Current Errata for Electromagnetic Waves

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Jeffery Joseph
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1 Cuent Eata fo Electomagnetic Waves Uman S. Inan and Aziz S. Inan August 8, 007 Peface Coections Page xiii: the section on Recommended Couse Content should have no section numbe. Chapte Coections Page 56: In the second line, the text infoms the eade that in typical tansmission lines e.g. a coaxial line, electical powe popagates in the dielectic egion between the conductos, typically at the velocity of light in fee space. Howeve, this discussion unfolds in Section 5. NOT Section.3. Page 94: Equation [.55] should contain an attenuation tem, as in equation [.37]: S av = ẑ [ C η c x + Cy ] cos φη e αz Page 97: The final expession in equation [.57] should ead Ex, y, z = E 0 e jβxx jβyy jβzz = [ˆxE 0x + ŷe 0y + ẑe 0z ]e jβxx jβyy jβzz Page 0: Two paagaphs into.6. Nonunifom Plane Waves, the wod panes should be spelled planes. Page 0: In section.7, the fist expession fo the unifom plane wave should not have ˆk as its diection. The diection can be abitay, but is not necessaily ˆk. Chapte 3 Coections Page 63: In the sixth line fom the bottom of the page, the expession fo λ z should ead: λ z = λ cos i 3 Page 84: In pat b. of the solution, the z-component of the electic field phaso should not be negative: E i y, z = η H 0 ŷ sin i + ẑ cos i e jβ cos i y sin i z 4

2 EE 4 - Electomagnetic Waves The following expessions [3.3] and [3.33] in the textbook ae coect, but we epint fo convenience: Γ = cos i j sin i ɛ cos i + j sin i ɛ = e jφ Γ = ɛ cos i j sin i ɛ ɛ cos i + j sin i ɛ = e jφ 5a 5b Page 9: Note that the φ and φ cuves in Figue 3.33 should intechange in the egion i > ic. The two cuves φ and π φ in Figue 3.33 must be flipped aound fo incidence angles geate than the citical angle. The sign of the phase angle φ must be positive fo an incidence angle geate than the citical angle, wheeas π φ must be negative. Page 9: Howeve, the two half-tangent equations should ead: φ sin i ɛ tan = cos i φ tan = ɛ cos i sin i ɛ 6a 6b Page 9: Because of these alteed half-tangent fomulae, equation [3.34] becomes: φ φ = φ φ tan = tan φ tan φ + tan φ tan φ = sin i cos i sin i ɛ 7 Page -- of 9

3 EE 4 - Electomagnetic Waves a Pependicula polaization b Paallel polaization Assuming that equations [3.3] and [3.33] ae veitable, we can deive the pope value of the phase change tangent: tan φ sin i ɛ = cos i = A tan φ = ɛ cos i sin i ɛ = ɛ A φ φ φ tan = tan φ = φ φ = A ɛ A + A ɛ A = A + ɛ A ɛ sin i ɛ φ cos tan = i + ɛ = sin i ɛ cos i sin i ɛ cos i ɛ cos i sin i ɛ ɛ φ sin i ɛ tan = cos i sin i ɛ ɛ φ sin i tan = cos i sin i ɛ Page 9: Theefoe, Figue 3.34 beas a diffeent appeaance. Page 93: In the solution to Example 3.4, equation [3.35] holds that φ max / Note that it is NOT φ max alone that equals Page 96: The exponential fo the tansmission coefficient T should be e j φ. Chapte 4 Coections Page 60: Equation [4.] lists m = 0 as a possible instantiation of the T E m mode, but, in actuality, no wave exists, so the equation eally should qualify m = ±, ±,... Page -3- of 9

