Description

Dane Johnson
Problem 1
Consider a downward propagating seismic wavefront generated by a source on the surface where the observed travel time in seconds, t, to a depth z meters below the surface can be modeled by
(1)
where s(z) denotes the vertical slowness (1/v(z) where v is the wavefront
velocity) and the kernel H is the Heaviside function.
(a) Estimating the integral in (1) for values of z = 0.2,0.4,…,20, using the midpoint rule we have the system:
t(0.2) ≈ t1 = s(0.1)∆ξ = s(0.1)(0.2) t(0.4) ≈ t2 = (s(0.1) + s(0.3))∆ξ = ((s(0.1) + s(0.3))(0.2) t(0.6) ≈ t3 = (s(0.1) + s(0.3) + s(0.5))∆ξ = ((s(0.1) + s(0.3) + s(0.5))(0.2)
…
t(20) ≈ t100 = (s(0.1) + s(0.3) + … + s(19.9))∆ξ = ((s(0.1) + s(0.3) + … + s(19.9))(0.2) These calculations can be collected in a matrix equation of the form:
.
That is, G is the lower triangular matrix where all entries on or below the main diagonal are 0.2. Note that although our equation includes s(0.1),s(0.3),…,s(19.9), these slowness values are not actually known to us yet. So we should stress that this matrix equation is an approximation and make no claim that this represents an analytical solution for s unless it can be verified that the model presented in part (b) is accurate (and if the model is indeed accurate all of our work in estimating slowness is unnecessary).
(b) For a seismic depth model having a linear depth gradient specifiedby ) we obtain ). We use this expression to calculate a vector strue in the accompanying matlab script for z = 0.1,0.3,…,19.9. Solving the integral in (1) analytically using this model we have:

In the matlab script we use this result to calculate the vector y for the sensor depths z = 0.2,0.4,…,20.0.
(c) The response to this part should be entirely satisfied in matlab.
(d) Adding even a small amount of noise (mean 0 and standard deviation 0.05 milliseconds) appears from plotting comparison to dramatically increase the error of the approximate solution from the model derived solution. See the plots at the end of the document for a visual comparison.
(e) Repeating the problem for 4 sensors (with hopefully not too manycopy/paste errors):
(a) Estimating the integral in (1) for values of z = 5,10,15,20, using the midpoint rule we have the system:
t(5) ≈ t1 = s(2.5)∆ξ = s(2.5)(5)
t(10) ≈ t2 = (s(2.5) + s(7.5))∆ξ = ((s(2.5) + s(7.5))(5) t(15) ≈ t3 = (s(2.5) + s(7.5) + s(12.5))∆ξ = (s(2.5) + s(7.5) + s(12.5))(5)
t(20) ≈ t4 = (s(2.5) + s(7.5) + s(12.5) + s(17.5))∆ξ = (s(2.5) + s(7.5) + s(12.5) + s(17.5))(5) These calculations can be collected in a matrix equation of the form:
.
(b) For a seismic depth model having a linear depth gradient specifiedby ) we obtain ). We use this expression to calculate a vector strue in the accompanying matlab script for z = 2.5,7.5,12.5,17.5. Solving the integral in (1) analytically using this model we have:

In the matlab script we use this result to calculate the vector y for the sensor depths z = 5,10,15,20.
(c) The response to this part should be entirely satisfied in matlab.
Conclusion
The recovery of the true solution when using n = 4 sensors is generally worse when looking only at residuals than using n = 100 sensors but adding noise in the latter case deteriorates our results significantly. We can compare the 2-norm of the difference between our approximation, s, and strue for
• Noiseless, n = 100 case: ||s − strue||2 ≈ 2.5951e − 08.
• Noisy*, n = 100 case: ||s − strue||2 ≈ 0.0037611.
• Noiseless, n = 4 case: ||s − strue||2 ≈ 3.1774e − 06.
• Noisy*, n = 4 case: ||s − strue||2 ≈ 2.1219e − 05.
*Since the noise is randomly generated these values will change a bit each time the script is run.
Using this metric for error shows that adding noise for 100 sensors is enough to result in a worse approximation than adding noise for even the 4 sensor case. In the script we can also see the relative increase in error for each case that results from introducing noise.
Another notable difference between performing this exercise with few sensor is that adding noise does not appear to drastically increase error beyond that seen without noise. That is, although using n = 4 sensors is worse that using n = 100 sensors in the absolute sense (with or without noise), adding noise does not increase the error beyond the noiseless case in the noticeable way we see with n = 100 sensors. The condition number of the matrix G in the n = 100 case is much higher (about 127.95) than the condition number of G when we set n = 4 (about 5.41). A higher condition number signifies that in solving the equation y = Gs for s, error in the solution for s will be more amplified by a perturbation in y.