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Linear Wire Antennas Ranga Rodrigo August 24, 2010 Lecture notes are fully based on Balanis [?]. Some diagrams and text are directly from the books. Contents 1 Inﬁnitesimal Dipole 1 2 Small Dipole 7 3 Finite-Length Dipole 9 • Wire antennas, linear or curved, are some of the oldest, simplest, cheap- est, and in many cases the most versatile for many applications. 1 Inﬁnitesimal Dipole • An inﬁnitesimal linear wire (l λ) is positioned symmetrically at the ori- gin of the coordinate system and oriented along the z axis. • The spatial variation of the current is assumed to be constant and given by ˆ I (z ) = a z I 0 where I 0 is a constant. 1 • The source only carries an electric current I e . I m and the potential func- tion F are zero. To ﬁnd A we write µ e − j kR A(x, y, z) = I e (x , y , z ) dl . 4π C R where (x, y, z) : Observation point coordinates. (x , y , z ) : Coordinates of the source. R : The distance from any point on the source to the observation point. C : Path along the length of the source. ˆ I e (x , y , z ) = a z I 0 . x = 0, y = 0, z = 0, for the inﬁnitesimal dipole. R= (x − x )2 + (y − y )2 + (z − z )2 = x2 + y 2 + z2 = r (constant). dl = dz . µI 0 − j kr l /2 µI 0 l − j kr ˆ A(x, y, z) = a z e ˆ d z = az e . 4π −l /2 4π z (x, y, z) θ (x , y , z ) φ y x 2 1 Next: H A : H A = ∇ × A, E A : ∇ × H A = J + j ω E A . Transformation from µ rectangular to spherical coordinates: sin θ cos φ sin θ sin φ cos θ Ar Ax A θ = cos θ cos φ cos θ sin φ − sin θ A y Aφ − sin φ cos φ 0 Az A x = 0, A y = 0, A z = 0. µI 0 l e − j kr A r = A z cos θ = cos θ. 4πr µI 0 l e − j kr A θ = −A z sin θ = − sin θ. 4πr A φ = 0. ˆ ar ∂ ∂ ˆ aθ 1 ∂ ∂ aφ ∂ ˆ ∂ ∇× A = (A φ sin θ) − Aθ + A r − (r A φ ) + (r A θ ) − Ar r sin θ ∂θ ∂φ r sin θ ∂φ ∂r r ∂r ∂θ 1 aφ ∂ ˆ ∂ H= (r A θ ) − Ar µ r ∂r ∂θ Hr = Hθ = 0. k I 0 l sin θ 1 Hφ = j 1+ e − j kr . 4πr j kr 1 E =EA = ∇× H. jω I 0 l cos θ 1 Er = η 1+ e − j kr . 2πr 2 j kr k I 0 l sin θ 1 1 Eθ = j η 1+ − e − j kr . 4πr j kr (kr )2 E φ = 0. The E and H components are valid everywhere except on the source itself. 3 z ˆ ar , E r , H r ˆ aφ , E φ , H φ θ ˆ aθ , E θ , H θ φ y x Power Density and Radiation Resistance • For a lossless antenna, the real part of the input impedance is designated as the radiation resistance, that power is transferred from the guided wave to the free space wave. 1 1 W = E × H ∗ = (a r E r + a θ E θ ) × a φ Hφ . ˆ ˆ ˆ ∗ 2 2 1 ∗ ∗ ˆ ˆ = a r E θ Hφ − a θ E r Hφ . 2 η I 0 l 2 sin2 θ 1 Wr = 1− j . 8 λ r 2 (kr )3 k|I 0 l |2 cos θ sin θ 1 Wθ = j η 2r 3 1+ j . 16π (kr )2 The complex power moving int eh radial direction 2π π P= W ·ds = (a r Wr + a θ Wθ ) · a r r 2 sin θd θd φ ˆ S 0 0 2π π = Wr r 2 sin θd θd φ 0 0 2 π I0l 1 =η 1− j . 3 λ (kr )3 4 The transverse component Wθ does not contribute to the integrals. Thus P does not represent the total complex power radiated by the antenna. Wθ is purely imaginary, and does not contribute to any real radiated power. It con- tributes to the imaginary (reactive) power. The reactive power density, which is most dominant for small values of kr , has both radial and transverse components. It merely changes between out- ward and inward directions to form a standing wave at a rate twice per cycle. It also moves in the transverse direction. Time average power radiated is 2 π I0l p rad = η . 3 λ For large values of kr (kr 1), the reactive power diminishes. For free space η 120π, 2 2 2π l l R rad = η = 80π2 . . 