Lithospheric Layering

Outline Receiver Function Method Mapping time to depth (Basic) Advanced applications a. Determining Vp/Vs and Moho depth b. Velocity modeling c. Determining layers of anisotropy and dip

Receiver functions Receiver functions are used to isolate the response function that describes P-wave to S-wave conversions at horizontal velocity interfaces (layers) in the earth below the receiver (hence the name receiver function)

P-wave to S-wave conversions  Frequency Domain h - horizontal v - vertical w - whitening

source * response = signal Forward Model (Convolution) Generation of recorded signal from Source and Earth Response source * response = signal = *

Inverse Model (Deconvolution ) Using the signal and source to get the Earth Response function signal / source = response = / CASE OF NO NOISE! EVEN SMALL % OF NOISE CAN CREATE UNSTABLE SOLUTION – INVERSE THEORY and REGULARIZATION TO SATBALIZE THE SOLUTION

Real Receiver function 3C Seismic Record: P-wave is the source (vertical) P-wave + converted S-wave are signal (isotropic-flat layer -> radial)

Outline Receiver Function Method Mapping time to depth (Basic) Advanced applications a. Determining Vp/Vs and Moho depth b. Velocity modeling c. Determining layers of anisotropy and dip

Move-out correction and Mapping time to depth P-wave Want to know the timing difference between the direct P arrival (ts) and the converted S arrival (ts) as a function of depth. This can be done if we know the velocity of the wave-front in the vertical and horizontal directions x S-wave z tp ts

Move-out correction and Mapping time to depth Just a geometry problem! Tpds = ts – tp 1/Vpx P-wave S-wave 1/Vpz tp ts

Move-out correction and Mapping time to depth Just a geometry problem! The horizontal velocity is known, the rayparmeter - ‘p’ We need to know the P-velocity as a function of depth: Vp(z) And the ratio between Vp and Vs (Poisson’s ratio).

Move-out correction and Mapping time to depth How much do errors in assumptions affect the time to depth mapping? Vp(z) – An avg velocity difference of 6.2 and 6.5 translates to ~ 3 km at 70 km depth ie if the Moho is at 70 km and the crust has an avg velocity of 6.5 km/s, we use 6.2 km/s and compute a depth of ~ 67 km The horizontal velocity is known, the rayparmeter - ‘p’ We need to know the P-velocity as a function of depth: Vp(z) And the ratio between Vp and Vs (Poisson’s ratio).

Move-out correction and Mapping time to depth How much do errors in assumptions affect the time to depth mapping? Vp/Vs ratio – An avg difference of 1.72 to 1.79 translates to ~ 7 km at 70 km depth ie if the Moho is at 70 km and the crust has an avg Vp/Vs of 1.79, we use 1.72 to compute a depth of ~ 63 km The horizontal velocity is known, the rayparmeter - ‘p’ We need to know the P-velocity as a function of depth: Vp(z) And the ratio between Vp and Vs (Poisson’s ratio).

Data Coverage 8/03 – 10/04 Teleseismic (blue) Distance 30-95 Deg Magnitude >=5.5 mb N - 179 Teleseismic Regional (Green) Distance <30 Deg Magnitude >=4.5 mb N - 571 Regional

70 km piercing pts across the array

Cross sections of stacked RF’s

Cross sections of stacked RF’s

Cross sections of stacked RF’s

Cross sections of stacked RF’s

Cross sections of stacked RF’s

Measurement of depth to Moho assuming Vp of 6.4 and Vp/Vs of 1.75

Outline Receiver Function Method Mapping time to depth (Basic) Advanced applications a. Determining Vp/Vs and Moho depth b. Velocity modeling c. Determining layers of anisotropy and dip

Moho depth and Vp/Vs ratio  If we assume Vp(z), we can write a function: H(Vp/Vs, D) = Tpms +Tppms + Tpsms which we can use to solve Vp/Vs and D Figures: Kennett, B

Example from station ES02 Moho depth 71 km Vp/Vs 1.77 Poisson’s 0.27 Pms + Ppms + Psms Pms + Ppms Depth below receiver (km) Vp/Vs ratio

Non-linear inversion for velocity and Vp/Vs

Identifying layers of Anisotropy and Dip

Identifying layers of Anisotropy and Dip

Identifying layers of Anisotropy and Dip

Thank You! The End…

Move-out correction and Mapping time to depth Just a geometry problem! Tppds = (ts – tp) + tp + tp = ts + tp 1/Vpx P-wave S-wave 1/Vpz tp ts

Move-out correction and Mapping time to depth Just a geometry problem! Tpsds = (ts – tp) + ts + tp = 2*ts 1/Vpx P-wave S-wave 1/Vpz

Examples from station ES34 Pms + Ppms + Psms Pms + Ppms Moho depth 65 km Vp/Vs 1.76 Poisson’s 0.26 Depth below receiver (km) Vp/Vs ratio