20.23. Problems 1041 20.8 For the Hansen window of Eq. (20.17.2), show that the encircled energy ratio and limiting directivity are given by the following closed-form expressions, 21 u0 | |2 2 2 − 2 + 2 2 − 2 F(u) udu J0 π u0 H J1 π u0 H 0 E(u0) = a = 1 − 2 2 2 − 2 (2a) |A(r⊥)| r⊥dr⊥ I0(πH) I1(πH) Aperture Antennas 0 a 2 2 2I1(πH) A(r⊥)2πr⊥dr⊥ D∞ = 0 = πH 2 a (ka) 2 2 2 − 2 πa |A(r⊥)| 2πr⊥dr⊥ I0(πH) I1(πH) 0 20.9 To prove Eq. (20.20.8) without invoking the Babinet principle, first derive the following inte- grals, where r⊥, r⊥ are two-dimensional transverse vectors with magnitudes r⊥ =|r⊥| and r⊥ =|r⊥|, ∞ ∞ jk 2 jk 2 2 kr⊥r −jk|r⊥−r⊥ | /2z 2 −jkr⊥/2z −jkr⊥ /2z ⊥ 21.1 Open-Ended Waveguides e d r⊥ = e e J0 2πr⊥dr⊥ = 1 2πz −∞ 2πz 0 z The aperture fields over an open-ended waveguide are not uniform over the aperture. then, using this result, show the relationship for any A(r⊥) and Ac(r⊥)= 1 − A(r⊥), The standard assumption is that they are equal to the fields that would exist if the guide ∞ 2 2 jk −jkr⊥/2z −jkr⊥ /2z kr⊥r⊥ were to be continued [1]. e A(r⊥)e J0 2πr⊥dr⊥ = 2πz 0 z Fig. 21.1.1 shows a waveguide aperture of dimensions a>b. Putting the origin in ∞ 2 2 jk −jkr⊥/2z −jkr⊥ /2z kr⊥r⊥ the middle of the aperture, we assume that the tangential aperture fields Ea, Ha are = 1 − e Ac(r⊥)e J0 2πr⊥dr⊥ 2πz 0 z equal to those of the TE10 mode. We have from Eq. (9.4.3): 20.10 Computer Experiment. Carry out the same experiment using the same numerical values as in Example 20.22.4, but instead of Eq. (20.22.42), take the ideal focal spot within the field- of-view [−a, a] to be a delta function, Eb(x)= δ(x) , for |x|≤a = 0.6λ (20.23.2) Make similar plots as in Fig. 20.22.6. Hint: In this case the expansion coefficients are simply, bn = ψn(a, k, 0)/λn, n = 0, 1,...,M. Fig. 21.1.1 Electric field over a waveguide aperture. πx 1 πx Ey(x )= E0 cos ,Hx(x )=− E0 cos (21.1.1) a ηTE a 2 2 2 where ηTE = η/K with K = 1 − ωc/ω = 1 − (λ/2a) . Note that the boundary conditions are satisfied at the left and right walls, x =±a/2. For larger apertures, such as a>2λ, we may set K 1. For smaller apertures, such as 0.5λ ≤ a ≤ 2λ, we will work with the generalized Huygens source condition (18.5.7). The radiated fields are given by Eq. (18.5.5), with fx = 0: e−jkr Eθ = jk cθ fy(θ, φ)sin φ 2πr (21.1.2) e−jkr Eφ = jk cφ fy(θ, φ)cos φ 2πr 21.1. Open-Ended Waveguides 1043 1044 21. Aperture Antennas − 2 where fy(θ, φ) is the aperture Fourier transform of Ey(x ), that is, The function cos(πvx)/(1 4vx) determines the essential properties of the H-plane pattern. It is essentially a double-sinc function, as can be seen from the identity: a/2 b/2 + = jkxx jkyy ⎡ ⎤ fy(θ, φ) Ey(x )e dx dy 1 1 −a/2 −b/2 ⎢ sin π vx + sin π vx − ⎥ cos(πvx) π ⎢ 2 2 ⎥ a/2 b/2 = ⎣ + ⎦ (21.1.10) πx 2 = jkxx · jkyy 1 − 4vx 4 1 1 E0 cos e dx e dy π vx + π vx − −a/2 a −b/2 2 2 The y -integration is the same as that for a uniform line aperture. For the x -integration, It can be evaluated with the help of the MATLAB function dsinc, with usage: we use the definite integral: cos(π x) π y = dsinc(x); % the double-sinc function = sinc(x + 0.5) + sinc(x − 0.5) a/2 1 − 4x2 4 πx 2a cos(kxa/2) cos ejkxx dx = 2 −a/2 a π 1 − (kxa/π) The 3-dB width of the E-plane pattern is the same as for the uniform rectangular aperture, Δθ = 0.886 λ/b. The dsinc function has the value π/4atv = 1/2. Its 3-dB It follows that: y x point is at vx = 0.5945, its first null at vx = 1.5, and its first sidelobe at vx = 1.8894 and 2ab cos(πv ) sin(πv ) = x y has height 0.0708 or 23 dB down from