Basics of Wave Propagation

Transcription

1 Basics of Wave Propagation S. R. Zinka Department of Electrical & Electronics Engineering BITS Pilani, Hyderbad Campus May 7, 2015

2 Outline 1 Time Harmonic Fields 2 Helmholtz Wave Equation 3 Lossy Materials 4 Poynting Vector 5 Reflection 6 Summary

3 Outline 1 Time Harmonic Fields 2 Helmholtz Wave Equation 3 Lossy Materials 4 Poynting Vector 5 Reflection 6 Summary

4 Complex Notation Any time-varying field such as F = F (x, y, z) cos (ωt + ψ) â can be written using of Euler s identity as, F = Re [F (x, y, z) e j(ωt+ψ)] [ â = Re F (x, y, z) e jψ e jωt] [ â = Re F s e jωt] â.

5 Let s Re-write Maxwell s Equations in Complex Form ( Ds e jωt) = ρ e,s e jωt ( Bs e jωt) = ρ m,s e jωt ( ) ( Hs e jωt) = Je,s e jωt + Ds e jωt ( ) = Je,s + jω Ds e jωt t ( ) ( Es e jωt) = Jm,s e jωt Bs e jωt ) = ( Jm,s jω Bs e jωt t

6 So, Maxwell s Equations in Complex Form are... Ds = ρ e,s Bs = ρ m,s Hs = Je,s + jω Ds Es = Jm,s jω Bs

7 Outline 1 Time Harmonic Fields 2 Helmholtz Wave Equation 3 Lossy Materials 4 Poynting Vector 5 Reflection 6 Summary

8 Wave The wave shown in the above diagram can be represented as F (x, t) = sin (βx βvt) = sin (βx ωt) (1) where, ω = 2πf = βv. (2)

9 Wave Equation Simple 1 - dimensional wave equation is given as 2 F x = 1 2 F 2 v 2 t 2 Using the complex notation, the above equation can be simplified as 2 F s x 2 = ( ) β 2 (jω) 2 F s = β 2 F s ω 2 F s x 2 + β2 F s = 0 (3) Using the theory of linear differential equations, solution for the above equation is given as F s = Ae jβx + Be jβx [( F = Re Ae jβx + Be jβx) e jωt] [ = Re Ae j(ωt+βx) + Be j(ωt βx)]. (4)

10 Now, let s prove that time harmonic electromagnetic fields exist in the form of waves...

11 Helmholtz Wave Equation In a source-less dielectric medium, Taking curl of (6) gives Ds = 0 Bs = 0 Hs = jω Ds = jωε Es (5) Es = jω Bs = jωµ Hs (6) ) ) ( Es = ( jωµ Hs ) ) ( Es 2 Es = jωµ ( Hs ) ) ( Es 2 Es = jωµ (jωε Es ) 2 Es = ( Es ω 2 µε Es 2 Es = 0 ω 2 µε Es (7) Similarly, it can be proved that 2 Hs = ω 2 µε Hs. (8)

12 Finally, Let s Analyze the Helmholtz Wave Equation Let s compare general wave equation (3) and Helmholtz wave equation (7). 2 Fs x 2 + β2 F s = 0 2 Es + ω 2 µε Es = 0 From the above comparison, we get, But, we already knew that β = ω µε. (9) v = ω β. So, from the above equations, we get v = 1 µε = 1 µr ε r c (10) where c is the light velocity.

