10 (4π 10 7 ) 2σ 2( = (1 + j)(.0104) = = j.0001 η c + η j.0104

Transcription

1 CHAPTER A uniform plane wave in air, E + x1 E+ x10 cos(1010 t βz)v/m, is normally-incident on a copper surface at z 0. What percentage of the incident power density is transmitted into the copper? We need to find the reflection coefficient. The intrinsic impedance of copper (a good conductor) is η c jωµ σ ωµ 10 (1 + j) (1 + j) 10 (4π 10 7 ) σ ( (1 + j)(.0104) ) Note that the accuracy here is questionable, since we know the conductivity to only two significant figures. We nevertheless proceed: Using η ohms, we write Ɣ η c η j j.0001 η c + η j.0104 Now Ɣ.9999, and so the transmitted power fraction is 1 Ɣ.0001, or about 0.01% is transmitted. 1.. The plane y 0 defines the boundary between two different dielectrics. For y<0, ɛ R1 1, µ 1 µ 0, and ɛ R1 0; and for y>0, ɛ R 5, µ µ 0, and ɛ R 0. Let E+ z1 150 cos(ωt 8y) V/m, and find a) ω: Haveβ 8 ω/c ω 8c sec 1. b) H 1 + : With E in the z direction, and propagation in the forward y direction, H will lie in the positive x direction, and its amplitude will be H x E y /η 0 in region 1. Thus H 1 + (150/η 0) cos(ωt 8y)a x 0.40 cos( t 8y)a x A/m. c) H1 : First, Ez1 ƔE+ z1 η 0/ 5 η 0 /1 η 0 / 5 + η 0 / E+ z1 0.38E+ z1 Then Hx1 +(0.38/η 0)E z (150) cos(ωt + 8y) 377 So finally, Hx cos( t + 8y)a x A/m A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and. If ɛ 1 ɛ 0, while ɛ R1 µ3 R1 and ɛ R µ3 R, find the ratio ɛ R /ɛ R1 if 0% of the energy in the incident wave is reflected at the boundary. There are two possible answers. First, since Ɣ.0, and since both permittivities and permeabilities are real, Ɣ ± we then set up Ɣ ±0.447 η η 1 η + η 1 η 0 (µ R /ɛ R ) η 0 (µ R1 /ɛ R1 ) (µ R1 /ɛ R1 ) η 0 (µ R /ɛ R ) + η 0 (µ R /µ 3 R ) (µ R1 /µ 3 R1 (µ ) µ R1 µ R R /µ 3 R ) + (µ R1 /µ 3 R1 ) µ R1 + µ R 00

2 1.3. (continued) Therefore µ R µ R1 1 ± (0.38,.6) ɛ R ɛ R1 ( µr µ R1 ) 3 (0.056, 17.9) 1.4. The magnetic field intensity in a region where ɛ 0 is given as H 5 cos ωt cos βza y A/m, where ω 5 Grad/s and β 30 rad/m. If the amplitude of the associated electric field intensity is kv/m, find a) µ and ɛ for the medium: In phasor form, the magnetic field is H ys H 0 e jβz + H 0 e +βz 5 cos βz H 0.5. The electric field will be x directed, and is E xs η(.5)e jβz η(.5)e +jβz (j)η(.5) sin βz. Given the electric field amplitude of kv/m, we write η, orη 400. Nowη 400 η 0 µ r /ɛ R µ and we also have β 30 (ω/c) R ɛ R.We solve these two equations simultaneously for µ R and ɛ R to find µ R 1.91 and ɛ R Therefore µ π µh/m and ɛ pf/m. b) E: From part a, electric field in phasor form is E xs j sin βz kv/m, and so, in real form: E(z, t) Re(E xs e jωt )a x sin βzsin ωt a x kv/m with ω and β as given The region z<0is characterized by ɛ R µ R 1 and ɛ R 0. The total E field here is given as the sum of the two uniform plane waves, E s 150e j10z a x + (50 0 )e j10z a x V/m. a) What is the operating frequency? In free space, β k 0 10 ω/c ω/ Thus, ω s 1,orf ω/π Hz. b) Specify the intrinsic impedance of the region z>0 that would provide the appropriate reflected wave: Use Now Ɣ E r 50ej0 E inc ej j0.11 η η 0 η + η 0 ( ) ( ) 1 + Ɣ j0.11 η η j177 1 Ɣ j0.31 c) At what value of z( 10 cm <z<0) is the total electric field intensity a maximum amplitude? We found the phase of the reflection coefficient to be φ 0.349rad, and we use z max φ β m 1.7cm Region 1, z<0, and region, z>0, are described by the following parameters: ɛ pf/m, µ 1 5 µh/m, ɛ 1 0, ɛ 00 pf/m, µ 50 µh/m, and ɛ /ɛ 0.5. If E e α1z cos( t β 1 z)a x V/m, find: a) α 1 : From Eq. (35), Chapter 11, we note that since ɛ 1 0, it follows that α 1 0. b) β 1 : β 1 ω µ 1 ɛ 1 ( ) ( )( ) rad/m. c) E s e j z a x V/m. d) Es1 : To find this, we need to evaluate the reflection coefficient, which means that we first need the two intrinsic impedances. First, η 1 µ 1 /ɛ 1 ( )/( )

