Option 3 : 10,000 watts

__Concept__:

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 log_{10} (G)

30 = 10 log_{10} (G)

3 = log_{10} (G)

∴ G = 10^{3} = 1000

∴ ERP = G ⋅ P_{t}

= 1000 × 10

= 10,000 wattsOption 3 : 66.66%

__Concept:__

The antenna efficiency of an antenna is given as:

\(\rm{\eta = \frac{{{R_r}}}{{{R_r} + {R_a}}}}\)

Rr : Radiation resistance

Ra : Effective Antenna resistance

__Calculation:__

Given:

Radiation resistance (Rr) = 60 Ω

Antenna resistance (Ra) = 30 Ω

\(\rm{\eta \% = \frac{{60}}{{60 + 30}} \times 100\%}\)

\(\rm{\eta \%= 66.66\%}\)

Option 3 : Gain

- Gain of an antenna is the ratio of the radiation intensity in a given direction to the radiation intensity that would be obtained if the power were radiated isotropically.
- The
**gain of an antenna takes the directivity into account.** - The term antenna gain describes how much power is transmitted in the direction of peak radiation to that of an isotropic source.

Mathematically, the gain is defined as:

G = η_{e}D

η_{e} = Antennas efficiency

D = Directivity of the antenna

** Directivity: **The ratio of the radiation intensity in a given direction from an antenna to the radiation intensity averaged over all directions is termed as directivity.

** Antenna's Efficiency: **Antenna's Efficiency is the ratio of the radiated power of the antenna to the input power accepted by the antenna.

Option 1 : 0.05

__Concept__:

- An antenna receives electric signals from the transmitter circuits and converts them into electromagnetic waves.
- The efficiency of the antenna in performing this conversion, known as antenna radiation efficiency, is defined as the ratio of the power dissipated into space to the net power delivered to the antenna by the transmitter circuits.
- It is denoted by η.

Mathematically, the radiation efficiency is calculated as:

\(η=\frac{P_{dissipated}}{P_{input}}\)

__Calculation__:

With P_{in} = 100 W, and P_{dissipated} = 5 W

\(η=\frac{5}{100}\)

**η = 0.05**

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

The output of a three-element co-linear antenna array operating in a free space environment is combined (after appropriate phase shifting) to maximize the signal received from a particular direction as shown in the figure:

If the inter-element spacing is half of the signal wavelength and direction of maximum response is 30° from a line perpendicular to the array, what are the phases to be applied to each element? Consider the first element as the reference.

Option 3 : \(\left[ {0\;\;\frac{\pi }{2}\;\;\;\pi } \right]\)

__Concept__**:**

To get the maximum output, the phases applied (leading phase) to each element must be given as the phase difference between the n^{th} element and the 1^{st} element, i.e.

\(= \frac{{2\pi }}{\lambda }{d_{1n}}cos\left( {90 - \phi } \right)\) which is a lagging phase.

d_{1n} = distance between 1^{st} and n^{th} elements.

ϕ is the angle between the direction of maximum response from the line perpendicular to the array.

__Calculation__**:**

ϕ = 30°

\({d_{12}} = \frac{\lambda }{2},\;{d_{13}} = \lambda \)

The phase difference between the 1^{st} and 2^{nd} elements will be:

\( = \frac{{2\pi }}{\lambda } \times \frac{\lambda }{2}cos\left( {90^\circ - 30^\circ } \right) = \frac{\pi }{2}\)

The phase difference between the 1^{st} and 3^{rd} elements will be:

Option 2 : increases the effective area of the antenna

__ Directive gain (G_{s})__: It is the ratio of radiation intensity to the average value of radiation intensity.

**Directivity (D**)**:**

1) It has the maximum value of G_{d}.

2) If we add more antenna elements in an array, the directive gain (G_{d}) will increase. (This is analogous to Cascading of systems with gain A)

i.e. **if G _{d} increases, D also increases.**

__Effective Area (A _{c})__:

It describes how has much power the antenna can capture from a given plane wave.

