Unit 6| Radiation P a g e | 1

Syllabus: Retarded potential, Electric and magnetic fields due to oscillating dipole (alternating current element), power radiated and radiation resistance, application to short monopole and dipole. Antenna Efficiency, Beam-width, Radiation Intensity, Directive Gain Power Gain & Front to Back Ratio. Advance topics on the subject.

RETARDED POTENTIAL

Explain the term Retarded Potential (S-14/4m) (W-15/5m) Time varying fields, produced by time varying sources, propagate point to point from source towards field point at a finite velocity. The velocity of propagation of time varying fields in free space is 3 × 10 /. Therefore, the time varying fields take a time Δ = / to propagate from source to field point. Therefore, the fields at instant at the field point correspond to the source at instant − Δ at the source point. For the source at instant , the corresponding field at the field point is obtained at instant + Δ. Therefore, the fields at the field point lag their sources in time, by an amount of the propagation delay time Δ = / seconds. These time lagging fields at field point are called as retarded fields. Therefore, the time varying potentials at field point are called as retarded potentials. For the practical distribution of a charge in a volume, the potential at a field point P is obtained as

= 4 The retarded potential at field point P due to time varying source is obtained as

() (/) () = = 4 4

RETARDED MAGNETIC VECTOR POTENTIAL

Write short note on Retarded Magnetic Vector Potential (W-16/5m) (W-14/6m) (S-15/5m) The concept of retarded vector magnetic potential ̅ is very useful to derive radiation field of antenna elements including current element.

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The vector magnetic potential ̅ at a field point (, , ) in space is given as ̅ ( − ⁄) ̅ = = (̅ − ⁄) 4 4

As the current is flowing in direction, the direction of current density ̅ will also be . Thus direction of ̅ will be . Such that, ̅ = Volume integration of current density is given as

= since is a vector quantity . . ̅ and its direction is , we have,

̅ = Since current flowing in the conductor is time varying, so, current density will be a function of time,

(̅ ) = cos

Where, is a differential current element in direction, it is given as . At a distance the delay of Δ will take place. Thus cos at a distant point corresponding to the current will be cos ( − Δ). But Δ = ⁄

̅ ∴ ( − ⁄) = cos ( − ⁄) The vector magnetic potential at a field point, which is at a distance from the source point is given by ̅ = cos ( − /) 4 Thus,

cos ( − /) = 4 This equation shows retarded vector magnetic potential.

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DEFINATION OF IMPORTANT TERMS: Antenna Gain Define Antenna Gain (S-16) The antenna gain is defined as the ratio of radiation intensity in a direction i.e. 4(, ) to the radiation intensity that would be obtained if the power fed to the antenna is radiated isotropically1. It is given as, (, ) (, ) = 4 It is also defined as the product of antenna efficiency and directivity , it is given as = It is also defined as the ratio of signal transmitted by an antenna in the maximum direction to that of a standard or reference antenna2. Antenna Bandwidth Antenna bandwidth is defined as the range of frequencies in which the antenna system maintains few important characteristics like gain, front-to-back ratio (FTB), standing wave ratio (SWR), impedance, polarization, etc. The antenna bandwidth depends upon two parameters: radiation pattern and impedance. At low frequency, the bandwidth of the antenna is inversly proportional to Q factor of antenna. It can be expressed as:

ℎ = = Δ = − = ∵ = 2

∴ Δ = − =

Where, is the center frequency or design frequency or resonant frequency, and Q factor is given as = 2 Therefore, for antennas with lower Q factor, the antenna bandwidth is very high and vice versa. Radiation Intensity Define Radiation Intensity (S-14) (S-15) (S-16) (W-14) (W-15) (W-16) pg600/601 The radiation intensity of an antenna does not depend on the distance from the radiator or antenna. It is denoted by U. The radiation intensity is defined as power per unit solid angle. It is expressed in Watts per Steradian (i.e. W/sr). It is given as,

(, ) = (, ) The average value of radiation intensity is given as

1 Isotropic means identical in all direction 2 Reference antenna are dipole antenna and isotropic radiator.

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= 4 Beam Width Define Beam Width (S-15) (S-16) (W-14) (W-15) (W-16) It is the measure of directivity of the antenna. The antenna beam width is an angular width in degrees. It is measured on a radiation pattern on major lobe. It is defined as the angular width in degrees between the two points on a major lobe of a radiation pattern where the radiated power decreases to half of its maximum value. Because of this, it is also called as Half Power Beam Width (HPBW). The beam width is also called as 3-dB Beam Width as reduction of power to half of its maximum value corresponds to the reduction of power by 3-dB.

