Engineering electromagnetics hayt 6th edition pdf

 

    The McGraw-Hill Companies. Engineering Electromagnetics. Sixth Edition. William H. Hayt, Jr.. John A. Buck. Textbook Table of Contents. The Textbook Table. Engineering Electromagnetics Sixth Edition William H. Hayt, Jr.. Leslie_Kaminoff,_Amy_Matthews_Yoga_Anatomy-2nd_E( zlibraryexau2g3p_onion).pdf Yoga. Hayt,Buck Engineering Electromagnetics 6e Pdf previous post Harrison Advanced Engineering Dynamics Pdf. next post Heat Exchanger.

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    Engineering Electromagnetics Hayt 6th Edition Pdf

    Solutions of engineering electromagnetics 6th edition william h. hayt, john a. neogosynchpromath.gq, Past Exams for Electromagnetic Engineering. University. Engineering Electromagnetics - 6th Edition [William H. Hayt] text book. :6 26EE <04 2<56 4:A= 0NI3=6E0E. Title: Solutions of engineering electromagnetics 6th edition william h hayt, john a buck pdf, Author: Erwin Aguilar, Name: Solutions of engineering.

    Textbook Table of Contents The Textbook Table of Contents is your starting point for accessing pages within the chapter. Copyright The McGraw Companies. All rights reserved. Any use is subject to the Terms of Use and Privacy Policy. If you have a question or a suggestion about a specific book or product, please fill out our User Feedback Form accessible from the main menu or contact our customer service line at C -1 A32 HP -2 32 34 E7 32 3?. M-0 -, 0: B ,0" 5C0 1-,? G K0 ; 60 M:

    The waveguide sections are all new. As is evident. Apart from the first chapter on vector analysis. It may also serve as a bridge between the basic course and more advanced courses that follow it. The book contains more than enough material for a one-semester course. The theme of the text has not changed since the first edition of The approach taken in the new material. Joel T. Hussain Al-Rizzo. Numerous examples. Pennsylvania State University.

    University of Florida. Answers to selected end-of-chapter problems can be found on the internet at www. South Dakota State University. Vladimir A. In a course that places more emphasis on dynamics. These include Madeleine Andrawis. Georgia Tech. Michelle Flomenhoft. University of Virginia. Juri Silmberg. My editors at McGraw-Hill. Weikle II. Buck Atlanta. Catherine Fields.

    My see- mingly odd conception of the cover illustration was brought into reality through the graphics talents of Ms Diana Fouts at Georgia Tech. Ryerson Polytechnic University and Robert M.

    Maxwell's First Equation Electrostatics 73 3. Scalars and Vectors 2 1. Applications of Gauss' Law: Some Symmetrical Charge Distributions 62 3. Vector Algebra 3 1. The Line Integral 85 4. Other Coordinate Systems: Circular Cylindrical Coordinates 15 1.

    Field of a Sheet Charge 44 2. The Experimental Law of Coulomb 28 2. Vector Components and Unit Vectors 6 1. The Cross Product 13 1. The Vector Field 9 1. Electric Flux Density 54 3. Gauss' Law 57 3. Divergence 70 3. Electric Field Intensity 31 2. Application of Gauss' Law: Differential Volume Element 67 3. The Dot Product 10 1. Field of a Line Charge 38 2. The Cartesian Coordinate System 4 1. Curl 8. Example of the Solution of Poisson's Equation 7.

    Potential Gradient 99 4. Continuity of Current 5. Semiconductors 5. Current Analogies 6.

    Engineering Electromagnetics (6th Edition, 2001) - Hayt & Buck + Solution Manual

    Curvilinear Squares 6. Examples of the Solution of Laplace's Equation 7. The Iteration Method 6. Boundary Conditions for Perfect Dielectric Materials 5. The Method of Images 5. Several Capacitance Examples 5.

