive negative charge on one conductor repels electrons on the dielectric toward the positive conductor and thus distorts the orbits of the electrons. A change in the path of electrons requires more energy, introducing a power loss. The atomic structure of rubber is more difficult to distort than the structure of some other dielectric materials. The atoms of materials, such as polyethylene, distort easily. Therefore, polyethylene is often used as a dielectric because less power is consumed when its electron orbits are distorted. Radiation and Induction Losses RADIAION and INDUCTION LOSSES are similar in that both are caused by the fields surrounding the conductors. Induction losses occur when the electromagnetic field about a conductor cuts through any nearby metallic object and a current is induced in that object. As a result, power is dissipated in the object and is lost.
Radiation losses occur because some magnetic lines of force about a conductor do not return to the conductor when the cycle alternates. These lines of force are projected into space as radiation, and this results in power losses. That is, power is supplied by the source, but is not available to the load.
Lecturer-16 Numerical Problems & Solutions 1. an r.m.s value of 10 microvolts/m. Calculate (a) the magnetic field strength; (b) The electric field strength at a receiving station is measured and found to have the amount of power incident on a receiving aerial with an effective area of 5 m2. Given: Electric field strength = 10 microvolts/m. Required: (a) Magnetic field strength, (b) incident power on a receiving aerial with effective area of 5 m2. Solution. The ratio of the field strengths is always the same and is given by E rms (V/m) / H rms (V/m) This ratio is called the free-space wave impedance. It is analogous to the characteristic impedance of a transmission line. (a) Hrms = 10 µv/m/377 Ω = 2.65 10 8 A/m (b) Power density is given by Erms Hrms = 10 10 6 2.65 10 8 W/m2 = 2.65 10 13 W/m2 This is the amount of power incident on a surface of area 1 m2. For an aerial with area 5 m2, the total incident power will be P = 2.65 10 13 W/m2 5 m2 = 1.33 pw 2. If the data in Question 1 applies to a receiver located 10 km from the transmitter, what will be the values of Erms and Hrms at a distance of 100 km? Given: Data of Example 1.1 applied to a receiver at 10 km from transmitter. Required: (a) Erms at 100 km, (b) Hrms at 100 km. Solution. Using Equation P D2 / P D1 = [D 1 / D 1 ] 2 where PD1, PD2 = power densities at distances D1 and D2 respectively. Using Equation above equation at a distance of 100 km, the power density will be reduced by a factor (10/100)2 = 0.01, so power density = 2.65 10 15 W/m2. Now, power density = Erms Hrms and since Hrms = Erms/377 Erms = 1µV /m Hrms = 2.65 10 9 A/m 3. A transmitter with an output resistance of 72 W and an r.m.s. output of 100 V is connected via a matched line to an antenna whose input resistance is 72 W. Its radiation resistance is also 72 W. Assuming that the antenna is 100% efficient at the operating frequency, how much power will be transmitted into free space? Given: Transmitter output = 100 V, transmitter output impedance = 72 Ω, antenna input impedance = 72 Ω, radiation resistance = 72 Ω, antenna efficiency = 100%.
Required: Power radiated into free space. Solution. The antenna has an input impedance Zin = 72 Ω and provides a matched termination to the 72 Ω line. The r.f. generator then sees an impedance of 72 Ω, so the r.m.s. voltage applied to the line will be 100/2 = 50 V. The amount of power radiated is calculated using radiated power = 50 2 / R where R = 72 Ω is the radiation resistance. The radiated power is therefore 34.7W. Notice that, because in this case R = Zin, maximum power is radiated into free space. 