\({A_c} = \frac{{{\lambda ^2}}}{{4\pi }}D\) From the above, it is clear that **if G _{d }increases then D also increases. This indicates that A_{c }also increases.**

Option 4 : 500 mm

**Calculation**:

The general relation for the effective aperture in terms of the peak antenna gain (G) of any antenna is given by:

\({{A}_{e}}=\frac{{{\lambda }^{2}}}{4\pi }.G\)

G = 1.5, for loop antenna.

And \({{A}_{e}}=\frac{3}{32\pi }{{m}^{2}}\) (Given)

\(So,\frac{3}{32\pi }=\frac{{{\lambda }^{2}}}{4\pi }.1.5\)

\({{\lambda }^{2}}=\frac{3}{8}\times 1.5=\frac{1}{4}\)

\(\lambda =\frac{1}{2}m\)

= 500 mmOption 1 : 27 dBi

Given:

Directivity (D) = 30 dBi, η = 50% = 0.5

In dB (η) = 10 log (0.5) = -3 dB

We know that,

\(\eta = \frac{G}{D}\)

In dB: (η)_{dB} = (G)_{dB} – (D)_{dB}

(G)_{dB} = (η)_{dB} + (D)_{dB}

= -3 dB + 30 dB

= 27 dB

Option 1 : No signal

__Polarization of wave:__

- It is defined as the time-varying behavior of the electric field strength vector at some fixed point in the space.
- There are three main types of polarisation namely, linear polarisation, circular polarisation, and elliptical polarisation.
- The polarization of an antenna is based on the E-plane orientation of the electromagnetic energy radiated/received by the antenna.
- Most antennas are typically either linearly (horizontal and vertical), or circularly polarized.
**It is important to match the polarization of the RF antenna to that of the incoming signal for maximum signal reception. Otherwise, there is a corresponding decrease in the level of the signal.**- If they were at right angles to one another (i.e. cross-polarised) then, at least in theory, no signal would be received.

Option 2 : 17.4

__Directive Gain (D):__

It is defined as the ratio of radiation intensity due to the test antenna to isotropic antenna (hypothetical antenna that radiates uniformly in all direction)

\(D = \frac{U}{U_0}=\frac{4\pi U}{P_{rad}}\)

Where,

U = radiation intensity due to test antenna, in watts per unit solid angle

U0 = radiation intensity due to an isotropic antenna, in watts per unit solid angle

Prad =total power radiated in watts

Since U is a directional dependent quantity, the directive gain of an antenna depends on the angle θ and Φ. If the radiation intensity assumes its maximum value then the directive gain is called the Directivity (D_{0}).

\(D_0=\frac{U_{max}}{U_0}=\frac{4\pi U_{max}}{P_{in}}\)

__Calculation__:

Given:

Efficiency, η = 0.90, U_{max} = 0.5 W/sr

Also, with Pin = 0.4 W, the radiated power will be:

P_{rad} = 0.4 × 0.9 = 0.36 W

∴ The directivity will be:

\(D= \frac{{4\pi \; \times \;0.5}}{{0.36}}\)

**D = 17.46**

For an infinitesimally small dipole in free space, the electric field E_{θ} in the far-field is proportional to (e^{-jkr}/r)sin θ, where k = 2π/λ. A vertical infinitesimally small electric dipole (δl ≪ λ) is placed at a distance h (h > 0) above an infinite ideal conducting plane, as shown in the figure. The minimum value of h, for which one of the maxima in the far-field radiation pattern occurs at θ = 60°, is

Option 1 : λ

__Concept:__

According to **image theory** Whenever we place a small electric dipole (δl) at a distance h (h > 0) above an infinite ideal conducting plane, the image of the small electric dipole will be formed at the same distance under the conducting plane. Thus it forms a two-element array.