3 λ λ Near-Field Region kr 1 I 0 l e − j kr Er −jη cos θ. 2πkr 3 I 0 l e − j kr Eθ − j η sin θ. 4πkr 3 E φ = Hr = Hθ = 0. I 0 l e − j kr Hφ sin θ. 4πr E r and E θ are in time-phase. E r and E θ are in time-phase quadrature with Hφ . Therefore, there is no time-average power ﬂow associated with them. Intermediate-Field Region kr > 1 I 0 l e − j kr Er η cos θ. 2πkr 2 k I 0 l e − j kr Eθ j η sin θ. 4πkr E φ = Hr = Hθ = 0. k I 0 l e − j kr Hφ j sin θ. 4πr 5 E r and E θ approach time-phase quadrature. They form a rotating vector whose tip traces and ellipse in a plane parallel to the direction of propagation: cross ﬁeld. Far Field kr 1 k I 0 l e − j kr Eθ jη sin θ. 4πkr Er E φ = Hr = Hθ = 0. k I 0 l e − j kr Hφ j sin θ. 4πr The ratio of E θ to Hφ is equal to Eθ Zw = η. Hφ where Z w is the wave impedance and η is the intrinsic impedance (377 120πΩ for free-space.) E - and H -ﬁeld components are perpendicular to each other, transverse to the direction of propagation, and r variations are separable from those of θ and φ. This relationship is applicable in the far-ﬁeld region of all antennas of ﬁnite dimensions. Directivity The average power density 2 1 1 η k I0l sin2 θ W av = Re E × H ∗ = a r ˆ |E θ |2 = a r ˆ . 2 2η 2 4π r2 The radiation intensity 2 η k I0l r2 2 U = r 2Wav = sin2 θ = E θ (r, θ, φ) . 2 4π 2η The maximum value occurs at θ = π/2: 2 η k I0l Umax = . 2 4π Umax 3 D 0 = 4π = . P rad 2 6 The maximum effective aperture λ2 3λ2 A em = D0 = . 4π 8π 1 P rad = |I 0 |2 R rad . 2 2 2π l R rad = η . 3 λ 2 Small Dipole The current distribution of the inﬁnitesimal dipole (l < λ/50) is I 0 , a constant. For a small dipole (λ/50 ≤ l ≤ λ/10) the triangular current distribution approx- imation must be used. z I0 |I | The current distribution is l a z I0 1 − 2 z , 0 ≤ z ≤ 2 , ˆ l I e (x , y , z ) = l a z I 0 1 + 2 z , − 2 ≤ z ≤ 0. ˆ l µ 0 2 e − j kR l /2 2 e − j kR A(x, y, z) = ˆ az I0 1 + z ˆ d z + az I0 1 − z dz . 4π −l /2 l R 0 l R Because the overall length of the dipole is small, the value of R for different values of z along the length of the wire are not much different from r . 7 z P (r, θ, φ) θ R l /2 dz θ r z l /2 φ=φ y x kl π Maximum phase error due to the assumption R r is 2 = 10 = 18◦ for λ/10. 1 µI 0 e − j kr ˆ ˆ A = a z Az = a z , 2 4πr which is one half of that obtained in the previous section for the inﬁnitesimal dipole. Far-Zone Fields, kr 1 k I 0 l e − j kr Eθ jη sin θ. 8πkr Er E φ = Hr = Hθ = 0. k I 0 l e − j kr Hφ j sin θ. 8πr Directivity and the maximum effective area are the same as for the inﬁnitesimal dipole. 2 2P rad 2 l R rad = = 20π |I 0 |2 λ which is 1/4 of the value for the inﬁnitesimal dipole. 8 3 Finite-Length Dipole We can analyze the radiation characteristics of a dipole with any length us- ing magnetic vector potential A. For a thin, center-fed ﬁnite-length dipole (l ≥ λ/10, d λ), the approximate current distribution can be written as a z I 0 sin k l − z , 0 ≤ z ≤ l , ˆ 2 2 I e (x = 0, y = 0, z ) = a z I 0 sin k l + z , − l ≤ z ≤ 0. ˆ 2 2 In the far ﬁeld, we have, z r, θ θ . For amplitude: R r . For phase: R r − z cos θ. z z P (r, θ, φ) P (r, θ, φ) θ R R θ l /2 dz dz θ r z z θ r l /2 y y φ=φ φ=φ x x kl π Maximum phase error due to the assumption R r is 2 = 10 = 18◦ for λ/10. µ e − j kR A(x, y, z) = I e (x , y , z ) dl . 4π C R µe − j kR A(x, y, z) = I e (x , y , z )e j kz cos θ d z . 4πr C The ﬁnite dipole antenna is subdivided into a number of inﬁnitesimal dipoles of length δz . For an inﬁnitesimal dipole of length d z positioned along the z-axis at z j ηk I e (x , y , z )e − j kR d Eθ sin θd z . 4πR d Er d E φ = d Hr = d Hθ = 0 j k I e (x , y , z )e − j kR d Hφ sin θd z . 4πR 9 Using the far ﬁeld approximation j ηk I e (x , y , z )e − j kr cos θ d Eθ sin θe j kz dz . 4πr Summing the contribution from all the inﬁnitesimal elements −l /2 ke − j kr −l /2 cos θ Eθ = d Eθ = j η sin θ I e (x , y , z )e j kz dz −l /2 4πr −l /2 Simplifying kl kl I 0 e − j kr cos 2 cos θ − cos 2 Eθ jη . 2πr sin θ cos kl cos θ − cos kl − j kr I0e 2 2 Hφ j . 2πr sin θ Power Density, Radiation Intensity, and Radiation Resistance 1 1 ∗ W av = Re E × H ∗ = Re a θ E θ × a φ Hφ ˆ ˆ 2 2 1 ˆ ˆ ∗ = Re a θ E θ × a φ E θ /η 2 10

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