    Biot-Savart Law 8. Uniqueness Theorem 7. Current and Current Density 5. The Nature of Dielectric Materials 5. Conductor Properties and Boundary Conditions 5. The Scalar and Vector Magnetic Potentials 8. The Dipole 4. Magnetic Flux and Magnetic Flux Density 8. Capacitance 5. Metallic Conductors 5. Stokes' Theorem 8. The Potential Field of a System of Charges: Conservative Property 95 4.

    Ampere's Circuital Law 8. Faraday's Law Materials and Inductance 9. The Nature of Magnetic Materials 9. Magnetic Boundary Conditions 9. Displacement Current Plane Wave Propagation in General Directions Propagation in Good Conductors: Skin Effect Force on a Moving Charge 9. Maxwell's Equations in Integral Form Potential Energy and Forces on Magnetic Materials 9. The Transmission-Line Equations Magnetization and Permeability 9.

    Several Practical Problems Transmission-Line Parameters The Magnetic Circuit 9. The Poynting Vector and Power Considerations Standing Wave Ratio Force on a Differential Current Element 9. Wave Propagation in Free Space Some Transmission-Line Examples Force Between Differential Current Elements 9. Wave Propagation in Dielectrics Graphical Methods Force and Torque on a Closed Circuit 9. Maxwell's Equations in Point Form Basic Waveguide Operation Rectangular Waveguides Dielectric Waveguides Flag for inappropriate content.

    This will be. Then The capacitance will be Qnet Next, apply the boundary conditions: Find the capacitance between them: The region between the spheres is filled with a perfect dielectric.

    If the inner sphere is at V and the outer sphere at 0 V: Concentric conducting spheres have radii of 1 and 5 cm. The potential of the inner sphere is 2V and that of the outer is -2V. We use the general expression derived in Problem 7. Two coaxial conducting cones have their vertices at the origin and the z axis as their axis.

    Cone A has the point A 1, 0, 2 on its surface, while cone B has the point B 0, 3, 2 on its surface. Integrate again to find: Let the volume charge density in Fig. These two equations can be unified to cover the entire range of x; the final expression for the electric field becomes: Use a development similar to that of Sec. This situation is the same as that of Fig.

    The solution is found from Eq. We will use the first three terms to evaluate the potential at 3,4: Using thirteen terms, and. The series converges rapidly enough so that terms after the sixth one produce no change in the third digit. Thus, quoting three significant figures, The four sides of a square trough are held at potentials of 0, 20, , and 60 V; the highest and lowest potentials are on opposite sides.

    Find the potential at the center of the trough: Here we can make good use of symmetry. The solution for a single potential on the right side, for example, with all other sides at 0V is given by Eq.

    Since we want V at the center of a square trough, it no longer matters on what boundary each of the given potentials is, and we can simply write: Solve for the potential at the center of the trough: The series solution will be of the form: From part a, the radial equation is: Functions of this form are called circular harmonics. Find H in cartesian components at P 2, 3, 4 if there is a current filament on the z axis carrying 8 mA in the az direction: Find H if both filaments are present: The Biot- Savart method was used here for the sake of illustration.

    A current filament of 3ax A lies along the x axis. Each carries a current I in the az direction. Find H at the origin: We use the Biot-Savart law, which in this case becomes: Again, find H at the origin: The parallel filamentary conductors shown in Fig.

    We need an expression for H in cartesian coordinates. We can start with the known H in cylindrical for an infinite filament along the z axis: The results of part a should help: For the finite-length current element on the z axis, as shown in Fig. The Biot-Savart law reads: Calculate H at P 0, 0, 3: Since the limits are symmetric, the integral of the z component over y is zero. Let a filamentary current of 5 mA be directed from infinity to the origin on the positive z axis and then back out to infinity on the positive x axis.

    The Biot-Savart law is applied to the two wire segments using the following setup: Three uniform cylindrical current sheets are also present: The problem asks you to find H at various positions. Before continuing, we need to know how to find H for this type of current configuration. The sketch below shows one of the slabs of thickness D oriented with the current coming out of the page.