For 2 element array, the array factor is given by:

\(\left| {A.F.} \right| = \frac{{sin\left( {N\frac{\Psi }{2}} \right)}}{{\sin \left( {\frac{\Psi }{2}} \right)}}\)

\(\Psi = Kdcos\theta\) ...1)

Where,

K: 2π/λ

θ: angle from antenna array axis at point ‘p’.

__Analysis: __

Given: A infinite small dipole (δl ≪ λ)

E_{θ} ∝ (e^{-jkr}/r) sin θ

We have to determine the minimum value of “h” for which one of the maxima in the far-field radiation pattern occurs at θ = 60^{0}.

For 2 element array (N = 2), the array factor will be:

\(\left| {A.F.} \right| = \frac{{sin\left( {N\frac{\Psi }{2}} \right)}}{{\sin \left( {\frac{\Psi }{2}} \right)}}\)

\(\left| {A.F.} \right| = \frac{{2sin\left( {\frac{\Psi }{2}} \right)cos\left( {\frac{\Psi }{2}} \right)}}{{\sin \left( {\frac{\Psi }{2}} \right)}}\)

\(\left| {A.F.} \right| = 2cos\left( {\frac{\Psi }{2}} \right)\)

\(|A.F.{|_{max}} = \;2\)

\(\left| {A.F.{|_N} = \left| {A.F.} \right|/} \right|A.F.{|_{max}}\)

\(|A.F.{|_N} = \;cos\left( {\frac{\Psi }{2}} \right)\) ...2)

From equation 1,

\(\Psi = Kdcos\theta\)

\(\Psi = \frac{{2{\rm{\pi }}}}{{\rm{\lambda }}}.2h.\cos 60\)

\(\Psi = \frac{{2{\rm{\pi }}}}{{\rm{\lambda }}}.h\)

Put this value in equation 2,

\(|A.F.{|_N} = \;cos\left( {\frac{{2{\rm{\pi }}}}{{2.{\rm{\lambda }}}}.h} \right)\)

\(|A.F.{|_N} = \;cos\left( {\frac{{\rm{\pi }}}{{\rm{\lambda }}}.h} \right)\)

|A.F.|_{N} will be maximum if

\(\frac{{\rm{\pi }}}{{\rm{\lambda }}}.h = n\pi\) where n = 0, 1, 2, 3....

For h_{min}, n = 1

\(\frac{{\rm{\pi }}}{{\rm{\lambda }}}.h = \pi \)

\(h = \lambda\)Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link

Option 3 : remains unchanged

__Concept__:

The capacity of a wireless link is given by Shanon Hartley theorem as:

\(C = B~lo{g_2}\left( {1 + \frac{S}{N}} \right)\)

Channel capacity depends on \(\left( {\frac{S}{N}} \right)\), and hence on S.

S = Received power at the receiver = P_{r}

\(S = {P_r} = \frac{{{P_T}{G_t}{G_r}}}{{{{\left( {\frac{{4\;{\bf{\pi }}R}}{\lambda }} \right)}^2}}}\)

\(= \frac{{{P_t}\left( {\frac{{{G_\pi }}}{{{\lambda ^2}}}{A_{te}}} \right)\left( {\frac{{4\pi }}{{{\lambda ^2}}}{A_{re}}} \right)}}{{{{\left( {\frac{{4\pi R}}{\lambda }} \right)}^2}}}\)

__Calculation__:

It is given,

A_{te' }= 2A_{te }

A_{re' }= 2 A_{re}

R' = 2 R

\({P_{r'}} = \frac{{{P_t}\left( {\frac{{4\pi \;{A_{te}}}}{{{\lambda ^2}}}} \right)\left( {\frac{{4\pi }}{{{\lambda ^2}}}{A_{rc}}} \right)2 \times 2}}{{{{\left( {\frac{{4\pi R}}{\lambda }} \right)}^2}{{\left( 2 \right)}^2}}}\)

P_{r}' = P_{r}

S' = S

\(\frac{{S'}}{N} = \frac{S}{N}\)

⇒ C' = C, i.e.