4. A 75 W aerial system is used to supply signals to two receivers. Each receiver has an input impedance of 75 W. What is the required value of the matching resistor? Given: 75 W aerial system, input impedance of each receiver = 75 Ω, no. of receivers = 2. Required: Value of matching resistor. Solution. Using Equation with n = 2, we obtain R = Z 0 = 25 W 5. A broadcast signal induces an open-circuit voltage of 100 µv into a rod aerial. The aerial system has a characteristic impedance of 50 Ω and it is used to supply signal to three identical receivers each of which has an input impedance of 50 Ω. If the matching network type is used, calculate (a) the value of the resistance (R) required for the matching network and (b) the terminated voltage appearing across the input terminals of the receiver. Given: 50 Ω aerial system, input impedance of each receiver = 50 Ω, no. of receivers = 3, open-circuit voltage in aerial = 100 µv. Required: (a) Value of matching resistor, (b) terminated voltage at receiver input terminal. Solution (a) For the matching network R = Z 0 = 25 W (b) Using equation V receiver = (1/2n) V antenna = (1/6) 100 µv = 16.67 µv 6. You will often find two types of flexible coaxial cables: one with a characteristic impedance Z0 of 50 W which is used mainly for r.f. instrumentation and the other has a characteristic impedance of 75 W used mainly for antennas. The inner diameter of the outer conductor is the same in both cables. How would you distinguish the impedance of the two cables using only your eye? Solution. In general, to save money, both cables are normally made with the same outer diameter. This is even more evident when the cables are terminated in a type of r.f. connector known as BNC.3
Since these connectors have the same outer diameter, by using Equation Z 0 = where d = outer diameter of the inner conductor D = inner diameter of the outer conductor e = dielectric constant of the space between inner and outer conductor (e = 1 for air) we can deduce that for Z0 = 75 W, the inner conductor will be smaller than that of the 50 W cable. In practice, you will be able to recognise this distinction quite easily. 7. The twin parallel transmission line shown in Figure 2.6 is separated by a distance (D) of 300 mm between the centre lines of the conductors. The diameter (d) of the identical conductors is 4 mm. What is the characteristic impedance (Z 0 ) of the line? Assume that the transmission line is suspended in free space, i.e. e = 1. Given: D = 300 mm, d = 4 mm, e = 1. Required: Z 0. Solution. Using Equation Z 0 = = = 600 Ω 8. Two microstrip lines are printed on the same dielectric substrate. One line has a wider centre strip than the other. Which line has the lower characteristic impedance? Assume that there is no coupling between the two lines. Solution. We know that Z 0 varies as a function of h/w. Therefore, the line with the lower characteristic impedance will have a wider centre conductor. 9. A transmission line has the following primary constants: R = 23 W km 1, G = 4 ms km 1, L = 125 µh km 1 and C = 48 nf km 1. Calculate the characteristic impedance, Z 0, of the line at a frequency of (a) 100 Hz, (b) 500 Hz, (c) 15 khz, (d) 5 MHz and (e) 10 MHz. Given: R = 23 W km 1, G = 4 ms km 1, L = 125 µh km 1 and C = 48 nf km 1. Required: Z0 at (a) 100 Hz, (b) 500 Hz, (c) 15 khz, (d) 5 MHz and (e) 10 MHz. Solution. Use Equation 2.17 in the calculations that follow. 1. At 100 Hz, R + jwl = (23 + j0.08) W km 1, G + jwc = (4 + j0.030) ms km 1 Z 0 = = 75.83 Ω -2.06 X 10-3 rad 2. At 500 Hz, R + jwl = (23 + j0.39) W km 1, G + jwc = (4 + j0.15) ms km 1 Z 0 = = 75.81 Ω -10.03 X 10-3 rad 3. At 15 khz, R + jwl = (23 + j11.78) W km 1, G + jwc = 4 + j4.52 ms km 1 Z 0 = = 65.42 Ω -0.19 rad (d)at 5 MHz,R + jwl = (23 + j3926.99) W km 1, G + jwc = (4 + j1508) ms km 1
Z 0 = = 50.03 Ω / -0.00 rad (e) At 10 MHz,R + jwl = (23 + j7853.98)w km 1, G + jwc = (4 + j3016) ms km-1 Z 0 = = 50.03 Ω -0.00 rad 10. The following measurements have been made on a line at 1.6 MHz where Zoc = 900 W 30 and Zsc = 400 W 10. What is the characteristic impedance (Z 0 ) of the line at 1.6 MHz? Given: f = 1.6 MHz, Zoc = 900 Ω 30, Zsc = 400 Ω 10. Required: Z 0 at 1.6 MHz. Solution. Using Equation Z 0 = = (900 Ω 30 Х 400 Ω 10 ) = 600 Ω 20. 11. A transmission has a loss of two nepers per kilometre. What is the loss in db for a length of 10 kilometres? Given: Attenuation constant (α) = 2 nepers per km. Required: Loss in db for a length of 10 km. Solution. If 1 km represents a loss of 2 nepers, then 10 km = 10 2 = 20 nepers. Therefore loss = 8.686 20 = 173.72 db 12. Calculate the voltage reflection coefficient for the case where Z L = (80 j10) Ω and Z 0 = 50 Ω. Given: Z L = (80 j10), Z 0 = 50Ω Required: Γ v Solution. Using Equation Γ v = = 0.24 14.03 13. Calculate the voltage reflection coefficients at the terminating end of a transmission line with a characteristic impedance of 50 Ω when it is terminated by (a) a 50 Ω termination, (b) an open-circuit termination, (c) a short-circuit termination and (d) a 75 Ω termination. Given: Z0 = 50 Ω, ZL = (a) 50 Ω, (b) open-circuit =, (c) short-circuit = 0 Ω, (d) = 75Ω. Required: Γv for (a), (b), (c), (d). Solution. Use Equation Γ v = (a) with ZL = 50 Ω, Γ v = 0 0 (b) with ZL = open circuit = infinite Ω, Γ v = 1 0 (c) with ZL = short circuit = 0 Ω, Γ v = -1 0 or 1 180 (d) with ZL = 75 Ω, Γ v = 0.2 0