3 1.6d) (continued) Next, using Eq. (39), Chapter 11, Then η µ ɛ 1 1 j(ɛ /ɛ ) j109 1 j0.5 Ɣ η η j η + η j j e j104 Now we multiply E s1 + by Ɣ and reverse the propagation direction to obtain E s1 71.8ej104 e j.5 103z V/m e) E s + : This wave will experience loss in region, along with a different phase constant. We need to evaluate α and β. First, using Eq. (35), Chapter 11, α ω µ ɛ ( ) 1 + ( ɛ ɛ Then, using Eq. (36), Chapter 11, β ω ) 1 1/ ( )( ) µ ɛ 1 + Then, the transmission coefficient will be ( ɛ ɛ [ ] 1/ 1 + (0.5) Np/m ) + 1 1/ rad/m τ 1 + Ɣ j e j7 The complex amplitude of E s + is then found by multiplying the amplitude of E+ s1 by τ. The field in region is then constructed by using the resulting amplitude, along with the attenuation and phase constants that are appropriate for region. The result is E + s 587e z e j7 e j z V/m 1.7. The semi-infinite regions z<0and z>1mare free space. For 0 <z<1m,ɛ R 4, µ R 1, and ɛ R 0. A uniform plane wave with ω rad/s is travelling in the a z direction toward the interface at z 0. a) Find the standing wave ratio in each of the three regions: First we find the phase constant in the middle region, β ω ɛ R c (4 108 ) rad/m 0

4 1.7a. (continued) Then, with the middle layer thickness of 1 m, β d.67 rad. Also, the intrinsic impedance of the middle layer is η η 0 / ɛ R η 0/. We now find the input impedance: [ ] η0 cos(β d) + jη sin(β d) η in η 377 η cos(β d) + jη 0 sin(β d) [ ] cos(.67) + j sin(.67) 31 + j141 cos(.67) + j sin(.67) Now, at the first interface, Ɣ 1 η in η j η in + η j j The standing wave ratio measured in region 1 is thus s Ɣ 1 1 Ɣ In region the standing wave ratio is found by considering the reflection coefficient for waves incident from region on the second interface: Then Ɣ 3 η 0 η 0 / η 0 + η 0 / 1 1/ 1 + 1/ 1 3 s 1 + 1/3 1 1/3 Finally, s 3 1, since no reflected waves exist in region 3. b) Find the location of the maximum E for z<0 that is nearest to z 0. We note that the phase of Ɣ 1 is φ rad. Thus z max φ β.15 (4/3).81 m 1.8. A wave starts at point a, propagates 100m through a lossy dielectric for which α 0.5 Np/m, reflects at normal incidence at a boundary at which Ɣ j0.4, and then returns to point a. Calculate the ratio of the final power to the incident power after this round trip: Final power, P f, and incident power, P i, are related through Try measuring that. P f P i e αl Ɣ e αl P f P i j0.4 e (0.5) (!) 1.9. Region 1, z<0, and region, z>0, are both perfect dielectrics (µ µ 0, ɛ 0). A uniform plane wave traveling in the a z direction has a radian frequency of rad/s. Its wavelengths in the two regions are λ 1 5cmandλ 3 cm. What percentage of the energy incident on the boundary is a) reflected; We first note that ( ) πc ( ) πc ɛ R1 and ɛ R λ 1 ω λ ω 03