4 EE 4 - Electomagnetic Waves Page 66: In the solution to pat b. of Example 4-, the value of v p should be 5c 5c 3, not 4. Page 69: In the cental paagaph, the expession fo popagation diection should be ˆk = β xˆx + β z ẑ β = ˆx cos im + ẑ sin im 8 Page 70: The incidence angles pictued in Figue 4.8 should be im instead of i. Page 76: Equation [4.] fails to sustain the facto of in its denominato. Its coection: α c = Powe lost pe unit length Powe tansmitted = P loss P av 9 This alteation will not affect any of the subsequent equations fo specific α c because equations [4.] and [4.4] aleady conside the dividend of. Page 90: In equations [4.39] fo Fee space below the slab, the exponentials should not be negatively signed. Each exponential should instead ead, fo x d/:...e +αxx+d/ Page 90: Futhemoe, seveal of the equations [4.39] lack a negative sign. We epint coect enditions of all nine equations hee fo convenience: Fo fee space above the slab x d/: E 0 z x = [ Exx 0 = j β [ α x H 0 y x = jωɛ 0 α x Fo the dielectic egion x d/: [ ] e αxx d/ ] e αxx d/ ] e αxx d/ 0a 0b 0c E 0 z x = β x x E 0 xx = j β β x C 0 cos β x x a b H 0 y x = jωɛ d β x C 0 cos β x x c Fo fee space below the slab x d/: [ ] Ez 0 x = e αxx+d/ Exx 0 = j β [ α x H 0 y x = jωɛ 0 α x [ ] e αxx+d/ ] e αxx+d/ a b c Page -4- of 9

5 EE 4 - Electomagnetic Waves Page 94: In the solutions to Example 4-6, cetain numeical answes equie evision: T M : β m 8.3 ad/cm, β m β 0.3 T M : β m 6.66 ad/cm, β m β 0.06 Page 96: In Figue 4.8, the H-field lines ae othogonal to the popagation diection, so the mode is TM, as witten. Howeve, the E z component, deived fom equation [4.39], coected above, indicates that the mode is, contay to the caption, the odd TM mode. Page 300: In equation [4.5], the pemittivities in the last equality should be ecipocated: α x d [ ] [ ɛ 0 ω ɛd µ d µd µ 0 ɛ 0 = πβ ɛ ] 0 d 3 ɛ d ɛ 0 µ 0 µ 0 ɛ d λ Page 30: In the solution to pat b. Example 4-8, the second line s β is missing an exponent: αx + βx = β n d n c 4 The answes also equie slight evision: Page 36: In the fifth line fom the top, d β dω β x ad/µm α x 4.07 np/µm β.703 ad/µm should be dω d β. Pages 34-35: Poblem 4-4 is ill-defined. Because the tansvese field E y is cosinusoidal, we know that the field H z is sinusoidal, evincing the odd TE mode. Howeve, computing βxd = 5, which lies between 3π and π, we begin to suspect that ou wave does not popagate in the odd TE mode. Accoding to Figue 4.9 on page 97, the intesections that occu between 3π and π belong to the even T E 4 mode. Similaly, when we compute α x = β x tan βxd = 6, , we obtain an inconsistent negative numbe. Thus, one of the paametes in the poblem is inconsistent with the given odd TE field. The slab thickness d = mm needs e-specification. Chapte 5 Coections Page 346: The tue uppe bound fo the dimension b is b.874 cm. Page 350: The second mathematical expession, concening [P loss ] y=0 leads to an eoneous ightmost expession, which should be a [P loss ] y=0 =... = C R s 4 + β 0 a3 4π 6 Page 359: In equation [5.48], the expession fo E φ should not have a leading negative sign: E φ = jaωµ C n J snl n s nl a cos nφ e j β nl z Page -5- of 9 7