14. The incident voltage measured along the transmission line is 100 V and the reflected voltage measured on the same line is 10 V. What is its VSWR? (e) Soloution. Using Equation VSWR = = 1.22 15. What is the VSWR of a transmission system if its reflection coefficient Γv is 0.1? Given: Γv = 0.1 Required: VSWR (a) Solution. Using Equation VSWR = = 1.22 16. A manufacturer quotes a maximum VSWR of 1.07 for a resistive load when it is used to terminate a 50 Ω transmission line. Calculate the reflected power as a percentage of the incident power. Given: VSWR = 1.07, Z0 = 50 W Required: Reflected power as a percentage of incident power Solution. Using Equation Γ = = 0.034 Since power is proportional to V 2 Pref = (0.034)2 Pinc = 0.001 Pinc = 0.1% of Pinc 17. A transmission line has the following primary constants: R = 23 Ω km 1, G = 4 ms km 1, L = 125 µh km 1 and C = 48 nf km 1. Calculate the propagation constant γ of the line, and the characteristic impedance Z0 of the line at a frequency of (a) 100 Hz, (b) 500 Hz, (c) 15 khz, (d) 5 MHz and (e) 10 MHz. Solution. The characteristic impedance Z0 will not be calculated here because it has already been carried out in Example 2.4. However, the results will be copied to allow easy comparison with the propagation results calculated here for the discussion that follows after this answer. Equation 2.41 will be used to calculate the propagation constant γ, and Equation 2.42 will be used to derive the attenuation constant α and the phase constant β in all the calculations that follow. (a) At 100 Hz, R + jωl = (23 + j(2π 100 125 µh)) = (23 + j0.08) Ω km 1 And G + jwc = (4 ms + j(2π 100 48 nf)) = (4 + j0.030) ms km 1 γ = j j = 0.30 nepers + 1.33 Х 10-3 rad km -1 Z 0 = 75.83 Ω -2.06 10-3 rad (b) At 500 Hz R + jwl = (23 + j(2π 500 125 µh)) = (23 + j0.39) Ω km 1 and G + jωc = (4 ms + j(2π 500 48 nf)) = (4 + j0.15) ms km 1
γ = j j = 0.30 nepers + 8.31 Х 10-3 rad km -1 Z 0 = 75.82 Ω -10.30 10-3 rad (c) At 15 khz R + jωl = (23 + j(2π 15 103 125 µh)) = (23 + j11.78) Ω km 1 and G + jωc =(4 ms + j(2π 15 103 48 nf)) = (4 + j4.52 10 3) ms km 1 γ = j j = 0.31 nepers + 242 Х 10-3 rad km -1 Z 0 = 65.42 Ω -0.19 rad (d) At 5 MHz R + jωl = (23 + j(2π 5 106 125 µh)) = (23 + j3926.99) Ω km 1 and G + jωc = (4 ms + j(2π 5 106 48 nf)) = (4 + j1508) ms km 1 γ = j j = 0.33 nepers + 76.95 rad km -1 Z 0 = 50.03 Ω -0.00 rad (e) At 10 MHz R + jωl = (23 + j(2π 10 106 125 µh) = (23 + j7853.98) Ω km 1 and G + jωc = (4 ms + j(2π 10 106 48 nf)) = (4 + j3016) ms km 1 γ = j j = 0.33 nepers + j153.9 rad km -1 Z 0 = 50.03 Ω -0.00 rad 18. The following measurements have been made on a line at 1.6 MHz where Zoc = 900 W/ 30 and Zsc = 400 W / 10. What is the characteristic impedance (Z0) of the line at MHz? Given: f = 1.6 MHz, Zoc = 900 W / 30, Zsc = 400 W / 10. Required: Z0 at 1.6 MHz.
Solution. Using Equation Z 0 = = 600Ω -20 0 19. A transmission has a loss of two nepers per kilometre. What is the loss in db for a length of 10 kilometres? Given: Attenuation constant (α) = 2 nepers per km. Required: Loss in db for a length of 10 km. Solution. If 1 km represents a loss of 2 nepers, then 10 km = 10 2 = 20 nepers. Therefore loss = 8.686 20 = 173.72 db 20. A 377 W transmission line is terminated by a short circuit at one end. Its electrical length is l/7. Calculate its input impedance at the other end. Ans - Zin = jz0 tan (2π l / λ ) =j377 1.254 = j472.8 W Similar reactive effects can also be produced by using an open-circuited load18 and applying it to Equation to produce inductive and capacitive reactances: Zin =- jz0 tan (2π l / λ ) Above Equation follows a cotangent curve and will therefore also produce positive and negative impedances. Adjustment of Z0 and line length will set the required reactance. 21. A transmission line has a characteristic impedance (Z0) of 90 W. Its electrical length is l/4 and it is terminated by a load impedance (ZL) of 20 W. Calculate the input impedance (Zin) presented by the line. Given: Z0 = 90 W, ZL = 20 W, l = l/4 Required: Zin Solution. Using Equation Zin = Z 0 2 / Z L = 405 Ω