cos kd kd 2 cosθ = π 2 ± nπ d λ cosθ = 1 2 ± n N db

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Augustus Griffith
5 years ago
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1 . (Balanis 6.43) You can confim tat AF = e j kd cosθ + e j kd cosθ N = cos kd cosθ gives te same esult as (6-59) and (6-6), fo a binomial aay wit te coefficients cosen as in section Tis single expession woks fo bot odd and even numbes of elements, N, and is a closed fom vesion witout a summation. N Te nulls occu wen cos kd cosθ N = cos kd cosθ = kd cosθ = π ± nπ d λ cosθ = ± n wee n is an intege. Fom tis one can see tat only fo d = λ/ will you ave nulls only at θ = and π. Tus te spacing must be λ/ to satisfy te design equiements. Fom te binomial seies o by expanding te expession above, te amplitude excitation coefficients ae z amplitude -λ -λ/ λ/ λ Since te expessions (6-64) toug (6-65b) ae valid fo ou spacing we can use tem. HPBW deg N D max.77 N db O, you can wok it out fom te aay facto above, fo example te HPBW is cos kd ( N ) cosθ = cos kd cos π ( N ) cos kλ ( N ) 4 cosθ = ( N ) cos π cosθ = / N π cosθ = cos / N θ = cos π cos / N HPBW = π cos π cos / N wic gives essentially te same answe.

2 . (Balanis.5) Fom te figue, te incident electic field is In te apetue, wee x = E = E i ẑe jk = y ŷ + z ẑ k = k cosφ ˆx sinφ ŷ E = E i ẑe jk sinφ y wee k = k( cosφ ˆx sinφ ŷ) ( y ŷ + z ẑ) = k sinφ y Te magnetic suface cuent associated wit tis electic field is M s = n E = ˆx E i ẑe jk sinφ y = ŷe i e jk sinφ y Fom (-5a) te pat diffeence is and te adiation integals (-c) and (-d) ae cosψ = y sinθ sinφ + z cosθ L θ e jk sinφ y cosθ sinφe jk ( y sinθ sinφ + z cosθ) d z cosθ sinφ e jk z cosθ d z e jk y ( sinθ sinφ +sinφ ) L φ e jk sinφ y cosφe jk ( y sinθ sinφ + z cosθ) d z cosφ e jk z cosθ d z e jk y ( sinθ sinφ +sinφ ) Pefoming te integals sin sin sinφ + sinφ ( L θ cosθ sinφ k cosθ k( sinθ sinφ + sinφ sin sin sinφ + sinφ ( L φ cosφ k cosθ k( sinθ sinφ + sinφ Te fields fom (-) ae E E θ j E i πk E φ j E i πk H H θ E φ H φ E φ cosφ cosθ sinθ sinφ + sinφ = 8E i k = 8E i k sin sinφ sin sinθ sinφ + sinφ sin ka ( sinθ sinφ + sinφ cosφ sin cosθ sinθ sinφ + sinφ sin ka ( sinθ sinφ + sinφ sin ka ( sinθ sinφ + sinφ sinφ sin sinθ sinφ + sinφ sin ka ( sinθ sinφ + sinφ

3 3. (Balanis.34) As descibed in section.5. fo ectangula apetues, te apetue fields fo te cicula apetue in space ae E = ŷe H = ˆx E η in te apetue and zeo elsewee. Note te coice of magnetic field is fo a plane wave emeging fom te apetue, in te positive z-diection, te diection of adiation. Te coesponding suface cuents ae M s = ˆn E = ẑ ŷe = ˆxE J s = ˆn H = ẑ ˆx E η = ŷ E η Te adiation fields fo due to te magnetic suface cuent ae aleady woked out in section.6.. We just need to divide by two. We find adiation integals due to te electic suface cuent. Fom (-43a) te pat diffeence is cosψ = ρ sinθ cos φ φ Te adiation integals ae a π N θ = E η cosθ sinφ e jk a π N φ = E η cosφ e jk ρ sinθ cos ( φ φ ) ρ sinθ cos ( φ φ ) Following (-48) toug (-5) fo te integation ρ d φ d ρ = E cosθ sinφ η ρ e jk ρ sinθ cos ( φ φ ) d φ d ρ a a π ρ d φ d ρ = E η cosφ ρ e jk ρ sinθ cos ( φ φ ) d φ d ρ E N θ = πa η cosθ sinφ J kasinθ E N φ = πa η cosφ J kasinθ π Te adiation integals fo te magnetic cuent ae, fom (-5) and (-5), (divided by two) Te total adiation fields ae L θ = πa E cosθ cosφ J kasinθ L φ = πa E sinφ J ( kasinθ ) E H E θ jka E E φ jka E H θ E φ H φ E θ J J ( kasinθ ) kasinθ sinφ ( cosθ + ) cosφ( cosθ + )

4 4. (Balanis 4.) Calculating te effective dielectic constant fom (4-) w / =. ε ε eff = ε + + ε + W / 6.86 You can calculate te caacteistic impedance fom (4-9) o fom you favoite micostip calculato, te esult is 47. Ω Tis is not substantially diffeent fom 5 Ω. Te eflection coefficient of a 5 Ω tansmission line attaced to a 47. Ω tansmission line is ρ = Z Z Z + Z.3

5 5. (Balanis 4.7) Following te design pocedue of section 4...C, tis widt is appaently good (4-6) v = m/s f =.6 GHz W = v 3.96 cm f ε + ε =. If you want te TM mode to ave lowest esonant fequency and be te dominant mode ten you will need to coose a naowe widt ( W <.9 cm) so tat L > W >, as we sall see below. Tis widt lage widt is fine since you can ely on te excitation fequency selection and te feed symmety to ensue tat only desied mode as significant amplitude. Te micostip feed line will usually be used wit a symmetic mode in z-diection. Te TM mode will be anti-symmetic in te z-diection and can not be excited. Te effective dielectic constant is (4.) =.7 m ε eff = ε + Te lengt extension is fom te mysteious fomula (4-) + ε + W / 9.5 ( ε eff +.3) W +.64 ΔL =.4 ( ε eff.58) W m and te lengt is fom (4.7) L = v f ε eff ΔL.93 cm Te single slot esonant input impedance is fom (4-) and (4-6) Te mutual conductance is (4-8) sin k W cosθ π G = π sin 3 θdθ cosθ k = π f R in = 38 Ω v G

6 G = π π sin k W cosθ cosθ J ( k L sinθ )sin 3 θdθ.393 Ω Te esonant input impedance wit mutual coupling included (4-7), wit te plus sign is cosen fo ou antisymmetic mode is We need to know B to find y. Since we can use (4-8b) Fo ou Zc = 75 Ω feed line we ave R in = (G + G ) 57 Ω.7.68 <. λ.87 B = W λ.636ln( k ).53 Ω G Y c = G =.4 B Y c = B =.4 Actually, te second condition is a little maginal, but I am tied of woking on tis poblem, so I will go aead wit te simple fomula (4-a) R in ( y R in = π cos y = L π cos L y R in. cm

Physics Courseware Electromagnetism

Physics Courseware Electromagnetism Pysics Cousewae lectomagnetism lectic field Poblem.- a) Find te electic field at point P poduced by te wie sown in te figue. Conside tat te wie as a unifom linea cage distibution of λ.5µ C / m b) Find