Electromagnetics william hayt free download

If so what are they? At the input end of the line, a DC voltage source, V0 , is connected. In a circuit in which a sinusoidal voltage source drives its internal impedance in series with a load impedance, it is known that maximum power transfer to the load occurs when the source and load impedances form a complex conjugate pair.

The condition of maximum power transfer will be met if the input impedance to the line is the conjugate of the internal impedance.

What average power is delivered to each load resistor? First, we need the input impedance. The parallel resistors give a net load impedance of 20 ohms. 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 A lossless transmission line is 50 cm in length and operating at a frequency of MHz.

Determine s on the transmission line of Fig. To achieve this, the imaginary part of the total impedance of part c must be reduced to zero so we need an inductor. 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. This is close to the value of the VSWR, as we found earlier. Problem 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. 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.

On the WTG scale, we read the zL location as 0. The distance is then 0. 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. 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.

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. 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. 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 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. We note a reading on that scale of about 0.

To this we add 0. A line drawn from the 0. This is at the zero position on the WTL scale. The load is at the approximate 0. The wavelength on a certain lossless line is 10cm. We read the input location as slightly more than 0.

The line length of 12cm corresponds to 1. Thus, to transform to the load, we go counter-clockwise twice around the chart, plus 0. A line is drawn to the origin from that position, and the compass with its previous setting is scribed through the line. A standing wave ratio of 2.

Probe measurements locate a voltage minimum on the line whose location is marked by a small scratch on the line. When the load is replaced by a short circuit, the minima are 25 cm apart, and one minimum is located at a point 7 cm toward the source from the scratch. This is a possible location for the scratch, which would otherwise occur at multiples of a half-wavelength farther away from that point, toward the generator.

As a check, I will do the problem analytically. At the point X, indicated by the arrow in Fig. With the short circuit removed, a voltage minimum is found 5cm to the left of X, and a voltage maximum is located that is 3 times voltage of the minimum.

This point is then transformed, using the compass, to the negative real axis, which corresponds to the location of a voltage minimum. On the chart, we now move this distance from the Vmin location toward the load, using the WTL scale. A line is drawn from the origin through the 0. With a short circuit replacing the load, a minimum is found at a point on the line marked by a small spot of puce paint.

The 1m distance is therefore 3. Therefore, with the actual load installed, the Vmin position as stated would be 3. This being the case, the normalized load impedance will lie on the positive real axis of the Smith chart, and will be equal to the standing wave ratio. This point is to be transformed to a location at which the real part of the normalized admittance is unity.

The stub is connected at either of these two points. The stub input admittance must cancel the imaginary part of the line admittance at that point. This point is marked on the outer circle and occurs at 0. The length of the stub is found by computing the distance between its input, found above, and the short-circuit position stub load end , marked as Psc. The length of the main line between its load and the stub attachment point is found on the chart by measuring the distance between yL and yin2 , in moving clockwise toward generator.

This occurs at 0. The attachment point is found by transforming yL to yin1 , where the former point is located at 0. The lossless line shown in Fig. For the line to be matched, it is required that the sum of the normalized input admittances of the shorted stub and the main line at the point where the stub is connected be unity.

So the input susceptances of the two lines must cancel. This line is one-quarter wavelength long, so the normalized load impedance is equal to the normalized input admittance. To cancel the input normalized susceptance of We therefore write 2.

The two-wire lines shown in Fig. In this case, we have a series combination of the loaded line section and the shorted stub, so we use impedances and the Smith chart as an impedance diagram. The requirement for matching is that the total normalized impedance at the junction consisting of the sum of the input impedances to the stub and main loaded section is unity.

In the transmission line of Fig. First, the load voltage is found by adding voltages along the right side of the voltage diagram at the indicated times. The load voltage as a function of time is found by accumulating voltage values as they are read moving up along the right hand boundary of the chart.

In the charged line of Fig. This problem accompanies Example Plots of the voltage and current at the resistor are then found by accumulating values from the left sides of the two charts, producing the plots as shown. A simple frozen wave generator is shown in Fig.

Determine and plot the load voltage as a function of time: Closing the switches sets up a total of four voltage waves as shown in the diagram below. Note that in Problem Given, a MHz uniform plane wave in a medium known to be a good dielectric. Also, the specified distance in part f should be 10m, not 1km. We use the good dielectric approximations, Eqs. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. The external and internal regions are non-conducting.

The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respec- tively. The dielectric is lossless and the operating frequency is MHz. 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.

Most microwave ovens operate at 2. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0. The outer conductor thickness is 0.

Use information from Secs. The result is squared, terms collected, and the square root taken. Consider a left-circularly polarized wave in free space that propagates in the forward z direc- tion.

Since the wave propagates in the positive y direction and has equal x and z amplitudes, we identify the polarization as left circular. With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the out- put. Suppose that the length of the medium of Problem Given the general elliptically-polarized wave as per Eq.

What percentage of the incident power density is transmitted into the copper? Next we apply Eq. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2. There are two possible answers. Reason: the limit of the area shrinking to zero must be taken before the results will be equal. The value of H at each point is given.

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. If so, what is its value? 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. 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. A long straight non-magnetic conductor of 0. A solid nonmagnetic conductor of circular cross-section has a radius of 2mm. All surfaces must carry equal currents. Use an expansion in cartesian coordinates to show that the curl of the gradient of any scalar field G is identically equal to zero. Thus, using the result of Section 8. The solenoid shown in Fig. Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector magnetic potential is shown in Fig.

By expanding Eq. Use Eq. The initial velocity in x is constant, and so no force is applied in that direction. Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7. 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. 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. 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: a filament by the current strip: 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.

The rectangular loop of Prob. Find the vector torque on the loop, referred to an origin: a at 0,0,0 : The fields from both current sheets, at the loop location, will be negative x-directed. 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. 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. So we must use the given origin. 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: a the magnetic flux density is 0. 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: I 1. Point P 2, 3, 1 lies on the planar boundary boundary separating region 1 from region 2.

The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg. We now have mmf 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. The reluctance of each gap is now 0. 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. This is d 0. I will leave the answer at that, considering the lack of fine resolution in Fig.

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. The currents return on a spherical conducting surface of 0. 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 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: a the solenoid of Fig.

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 location of the sliding bar in Fig. The rails in Fig. 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. Then D 1. Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance Sec.

The parallel plate transmission line shown in Fig. Neglect fields outside the dielectric. The equation is thus not valid with these fields. A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region.

Then the power factor is P. Note that in Problem Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. The external and internal regions are non-conducting. 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. 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.

Most microwave ovens operate at 2. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0. 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. 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. Similarly, a positive z component for E requires a negative y component for H.

Therefore, 10! 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 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.

A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2.

There are two possible answers. 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 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. A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean.

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. Determine the standing wave ratio in front of the plate. Repeat Problem A uniform plane wave is normally incident from the left, as shown. 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 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 Therefore, for s polarization,. The fraction transmitted is then 0. 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.

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. Hayt available in Hardcover on , also read synopsis and reviews.

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