5 1.9a. (continued) Therefore ɛ R1 /ɛ R (λ /λ 1 ). Then with µ µ 0 in both regions, we find Ɣ η η 1 η 0 1/ɛ R η 0 1/ɛ R1 ɛ R1 /ɛ R 1 η + η 1 η 0 1/ɛ (λ /λ 1 ) 1 R + η 0 1/ɛ R1 ɛ R1 /ɛ R + 1 (λ /λ 1 ) + 1 λ λ λ + λ The fraction of the incident energy that is reflected is then Ɣ 1/ b) transmitted? We use part a and find the transmitted fraction to be 1 Ɣ 15/ c) What is the standing wave ratio in region 1? Use s 1 + Ɣ 1 Ɣ 1 + 1/4 1 1/ In Fig. 1.1, let region be free space, while µ R1 1, ɛ R1 0, and ɛ R1 is unknown. Find ɛ R! if a) the amplitude of E1 is one-half that of E+ 1 : Since region is free space, the reflection coefficient is Ɣ E 1 E 1 + η 0 η 1 η 0 η 0 / ɛ R1 ɛ R1 1 η 0 + η 1 η 0 + η 0 / ɛ R1 ɛ 1 R1 + 1 ɛ R1 9. b) P1,avg is one-half of P 1,avg + : This time Ɣ ɛ R1 1 ɛ R1 + 1 c) E 1 min is one-half E 1 max : Use 1 ɛ R1 34 E 1 max E 1 min s 1 + Ɣ 1 Ɣ Ɣ Ɣ 1 3 ɛ R1 1 ɛ R1 + 1 ɛ R A 150 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 0.3 wavelengths in front of the interface. Determine the impedance of the unknown material: First, the field minimum is used to find the phase of the reflection coefficient, where z min 1 (φ + π) 0.3λ β φ 0.π where β π/λ has been used. Next, Ɣ s 1 s

6 1.11. (continued) So we now have We solve for η u to find Ɣ 0.5e j0.π η u η 0 η u + η 0 η u η 0 ( j1.33) j A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean. For seawater, σ 4 S/m, and ɛ R 78. a) Determine the fractions of the incident power that are reflected and transmitted: First we find the loss tangent: σ ωɛ 4 π( )(78)( ) 18.4 This value is sufficiently greater than 1 to enable seawater to be considered a good conductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 11), the intrinsic impedance is η s πf µ/σ (1 + j), and the reflection coefficient becomes Ɣ πf µ/σ (1 + j) η0 πf µ/σ (1 + j)+ η0 where πf µ/σ π( )(4π 10 7 )/ The fraction of the power reflected is P r Ɣ [ πf µ/σ η0 ] + πf µ/σ P i [ πf µ/σ + η 0 ] + πf µ/σ [ ] [ ] The transmitted fraction is then P t P i 1 Ɣ b) Qualitatively, how will these answers change (if at all) as the frequency is increased? 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 A right-circularly-polarized plane wave is normally incident from air onto a semi-infinite slab of plexiglas (ɛ R 3.45, ɛ R 0). Calculate the fractions of the incident power that are reflected and transmitted. Also, describe the polarizations of the reflected and transmitted waves. First, the impedance of the plexiglas will be η η 0 / Then Ɣ The reflected power fraction is thus Ɣ 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 power fraction is now 1 Ɣ The transmitted field will be right circularly polarized (as the incident field) for the same reasons. 05

7 1.14. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect conductor. a) Construct the superposition of the incident and reflected waves in phasor form: Assume positive z travel for the incident electric field. Then, with reflection coefficient, Ɣ 1, the incident and reflected fields will add to give the total field: E tot E i + E r E 0 (a x + ja y )e jβz E 0 (a x + ja y )e +jβz E 0 (e jβz e jβz) a x + j (e jβz e jβz) a y E 0 sin(βz) [ ] a y ja x } {{ } j sin(βz) } {{ } j sin(βz) b) Determine the real instantaneous form of the result of part a: E(z, t) Re {E tot e jωt} E 0 sin(βz) [ ] cos(ωt)a y + sin(ωt)a x c) Describe the wave that is formed: This is a standing wave exhibiting circular polarization in time. At each location along the z axis, the field vector rotates clockwise in the xy plane, and has amplitude (constant with time) given by E 0 sin(βz) Consider these regions in which ɛ 0: region 1, z<0, µ 1 4 µh/m and ɛ 1 10 pf/m; region, 0 <z<6cm, µ µh/m, ɛ 5 pf/m; region 3, z>6cm, µ 3 µ 1 and ɛ 3 ɛ 1. a) What is the lowest frequency at which a uniform plane wave incident from region 1 onto the boundary at z 0 will have no reflection? This frequency gives the condition β d π, where d 6 cm, and β ω µ ɛ Therefore π β d π ω (.06) 1 µ ɛ f GHz ( 10 6 )( ) b) If f 50 MHz, what will the standing wave ratio be in region 1? At the given frequency, β (π ) ( 10 6 )( ). rad/m. Thus β d.(.06) The intrinsic impedance of regions 1 and 3 is η 1 η 3 ( )/(10 11 ) 63. The input impedance at the first interface is now [ ] 63 cos(.133) + j83 sin(.133) η in j cos(.133) + j 63 sin(.133) The reflection coefficient is now The standing wave ratio is now Ɣ η in η 1 η in + η j j s 1 + Ɣ 1 Ɣ A uniform plane wave in air is normally-incident onto a lossless dielectric plate of thickness λ/8, and of intrinsic impedance η 60. Determine the standing wave ratio in front of the plate. Also find the fraction of the incident power that is transmitted to the other side of the plate: With the a thickness of λ/8, we have βd π/4, and so cos(βd) sin(βd) 1. The input impedance thus becomes [ ] j60 η in j j377 06

