Module I: Electromagnetic waves

Transcription

1 Module I: Electromagnetic waves Lecture 9: EM radiation Amol Dighe

2 Outline 1 Electric and magnetic fields: radiation components 2 Energy carried by radiation 3 Radiation from antennas

3 Coming up... 1 Electric and magnetic fields: radiation components 2 Energy carried by radiation 3 Radiation from antennas

4 E and B from A and φ Now that we know φ( x, t) = 1 4πɛ 0 ρ( x, t ) x x d 3 x, A( x, t) = µ 0 4π J( x, t ) x x d 3 x We can calculate E = φ A/ t and B = A to obtain E( x, t) and B( x, t). However note the subtle point: = ˆx i x i, t t = t, (2) x whereas ρ( x, t ) and J( x, t ) are specified in terms of the retarded time t. Hence, we ll have to use ρ( x, t ) x i = ρ( x, t ) t t J( x, t ) x i = J( x, t ) t t t x i t x i t (1) (3) t

5 Intermediate derivatives: gradient of φ Defining r x x = (r 1, r 2, r 3 ) and r r, we have ρ( x, t ) x i r = ρ( x, t ) ( ) 1 t t x i + 1 r r x i ρ( x, t ) (4) t Further using we get φ = x i ( 1 r ) t = r i r 3, x i ρ( x, t ) = ρ t r t t r dx i = ρ( x, t ) 1 t t c = = [ 1 ρ( x, t )ˆx i 4πɛ 0 [ 1 ρ( x, t ) 4πɛ 0 r 2 1 4πɛ 0 ( ) 1 t x i + 1 r r ˆxi ˆr ˆr cr [ [ρ( x )] ˆr r 2 [ ρ( x )] cr ρ( x, t ) t r i r, (5) ] x i ρ( x, t ) d 3 x t ] d 3 x ] d 3 x (6)

6 Intermediate derivatives: curl of A Similar to the case of derivatives of ρ, we have x i J( x, t ) = J t r t t r dx i = J( x, t ) 1 r i t c r, (7) t And then A = µ 0 4π [ ( ) ˆx i 1 x i r J( x, t ) + 1 r ˆxi = µ [ 0 J( x, t ) 4π r 2 ˆr + 1 ] J( x, t ) ˆr d 3 x cr t = µ [ 0 [ J( x )] ˆr 4π r 2 + [ ] J( x )] ˆr cr ] dx i J( x, t ) t d 3 x d 3 x (8)

7 Electric and magnetic fields The electric field can now be calculated to be E( x, t) = φ A t ( 1 [ρ( x E( x, )] t) = 4πɛ 0 The magnetic field is ˆr r 2 + [ ρ( x )] cr ˆr [ ) J( ~x )] c 2 d 3 x (9) r B( x, t) = A B( x, t) = µ ( 0 [ J( x )] 4π r 2 ˆr + [ ) J( ~x )] ˆr d 3 x (10) cr

8 E( x, t) and B( x, t): behaviour at large x In both, E( x, t) and B( x, t), there are terms that behave as 1/r 2 and there are terms that behave as 1/r. The former are proportional to the sources, the latter are proportional to the rate of change of sources. When the sources are confined to a small region x < d, then for x >> d, the 1/r terms dominate over the others. These are the radiative components of the fields. The radiative E and B fields are then E rad ( x, t) ( 1 [ ρ( x )] = ˆr [ ) J( x )] 4πɛ 0 cr c 2 d 3 x (11) r B rad ( x, t) = µ ( 0 [ ) J( x )] ˆr d 3 x (12) 4π cr For sufficiently large r, we can take r x.

9 For monochromatic sources Monochromatic sources correspond to ρ( x, t) = ρ 0 ( x )e iωt and J( x, t) = J 0 ( x )e iωt. ( Real part always implicit) This dependence implies [ ρ( x ) ] = ρ( x, t ) = ρ 0 ( x )e i(kr ωt), (13) [ J( x )] = J( x, t ) = J 0 ( x )e i(kr ωt). (14) And hence 1 φ( x, t) = 4πɛ 0 A( x, t) = µ 0 4π ρ 0 ( x ) ei(kr ωt) d 3 x, r J0 ( x ) ei(kr ωt) d 3 x (15) r The potentials also then have the same frequency ω.

10 E rad and B rad in terms of J only Given the continuity equation J( x, t) + ρ( x, t) = 0, (16) it is clear that if [ J( x )] is known everywhere, so is [ ρ( x )], and the radiative E and B fields can be written in terms of J only. For monochromatic sources, Some algebraic manipulation using the above result yields (See Panofsky-Phillips) B rad ( x, t) = E rad ( x, t) 1 4πɛ 0 c 3 1 4πɛ 0 c 2 [ J( x )] ˆr d 3 x r (17) ([ J( x )] ˆr) ˆr d 3 x r (18) Note that µ 0 /(4π) = 1/(4πɛ 0 c 2 ), one can use any combination.

11 Frequency components of radiation fields Fourier components of B rad ( x, t) and E rad ( x, t) give the radiation fields in terms of their frequency components: B rad ω ( x) = E rad ω ( x) = i ( Jω 4πɛc 2 ( x ) e ikr k) d 3 x (19) r i ( ( 4πɛ 0 c J ω ( x ) e ikr k) ˆr) d 3 x (20) r If the sources are monochromatic, the above directly give the corresponding E and B fields. We shall use this in calculating the power radiated by periodically time-varying charges and currents.