13 Solution of Helmholtz Equation Vector Helmholtz equation can be decomposed as shown below: 2 E xs + ω 2 µεe xs = 0 2 Es + ω 2 µε Es = 0 2 E ys + ω 2 µεe ys = 0 2 E ys + ω 2 µεe ys = 0 Since all the differential equations are similar, let s solve just one equation using variable-separable method. If E xs can be decomposed into E xs = A (x) B (y) C (z) then substituting the above equation into Helmholtz equation gives 2 E xs x E xs y 2 2 E xs + ω 2 µεe xs = E xs z 2 + ω 2 µεe xs = 0 B (y) C (z) 2 A x + 2 A (x) C (z) 2 B y + 2 A (x) B (y) 2 C z + 2 ω2 µεa (x) B (y) C (z) = A A (x) x B 2 B (y) y C 2 C (z) z 2 γ2 = 0

14 Solution of Helmholtz Equation... Cont d 1 2 A A (x) x B 2 B (y) y C 2 C (z) z 2 γ2 = A A (x) x B 2 B (y) y C 2 C (z) z 2 γ2 x γy γ 2 z 2 = 0 (11) The above equation can be decomposed into 3 separate equations: 1 2 A A (x) x 2 γ2 x = B B (y) y 2 γ2 y = C C (z) z 2 γ2 z = 0 It is sufficient to solve only one of the above equations and it s solution is given as 2 A x 2 γ2 xa (x) = 0 A (x) = L 1 e γxx + L 2 e γxx = L e γxx + L + e γxx (12)

15 Solution of Helmholtz Equation... Cont d So, finally E xs is given as E xs = ( L e γxx + L + e γxx) ( M e γyy + M + e γyy) ( N e γzz + N + e γzz) (13) E x = Re [ (L e γxx + L + e γxx) ( M e γyy + M + e γyy) ( N e γzz + N + e γzz) e jωt] (14) with the condition γ 2 x + γ 2 y + γ 2 z = γ 2. (15)

16 Outline 1 Time Harmonic Fields 2 Helmholtz Wave Equation 3 Lossy Materials 4 Poynting Vector 5 Reflection 6 Summary

17 Complex Permittivity Ds = ρ e,s Bs = ρ m,s Hs = Je,s + jω Ds = σ Es + jωε Es = jω (1 }{{}}{{} ε j σ ) ωε E s }{{} conduction current displacement current εs Es = Jm,s jω Bs From now onwards, we will be using complex permitivity ɛ s = ε ( 1 j σ ωε ) instead of simple ɛ when ever possible. Another important definition is that of loss tangent: tan θ = σ ωε

18 Propagation Constant So, from the previous slide, for lossy dielectrics, ε s is a complex number and is given as ( ε s = ε 1 j σ ). ωε Then propagation constant γ is given from the equation γ 2 = ω 2 µε s ( = ω 2 µε 1 j σ ) ωε = ω 2 µε + jωµσ. (16) From the above equation, γ can be written as γ = α + jβ. Now, we need to find out the values of α and β. We have γ 2 = (α + jβ) 2 = ( α 2 β 2) + j (2αβ) (17) Comparing (16) and (17) we get, α 2 β 2 = ω 2 µε β = ωµσ 2α. (18)

19 Propagation Constant... Contd Solving the set of equations (18) gives ( ωµσ ) 2 α 2 = ω 2 µε 2α 4α 4 ω 2 µ 2 σ 2 = 4α 2 ω 2 µε 4α 4 + 4α 2 ω 2 µε ω 2 µ 2 σ 2 = 0 4ξ 2 + 4ξω 2 µε ω 2 µ 2 σ 2 = 0 [ ] ξ = ω2 µε ( σ ) 2 ± ωε α = [ ] ξ = ω µε ( σ ) (19) 2 ωε Similarly it can be proved that [ ] β = ω µε ( σ ) ωε (20)

20 Skin Depth Skin Depth: The distance δ, through which the wave amplitude decreases by a factor 1 e is called skin depth or penetration depth of the medium, that is From the above equation, E 0 e αδ = E 0 e. δ = 1 α (21)

21 Outline 1 Time Harmonic Fields 2 Helmholtz Wave Equation 3 Lossy Materials 4 Poynting Vector 5 Reflection 6 Summary

22 Poynting Vector One Vector Identity: ( ) ) ) A B = B ( A A ( B From the above vector identity, ( ) ) ) E H = H ( E E ( H ( = H = = ) ( Jm B E t ) ( ( H 0 B H t ( µ H H t ) ( σ E + D t ) ) σ E E + D E t ) σ E E + ε E E t ( ) H = µ H t + ε E E σ E E = µ 1 t 2 = ( µ t 2 H 2 + ε 2 E 2) σ E 2. ( ) H H t ( ) + ε 1 E E σ E E 2 t

