Basics of Antenna MCQ || Antenna Questions and Answers

Answer.1. Capable of a larger duty cycle

Explanation:-

• TWT (Travelling Wave Tube) is an electron tube used for amplification at microwave frequencies- generally identified as frequencies between 500 MHz and 300 Hz._. 
• Its power generation capability range from watts to megawatts.
• For helix TWTs, bandwidths may be as high as two octaves or more and power levels of tens to hundreds of watts.
• The duty or cycle time of the TWT is the time the TWT is ON divided by the time base. TWT has the advantage of producing a high duty cycle.

Answer.3. 66.66%

Explanation:-

Given:

Radiation resistance (Rr) = 60 Ω

Antenna resistance (Ra) = 30 Ω

The antenna efficiency of an antenna is given as

η = Rr/(Rr + Ra)

where

Rr : Radiation resistance

Ra : Effective Antenna resistance

η% = 60/(60 + 30) × 100%

η% = 66%

Answer.1. Monopulse

Explanation:-

• In monopulse antenna, simultaneously two or more lobes of the antenna are used to track a target hence giving better tracking than other methods.
• Monopulse radar is a radar system that uses additional encoding of the radio signal to provide accurate directional information.
• The name refers to its ability to extract range and direction from a single signal pulse.
• Monopulse radar avoids problems seen in conical scanning radar systems, which can be confused by rapid changes in signal strength.
• This system also makes jamming more difficult.
• Most radars designed since the 1960s are monopulse systems.
• The monopulse method is also used in passive systems, such as electronic support measures and radio astronomy.
• Monopulse radar systems can be constructed with reflector antennas, lens antennas, or array antennas.
• Historically, monopulse systems have been classified as either phase-comparison monopulse or amplitude monopulse. This is because some common implementations have been based on a phase comparison or an amplitude comparison.
• Modern systems determine the direction from the monopulse ratio, which contains both amplitude and phase information.
• The monopulse method does not require that the measured signals are pulsed.

Answer.3. 10,000 watts

Explanation:-

Effective radiated power (ERP) is defined as:

ERP = G ⋅ P_t watts

Where, G = Gain of the Antenna

P_t = transmitted power

Calculation:

Given that, G = 30 dB and P_t = 10 W

Since, G(in dB) = 10 log₁₀ (G)

30 = 10 log₁₀ (G)

3 = log₁₀ (G)

∴ G = 10³ = 1000

∴ ERP = G ⋅ P_t

= 1000 × 10

= 10,000 watts

Answer.3. 0.96

Explanation:-

For two lines connected with different impedances, the magnitude of the reflection coefficient is given by:

$\Gamma = |\frac{{{Z_L} – {Z_0}}}{{{Z_L} + {Z_0}}}|$

The efficiency of reflection coefficient can be written as:

η = 1 – Γ²

Calculation:

With Z_L = 50 Ω and Z₀ = 73 Ω, the reflection coefficient will be:

$\Gamma = |\frac{{50 – 73}}{{50 + 73}}|$

Γ = 0.186

The efficiency will now be:

η = 1 – (0.186)²

η = 0.96

Answer.2. 16

Explanation:-

The range of the radar is given as:

$R = {\left[ {\frac{{{P_t}{G^2}\sigma {A_e}}}{{{{\left( {4\pi } \right)}^2}{P_r}}}} \right]^{\frac{1}{4}}}$

G= Antenna Gain

σ = Radar Cross-Section

Pr = Reflected Power

Ae = Effective Area of the Antenna

Observation:

We observe that, R ∝ (Pt)1/4

Now, to double the range R, Pt has to be increased by a factor of 16

Answer.1. 0.5 of the wave length

Explanation:-

The T.V antenna is also known as the Yagi antenna. It has a folded dipole as the main radiating or driven element and ‘parasitic’ elements that pickup radiated power from the dipole and reradiated it. This improves the directivity and gain of the antenna. In standard TV receiving antenna, the dipole element is 0.5 of the wavelength.