The maximum capacity of the channel remains unchanged.

Option 3 : 90°

The radiation pattern of a Hertzian dipole is shown below:

Let us consider, \(l\left( {OB} \right) = x\)

∴ \(l\left( {OA} \right) = \frac{x}{2} \times \sqrt 2 = \frac{x}{{\sqrt 2 }}\;\)

Point A is \(\left( {\frac{1}{{\sqrt 2 }}} \right)\) times electric field at point B

It is a half-power bandwidth point.

∠ AOD = Half power bandwidth is

HPBW=90°

Consider the following statements:

1. The antennas radiate energy.

2. An antenna is a transition device or transducer between a guided wave and a free space wave or vice versa.

3. The resonators and transmission lines store energy.

4. An antenna converts electromagnetic signals to currents or vice versa.

Which of these statements are correct?

Option 1 : 1, 2 and 4 only

**Explanation:**

**An antenna is a transducer; that is, a device that converts signals in one form into another form. **

In the case of an antenna, these two forms are:

(1) conductor-bound voltage and current signals and

(2) electromagnetic waves.

- The Antenna radiates energy
**because in operation they all carry time-varying currents and, consequently, accelerating electrons.**The dipole antenna is an example of a distributed circuit that owes its existence to the fact that it radiates well. - It is designed for the efficient conversion of electrical energy into radio waves.
- But any system of conductors that carry varying currents behaves in accordance with the principles.
- The same processes, including radiation, take place whether we call the system an antenna or something else. For example, what we generally refer to as reflection from a conducting surface (the radar dish, the ground) is actually radiation from free electrons set in motion by incident electric fields.

From the above explanation statements, **1,2, and 4 are satisfied hence they are true**.

- When one end of the waveguide is terminated in a shorting plate then reflections will occur and when another shorting plate is kept at a distance of a multiple of λ/2, then hollow space so formed can support a signal which bounces back and forth between the two shorting plates.
**This results in resonance and the hollow space is called cavity and the resonator as the cavity resonator.** - An electrical circuit composed of discrete components can act as a resonator when both an inductor and capacitor are included.
- When a source of radio waves at one of the cavity's resonant frequencies is applied
**, the oppositely-moving waves form standing waves and the cavity stores electromagnetic energy.****** **Whereas**when a sinusoidal source is applied to the input of an ideal twin lead transmission line.**The spacing between conductors is much less than a wavelength, and the output of the transmission line is terminated into an open circuit.**- Hence no standing waves are formed and due to this transmission lines don't store energy.

**So resonators store energy but transmission lines do not**.

Hence statement 3 is false.

**The solution is option 1.**

Option 1 : 39 dB

**Concept:**

The efficiency of dish antenna is given by:

\(Gain \ = \ \frac{η \ 4 \ \pi \ A_e}{λ^2}\) ---(1)

Where,

η = efficiency

A_{e} = Area of parabolic antenna

λ = Wavelength

**Calculation:**

η = 80 %

ν = 9.5 GHz

D = 1 m

As c = νλ

\(\lambda=\frac{3 \ \times \ 10^8}{9.5 \ \times \ 10^9}=0.0315\)

\(A_e=\frac{\pi \ D^2}{4}=\frac{\pi}{4}\)

Putting all values in equation (1)

\(Gain = \frac{0.80 \ \times 4\pi^2}{4 \ (0.0315)^2}=7912.62\)

**Gain in db = 10log _{10}(7917.62) = 39 db.**

__Important Points__

1. The parabolic reflector antenna is also known as the satellite dish antenna.

2. They have very high gain, bandwidth, and low cross-polarization.

3. They operate in the VHF region of frequency range from 30 - 300 MHz

Option 4 : Isotropic antenna

__Concept:__

- The gain of an antenna is the ratio of the radiation intensity in a given direction to the radiation intensity that would be obtained if the power were radiated isotropically.
- The gain of an antenna takes the directivity into account.
- The term antenna gain describes how much power is transmitted in the direction of peak radiation to that of an isotropic source.