    The problem statement implies that both slabs are of infinite length and width. For example, if the sketch below shows the upper slab in Fig. Thus H will be in the positive x direction above the slab midpoint, and will be in the negative x direction below the midpoint. Reverse the current, and the fields, of course, reverse direction.

    We are now in a position to solve the problem. Find H at: This point lies within the lower slab above its midpoint. Thus the field will be oriented in the negative x direction. Referring to Fig. The total field will be this field plus the contribution from the upper slab current: Here the fields from both slabs will add constructively in the negative x direction: Since 0. The field from the lower slab will be negative x-directed as well, leading to: This point lies above both slabs, where again fields cancel completely: Consider this situation as illustrated in Fig.

    There sec. The only way to enclose current is to set up the loop which we choose to be rectangular such that it is oriented with two parallel opposing segments lying in the z direction; one of these lies inside the cylinder, the other outside.

    The loop is now cut by the current sheet, and if we assume a length of the loop in z of d, then the enclosed current will be given by Kd A. If we assume an infinite cylinder length, there will be no z dependence in the field, since as we lengthen the loop in the z direction, the path length over which the integral is taken increases, but then so does the enclosed current — by the same factor. Thus H would not change with z.

    There would also be no change if the loop was simply moved along the z direction. Again, using the Biot-Savart law, we note that radial field components will be produced by individual current elements, but such components will cancel from two elements that lie at symmetric distances in z on either side of the observation point.

    Suppose the rectangular loop was drawn such that the outside z-directed segment is moved further and further away from the cylinder. We would expect Hz outside to decrease as the Biot-Savart law would imply but the same amount of current is always enclosed no matter how far away the outer segment is. We therefore must conclude that the field outside is zero.

    With our rectangular path set up as in part a, we have no path integral contributions from the two radial segments, and no contribution from the outside z-directed segment. Find H everywhere: Between the cylinders, we are outside the inner one, so its field will not contribute. A toroid having a cross section of rectangular shape is defined by the following surfaces: The construction is similar to that of the toroid of round cross section as done on p.

    Inner and outer currents have the same magnitude. We can now proceed with what is requested: We obtain 2. A current filament on the z axis carries a current of 7 mA in the az direction, and current sheets of 0. Calculate H at: Here, we are either just inside or just outside the first current sheet, so both we will calculate H for both cases. We require that the total enclosed current be zero, and so the net current in the proposed cylinder at 4 cm must be negative the right hand side of the first equation in part b.

    Current density is distributed as follows: Symmetry does help significantly in this problem. As a consequence of this, we find that the net current in region 1, I1 see the diagram on the next page , is equal and opposite to the net current in region 4, I4.

    Also, I2 is equal and opposite to I3. H from all sources should completely cancel along the two vertical paths, as well as along the two horizontal paths. Assuming the height of the path is. Therefore, H will be in the opposite direction from that of the right vertical path, which is the positive x direction. We can use the results for regions 1 and 2 to construct the field everywhere: The value of H at each point is given.

    We use the approximation: Each curl component is found by integrating H over a square path that is normal to the component in question. Along each segment, the field is assumed constant, and so the integral is evaluated by summing the products of the field and segment length 4 mm over the four segments.

    The x component of the curl is thus: If so, what is its value? So, YES. The integral is: Let the surface have the ar direction: Their centers are at the origin. This problem was discovered to be flawed — I will proceed with it and show how. The reader is invited to explore this further. When evaluating the curl of G using the formula in spherical coordinates, only one of the six terms survives: Integrals over x, to complete the loop, do not exist since there is no x component of H.

    The path direction is chosen to be clockwise looking down on the xy plane. We first find the current density through the curl of the magnetic field, where three of the six terms in the spherical coordinate formula survive: A long straight non-magnetic conductor of 0.

    Here we use H outside the conductor and write: A solid nonmagnetic conductor of circular cross-section has a radius of 2mm. If the conductor is 1m in length and has a voltage of 1mV between its ends, find: We first need to find J, H, and B: The current density will be: This is part a over again, except we change the upper limit of the radial integration: This is entirely outside the current distribution, so we need B there: All surfaces must carry equal currents.

