Electromagnetics for engineers solution manual pdf

Though the junkie might howl his protest all night long. Mud in the boots, mud in the trousers, mud flecking their hair, mud coating their faces. It bit into their clothes, cut into the tops of their heads, attacked their eyes so they could barely open them. In the middle of the mud fight they stopped battling each other and started battling the rain.

He shudders convulsively, sensing the prickle of an immensity up in the sky, staring down at him. A rain of shooting stars, and from this conflagration the universe opening up before him.

For an instant, Dradin can sense every throbbing artery and arrhythmic heart in the city below him - every darting quicksilver thought of hope, of pain, of hatred, of love. View step-by-step homework solutions for your homework. Ask our subject experts for help answering any of your homework questions!

He passed half way around the city to the north side, which was less well guarded than the south. He chose a place far from the north gate, and crept toward the city on his belly through the garden stuff growing in the cultivated fields.

He stopped often to look and listen, but he saw no sign of life on the city wall. A shedlike building abutted against the wall, and beyond this was a narrow street. The fumes in the room were choking him. I expected you to be more phlegmatic, even a little disinterested in the justice or the injustice of all this, instead you get whipped with your own pistol. He closed his eyes and focused on getting the right words out.

It would be password-protected, and he did not have time to hack. They included two short padlock keys and two long, old-fashioned rusting keys that might be used for a garden door.

If Di Tivoli had a place outside Rome, he could go there, lie low for a day, while the investigators concentrated on questioning Innocenzi. He switched on the massive TV and channel-surfed for a bit, familiarizing himself with the large remote control. He checked the cables at the back and saw the screen was hooked up to a small-format computer.

Steve M. Hayt, Jr. Engineering Fuel proextender manual Finally they dragged them to the doorway and slid them down the ladder one by one, letting them fall most of the way to the ground.

They tried to rise, and finally succeeded. Before he could employ any more of his undressing skills, I pulled my shirt over my head in a single sweeping motion. Recently he had purchased a love charm from Sobito.

He was wondering now if he had thrown away, uselessly, the little treasure he had paid for it. Access Free Engineering Circuit Analysis 7th Edition Hayt Solution Manual Getting the books engineering circuit analysis 7th edition hayt solution manual now is not type of inspiring means.

You could not lonely going like book store or library or borrowing from your contacts to entre them. At last I have an opportunity to thank Marcel Duret for his promotion of Haitian culture in Japan and the enjoyable and informative trips there. My thanks also to Daniel Morel for his time and his work. My deepest gratitude to the John D. Thanks lastly, to Pascalle Monnin for the art used on the book cover.

For a long time Ozone Eddy worked as a stall for a bunch of street dips in the Quarter, then upgraded as a money washer at the track, which cost him an ice pick through both kneecaps.

On his last bust, the judge took mercy on him and gave him probation, contingent on his attendance at twelve-step meetings. Fire in the sky to the northeast, the direction of Fort Bowie.

And Cochise, with over two hundred Chiricahuas, was on the warpath in that general direction. It was easy for Chee to abandon one ambushed troop for a chance to assist in the sack of a whole garrison.

And when Chee discovered his error it was too late. Kenneth Ligutom. Academic year. Hayt is an author specialised in engineering. 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. 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.

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. 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.

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. If the velocity on the line is 2. 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.

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. The reflection coefficient encountered by waves incident on ZL1 from line 1 can now be found, along with the standing wave ratio: 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.

A lossless transmission line is 50 cm in length and operating at a frequency of MHz. 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: a 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.

Use analytic methods or the Smith chart or both to find: a s; b ZL if the line is 1 m long; c the distance from the load to the nearest voltage maximum: 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. 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. 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.

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. 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.

Suppose that the scratch locates the first voltage minimum. With the short in place, the first minimum occurs at the load, and the second at 25 cm in front of the load. The effect of replacing the short with the load is to move the minimum at 25 cm to a new location 7 cm toward the load, or at 18 cm. 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. Using the Smith chart see below we first draw a line from the origin through the 0.

As a check, I will do the problem analytically. A 2-wire line, constructed of lossless wire of circular cross-section is gradually flared into a coupling loop that looks like an egg beater.

At the point X, indicated by the arrow in Fig. A probe is moved along the line and indicates that the first voltage minimum to the left of X is 16cm from X. 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. Use the Smith chart to determine: a f : No Smith chart is needed to find f , since we know that the first voltage minimum in front of a short circuit is one-half wavelength away.

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. We find the stub length by moving from Poc to the point at which the admittance is j 0. 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. To find the stub input susceptance, use the Smith chart to transform the short circuit point 0.

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. Determine and plot the voltage at the load resistor and the current in the battery as functions of time by constructing appropriate voltage and current reflection diagrams: Referring to the figure, closing the switch launches a voltage wave whose value is given by Eq.

So the voltage wave traverses the line and does not reflect. The voltage reflection diagram would be that shown in Fig. Likewise, the current reflection diagram is that of Fig. First, the load voltage is found by adding voltages along the right side of the voltage diagram at the indicated times.

Second, the current through the battery is found by adding currents along the left side of the current reflection diagram. Both plots are shown below, where currents and voltages are expressed to three significant figures. Note also that when the switch is opened, the reflection coefficient at the generator end of the line becomes unity.

The reflection diagram is now constructed in the usual manner, and is shown on the next page. 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.

The resulting function, plotted just below the reflection diagram, is found to be a sequence of pulses that alternate signs. 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.

No reflection occurs at the load end, since the load is matched to the line. The reflection diagram and load voltage plot are shown below.

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.

The reflection diagram is drawn and is used to construct the load voltage with time by accumulating voltages up the right hand vertical axis. How many modes propagate? For the first mode, we are given 2nd 0.

If the operating frequency is 32 GHz, which modes will propagate? For the guide of Problem Assume a propagation distance of 10 cm: From Problem A parallel-plate guide is partially filled with two lossless dielectrics Fig. At a certain frequency, it is found that the TM1 mode propagates through the guide without suffering any reflective loss at the dielectric interface.

The So the answer is yes. In the guide of Problem To summarize, as frequency is lowered, the ray angle in guide 1 decreases, which leads to the incident angle at the interface increasing to eventually reach and surpass the critical angle. This defines the cutoff condition in guide 2. The cutoff frequency for mode mp is, using Eq. We require that the frequency lie between the cutoff frequencies of the T E10 and T E01 modes.

We note first that f must be greater than fc01 to support both modes, but must be less than the cutoff frequency for the next higher order mode. Two rectangular waveguides are joined end-to-end. Solution: a y. E Indicates E is infinite. Sections and Coordinate Systems Problem 2. Solution: Use the coordinate variables column in Table Note that in both the cylindrical and spherical coordinates, is in Quadrant I. Note that in both the cylindrical and spherical coordinates, is arbitrary and may take any value.

Note that in both the cylindrical and spherical coordinates, is in Quadrant II. Solution: a. Also sketch the outlines of each of the surfaces. Also sketch the outline of each volume. Solution: z z. Find: a the surface area of the spherical section, b the enclosed volume. Also sketch the outline of the section. Solution: z. At point P, find: a the vector component of E perpendicular to the cylinder, b the vector component of E tangential to the cylinder. Find: a the scalar component, or projection, of B in the direction of A, b the vector component of B in the direction of A, c the vector component of B perpendicular to A.

Solution: It doesnt matter whether the vectors are evaluated before vector products are calculated, or if the vector products are directly calculated and the general results are evaluated at the specific point in question. From Eq. Solution: From Table a.

Solution: The cross product of two vectors produces a third vector which is perpendicular to both of the original vectors. Two vectors exist that satisfy the stated. Open navigation menu. Close suggestions Search Search. User Settings. Skip carousel. Carousel Previous. Carousel Next. What is Scribd? Explore Ebooks. Bestsellers Editors' Picks All Ebooks. Explore Audiobooks.

Bestsellers Editors' Picks All audiobooks. Explore Magazines. Editors' Picks All magazines.