Mostly, the antenna pattern is expressed in terms of the angular width between first side lobes or first nulls. Then such angular beam width is called as First Nulls Beam Width (FNBW).

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Radiation Resistance Define Radiation Resistance (S-15) Radiation Resistance is defined as a fictitious resistance such that when it is connected in series with antenna dissipates same power as the antenna radiates. It is denoted by . But practically, the energy supplied to the antenna is not completely radiated in the form of electromagnetic waves, but there are certain radiation losses due to the loss resistance denoted by . Thus, the total power is given by, = ℎ + = + = ( + ) The radiation resistance of antenna depends on antenna configuration, ratio of length and diameter of conductor used, location of the antenna with respect to ground and other objects. Directivity or Directive Gain Define Directivity (S-15) (S-16) Define Directive Gain (W-15) (W-16) (S-14) (S-16)

The directive gain is defined as the ratio of the power density (,) to the average power radiated. For isotropic antenna, the value of the directivity gain is unity. The directive gain is given as,

(,) (, ) = The average power in terms of radiated power is given as,

= / 4 (,) (, ) ∴ (, ) = = 4 4 The numerator in the above ratio is the radiation intensity while the denomenator is the average value of the radiation intensity. Therefore, (, ) (, ) = Thus the directive gain can be defined as a measure of the concentration of the radiated power in a particular direction (, ). The ratio of maximum power density to the average power radiated ( ) is called Maximum directive gain or Directivity of the antenna. It is given by,

= = = 4 The directivity of an antenna is a dimensionless quantity.

The directivity can be expressed in terms of beam solid angle Ω or Beam Area as,

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4 4 = = Ω Antenna efficiency (or Radiation Efficiency) and Power Gain Define Antenna Efficiency pg.602 (W-14) (W-15) (W-16) & Define Power Gain (S-14) (W-15) The practical antenna is made up of a conductor having finite conductivity and resistivity, which incur ohmic power loss in the antenna. It is given by = . Thus, radiated power becomes less than input power . The Radiation efficiency or Antenna efficiency is given as

=

As = +

= + ∵ = & = = = + ( + )

= + The ratio of the power radiated in a particular direction (, ) to the actual power input to the antenna is called Power gain of antenna. It is denoted by (, ) and is given by

(, ) (, ) =

The maximum power gain is defined as the ratio of the maximum radiation intensity to the radiation intensity due to isotropic lossless antenna. It is given as,

= 4

∵ = × 4 × 4 = 4 × × 4 = 4

=

Generally, both power gain and directional gain (directivity) are expressed in decibles (dB).

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Front to back ratio Define Front to back ratio (W-14) (W-15) (W-16) (S-14) (S-16) Front to Back Ratio (FBR) is defined as the ratio of the power radiated in the desired direction to the power radiated in the opposite direction. Ideally, power radiated in desired direction should be 100% and power radiated in opposite direction should be 0%. Thus, ideally FBR must be infinity. But practically, it must be as large as possible, i.e. power radiated in desired direction must be very high as compared to the power radiated in opposite direction. FBR depends upon frequency of operation. So, when frequency of antenna changes, the FBR also changes. Similarly, FBR depends on the spacing between the antenna elements. So, if the spacing between antenna elements increases, the FBR decreases. The FBR also depends upon the electrical length of the parasitic elements of the antenna. The FBR can be raised by diverting the gain of backward direction to the forward (desired) direction by adjusting the length of parasitic elements of antennas. This is known as tuning.

Effective Area or Effective aperture Define Effective Area (W-16) It is defined as the ratio of power received by the load to the average power density produced at a point. It is given as,

= It is measured as . Thus, it is an area which extracts energy from the electromagnetic wave, out of the total area of antenna. The maximum effective aperture is obtained when power received is maximum. Effective Length Define Effective Length (S-16)

The effective length of an antenna carrying peak current Im is defined as the length of an imaginary linear antenna with a uniform distributed current, such that both the antenna have the same far field in q = p/2 plane. It is represented by Leff. The effective length of a receiving antenna is defined as the ratio of the open circuit voltage VOC induced at the open terminal of an antenna to the incident electric field intensity Ei producing VOC. It is given by

The effective length of an antenna used for transmitting is same as that for the same antenna used for receiving.