    With this requirement, we find: Use an expansion in cartesian coordinates to show that the curl of the gradient of any scalar field G is identically equal to zero. We proceed as follows: Thus, using the result of Section 8. We can then write: The solenoid shown in Fig. Since H is only in the z direction, Vm should vary with z only. Since we have parallel current sheets carrying equal and opposite currents, we use Eq.

    Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector magnetic potential is shown in Fig. Select the proper zero reference and sketch the results on the figure: By expanding Eq. Use Eq. The initial velocity in x is constant, and so no force is applied in that direction. We integrate once: Integrating a second time yields the z coordinate: Have 1 1 K. Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7.

    We begin by visualizing the problem. We can construct the differential equations for the forces in x and in y as follows: The positions are then found by integrating vx and vy over time: It establishes a negative x-directed velocity as it leaves the field, given by the acceleration times the transit time, tt: A rectangular loop of wire in free space joins points A 1, 0, 1 to B 3, 0, 1 to C 3, 0, 4 to D 1, 0, 4 to A.

    The wire carries a current of 6 mA, flowing in the az direction from B to C. A filamentary current of 15 A flows along the entire z axis in the az direction. The field from the long wire now varies with position along the loop segment. This will be the vector sum of the forces on the four sides. Note that by symmetry, the forces on sides AB and CD will be equal and opposite, and so will cancel. Find the total force on the rectangular loop shown in Fig.

    Uniform current sheets are located in free space as follows: Find the vector force per meter length exerted on a current filament carrying 7 mA in the aL direction if the filament is located at: Find the vector force on a 1-m length of the 1-mA filament and plot F versus k: The total B field arising from the two 25A filaments evaluated at the location of the 1-mA filament is, in cartesian components: Find the magnitude of the total force acting to split the outer cylinder apart along its length: We wish to find the force acting to split the outer cylinder, which means we need to evaluate the net force in one cartesian direction on one half of the cylinder.

    Since the outer cylinder is a two-dimensional current sheet, its field exists only just outside the cylinder, and so no force exists. If this cylinder possessed a finite thickness, then we would need to include its self-force, since there would be an interior field and a volume current density that would spatially overlap.

    Two infinitely-long parallel filaments each carry 50 A in the az direction. Find the force exerted on the: We first need to find the field from the current strip at the filament location. A current of 6A flows from M 2, 0, 5 to N 5, 0, 5 in a straight solid conductor in free space. An infinite current filament lies along the z axis and carries 50A in the az direction.

    Compute the vector torque on the wire segment using: The rectangular loop of Prob. Find the vector torque on the loop, referred to an origin: The fields from both current sheets, at the loop location, will be negative x-directed.

    They will add together to give, in the loop plane: This filament carries a current of 3 A in the ax direction. An infinite filament on the z axis carries 5 A in the az direction. Obtain an expression for the torque exerted on the finite conductor about an origin located at 0, 2, 0: Assume that an electron is describing a circular orbit of radius a about a positively-charged nucleus.

    The hydrogen atom described in Problem 16 is now subjected to a magnetic field having the same direction as that of the atom. What are these decreases for the hydrogen atom in parts per million for an external magnetic flux density of 0. We first write down all forces on the electron, in which we equate its coulomb force toward the nucleus to the sum of the centrifugal force and the force associated with the applied B field. With the field applied in the same direction as that of the atom, this would yield a Lorentz force that is radially outward — in the same direction as the centrifugal force.

    Finally, 1m e2 a 2 B 2. Calculate the vector torque on the square loop shown in Fig. The field is uniform and so does not produce any translation of the loop.

    We observe two things here: So we must use the given origin. Find H in a material where: Then M 0. At radii between the currents the path integral will enclose only the inner current so, 3. Find the magnitude of the magnetization in a material for which: Three current sheets are located as follows: Find B everywhere: With this in mind, we can construct the following expressions for the B field in all four regions: Compute for: Let its center lie on the z axis and let a dc current I flow in the az direction in the center conductor.