8 1.16. (continued) The reflection coefficient is then Ɣ (43 j9) 377 (43 j9) j rad Therefore s and 1 Ɣ 1 (.6) Repeat Problem 1.16 for the cases in which the frequency is a) doubled: If this is true, then d λ/4, and thus η in (60) / The reflection coefficient becomes Ɣ 0.36 s Then 1 Ɣ 1 (.36) b) quadrupled: Now, d λ/, and so we have a half-wave section surrounded by air. Transmission will be total, and so s 1 and 1 Ɣ In Fig. 1.6, let η 1 η 3 377, and η 0.4η 1. 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: a) as a function of l if f.5ghz: With η 1 η 3 η 0 and with η 0.4η 0, Eq. (41) becomes [ cos(βl) + j0.4 sin(βl) η in 0.4η cos(βl) + j sin(βl) [ ] 1 j1.05 sin(βl) η 0 cos (βl) sin (βl) Then Ɣ (η in η 0 )/(η in + η 0 ), from which we find Ɣ ƔƔ ] [ ] 0.4 cos(βl) j sin(βl) 0.4 cos(βl) j sin(βl) [[ 1 cos (βl) 6.5 sin (βl) ] + (1.05) sin (βl) [ 1 + cos (βl) sin (βl) ] + (1.05) sin (βl) ] 1/ Then s (1 + Ɣ )/(1 Ɣ ). Now for a uniform plane wave, β ω µɛ nω/c. Given that η 0.4η 0 η 0 /n,wefindn.5 (assuming µ µ 0 ). Thus, at.5 GHz, βl nω c l (.5)(π)( ) l 1.95 l(lin m) l(lin cm) Using this in the expression for Ɣ, and calculating s as a function of l in cm leads to the first plot shown on the next page. b) as a function of frequency if l cm. In this case we use βl (.5)(π)(0.0) f f(fin Hz) f(fin GHz) Using this in the expression for Ɣ, and calculating s as a function of f in GHz leads to the second plot shown on the next page. MathCad was used in both cases. 07

9 1.18 (continued) Plots for parts a and b You are given four slabs of lossless dielectric, all with the same intrinsic impedance, η, known to be different from that of free space. The thickness of each slab is λ/4, where λ is the wavelength as measured in the slab material. 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 halfwave sections. Let the inter-slab distances be d 1, d, and d 3 (from left to right). Two possibilities are i.) d 1 d d 3 0, thus creating a single section of thickness λ, or ii.) d 1 d 3 0, d λ/, thus yielding two half-wave sections separated by a half-wavelength. b) the stack presents the highest reflectivity to the incident wave: The best choice here is to make d 1 d d 3 λ/4. Thus every thickness is one-quarter wavelength. The impedances transform as follows: First, the input impedance at the front surface of the last slab (slab 4) is η in,1 η /η 0. We transform this back to the back surface of slab 3, moving through a distance of λ/4 in free space: η in, η0 /η in,1 η0 3/η. We next transform this impedance to the front surface of slab 3, producing η in,3 η /η in, η 4 /η0 3. We continue in this manner until reaching the front surface of slab 1, where we find η in,7 η 8 /η0 7. Assuming η<η 0, the ratio η n /η0 n 1 becomes smaller as n increases (as the number of slabs increases). 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 1.1 is incident onto the ocean surface at an angle to the normal of 60. Determine the fractions of the incident power that are reflected and transmitted for a) s polarization: To review Problem 1, we first we find the loss tangent: σ ωɛ 4 π( )(78)( ) 18.4 This value is sufficiently greater than 1 to enable seawater to be considered a good conductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 11), and with µ µ 0, the intrinsic impedance is η s πf µ/σ (1 + j) 7.0(1 + j). 08