Isotropic source: It radiates equally in all directions.

Mathematically, the gain is defined as:

G = ηeD

ηe = Antennas efficiency

D = Directivity of the antenna

Since the value of ηe lies between 0 and 1, the directive gain of the antenna lies between 0 and the directivity of the antenna.

Directivity: The ratio of the radiation intensity in a given direction from an antenna to the radiation intensity averaged over all directions is termed as directivity.

Antenna's Efficiency: Antenna's Efficiency is the ratio of the radiated power of the antenna to the input power accepted by the antenna.

Option 2 : Both antennas radiate same power

__Concept:__

P_{rad} = I_{rms}^{2} R_{rad}

Where

I_{rms} = RMS current

R_{rad} = radiated resistance

P_{rad} = radiated power

So P_{rad} ∝ R_{rad}

and \({R_{rad}} = 80\;{\pi ^2}{\left( {\frac{l}{\lambda }} \right)^2}\)

l = length of antenna

λ = wavelength

Since \(\lambda = \frac{c}{f}\)

So R_{rad} ∝ l^{2}f^{2}

__Solution:__

Given:

dioples are fed with same current (I_{1 }= I_{2})

l_{1} = 1.5 m, f_{1} = 100 MHz = 100 × 10^{6} Hz

l_{2} = 15 m, f_{2} = 10 MHz = 10 × 10^{6} Hz

So, \(\frac{{{P_{rad}}_1}}{{{P_{rad}}_2}} = \frac{{l_1^2f_1^2}}{{l_2^2f_2^2}}\)

\(\frac{{{P_{rad}}_1}}{{{P_{rad}}_2}} = {\left[ {\frac{{1.5 \;\times \;100\; \times \;{{10}^6}}}{{15 \;\times \;10\; \times \;{{10}^6}}}} \right]^2}\)

= \({\left[ {\frac{1}{1}} \right]^2}\)

So, P_{rad1}= P_{rad2}

**Statement (I****):** Slot antenna is essentially a wave-guide with a slot length of one half lengths with a radiation pattern similar to that of a dipole with plane reflection.

Option 3 : Statement (I) is true but Statement (II) is false

__Slot antenna____:__

The slot antenna is a very efficient radiator.

A λ/2 slot is cut in a flat metal sheet in such a way that the width of the slot is small i.e. w < λ. (Statement I is correct)

Therefore the currents are not confined to the edges of the slot but spread out over the sheet.

The gain of the slot antenna is in the range of 10 – 15 dBi and Horn is in the range of 25 dBi. **(Statement II is Incorrect)**

Hence, Statement (I) is true but Statement (II) is false.

Option 4 : none of these

__Concept:__

The gain of an antenna is given by the product of the radiation efficiency of the antenna and the directivity of the antenna.

Product of directivity and efficiency.

G = η D (Power Gain)

\(G = (\frac{\pi d}{\lambda})^2 \times aperture ~efficiency\)

Where;

η = efficiency and ‘D’ is the Directivity

**Analysis:**

D = 30 dB = 1000

G = 0.55 × 1000 = 550

\(G = (\frac{\pi d}{\lambda})^2 \times aperture ~efficiency\)

Solving the above, we get

d = 10 m

The half-power beamwidth is given as:

\(\frac{70~\lambda}{d}= 7\)°

Option 1 : Its length is an integral multiple of \(\frac{{n\lambda }}{2}\)

__Analysis__:

Mathematically, it has been obtained that when βl = nπ, the antenna behaves as a short circuit.

βl = phase length

Since one complete time period or wavelength of a sinusoidal sin wave covers 2π radians, i.e.

λ = 2π radians

\(\pi =\frac{\lambda}{2}\)

The expression for resonance can be written as:

\(β l=\frac{{n\lambda }}{2}\)

∴ An antenna behaves as a Resonant Circuit if Its length is an integral multiple of \(\frac{{n\lambda }}{2}\).