0
11kviews
Derive radiation resistance of the infinitesimal dipole. Explain its significance
1 Answer
0
601views

The input impedance of an antenna consists of real and imaginary parts. For a lossless antenna, the real part of the input impedance was designated as radiation resistance. It is through the mechanism of the radiation resistance that power is transferred from the guided wave to the free-space wave. To find the input resistance for a lossless antenna, the Poynting vector is formed in terms of the E- and H-fields radiated by the antenna. By integrating the Poynting vector over a closed surface (usually a sphere of constant radius), the total power radiated by the source is found. The real part of it is related to the input resistance. Fields radiated by the current carrying elements are given by

$$H_r = H_0 = 0$$ $$H_ϕ = j\frac{kI_0l sin θ}{4πr}\left[1 + \frac{1}{jkr}\right]e^{-jkr} .........(1)$$ $$E_r = η\frac{I_0lcos θ}{2πr^2}\left[1+\frac{1}{jkr}\right]e^{-jkr}$$ $$E_θ = jη\frac{kI_0l sin θ}{4πr}\left[1 + \frac{1}{jkr} - \frac{1}{(kr)^2}\right]e^{-jkr}$$ $$E_ϕ = 0........... (2)$$

For the infinitesimal dipole, the complex Poynting vector can be written using (1) and (2)

$$W = \frac{1}{2}(E × H^*) = \frac{1}{2}(\hat{a}_r E_r + \hat{a}_θE_θ) × (\hat{a}_ϕH_ϕ^*)$$ $$= \frac{1}{2}(\hat{a}_rE_θH_ϕ^* - \hat{a}_θE_rH_ϕ^*).........(3)$$

Whose radial $W_r$ and transverse $W_θ$ components are given, respectively, by

$$W_r = \frac{η}{8}\left|\frac{I_0l}{λ}\right|^2\frac{sin^2θ}{r^2}\left[1 - j\frac{1}{(kr)^3}\right]$$ $$W_θ = jη\frac{k|I_0l|^2 cos θ sin θ}{16π^2r^3}\left[1 + \frac{1}{(kr)^2}\right]......... (4)$$

The complex power moving in the radial direction is obtained by integrating (3) (4) over a closed sphere of radius r. Thus it can be written as

$$P = \unicode{x222F}_s W . ds = \int_0^{2π}\int_0^{π} (\hat{a}_rW_r + \hat{a}_θW_θ). \hat{a}_r r^2 sin θ dθ dϕ...............(5)$$

Which reduces to

$$P = \int_0^{2π}\int_0^{π}W_r r^2 sin θ dθ dϕ = η\frac{π}{3}\left|\frac{I_0l}{λ}\right|^2 \left[1 - j\frac{1}{(kr)^3}\right].......... (6)$$

The transverse component Wθ of the power density does not contribute to the integral. Thus (6) does not represent the total complex power radiated by the antenna. Since Wθ, as given by (4), is purely imaginary, it will not contribute to any real radiated power. However, it does contribute to the imaginary (reactive) power which along with the second term of (6) can be used to determine the total reactive power of the antenna. The reactive power density, which is most dominant for small values ofkr, has both radial and transverse components. It merely changes between outward and inward directions to form a standing wave at a rate of twice per cycle. It also moves in the transverse direction as suggested by (4).Equation (5), which gives the real and imaginary power that is moving outwardly, can also be written as

$$P = \frac{1}{2}\iint_s E × H^* . ds = η\frac{π}{3}\left|\frac{I_0l}{λ}\right|^2 \left[1 - j\frac{1}{(kr)^3}\right]$$

$$= P_{rad} + j2ω (\hat{W}_m - \hat{W}_e).............. (7)$$

Where

P = power (in radial direction)

$P_rad$= time-average power radiated

$W_m$ = time-average magnetic energy density (in radial direction).

$W_e$ = time-average electric energy density (in radial direction)

$2ω( W_m − W_e)$ = time-average imaginary (reactive) power (in radial direction)

From (7)

$$P_{rad} = η\frac{π}{3}\left|\frac{I_0l}{λ}\right|^2$$

$$2ω (\hat{W}_m - \hat{W}_e) = η\frac{π}{3}\left|\frac{I_0l}{λ}\right|^2 \frac{1}{(kr)^3}...............(8)$$

It is clear from (8) that the radial electric energy must be larger than the radial magnetic energy. For large values of kr (kr >>1 or r >>λ), the reactive power diminishes and vanishes when kr =∞. Since the antenna radiates its real power through the radiation resistance, for the infinitesimal dipole it is found by equating (8) to

$$\boxed{P_{rad} =η\frac{π}{3}\left|\frac{I_0l}{λ}\right|^2 = \frac{1}{2}|I_0|^2R_r}$$

Where R_r is the radiation resistance. Equation (4 - 18) reduces to

$$R_r = η\frac{2π}{3}\left|\frac{l}{λ}\right|^2 = 80π^2 \left(\frac{1}{λ}\right)^2$$

For a free-space medium (η=120π). It should be pointed out that the radiation resistance of (9) represents the total radiation resistance since (4) does not contribute to it. For a wire antenna to be classified as an infinitesimal dipole, its overall length must be very small (usually l ≤ λ/50).

Please log in to add an answer.