INDUCTION FIELD AND RADIATION FIELD

Explain what is Induction field and Radiation field. (W-15/4m) The two distinct electromagnetic fields associated with an antenna are Induction field and

Radiation field. The expression for field component in direction

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---(1) The induction field consists of the lines of force, which are setup by the voltage and current in the antenna and which collapse back into antenna twice each cycle. It contains only reactive energy. The electric and magnetic fields are 900 out of phase with each other. Above equation (1) shows that the intensity of induction field is inversely proportional to the square of the distance. Thus, with the increase in distance from antenna, the induction field decreases very rapidly. In the distance very close to antenna i.e. 1/8 of wavelength or less, the induction field is very strong and may cause the induction of large voltages and currents in nearby conductors. The radiation field consists of lines of force, which have become detached from the antenna and are moving on out into space as an electromagnetic wave. It contains real power. The electric and magnetic field are in phase with each other, so that actual power is removed from the antenna and carried away by this field. Above equation (1) shows that, the intensity of the radiation field is inversely proportional to the distance from the antenna. Thus, with the increase in distance from antenna, the induction field decreases not as rapidly as radiation field. At a distance of 1/2p wavelength, the induction and radiation fields are equal in intensity. At distances shorter than this (I.e. 1/2p wavelength), the induction field is stronger than radiation field. At distances larger than this (I.e. 1/2p wavelength), the radiation field is stronger than induction field. For communication, radiation field is thus of greater importance. This field extends to larger distances with sufficient intensity that is useful for transmitting information. This is also called as “Radio Signal”. The Radiation field term in equation (1) shows flow of energy away from the current element while the Induction field term shows the energy stored in the field during one quarter of the cycle, which is returned back during next cycle. The condition at which the amplitude of both induction and radiation fields are equal is given by

RADIATION RESISTANCE

( ) Given that average power radiated by a current carrying element is . Find the expression for the total radiated power. Also, find radiation resistance. (S-15/8m) (S-16/8m)

Show that the radiation resistance of current element is given by = ohms. (S- 14/9m) (W-16/14m) (W-14/13m) The electric and magnetic fields of alternating current element for any general direction of propagation are given as

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sin − sin{( − /)} cos{( − /)} = + 4 2 cos cos{( − /)} sin{( − /)} = + 4 sin − sin{( − /)} cos{( − /)} sin{( − /)} = + + 4

The component of Poynting’s vector in direction is

= − 2 cos cos{( − /)} sin{( − /)} sin − sin{( − /)} = − + 4 4 cos{( − /)} + 2 sin cos cos{2( − /)} sin{2( − /)} sin{2( − /)} = − − − + 16 2 2 sin 2 cos{2( − /)} sin{2( − /)} sin{2( − /)} = − − + 16 2 2

The component of Poynting’s vector in direction is

= sin − sin{( − /)} cos{( − /)} = + 4 sin{( − /)} sin − sin{( − /)} cos{( − /)} + + 4 sin sin{( − /)} cos{( − /)} sin{( − /)} = − 16 sin{( − /)} cos{( − /)} cos{( − /)} − + sin{( − /)} sin{( − /)} cos{( − /)} − + sin sin{( − /)} × 2 sin{( − /)} cos{( − /)} = − 16 cos{( − /)} − sin{( − /)} sin{( − /)} cos{( − /)} + + sin (1 − 2{( − /)}) sin 2{( − /)} 2{( − /)} = − + 16 2 sin 2{( − /)} + 2

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sin 2{( − /)} sin 2{( − /)} 2{( − /)} = − − + 16 2 2 sin 2{( − /)} + 2 The above equation consists of sine and cosine terms. The average value of sine and cosine terms becomes zero. Thus, if we take average of the sine and cosine terms will become zero. sin ∴ = − 0 − 0 + 0 + 0 16 2

sin 1 sin 1 sin = × = = 1 16 2 2 4 2 4 √ √ sin / sin = = 2√√ 4 2 4

= Above equation gives an expression for average power radiated by a current carrying element. The total radiated power can be obtained by integrating above equation over a complete range of sphere of radius r.

sin sin = = = sin 2 4 2 4

sin sin = sin = × sin 2 4 2 4

3 sin − sin 3 = × sin = × 2 4 2 4 4 3 sin − sin 3 = × 2 4 4

3 ∫ sin − ∫ sin 3 = × 2 4 4

cos 3 [ ] 3 − cos − − 3 = × [] 2 4 4

cos 3 cos 0 [ ] 3 − cos + cos 0 − − 3 + 3 = × [2] 2 4 4

(−1) 1 3[−(−1) + 1] − − + 3 3 = × [2] 2 4 4

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2 6 − 18 − 2 16 = × 3 [2] = × [2] = × [2] 2 4 4 2 4 12 2 4 12

1 = × 1 12 = 12 Above expression gives Power radiated by constant current antenna.