    Find a H everywhere: This result will depend on the current and not the materials, and is: Point P 2, 3, 1 lies on the planar boundary boundary separating region 1 from region 2. Find H2: The normal component of H1 will now be: The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg.

    Find B in the: We now have mmf The air gap reluctance adds to the total reluctance already calculated, where 0. The flux in the center leg is now In Problem 9. Using this value of B and the magnetization curve for silicon steel,. Using Fig. A toroidal core has a circular cross section of 4 cm2 area. The mean radius of the toroid is 6 cm.

    There is a 4mm air gap at each of the two joints, and the core is wrapped by a turn coil carrying a dc current I1. I will use the reluctance method here. The reluctance of each gap is now 0. We are not sure what to use for the permittivity of steel in this case, so we use the iterative approach. From Fig. Then, in the linear material, 1.

    This is still larger than the given value of. The result of 0. A toroid is constructed of a magnetic material having a cross-sectional area of 2. There is also a short air gap 0. Calculate the total flux in the toroid if: This is d 0.

    In this case we use the magnetization curve, Fig. We can begin with the value of 6 found in part a, assuming infinite permeability: I will leave the answer at that, considering the lack of fine resolution in Fig. Determine the total energy stored in a spherical region 1cm in radius, centered at the origin in free space, in the uniform field: First we find the energy density: This field differs from H2 only by the negative x component, which is a non-issue since the component is squared when finding the energy density.

    A toroidal core has a square cross section, 2. First, the magnetic field strength will be the same as in part a, since the calculation is material-independent.

    Three planar current sheets are located in free space as follows: The currents return on a spherical conducting surface of 0.

    We thus perform the line integral of H over a circle, centered on the z axis, and parallel to the xy plane: Find the inductance of the cone-sphere configuration described in Problem 9. The inductance is that offered at the origin between the vertices of the cone: From Problem 9. Second method: Use the energy computation of Problem 9.

    The core material has a relative permeability of If the core is wound with a coil containing turns of wire, find its inductance: The stored energy within the specified volume will be: A coaxial cable has conductor dimensions of 1 and 5 mm. Find the inductance per meter length: The interfaces between media all occur along radial lines, normal to the direction of B and H in the coax line. B is therefore continuous and constant at constant radius around a circular loop centered on the z axis.

    A rectangular coil is composed of turns of a filamentary conductor. Find the mutual inductance in free space between this coil and an infinite straight filament on the z axis if the four corners of the coil are located at a 0,1,0 , 0,3,0 , 0,3,1 , and 0,1,1: In this case the coil lies in the yz plane.

    Find the mutual inductance of this conductor system in free space: We first find the magnetic field inside the conductor, then calculate the energy stored there. It may be assumed that the magnetic field produced by I t is negligible. The magnetic flux will be: The location of the sliding bar in Fig. Find the voltmeter reading at: Have 0.

    First the flux through the loop is evaluated, where the unit normal to the loop is az. The rails in Fig. In this case, there will be a contribution to the current from the right loop, which is now closed. Develop a function of time which expresses the ohmic power being delivered to the loop: First, since the field does not vary with y, the loop motion in the y direction does not produce any time-varying flux, and so this motion is immaterial. We can evaluate the flux at the original loop position to obtain: This will be Id 0.

    Then D 1. We set the given expression for Jd equal to the result of part c to obtain: Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance Sec. This we already found during the development in part a: We have. The parallel plate transmission line shown in Fig. Neglect fields outside the dielectric.

    We evaluate the flux integral of Jd over the given cross section: Note that B as stated is constant with position, and so will have zero curl. The equation is thus not valid with these fields. Nevertheless, we press on: We use the result of part a: The procedure here is similar to the development that leads to Eq. Begin by taking the curl of both sides of the Faraday law equation: In free space, the magnetic field of the uniform plane wave can be easily found using the intrinsic impedance: First, we note that if E at a given instant points in the negative x direction, while the wave propagates in the forward y direction, then H at that same position and time must point in the positive z direction.