the main lobe. It follows from v = a sin θ/λ fy(θ, φ) E0 2 (21.1.3) x π 1 − 4vx πvy that the 3-dB width in angle space will be Δθx = 2×0.5945 λ/a = 1.189 λ/a. Thus, the 3-dB widths are in radians and in degrees: where vx = kxa/2π and vy = kyb/2π, or, λ o λ λ o λ a b Δθx = 1.189 = 68.12 ,Δθy = 0.886 = 50.76 (21.1.11) v = sin θ cos φ, v = sin θ sin φ (21.1.4) a a b b x λ y λ Example 21.1.1: Fig. 21.1.2 shows the H- and E-plane patterns for a WR90 waveguide operating The obliquity factors can be chosen to be one of the three cases: (a) the PEC case, if at 10 GHz, so that λ = 3 cm. The guide dimensions are a = 2.282 cm, b = 1.016 cm. The the aperture is terminated in a ground plane, (b) the ordinary Huygens source case, if it typical MATLAB code for generating these graphs was: is radiating into free space, or (c) the modified Huygens source case. Thus, a = 2.282; b = 1.016; la = 3; c 1 1 1 + cos θ 1 1 + K cos θ θ = , , (21.1.5) th = (0:0.5:90) * pi/180; cφ cos θ 2 1 + cos θ 2 K + cos θ By normalizing all three cases to unity at θ = 0o, we may combine them into: vx = a/la * sin(th); vy = b/la * sin(th); 1 + K cos θ K + cos θ = = cE(θ)= ,cH(θ)= (21.1.6) K = sqrt(1 - (la/(2*a))^2); % alternatively, K 0, or, K 1 1 + K 1 + K cE = (1 + K*cos(th))/(K+1); % normalized obliquity factors where K is one of the three possible values: cH = (K + cos(th))/(K+1); 2 η λ gH = abs(cH .* dsinc(vx).^2); % uses dsinc K = 0 ,K= 1 ,K= = 1 − (21.1.7) gE = abs(cE .* sinc(vy)).^2; % uses sinc from SP toolbox ηTE 2a The normalized gains along the two principal planes are given as follows. For the xz-or figure; dbp(th,gH,45,12); dB gain polar plot o figure; dbp(th,gE,45,12); the H-plane, we set φ = 0 , which gives Eθ = 0: | |2 2 The three choices of obliquity factors have been plotted for comparison. We note that the Eφ(θ) 2 cos(πvx) a gH(θ)= = cH(θ) ,vx = sin θ (21.1.8) Huygens source cases, K = 1 and K = η/ηTE, differ very slightly. The H-plane pattern |E |2 1 − 4v2 λ φ max x vanishes at θ = 90o in the PEC case (K = 0), but not in the Huygens source cases. = And, for the yz-orE-plane, we set φ = 90o, which gives E = 0: The gain computed from Eq. (21.1.13) is G 2.62 or 4.19 dB, and computed from Eq. (21.1.14), φ 2 G = 2.67 or 4.28 dB, where K = η/ηTE = 0.75 and (K + 1) /4K = 1.02. 2 2 This waveguide is not a high-gain antenna. Increasing the dimensions a, b is impractical |Eθ(θ)| 2 sin(πvy) b g (θ)= = c (θ) ,v= sin θ (21.1.9) and also would allow the propagation of higher modes, making it very difficult to restrict E 2 E y |Eθ| πvy λ max operation to the TE10 mode. 21.1. Open-Ended Waveguides 1045 1046 21. Aperture Antennas H− Plane Pattern E− Plane Pattern o o θ 0 θ θ 0 θ 2 o o a/λ K (K + 1) /(4K) 45 45o 45 45o 0.6 0.5528 1.0905 0.8 0.7806 1.0154 1.0 0.8660 1.0052 −9 −6 −3 −9 −6 −3 90o 90o 90o 90o 1.5 0.9428 1.0009 dB dB 2.0 0.9682 1.0003 The gain-beamwidth product is from Eqs. (21.1.11) and (21.1.13), p = GΔθx Δθy = 2 2 o o = = 135o 135 135o 135 4π(0.81)(0.886)(1.189) 10.723 rad 35 202 deg . Thus, another instance of the general formula (16.3.14) is (with the angles given in radians and in degrees): 180o 180o 10.723 35 202 = = = = = G o o (21.1.15) Fig. 21.1.2 Solid line has K η/ηTE, dashed, K 1, and dash-dotted, K 0. Δθx Δθy Δθx Δθy Next, we derive an expression for the directivity and gain of the waveguide aperture. 21.2 Horn Antennas o The maximum intensity is obtained at θ = 0 . Because cθ(0)= cφ(0), we have: The only practical way to increase the directivity of a waveguide is to flare out its ends 2 into a horn.