6 EE 4 - Electomagnetic Waves Page 369: In the line afte the expession fo f, the expession fo h is incoect, since h = γ + ω µɛ. Page 380: In the unlabeled equation diectly above equation [5.76], all instantiations of z should be d due to substitution. Page 38: In the lowe-ight-hand diagam of Figue 5.0, b should be d. Page 390: In the solution to pat a. of Example 5-9, the T M mode has a lowe esonance fequency than the T M 0 mode, so f.57 GHz should supplant f 0.45 GHz. Page 39: The othe nonzeo field components listed in equation [5.90] can be deived though substitution into equations [5.3] NOT equations [5.6]. Page 396: The solution fo Example 5- assumes that d = a, but this assumption is spuious since the equation fo Q holds only fo d = λ T Enl. The poblem statement should eithe specify d = a, o the solution should compute d = λ T Enl. Chapte 6 Coections Page 48: The label above equation [6.4] should be Centifugal athe than Centifigal. Page 43: The fist integated equation lacks constant C : x = E 0 t + v 0 E 0 cos ω c t + C 8 B 0 ω c B 0 The value of C is zeo, but its epesentation should still appea in the equation. Page 43: At the bottom ight cone of the page, the solution to Example 6-5 efes to expessions [6.6], but the equations officially numbeed [6.6] have not yet been intoduced. The equations employed actually oiginate fom unnumbeed expessions fo ṽ x t and ṽ y t on page 430. Pages : The deivation fo maximal magnetic field has some flaws. The coect deivation follows: ˆ = cos ˆx + sin ŷ ˆ t = sin ˆx + t t cos ŷ ˆ t = ωˆ ˆ = sin ˆx + cos ŷ ˆ t = cos ˆx sin ŷ t t ˆ t = ωˆ v = v ˆ + v ˆ Page -6- of 9

7 EE 4 - Electomagnetic Waves v t = v t ωv v ˆ + t + ωv ˆ v t = v t v Letting a epesent the paticle acceleation... ˆ + v t + v v ˆ a = v t = F m = ω c v + A ˆ + ω c v ˆ Juxtaposing coefficients of v t and F m in ˆ, we obtain the equation: v t + v v = ω c v v t = v ω c v v t = v ω c v v t = v ω c v v v = v ω c v v = v = ω c v v + v = ω c v = e d ω c d + C e d ωc d + C v = ω c + C We know that v = a = 0, so... v = ω c + C v = a = aω c + C a = 0 C = a ω c Page -7- of 9

8 EE 4 - Electomagnetic Waves At the oute conducto edge, vb, = v ˆ: v = ω c a ω c Substituting ω c = qeb 0 m e v = b = bω c a ω c b m ev = [ ] b m eωc 4 a + a4 4b m ev = m eωc [ b 4 4b a b + a 4] m ev = m eωc [ b 4b a ] q e V 0 = m eω c into the expession above, 4b b a q e V 0 = q eb 0 b m e m e 4b a q eb 0 m e B 0 4b b a = q e V 0 m e b 4b a m e V 0 = B 0 = 8b m e V 0 q e b a q e This equation supplants the boxed equation on page 434. B 0 = b 8m e V 0 /q e b a 9 Page 443: In the seventh line of text below the timeline, the sentence should ead: Most applications of plasma physics ae concened with ionized gases. Page 47: Equation [6.55b] is mislabeled as [.55b]. Page 490: The second β in equation [6.76b] should be β 3 : Idl sin S = 4π ηβ [ j ] β Page 494: The poblem statement fo Example 6-7 should ave that λ = 0 m. Page -8- of 9

9 EE 4 - Electomagnetic Waves Page 498: The last line in equation [6.86] fails to eadicate the multiplie and the squae exponent in the denominato sine: E = j60i [ ] 0 cos βl cos cos βl e jβ sin It appeas that equation [6.87] coectly omits these supefluous factos. Appendix Coections Page App-37: The answes to Poblem 4.33a. should compise multiples of 666 MHz because, fo a dielectic on a gound plane, the thickness of the dielectic should be d/, making d =.5 cm in this poblem. Substitute into equation [4.50]. Page -9- of 9

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007 School of Electical and Compute Engineeing, Conell Univesity ECE 303: Electomagnetic Fields and Waves Fall 007 Homewok 8 Due on Oct. 19, 007 by 5:00 PM Reading Assignments: i) Review the lectue notes.