    We use the general formula, Eq. Using With Es in the positive y direction at a given time and propagating in the positive x direction, we would have a positive z component of Hs , at the same time.

    Find the distance a uniform plane wave can propagate through the material before: A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region.

    In a non-magnetic material, we would have: Find both the power factor and Q in terms of the loss tangent: First, the impedance will be: Then the power factor is P.

    Assume x-polarization for the electric field. At this point a flaw becomes evident in the problem statement, since solving this part in two different ways gives results that are not the same. I will demonstrate: We apply our equation to the result of part a: Note that in Problem A positive y component of E requires a posi- tive z component of H for propagation in the forward x direction. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. From part a, we have 4.

    This will be 4. The external and internal regions are non-conducting. We use J 1. Use The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respectively. The dielectric is lossless and the operating frequency is MHz.

    Calculate the resistance per meter length of the: Again, 70 applies but with a different conductor radius. A hollow tubular conductor is constructed from a type of brass having a conductivity of 1.

    The inner and outer radii are 9 mm and 10 mm respectively. Calculate the resistance per meter length at a frequency of a dc: In this case the current density is uniform over the entire tube cross-section.

    We write: Now the skin effect will limit the effective cross-section. Therefore we can approximate the resistance using the formula: Most microwave ovens operate at 2. Since the conductivity is high, we. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0. Assuming the conductor is non-magnetic, determine the frequency and the conductivity: The outer conductor thickness is 0.

    Use information from Secs. The coax is air-filled. The result is squared, terms collected, and the square root taken. For a good dielectric Teflon we use the approximations: Consider a left-circularly polarized wave in free space that propagates in the forward z direction. The electric field is given by the appropriate form of Eq. We find the two components of Hs separately, using the two components of Es. Specifically, the x component of Es is associated with a y component of Hs , and the y component of Es is associated with a negative x component of Hs.

    With the Poynting vector in the positive x direction, a positive y component for E requires a positive z component for H. Similarly, a positive z component for E requires a negative y component for H. Therefore, 10! This is most clearly seen by first converting the given field to real instantaneous form: With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the output.

    Suppose that the length of the medium of Problem Describe the polarization of the output wave in this case: With the wave propagating in the forward z direction, we find: Given the general elliptically-polarized wave as per Eq. What percentage of the incident power density is transmitted into the copper? We need to find the reflection coefficient. We nevertheless proceed: A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2.

    There are two possible answers. This wave will experience loss in region 2, along with a different phase constant. First, using Eq.

    The field in region 2 is then constructed by using the resulting amplitude, along with the attenuation and phase constants that are appropriate for region 2. Also, the intrinsic impedance We now find the input impedance: Calculate the ratio of the final power to the incident power after this round trip: Pi Try measuring that.

    A MHz uniform plane wave in normally-incident from air onto a material whose intrinsic impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of an electric field minimum at 0.

    Determine the impedance of the unknown material: A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean. First we find the loss tangent: Within the limits of our good conductor approximation loss tangent greater than about ten , the reflected power fraction, using the formula derived in part a, is found to decrease with increasing frequency.

    The transmitted power fraction thus increases. Calculate the fractions of the incident power that are reflected and trans- mitted. The total electric field in the plane of the interface must rotate in the same direction as the incident field, in order to continually satisfy the boundary condition of tangential electric field continuity across the interface.

    Therefore, the reflected wave will have to be left circularly polarized in order to make this happen. The transmitted field will be right circularly polarized as the incident field for the same reasons. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect conductor.

    Assume positive z travel for the incident electric field. This is a standing wave exhibiting circular polarization in time. Determine the standing wave ratio in front of the plate. Repeat Problem A uniform plane wave is normally incident from the left, as shown.

    Plot a curve of the standing wave ratio, s, in the region to the left: Thus, at 2. In this case we use 2. MathCad was used in both cases. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally-incident. The slabs are to be arranged such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness.

