An antenna is a key component of a wireless link which efficiently couples electromagnetic energy from the transmitter to free space and from free space to the receiver. An antenna is generally a bidirectional device, i.e, the power through the antenna can flow in both the directions, hence it works as a transmitting as well as a receiving antenna. An antenna acts as an interface between the radiated electromagnetic waves and the guided waves. It can be thought of as a mode transformer which transforms a guided wave field distribution into a radiated-wave field distribution.
Ultra High Frequency (UHF) spectrum is defined as :
300 MHz - 3 GHz
UHF runs from 300 MHz to 3 GHz — option 3.
The ITU band structure, which makes every answer derivable rather than memorised. Each band spans exactly one decade, from \(3\times10^{n}\) to \(3\times10^{n+1}\) Hz:
| Band | Name | Frequency | Wavelength |
|---|---|---|---|
| VLF | Very Low | 3–30 kHz | 100–10 km |
| LF | Low | 30–300 kHz | 10–1 km |
| MF | Medium | 300 kHz–3 MHz | 1000–100 m |
| HF | High | 3–30 MHz | 100–10 m |
| VHF | Very High | 30–300 MHz | 10–1 m |
| UHF | Ultra High | 300 MHz–3 GHz | 1 m–10 cm |
| SHF | Super High | 3–30 GHz | 10–1 cm |
| EHF | Extremely High | 30–300 GHz | 10–1 mm |
Counting up from HF — High, Very High, Ultra High — each step multiplies by ten, which places UHF two decades above HF's 3–30 MHz. Option 1 is LF, option 2 is HF and option 4 is EHF, so the distractors are simply neighbouring bands.
Why the wavelength matters here. UHF wavelengths run from 1 m down to 10 cm, so a half-wave dipole is between 50 cm and 5 cm long — small enough to fit inside a handset. That is precisely why mobile telephony, Wi-Fi, Bluetooth, GPS and television all sit in this band, and it connects directly to the passage's point that antenna size is set by wavelength.
How UHF propagates. It is essentially line-of-sight. The ionosphere no longer reflects at these frequencies — the sky-wave propagation that HF relies on stops working — so coverage is limited to the radio horizon,
\(d\approx4.12\left(\sqrt{h_t}+\sqrt{h_r}\right)\ \text{km}\)
with the heights in metres. This is why UHF services need many cell sites or a satellite rather than one distant transmitter.
The trade-off that makes UHF popular. Higher frequency buys wide bandwidth and small antennas at the cost of range and of penetration through buildings and foliage — a balance that suits short-range, high-capacity services exactly.
Hence, UHF is 300 MHz to 3 GHz.
Similar Questions
If the axial ratio of an electromagnetic wave is zero dB. Then, the wave is:
A plane wave is propagating in a uniform medium with propagation constant \((0.6\pi + j\,0.4\pi)\) its wavelength is
Given below are two statements :
Statement I : Phase velocity of an electromagnetic wave in a bounded medium can be greater than the speed of light in an unbounded medium.
Statement II : Frequency of electromagnetic wave is affected by the permittivity of the medium.
In the light of the above statements, choose the correct answer from the options given below :
Assertion (A) : The ionosphere is that region of the earth’s atmosphere in which the constituent gases are ionised by radiations from outer space.
Reason (R) : Throughout the ionosphere, there are several layers in which ionization density either reaches maximum or remains almost constant. It depend on the intensity of sun. It also depends upon the atmospheric pressure.
The free space wave number ‘k0’ is defined as
Assertion (A) : Light is electromagnetic wave in nature, these may be x-rays, radio waves, microwaves etc.
Reason (R) : The amount of energy depends on the intensity of the light rays. The energy associated with each photon is proportional to the frequency.
Select your answer using the codes given below :
Frequency in UHF range propagated by means of
Important Questions from Electromagnetic Wave Propagation
For sky waves, following statements are given:
(A) n > 1, this shows 81 \(\rm\frac{N}{f^2}\) positive
(B) n > 1, show 81 \(\rm\frac{N}{f^2}\) Negative
(C) n < 1 shows 81 \(\rm\frac{N}{f^2}\) < 1
(D) v g x v p= c 2
(E) n = 0 shows 81 \(\rm\frac{N}{f^2}\) = 1, f = f c
Choose the correct answer from the options given below:
If the Polarization vector is given as N and the Direction of propagation is given as K then which one of the following relations is correct?
The wave length (λ) in meters of an electromagnetic wave is related to its frequency (f) in MHz as:
Bending of light wave as it passes between material of different optical density
The wave impedance of a medium is equal to:
