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Reflection+ Transmission Coefficients for 2-Layer Problem

Reflection+ Transmission Coefficients for 2-Layer Problem. Outline. Two-Layer Problem: Reflection+Transmission Coefficients, Snell’s Law, Refractions, Critical angle, Post-critical Waves. 2-Layer Medium: Find R & T. 2 . 2 . i(k x - k z - wt) . k = w – k . D = e . z . x . z .

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Reflection+ Transmission Coefficients for 2-Layer Problem

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  1. Reflection+ Transmission Coefficients for 2-Layer Problem

  2. Outline • Two-Layer Problem: Reflection+Transmission Coefficients, Snell’s Law, Refractions, Critical angle, Post-critical Waves.

  3. 2-Layer Medium: Find R & T 2 2 i(k x - k z - wt) k = w – k D = e z x z x c 2 (2) { increase x, z - increase t { i(k x + k z - wt) U = R e z x z x r ’ i(k x - k’ z - wt) D = Te x z 2 2 k’ = w – k z x c’ 2 (2) r c c’

  4. 2-Layer Medium: Find R & T + i(k x + k z - wt) i(k x - k z - wt) e z P = U + D = x e z + R x z x - r ’ P = i(k x - k’ z - wt) Te x z r c (2) c’ Two unknowns R & T, need two equations of constraint

  5. + - P = P @z=0 i(k x - wt) i(k x - wt) = Te i(k x - wt) e x x e + R x @z=0 i( wt) Divide by e ik x ik x = Te ik x e x x e + R x Snell’s Law ’ r sin O = sin O’ Implies k = k x x c c’ c r O c’ O ik x x B.C. #1: Pressure Must Match at z=0

  6. Pressure must match at z=0 + - P = P @z=0 @z=0 .. .. + - w = w r r r ik x ik x = Te ik x e x x e + R x so w+ = w- @ z=0 ’ r .. ik x ik x = -ik Te ik x e x x -ik e + ik R x Recall w = -1 dP/dz z z z r O O ik x x B.C. #2: Particle Accel. Must Match at z=0 Vertical acceleration matches @ z=0 @z=0 c r z=0 c’

  7. @z=0 @z=0 , , , , r r r r r r r ik x ik x = Te ik x e x x e + R x c’ cos O - c cos O r R = c cos O + c’ cos O ’ r 2 r c’ cos O ik x T = ik x = -ik Te ik x e x x -ik e + ik R x c cos O + c’ cos O r z z z O O ik x x Pressure & accelration must match at z=0 c r c’

  8. Special Cases: Bright Spots O = 0: c’ - c r R = c’ + c r , , , , r r r r c’ cos O - c cos O r R = c cos O + c’ cos O r BRIGHT SPOT Positive Refle. Coeff when Model is low over hi velocity Negative Refle. Coeff when Model is hi over low velocity Gas sand has lower impedance than wet sand. Works well in GOM and young sedimentary basins

  9. Special Cases: Critical Angle c r O , , r r c’ cos O - c cos O r c’ R = O c cos O + c’ cos O ’ r r c’ c’ 1 < sin O = sin O’ c c O = critical angle:

  10. Special Cases: Post Critical Reflections , , r r c’ cos O - c cos O r R = c cos O + c’ cos O ’ r r c’ c’ 1 < sin O = sin O’ c c O = critical angle: c r O c’ O

  11. Special Cases: Post Critical Reflections , , r r c’ cos O - c cos O r R = c cos O + c’ cos O ’ r r c’ c’ 1 < sin O = sin O’ c c O = critical angle: c r O c’ O

  12. Special Cases: Post Critical Reflections , , r r c’ cos O - c cos O r R = c cos O + c’ cos O ’ r r c’ 1 < c O = critical angle: c r O c’ O c’ sin O = 1 c

  13. Special Cases: Post Critical Reflections , , r r c’ cos O - c cos O r R = c cos O + c’ cos O ’ r r c’ 1 < c O = post critical angle: c r O c’ O Total Energy Reflection Post Critical Phase Shift c’ sin O > 1 c

  14. Special Cases: Post Critical Reflection Coeff. A 2 , , = 1 – sin cos O O O r r { 2 c’sinO = 1 – > 1 c { sin = c’sinO { { c c c c 2 2 = i - 1 - 1 c’sinO c’sinO c’sinO { { ip/2 ip/2 ip/2 = e e e B c’ cos O - c cos O r R = c cos O + c’ cos O r

  15. Special Cases: Post Critical Reflection Coeff. A c cos O i(k x + k z - wt) U = R e z x , , , , r r r r c’ A - c B { c r 2 cos O = - 1 c’sinO R = { c’ A + c B r ip/2 ip/2 ip/2 ip/2 = e e e e B i a e c’ cos O - c cos O r R = c’ cos O r + Reflection coefficient is now a complex number Phase change in upgoing post critical reflections Havoc AVO and stacking!

  16. Post-critical reflections Layered Medium & Critical Angle CSG Model 0 0 3 km/s Sea floor Post-critical reflection ray Z (km) Time (s) 4.0 3.5 1.5 km/s 0 6 0 6 X (km) X (km)

  17. Special Cases: Pressure+Marine Free Surface 1 -1 , , - c cos O r r r = -1 R = c cos O r c’ cos O - c cos O r R = c cos O + c’ cos O r Free Surface r’c’=0: P = 1 + R = 1 – 1 = 0 ghosts

  18. Special Cases: Part. Vel. + Land Free Surface 1 +1 , , + c cos O r -r r = 1 R = c cos O r c’ cos O + c cos O r R = c cos O + c’ cos O r Part. Free Surface r’c’=0: Part. Part. w = 1 + R = 1 + 1 = 2 ghosts

  19. Special Cases: Part. Vel. + Land Free Surface r r r so w+ = w- @ z=0 .. - ik x ik x = -ik Te ik x e x x -ik e + ik R x Recall w = -1 dP/dz Downgoing particle-z velocity Down particle-z vel. Up particle-z vel. z z z r = D D - U R z z z R =-R part. press. Why is particle velocity R opposite polarity to pressure R? @z=0

  20. Special Cases: Impedance ik z P = e z dP/dz = - P dP/dz = P dP/dz = e ik z .. z ik ik ik ik where Therefore w z z z z .. r c r Recall w = -1 dP/dz .. . . r But w = -iww Therefore w dP/dz = P or P/w = iw r . Impedance

  21. Special Cases: Conservation of Energy Flow Previously we found that the Energy Density in deforming a cube was given by P 2 rc 2 Therefore the rate at which energy is flowing across a flat interface by a propagating plane wave is by P 2 c P 2 Energy/area/time = = rc rc 2 Conservation of energy demands energy flow of incident wave is same as the transmitted and reflected wave: 1 2 = R + T 2 2 rc rc r c’ ’ r c’ ’ 2 > 1 Note: rc r c’ ’ +

  22. Special Cases P = sinO1/v1 = sinO2/v2=sinO3/v3=sinO4/v4 Slowness = inverse apparent Vx v1 v2 v3 v4

  23. Time Slope = v5 v1 v2 v3 v4 v5

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