/(2) 376.7 × 4 × = = 12() 12 376.7 × 4 = × × = 40 × × 12 Where, is the peak value of current. The rms value (effective value) of current for complete cycle

. . is given as

= ! = √2 √2 ∴ = 2 ∴ = 2 × 376.7 × 4 = × × = 40 × × 2 × 12 = 80 × = × 80 The power is given as

= × Comparing above two equations, we get,

= 80 Ω

Above equation is an expression for Radiation Resistance

= Ω Hence, Proved. For z direction of propagation, = , we get,

= Ω

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SHORT ANTENNA The radiation resistance for practical short dipole antenna is one fourth of that of current element of same length. Therefore,

80 = = 20 ( ) 4 The radiation resistance of the monopole antenna of height ℎ = /2 is

20 2ℎ ℎ = = 10 = 10 = 40 () 2

Show that =

NUMERICALS

Calculate the power radiated and radiation resistance of a current element . long and carrying current of 5 Amp. (S-15/7m) Given: = 0.2; = 5 ; The radiation resistance is given by

0.2 = 80 = 80 = 80 (0.2)

= 31.58 Ω The power radiated is given as

5 = × = × = × 31.58 √2 √2 = 394.78

An antenna has a radiation resistance of , a loss resistance of and power gain of . Determine antenna efficiency and its directivity. (S-16/6m) (W-15/7m)

Given: = 72 Ω; = = 8 Ω; = () = 12

The antenna efficiency is given as × 72 = = = = = 0.9 + × + × + 72 + 8

Percentage antenna efficiency is

= 90% The directivity is given as

() 12 () = = 0.9

() = 13.33

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Note: If power gain was not given in dB then would have been converted to dB as

() = 10 log

A monopole antenna of height 10 cm operates at a frequency of 300 MHz and is situated above ground. Find its radiation resistance. (S-14/6m) (W-15/4m) Given: ℎ ℎ = 10 = 0.1; = 300; 3 × 10 = = = 1 300 × 10 The radiation resistance of the monopole antenna is

ℎ 0.1 = 40 = 40 = 40 × 0.01 () 1

() = 3.947 Ω

A transmitting antenna having an effective height of 100 meters has a current at the base 100 A at the frequency of 300 KHz. Calculate (i) The field strength at a distance of 100 km (ii) The value of radiation resistance (iii) The power radiated. (S-16/6m) Given: ℎ = = 100 ; = 100 ; = 300 ; 3 × 10 = = = 1 300 × 10 The radiation electric field amplitude at a distant field point P is given as:

sin = 4 For maximum radiation, consider the maximum value of sine i.e. sin = 1

∴ = 4 The given current is a peak current. The average of the peak current is approximately half of the peak current. i.e. = /2 Part (i): The field strength (i.e. amplitude of electric field) at a distance of 100 Km (i.e. r = 100 Km) is given as

(/2)(2) (100/2)(100 )(2 × 300 × 10 ) = = = 4 4(100 × 10 )(3 × 10 ) 4(100 × 10 )(3 × 10 ) = 9.41 × 10 / Part (ii): ℎ 100 1 = = 1000 10 ℎ = 10 Therefore, antenna acts as a short monopole. The radiation resistance of short monopole is given as

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ℎ 1 = 40 = 40 10

= 3.947 Ω

Part (iii): Power radiated by constant current antenna is given by As the given antenna is short monopole, the power radiated reduces to one-eighth of the power radiated by constant current antenna.

(2) × 100 × 100 1 1 = × = × ⎛ ⎞ 8 12 8 12(3 × 10) ⎝ ⎠ (2 × 300 × 10) × 100 × 100 1 = × ⎛ ⎞ 8 12(3 × 10) ⎝ ⎠ = 4934.8

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