10 1.0a. (continued) Next we need the angle of refraction, which means that we need to know the refractive index of seawater at 50MHz. For a uniform plane wave in a good conductor, the phase constant is β n sea ω c.. µσ πf µσ n sea c 4πf 6.8 Then, using Snell s law, the angle of refraction is found: sin θ n sea n 1 sin θ sin(60 ) θ 1.9 This angle is small enough so that cos θ. 1. Therefore, for s polarization, Ɣ s. η s η s1 7.0(1 + j) 377/ cos 60 η s + η s1 7.0(1 + j)+ 377/ cos j The fraction of the power reflected is now Ɣ s The fraction transmitted is then b) p polarization: Again, with the refracted angle close to zero, the relection coefficient for p polarization is Ɣ p. η p η p1 7.0(1 + j) 377 cos 60 η p + η p1 7.0(1 + j)+ 377 cos j The fraction of the power reflected is now Ɣ p The fraction transmitted is then A right-circularly polarized plane wave in air is incident at Brewster s angle onto a semi-infinite slab of plexiglas (ɛ R 3.45, ɛ R 0, µ µ 0). a) Determine the fractions of the incident power that are reflected and transmitted: In plexiglas, Brewster s angle is θ B θ 1 tan 1 (ɛ R /ɛ R1 ) tan 1 ( 3.45) Then the angle of refraction is θ 90 θ B (see Example 1.9), or θ 8.3. With incidence at Brewster s angle, all p-polarized power will be transmitted only s-polarized power will be reflected. This is found through Ɣ s η s η 1s.614η 0.11η η s + η 1s.614η η 0 where η 1s η 1 sec θ 1 η 0 sec(61.7 ).11η 0, and η s η sec θ (η 0 / 3.45) sec(8.3 ) 0.614η 0. Now, the reflected power fraction is Ɣ (.549).30. 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.30/ 0.15, or 15%. The fraction of the incident power that is transmitted is then the remainder, or 85%. b) Describe the polarizations of the reflected and transmitted waves: Since all the p-polarized component 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. 09

11 1.. A dielectric waveguide is shown in Fig with refractive indices as labeled. Incident light enters the guide at angle φ from the front surface normal as shown. Once inside, the light totally reflects at the upper n 1 n interface, where n 1 >n. All subsequent reflections from the upper an lower boundaries will be total as well, and so the light is confined to the guide. Express, in terms of n 1 and n, the maximum value of φ such that total confinement will occur, with n 0 1. The quantity sin φ is known as the numerical aperture of the guide. From the illustration we see that φ 1 maximizes when θ 1 is at its minimum value. This minimum will be the critical angle for the n 1 n interface, where sin θ c sin θ 1 n /n 1. Let the refracted angle to the right of the vertical interface (not shown) be φ, where n 0 sin φ 1 n 1 sin φ. Then we see that φ + θ 1 90, and so sin θ 1 cos φ. Now, the numerical aperture becomes sin φ 1max n 1 sin φ n 1 cos θ 1 n 1 1 sin θ 1 n 1 1 (n /n 1 ) n n 1 n 0 Finally, φ 1max sin 1 (n 1 n ) is the numerical aperture angle Suppose that φ 1 in Fig is Brewster s angle, and that θ 1 is the critical angle. Find n 0 in terms of n 1 and n : With the incoming ray at Brewster s angle, the refracted angle of this ray (measured from the inside normal to the front surface) will be 90 φ 1. Therefore, φ 1 θ 1, and thus sin φ 1 sin θ 1. Thus sin φ 1 n 1 n 0 + n 1 sin θ 1 n n 0 (n 1 /n ) n 1 n 1 n Alternatively, we could have used the result of Problem 1., in which it was found that sin φ 1 (1/n 0 ) n 1 n, which we then set equal to sin θ 1 n /n 1 to get the same result A Brewster prism is designed to pass p-polarized light without any reflective loss. The prism of Fig is made of glass (n 1.45), and is in air. Considering the light path shown, determine the apex angle, α: With entrance and exit rays at Brewster s angle (to eliminate reflective loss), the interior ray must be horizontal, or parallel to the bottom surface of the prism. From the geometry, the angle between the interior ray and the normal to the prism surfaces that it intersects is α/. Since this angle is also Brewster s angle, we may write: ( ) ( ) α sin 1 1 sin rad n 1 + (1.45) 1.5. In the Brewster prism of Fig. 1.19, determine for s-polarized light the fraction of the incident power that is transmitted through the prism: We use Ɣ s (η s η s1 )/(η s + η s1 ), where η s η cos(θ B ) η n/ 1 + n η 0 n 1 + n and η s1 η 1 cos(θ B1 ) η 1 1/ 1 + n η n 10