23 Poynting Vector... Cont d ( ) E H = ( µ t 2 ( ) ( E H dv = t ( E ) H ds } {{ }??? H 2 + ε 2 E 2) σ E 2 ( µ 2 H 2 + ε 2 E 2) σ E 2) dv = ( µ t 2 H 2 + ε 2 E 2) dv σ E 2 dv }{{}}{{} Rate of decrease of stored Energy Ohmic power dissipated

24 Poynting Vector... Physical Interpretation P = E H = E H sin θˆk (22)

25 Instantaneous & Time Average Power Instantaneous power corresponding to the above set of voltage & current is defined as P inst (t) = v 0 i 0 cos (ωt + φ 1) cos (ωt + φ 2) Time average power is defined as = v 0i 0 2 [cos (ωt + φ 1 + ωt + φ 2) + cos (ωt + φ 1 ωt φ 2)] = v 0i 0 2 [cos (2ωt + φ 1 + φ 2) + cos (φ 1 φ 2)] (23) P avg = 1 T 0 ˆ T0 0 P inst dt = v 0i 0 2 cos (φ 1 φ 2) (24)

26 Time Average Power - Complex Notation v real = v 0 cos (ωt + φ 1) v complex = V = v 0 e j(ωt+φ 1) i real = i 0 cos (ωt + φ 2) i complex = I = i 0 e j(ωt+φ 2) P avg = v 0 i 0 2 cos (φ 1 φ 2) P avg = 1 2 Re (VI ) = v 0 i 0 2 cos (φ 1 φ 2)

27 Instantaneous & Time Average Poynting Vector Pinst = Einst Hinst Pavg = Re [ 1 ( ) ] Es H 2 s

28 Plane Wave in Free Space... Poynting Vector

29 Free Space / Uniform Dielectric Medium Impedance In source-less medium, Es = jω Bs ( E 0 e γzz ˆx ) = jωµ Hs So, Hs = j [ ( E0 e γzz ˆx )] ωµ = j [ ( E0 e γzz ˆx )] ωµ = j [ ] Exs ωµ z ŷ = j ωµ z ( ) jγ0 = E xs ŷ = j ( jω ) µε s ωµ ωµ E xs µ jωµ = = H ys ε s σ + jωε. [ (E0 e γzz ) E xs ŷ ] ŷ For loss-less case (i.e., α = 0 and σ = 0), E xs = ωµ H ys β = ωµ µ ω µε = ε. Wave Propagation Direction

30 Outline 1 Time Harmonic Fields 2 Helmholtz Wave Equation 3 Lossy Materials 4 Poynting Vector 5 Reflection 6 Summary

31 Reflection of Plane Wave at Normal Incidence Electric fields on both sides are given as, Es,i = E i e γ1 z z ˆx, Es,t = E t e γ2 z z ˆx, and Es,r = E r e γ1 z z ˆx. Similarly, magnetic fields on both sides are given as, Hs,i = Es,t = E i η 1 e γ1 z z ŷ, E t η 2 e γ2 z z ŷ, and x Es,r = E r η 1 e γ1 z z ˆx. From the boundary conditions (at z = 0), E i + E r = E t, and E i η 1 E r η 1 = E t η 2. (25) From (25), Γ = E r E i = τ = E t E i = η 2 η 1 η 2 + η 1, (26) 2η 2 η 2 + η 1, and (27) 1 + Γ = τ. (28)

32 Outline 1 Time Harmonic Fields 2 Helmholtz Wave Equation 3 Lossy Materials 4 Poynting Vector 5 Reflection 6 Summary

33 Summary