1. Which of the following expression is correct for radiation efficiency?
a) \(\eta_r = \frac{P_r}{P_l}\)
b) \(\eta_r = \frac{P_r}{P_r-P_l}\)
c) \(\eta_r = \frac{P_r}{P_r+P_l}\)
d) \(\eta_r = \frac{P_l}{P_r+P_l}\)
Explanation: Radiation efficiency is defined as the ratio of power radiated to the total input power to the antenna. Total input power is the sum of the radiated power Pr and the ohmic losses Pl.
\(\eta_r = \frac{P_r}{P_{in}} = \frac{P_r}{P_r+P_l}\)
2. Which of the following represents the relation between maximum power gain and maximum directivity gain of the antenna?
a) Gpmax = ηrGdmax
b) Gpmax = ηr/Gdmax
c) ηr = \(\sqrt{(G_{pmax} G_{dmax})}\)
d) ηr = \(\frac{G_{dmax}+G_{pmax}}{G_{dmax}-G_{pmax}}\)
Explanation: Maximum power gain is obtained when there are no ohmic losses. Gpmax=\(\frac{U_{max}}{P_{in}/4π}\)
Maximum directive gain Gdmax=\(\frac{U_{max}}{P_r/4π}\, and\, \eta_r=\frac{P_r}{P_{in}}\)
Gpmax=ηr Gdmax
3. What is the maximum power gain when the radiation resistance is 72Ω, loss resistance is 8Ω and the maximum directive gain is 1.5?
a) 1.15
b) 1.35
c) 1.25
d) 1.53
Explanation: Maximum power gain Gpmax=ηr Gdmax
Radiation efficiency ηr=\(\frac{R_r}{R_r+R_l}=\frac{72}{72+8}=\frac{72}{80}=0.9\)
Now, Gpmax=ηr Gdmax=0.9×1.5=1.35
4. The radiation efficiency value is ______
a) 0
b) 1<η<∞
c) 0≤η≤1
d) ∞
Explanation: Radiation efficiency ηr=\(\frac{R_r}{R_r+R_l}\)
Rr+Rl>Rr So \(\frac{R_r}{R_r+R_l}\) < 1 and If Rl=0 then ηr=1
Therefore, the value of efficiency lies in the range 0 to 1.
5. The value of maximum power gain is always greater than or equal to the maximum directive gain.
a) True
b) False
Explanation: Since Gpmax=ηrGdmax and the value of radiation efficiency lies in range 0 to 1, The maximum power gain will be always less than or equal to the maximum directive gain of antenna.
6. What is the radiation resistance of an antenna if it radiates 1kW and current in it is Irms=10A?
a) 0.1Ω
b) 1Ω
c) 10Ω
d) 100Ω
Explanation: Power radiated Prad=Irms2 Rrad
⇨ Rrad=\(\frac{P_{rad}}{I_{rms}^2} = \frac{1000}{10×10}=10\Omega\)
7. What is the radiation resistance of a short dipole of length L?
a) 20π2 \((\frac{L}{\lambda})^2\)
b) 80π2 \((\frac{l}{\lambda})^2\)
c) 40π2 \((\frac{l}{\lambda})^2\)
d) 160π2 \((\frac{l}{\lambda})^2\)
Explanation: Radiation resistance of a Hertzian dipole of length l is given by R=80π2 \((\frac{l}{\lambda})^2\)
Then short dipole is of length l/2, so the radiation resistance is given by R=80π2\((\frac{L/2}{\lambda})^2=20\pi^2 (\frac{L}{\lambda})^2.\)
8. If the length of the dipole decreases, then the radiation resistance will________
a) increase
b) decrease
c) depends on current distribution
d) not change
Explanation: Since the radiation resistance of a Hertzian dipole is R=80π2 \((\frac{l}{\lambda})^2\), the radiation resistance will decrease as the length of the dipole decreases. It is directly proportional to the square of the length of the dipole.
9. For a half-wave dipole with length λ/12, what is the antenna efficiency if the Radiation resistance is 2Ω?
a) 0.73
b) 0.073
c) 0.37
d) 0.78
Explanation: Radiation resistance \(R_{rad}=80π^2(\frac{l}{\lambda})^2=80π^2 (\frac{\lambda/12}{\lambda})^2\)=5.48Ω
Antenna efficiency \(\eta=\frac{R_{rad}}{R_{rad}+R_{loss}}=\frac{5.48}{5.48+2}=0.73\)
10. Find the radiation resistance of a Hertzian dipole of length 1m and operating at a frequency 1MHz?
a) 0.08Ω
b) 8.8mΩ
c) 8.8Ω
d) 0.88Ω
Explanation: \(λ=\frac{c}{f}=\frac{3×10^8}{1×10^6}=300m\)
Radiation resistance of a Hertzian dipole is given by \(R=80π^2(\frac{l}{\lambda})^2=80π^2 (\frac{1}{300})^2\)=0.0088Ω