    Specify an arrangement of slabs and air spaces such that a the wave is totally transmitted through the stack: In this case, we look for a combination of half- wave sections. Let the inter-slab distances be d1 , d2 , and d3 from left to right. Two possibilities are i. Thus every thickness is one-quarter wavelength. The impedances transform as follows: The reflection coefficient for waves incident on the front slab thus gets close to unity, and approaches 1 as the number of slabs approaches infinity.

    The 50MHz plane wave of Problem Determine the fractions of the incident power that are reflected and transmitted for a s polarization: To review Problem 12, we first we find the loss tangent: Therefore, for s polarization,.

    The fraction transmitted is then 0. Again, with the refracted angle close to zero, the relection coefficient for p polar- ization is. Since the wave is circularly-polarized, the s-polarized component represents one-half the total incident wave power, and so the fraction of the total power that is reflected is.

    Since all the p-polarized com- ponent is transmitted, the reflected wave will be entirely s-polarized linear. The transmitted wave, while having all the incident p-polarized power, will have a reduced s-component, and so this wave will be right-elliptically polarized.

    A dielectric waveguide is shown in Fig. All subsequent reflections from the upper an lower boundaries will be total as well, and so the light is confined to the guide.

    Find n0 in terms of n1 and n2: A Brewster prism is designed to pass p-polarized light without any reflective loss. The prism of Fig. In the Brewster prism of Fig.

    The light is incident from air, and the returning beam also in air may be displaced sideways from the incident beam. More than one design is possible here. Using the result of Example For this to work, the Brewster angle must be greater than or equal to the critical angle.

    Using Eq. Assume conductivity does not vary with frequency: In a good conductor: Over a certain frequency range, the refractive index of a certain material varies approximately linearly. The pulse will broaden and will acquire a frequency sweep chirp that is precisely linear with time. Additionally, a pulse of a given bandwidth will broaden by the same amount, regardless of what carrier frequency is used.

    Describe the pulse at the output of the second channel and give a physical explanation for what hap- pened. In fact, we may write in general: Physically, the pulse acquires a positive linear chirp frequency increases with time over the pulse envelope during the first half of the channel.

    The pulse, if originally transform-limited at input, will emerge, again transform-limited, at its original width. We use the expression for input impedance Eq. Find and s: If the velocity on the line is 2.

    Solutions of engineering electromagnetics 6th edition william h. hayt, john a. neogosynchpromath.gq - Docsity

    I will do this using two different methods: The Hard Way: Find L, C, R, and G for the line: If the inner radius of the outer conductor is 4 mm, find the radius of the inner conductor so that assuming a lossless line: Two aluminum-clad steel conductors are used to construct a two-wire transmission line.

    The radius of the steel wire is 0. The dielectric is air, and the center-to-center wire separation is 4 in. The first question is whether we are in the high frequency or low frequency regime.

    Furthermore, the skin depth is considerably less than the aluminum layer thickness, so the bulk of the current resides in the aluminum, and we may neglect the steel. Each conductor of a two-wire transmission line has a radius of 0. Pertinent dimensions for the transmission line shown in Fig. The conductors and the dielectric are non-magnetic.

    A transmission line constructed from perfect conductors and an air dielectric is to have a maximum dimension of 8mm for its cross-section. The line is to be used at high frequencies.

    Specify its dimensions if it is: With the maximum dimension of 8mm, we have, using Find c L: Coaxial lines 1 and 2 have the following parameters: The reflection coefficient encountered by waves incident on ZL1 from line 1 can now be found, along with the standing wave ratio: The line 1 length now has a load impedance of We need to find its input impedance.

    In line 1, having a dielectric constant of 2. For the transmission line represented in Fig. A ohm transmission line is 0. The line is operating in air with a wavelength of 0. Determine the average power absorbed by each resistor in Fig.

    The next step is to determine the input impedance of the 2. The power dissipated by the ohm resistor is now 1 V 2 1 Find s on both sections 1 and 2: For section 2, we consider the propagation of one forward and one backward wave, comprising the superposition of all reflected waves from both ends of the section.

    We first need the input impedance of the.

    Engineering Electromagnetics (6th Edition, 2001) - Hayt & Buck + Solution Manual

    A lossless transmission line is 50 cm in length and operating at a frequency of MHz. We then find the input impedance to the shorted line section of length 20 cm putting this impedance at the location of ZL , so we can combine them: This problem was originally posed incorrectly.

    The corrected version should have an inductor in the input circuit instead of a capacitor. I will proceed with this replacement understood, and will change the wording as appropriate in parts c and d: Note that the dielectric is air: With the length of the line at 2. To achieve this, the imaginary part of the total impedance of part c must be reduced to zero so we need an inductor.

    Continuing, for this value of L, calculate the average power: The power delivered to the load will be the same as the power delivered to the input impedance. Use analytic methods or the Smith chart or both to find: I will first use the analytic approach. Using normalized impedances, Eq. A line drawn from the origin through this point intersects the outer chart boundary at the position 0. With a wavelength of 1. On the WTL scale, we add 0. A straight line is now drawn from the origin though the 0.

    A compass is then used to measure the distance between the origin and zin. With this distance set, the compass is then used to scribe off the same distance from the origin to the load impedance, along the line between the origin and the 0. The difference in imaginary parts arises from uncertainty in reading the chart in that region.

    This is close to the value of the VSWR, as we found earlier. Next, yL is inverted to find zL by transforming the point halfway around the chart, using the compass and a straight edge. This is close to the computed inverse of yL , which is 1. Now, the position of zL is read on the outer edge of the chart as 0.

    The point is now transformed through the line length distance of 1. The final reading on the WTG scale after the transformation is found through 0. Drawing a line between this mark on the WTG scale and the chart center, and scribing the compass arc length on this line, yields the normalized input impedance. I will specify answers in terms of wavelength. Make use of the Smith chart to find: Referring to the figure below, we start by marking the given zL on the chart and drawing a line from the origin through this point to the outer boundary.

    On the WTG scale, we read the zL location as 0. Moving from here toward the generator, we cross the positive R axis at which the impedance is purely real and greater than 1 at 0. The distance is then 0. Using a compass, we set its radius at the distance between the origin and zL. What is s on the remainder of the line? This will be just s for the line as it was before. This would return us to the original point, requiring a complete circle around the chart one-half wavelength distance.

    The distance from the resistor will therefore be: With the aid of the Smith chart, plot a curve of Zin vs. Then, using a compass, draw a circle beginning at zL and progressing clockwise to the positive real axis. On the chart, radial lines are drawn at positions corresponding to. The intersections of the lines and the circle give a total of 11 zin values.

    The table below summarizes the results. A fairly good comparison is obtained. Use the Smith chart to find ZL with the short circuit replaced by the load if the voltage readings are: We mark this on the positive real axis of the chart see next page. The load position is now 0. A line is drawn from the origin through this point on the chart, as shown.

    We then scribe this same distance along the line drawn through the. A line is drawn from the origin through this location on the chart. Use the Smith chart to find: Drawing a line from the chart center through this point yields its location at 0. Alternately, use the s scale at the bottom of the chart, setting the compass point at the center, and scribing the distance on the scale to the left. This distance is found by transforming the load impedance clockwise around the chart until the negative real axis is reached.

    This distance in wavelengths is just the load position on the WTL scale, since the starting point for this scale is the negative real axis. So the distance is 0. Transforming the load through this distance toward the generator involves revolution once around the chart 0. A line is drawn between this point and the chart center. This is plotted on the Smith chart below. We then set on the compass the distance between yL and the origin.

    The same distance is then scribed along the positive real axis, and the value of s is read as 2. First we draw a line from the origin through zL and note its intersection with the WTG scale on the chart outer boundary. We note a reading on that scale of about 0. To this we add 0. A line drawn from the 0.