12 1.5. (continued) Thus, at the first interface, Ɣ (1 n )/(1+n ). At the second interface, Ɣ will be equal but of opposite sign to the above value. The power transmission coefficient through each interface is 1 Ɣ, so that for both interfaces, we have, with n 1.45: P ( tr 1 Ɣ ) [1 P inc ( n ) ] 1 n Show how a single block of glass can be used to turn a p-polarized beam of iight through 180, with the light suffering, in principle, zero reflective loss. The light is incident from air, and the returning beam (also in air) may be displaced sideways from the incident beam. Specify all pertinent angles and use n 1.45 for glass. More than one design is possible here. The prism below is designed such that light enters at Brewster s angle, and once inside, is turned around using total reflection. Using the result of Example 1.9, we find that with glass, θ B 55.4, which, by the geometry, is also the incident angle for total reflection at the back of the prism. For this to work, the Brewster angle must be greater than or equal to the critical angle. This is in fact the case, since θ c sin 1 (n /n 1 ) sin 1 (1/1.45) Using Eq. (59) in Chapter 11 as a starting point, determine the ratio of the group and phase velocities of an electromagnetic wave in a good conductor. Assume conductivity does not vary with frequency: In a good conductor: β πf µσ ωµσ dβ dω 1 [ ωµσ ] 1/ µσ Thus ( ) dω dβ 1 dβ ω dω µσ v g and v p ω β ω ω ωµσ/ µσ Therefore v g /v p. 11

13 1.8. Over a certain frequency range, the refractive index of a certain material varies approximately linearly with frequency: n(ω). n a + n b (ω ω a ), where n a, n b, and ω a are constants. Using β nω/c: a) determine the group velocity as a function (or perhaps not a function) of frequency: v g (dβ/dω) 1, where dβ dω d [ na ω + n ] b(ω ω a )ω 1 dω c c c [n a + n b (ω ω a )] so that b) determine the group dispersion parameter, β : v g (ω) c [n a + n b (ω ω a )] 1 β d β ω0 dω d 1 dω c [n a + n b (ω ω a )] n b /c ω0 c) Discuss the implications of these results, if any, on pulse broadening: The point of this problem was to show that higher order terms (involving d 3 β/dω 3 and higher) in the Taylor series expansion, Eq. (89), do not exist if the refractive index varies linearly with ω. These higher order terms would be necessary in cases involving pulses of exremely large bandwidth, or in media exhibiting complicated variations in their ω-β curves over relatively small frequency ranges. With d β/dω constant, the three-term Taylor expansion of Eq. (89) describes the phase constant of this medium exactly. 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 A T 5 ps transform-limited pulse propagates in a dispersive channel for which β 10 ps /km. Over what distance will the pulse spread to twice its initial width? After propagation, the width is T T + ( τ) T. Thus τ 3T, where τ β z/t. Therefore β z T 3T 3T or z β 3(5ps) 10 ps /km 4.3 km A T 0 ps transform-limited pulse propagates through 10 km of a dispersive channel for which β 1 ps /km. The pulse then propagates through a second 10 km channel for which β 1 ps /km. Describe the pulse at the output of the second channel and give a physical explanation for what happened. Our theory of pulse spreading will allow for changes in β down the length of the channel. In fact, we may write in general: τ 1 L β (z) dz T 0 Having β change sign at the midpoint, yields a zero τ, and so the pulse emerges from the output unchanged! Physically, the pulse acquires a positive linear chirp (frequency increases with time over the pulse envelope) during the first half of the channel. When β switches sign, the pulse begins to acquire a negative chirp in the second half, which, over an equal distance, will completely eliminate the chirp acquired during the first half. The pulse, if originally transform-limited at input, will emerge, again transform-limited, at its original width. More generally, complete dispersion compensation is achieved using a two-segment channel when β L β L, assuming dispersion terms of higher order than β do not exist. 1