Length of Dipole = λ/2

Answer.1. 0.05

Explanation:-

Radiation efficiency is calculated as

η = P_Dissipated/P_input

With P_in = 100 W, and P_dissipated = 5 W

η = 5/100

η = 0.05

The percentage efficiency will be: 0.05 × 100 = 5 %

Answer.4. 2

Explanation:-

Maximum range of a radar system is given as

${R_{max}} = {\left( {\frac{{{P_t}{G^2}\sigma {A_e}}}{{{{\left( {4\pi } \right)}^2}{S_{min}}}}} \right)^{1/4}}$

Also,

Ae = πD²/4

Where,

• Rmax = Maximum range of the radar
• Smin = Minimum Detectable Signal Power
• P_t = Peak transmitted power
• Ae = Aperture Area
• D = Antenna diameter

R_max can now be written as:

${R_{max}} = {\left( {\frac{{{P_t}{G^2}\sigma \pi {D^2}}}{{4\left( {4{\pi ^2}} \right){S_{min}}}}} \right)^{1/4}}$

Rmax ∝ P_t1/4

So, If the peak transmitted power is increased by a factor of 16, then radar range will be increased by a factor of 2.

Answer.1. 2

Explanation:-

Maximum range of a radar system is given as

${R_{max}} = {\left( {\frac{{{P_t}{G^2}\sigma {A_e}}}{{{{\left( {4\pi } \right)}^2}{S_{min}}}}} \right)^{1/4}}$

Also,

Ae = πD²/4

Where,

• Rmax = Maximum range of the radar
• Smin = Minimum Detectable Signal Power
• P_t = Peak transmitted power
• Ae = Aperture Area
• D = Antenna diameter

R_max can now be written as:

${R_{max}} = {\left( {\frac{{{P_t}{G^2}\sigma \pi {D^2}}}{{4\left( {4{\pi ^2}} \right){S_{min}}}}} \right)^{1/4}}$

Rmax ∝ P_t1/4

So, If the antenna diameter is increased by a factor of 4, the radar range will be increased by a factor of 2.

Answer.2. Bandwidth of the transmitted pulse

Explanation:-

Range resolution is the ability of a radar to distinguish between two or more targets that are very close to each other.

The degree of range resolution depends on:

• the width of the transmitted pulse.
• the type and size of the target.
• on the efficiency of the receiver.
• Pulse width is the primary factor in range resolution. A well-designed radar system can distinguish targets separated by one-half the pulse-width time (τ).

There the theoretical range resolution of radar can be calculated as-

S_r(separation between targets) ≥ C_oτ/2

Where τ = Pulse width.

And Bandwidth of the pulse ∝ 1/τ

Hence Bandwidth of the transmitted pulse determines the range resolution of a radar.

Answer.3. 2%

Explanation:-

The radiation resistance of dipole is given by:

${R_{rad}} = 80{\pi ^2}{\left( {\frac{{dl}}{\lambda }} \right)^2}$

Taking log

log(R_rad) = log(80π²) + 2 log dl – 2 log λ

Differentiating the above, we get:

$\frac{{d{R_{rad}}}}{R} = \frac{{2dl}}{l}$

As λ = constant, its derivative will be 0

Multiply both the sides by 100, the above equation becomes:

$\frac{{d{R_{rad}}}}{R} \times 100 = 2\left( {\frac{{dl}}{l} \times 100} \right)$

% R_rad = 2 × 1%

= 2%

Answer.2. 73 W

Explanation:-

I_m = 0.5 Amp

Radiation Resistance, R_rad = 146 Ω

Calculation:

Power radiated by 1 half-wave dipole antenna.

${P_{rad}} = \frac{1}{2}\:I_m^2 \times {R_{rad}}$

= 18.25 watt

∴ Total power radiated by 4 half-wave dipole antenna.

P_T = 4 P_rad = 73 W

Answer.1. The duplexer

Explanation:-

• A duplexer is an electronic device that allows bi-directional (duplex) communication over a single path.
• In radar and radio communications systems, it isolates the receiver from the transmitter while permitting them to share a common antenna.
• Most radio repeater systems include a duplexer.
• Whenever a single antenna is used for both transmitting and receiving, as in the shown radar system, an electronic switch is used. Switching systems of this type are called duplexers.

Answer.1. 0.997

Explanation:-

For a lossless antenna, the radiation efficiency e_cd=1.

Overall efficiency of antenna is given by e_o =e_cd (1-∣γ²∣)=1 × (1 – (0.15²))=0.977.