12 Coming up... 1 Electric and magnetic fields: radiation components 2 Energy carried by radiation 3 Radiation from antennas

13 Energy from a radiation pulse The Poynting vector is N = E H, which gives the power radiated per unit area along it. Total energy radiated per unit area normal to N is N( x, t)dt = = = 2π E( x, t) H( x, t)dt (21) ω,ω,t= E rad ω 0 E rad ω e iωt dω H rad H rad ωdω + 2π ω e iω t dω dt 0 E rad ω H rad ω dω (22) But since E( x, t) and H( x, t) are real, E ω = E ω and H ω = H ω. Then N( x, t)dt = 2π 0 E rad ω ( H rad ω ) dω + c.c. (23)

14 Calculating radiated energy for a pulse Substituting the expressions for E rad ω earlier, after a bit of algebra, gives E rad ω ( H rad ω ) = 1 µ0 (4π) 2 ɛ 0 and H rad ω = B rad ω /µ 0 obtained ( Jω ( x ) ) e ikr 2 k d 3 x ˆr (24) r Total radiated energy across a surface ds = r 2 Ωˆr is U = 2π E rad ω U = 1 µ0 4π ɛ 0 ( H rad ω ) dω r 2 dωˆr + c.c. (25) ω ( Jω ( x ) ) e ikr 2 k d 3 x dωr 2 dω (26) r In other words, du ω dω = 1 µ0 4π ɛ 0 ( Jω ( x ) ) 2 k e ikr d 3 x (27)

15 Average power radiated by a monochromatic source Here we calculate the average radiated power /area over a cycle: 1 T N( x, t)dt = 1 T E( x, t) H( x, T 0 T t)dt (28) 0 where T is the periodicity of the wave. Since E rad ( x, t) = E rad 0 e iωt and H rad ( x, t) = H rad 0 e iωt, the averaging gives N = 1 2 E 0 rad H0 rad = (4π) 2 µ0 ɛ 0 ( J0 ( x ) ) e ikr 2 k d 3 x ˆr (29) r The average power radiated is then d P dω = (4π) 2 µ0 ɛ 0 ( J0 ( x ) ) 2 k e ikr d 3 x (30)

16 Long-distance approximation When x >> x, then we have kr = k x x k x k x (31) Then for a radiation pulse, du ω dω = 1 µ0 4π ɛ 0 And for a monochromatic source, d P dω = (4π) 2 µ0 ɛ 0 ( Jω ( x ) ) k e i k x 2 d 3 x ( J0 ( x ) ) k e i k x 2 d 3 x (32) (33) This can be used in many instances, for example for radiation from antennas, as will be seen in later chapters.

17 Coming up... 1 Electric and magnetic fields: radiation components 2 Energy carried by radiation 3 Radiation from antennas

18 Dipole antenna from a coaxial cable Feeding through a coaxial cable: An antenna may be represented as a monochromatic sinusoidal current with frequency ω: [ ( )] J0 ( x I 0 L ) = sin(kl/2) δ(x )δ(y ) sin k 2 z ẑ (34) (Note: direction of current is the same for z > 0 and z < 0)

19 Calculating power radiated We have x = (0, 0, z ). Using azimuthal symmetry, we choose ˆr = (sin θ, 0, cos θ), so that k = k(sin θ, 0, cos θ), so that J0 ( x ) k = I [ ( )] 0k sin θ L sin(kl/2) δ(x )δ(y ) sin k 2 z ŷ k x = kz cos θ (35) Then we get ( J 0 ( x ) k)e i k x d 3 x [ ( 2I 0 kl cos θ = cos sin θ sin(kl/2) 2 ) cos ( ) ] kl 2 (36) This can be used to calculate the radiation pattern.

20 Dependence of radiation pattern on kl In the large wavelength limit, kl 1 ( J( x ) k)e i k x d 3 x = I 0 kl sin θ (37) 2 The average power radiated is then d P dω = (4π) 2 = (4π) 2 Dipole radiation pattern! µ0 ( J0 ɛ 0 ( x ) ) k e i k x 2 d 3 x µ0 (I 0 kl) 2 sin 2 θ (38) ɛ 0 As kl increases, stronger angular dependences appear. Check CDF demo at

21 Antenna patterns: dipole ps7183/ps469/prod_white_paper0900aecd806a1a3e.html

22 Antenna patterns: patch (Note: scale in db) ps7183/ps469/prod_white_paper0900aecd806a1a3e.html

23 Antenna patterns: patch array (Note: scale in db) ps7183/ps469/prod_white_paper0900aecd806a1a3e.html

24 Antenna patterns: yagi (Note: scale in db) ps7183/ps469/prod_white_paper0900aecd806a1a3e.html

25 Antenna patterns: dish (Note: scale in db)

26 Recap of topics covered in this lecture E and B fields in the presence of moving sources Radiative components of E and B: the 1/r behaviour that dominates at large distances Poynting vector and power radiated by EM waves Long distance approximation for radiated power Radiation patterns in antennas