For a half wave dipole antenna A. The average value of current is 0.64 (unit). B. The ratio of electric field intensity just near to the antenna surface and the potential developed on the antenna surface after reception of signal, is equal to the effective height C. Effective aperture will remain same if antenna gain will be improved by some technique. D. The effective height will be 0.64 l, where l is the physical length of antenna. E. If length of antenna becomes l = 0.1λ, then its current distribution become triangular. Choose the correct answer from the options given below :
A, D and E only
Background. A half-wave dipole (l = λ/2) carries an approximately sinusoidal current: maximum at the centre feed point, falling to zero at the two open ends.
A — "The average value of current is 0.64." TRUE. Averaging a half-sine over its length gives
\(I_{avg}=\dfrac{2}{\pi}I_{max}=0.637\,I_{max}\approx 0.64\)
in normalised units. It is the same 2/π that appears as the average of a rectified sine wave.
D — "Effective height = 0.64 l." TRUE. The effective height (or effective length) is the length a uniform-current antenna would need in order to radiate the same field as the actual tapered-current antenna. Because the actual current averages 2/π of its peak,
\(h_{eff}=\dfrac{2}{\pi}\,l = 0.64\,l\)
For a half-wave dipole this works out to \(h_{eff}=0.64\times\lambda/2 = \lambda/\pi\).
E — "If l = 0.1λ the current distribution becomes triangular." TRUE. On a very short (Hertzian) dipole only a tiny portion of the sine curve is used, and near its peak a sine is almost linear. The distribution therefore degenerates into a straight taper from maximum at the feed to zero at each end — a triangular distribution — for which the effective height is \(h_{eff}=l/2\).
B — the garbled "ratio of electric field intensity near the surface to the potential developed … equals the effective height." FALSE. Effective height is properly defined either as \(h_{eff}=\dfrac{V_{oc}}{E}\) (open-circuit induced voltage per unit incident field, i.e. potential ÷ field, not field ÷ potential) or by the current-distribution integral above. As written the ratio is inverted, so the statement is not accepted.
C — "Effective aperture will remain the same if the gain is improved." FALSE. Gain and effective aperture are locked together at a fixed wavelength:
\(A_e=\dfrac{\lambda^{2}}{4\pi}G\)
so raising G necessarily raises Ae. (For a half-wave dipole G = 1.64, giving \(A_e = 0.13\lambda^{2}\).)
Hence, the correct answer is A, D and E only.
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The expression given below is :
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Which of the following antennas is the standard reference antenna for the directiveness?
Consider the following statements:
(a) Fiber optic cable is much lighter than copper cable
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(c) Optical transmission is inherently bidirectional.
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A. Number of dipoles of unequal size
B. Number of dipoles equally spaced
C. Collinear dipoles
D. Dipoles in phase
E. Dipoles are 90 out of phase
Choose the correct answer from the options given below:
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For an isotropic antenna P n(θ, φ) = 1, D = 1, for all θ and φ. The beam area for the isotropic antenna is given by:
