Finding Buried Targets Using Acoustic Excitation Zackary R. Kenz Advisor: Dr. H.T. Banks In collaboration with Dr. Shuhua Hu, Dr. Grace Kepler, Clay Thompson NCSU, L3 Communications team led by Dr. Jerrold Levine, and Dr. Richard Albanese Center for Research in Scientific Computation Department of Mathematics North Carolina State University October 7, 2010
1 Math Modeling and Differential Equations 2 1D Model Formulation 3 Simulation Setup and Results 4 Conclusion
Goals for Math Modeling Portion Introduction to math modeling Introduce differential equations Examine how changing parts of a differential equation can change resulting behavior
Goals for Target Detection Portion Acoustic Dynamics Given an impact to the soil, what sort of wave propagation dynamics are expected? How do changes in soil properties affect dynamics? Electromagnetic Signal Dynamics Given an arbitrarily moving target in the soil, what will a reflected radar signal look like? Device Development
1 Math Modeling and Differential Equations 2 1D Model Formulation 3 Simulation Setup and Results 4 Conclusion
Math Modeling Overview Modeling is developing equations to explain some phenomenon, and then using the equations to make predictions or answer questions about the phenomenon. Validating the equations with real life data is of particular interest.
Modeling Cycle The Iterative Modeling Process (iii) Abstraction or Mathematization resulting in a mathematical model (ii) Formalization of properties, relationships and mechanisms which result in a biological or physical model (iv) Formalization of Uncertainty/Variablity in model and data resulting in a statistical model (i) Empirical Observations (experiments and data collection) (v) Model Analysis (vii) Changes in understanding of mechanisms, etc., in the real system. (vi) Interpretation and Comparison (with the real system) Formation Stage: (i),(ii),(iii),(iv) Solution Stage: (v) Interpretation Stage: (vi), (vii)
Differential Equations and Modeling Modeling often uses differential equations to describe relationships between variables There are "ordinary" and "partial" differential equations Question: What are differential equations?
What is an ODE? A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation means the function of interest depends only on one independent variable. Instead of solving for numbers, like solving for x in x 2 1 = 0, we are solving for a function
ODE Example 1 dy(t) dt = 2 The left hand side is the rate of change of the solution y(t). The equation says that we want the rate of change of the unknown function y(t) to be a constant. What simple function has a constant rate of change (slope)?
Example 1 Solution dy(t) = 2 dt If we integrate both sides with respect to t, the solution is c is a "constant of integration" y(t) = 2t + c This could describe the motion of someone walking down the sidewalk at a constant rate.
Adding an Initial Condition Solution from previous slide: y(t) = 2t + c In a particular problem, we might want the person to start walking at position 5. Mathematically, we write y(0) = 5. Applying that condition to the problem: Solving gives c = 5. The full solution would then be y(0) = 2 0 + c = 5. y(t) = 2t + 5
Components of an ODE Problem A full ODE problem is both the main equation (which includes the derivatives) and also some conditions like our y(0) = 5. The number of conditions depends on the highest number of derivatives you have. Now we ll look at a slightly more complicated example.
ODE Example 2 dy(t) dt = 2y(t) This equation says that the rate of change of the solution is proportional to the value of the function at every point t. It s a little more complicated to get the solution for this case, so we won t solve it here but just state the answer: y(t) = e 2t + c. Changing the right hand side of the ODE from a constant to the unknown function changed the solution behavior.
ODE Example 3 dz(t) dt = 2z(t) By changing the constant to 2, the solution is now z(t) = e 2t + d.
Example 2 vs Example 3 We ll set the constants of integration equal to 0 (arbitrary). We can compare the two solutions y(t) = e 2t and z(t) = e 2t in the figure below. Function Value 6 4 2 0 y(t) z(t) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t
A Little More on ODEs These equations can get more complicated. For example, Higher derivatives can be in the equation Parameters can be in the model. The spring system is a common example: m d 2 y(t) dt 2 + c dy(t) dt + ky(t) = 0 m: mass c: damping k: stiffness
Example Solution for Spring System 2 Damped Spring System 1.5 Postion x(t) 1 0.5 0 0.5 0 2 4 6 8 10 Time t
Brief PDE Discussion What if we want the solution to depend on multiple independent variables, like the three spatial dimensions? A partial differential equation (PDE) seeks a solution with more than one independent variable. For example, we might want to track the temperature of a room over time. We would then be trying to find a function like T(x, y, z, t).
Brief PDE Discussion Derivatives are now "partial" derivatives (denoted with ), meaning with respect to one variable holding the others constant. For example, if we wanted the rate of change of temperature with respect to time, we d denote that quantity as T(x, y, z, t). t
The Wave Equation An equation that can be used to describe simple wave motion in one spatial dimension is 2 u(z, t) t 2 = c 2 2 u(z, t) z 2. Upon solving analytically, the constant c represents the speed of the wave. This particular form of the wave equation can be solved by hand, but more complicated forms are solved numerically. There is a shorthand notation for partial derivatives that turns the wave equation above into u tt = c 2 u zz.
Moving Forward With this brief introduction to modeling and differential equations, we can start discussing my research project of detecting buried targets. The project is an example of the modeling cycle.
1 Math Modeling and Differential Equations 2 1D Model Formulation 3 Simulation Setup and Results 4 Conclusion
Definitions Elastic material: returns to its original shape after applying stress forces (rubber band) Viscous material: Resists stresses (tar) Viscoelastic material: Has both properties (rubber)
Problem Situation Open field where we want to detect buried objects Thump in one area, study wave propagation outward from thumper Likely need a 2D or 3D model to capture full dynamics As a first approximation, we ll use a 1D model see which features of wave propagation we can capture
1D Problem One dimensional problem schematic: air. soil. soil Observations of the wave form at a particular depth will be taken at the z 10 position
1D Model: Key Assumptions Both soil and target are uniform in horizontal (i.e., x and y) directions The column is a continuum Dry soil behaves as a Kelvin-Voigt viscoelastic solid for small-amplitude vibrations 1 1 B.O. Hardin, The nature of damping in sands, J. Soil Mech. Found. Div., 91 (1965), 63-97.
1D Model: Notation u(z, t) denotes the displacement (units: m) in the z-direction at position z at time t Kelvin-Voigt stress: σ = κ u(z,t) z + η 2 u(z,t) t z κ: elastic modulus (i.e. stiffness) in kg m s 2 η: damping coefficient in kg m s ρ: soil density in kg m 3 = Pa
1D Model: Only Soil From the equations of motion for a continuum, we obtain ρ 2 u(z,t) = t 2 z (σ) = z ( κ u(z,t) z ) + η 2 u(z,t) t z Remember from the wave equation slide that the wave speed was the constant c in u tt = c 2 u zz. For our model above, if we neglect damping we get c = κ/ρ.
1D Model: Including Rigid Target Model the rigid target as a point mass in the column at z 10 : ( ) ρ 2 u(z,t) = t 2 z κ u(z,t) z + η 2 u(z,t) t z, z (z p0, z 10 ) (z 10, ) M 2 u(z 10,t) t 2 ( = S κ u(z+ 10,t) z ) + η 2 u(z + 10,t) t z ( ) S κ u(z 10,t) z + η 2 u(z 10,t) t z. M: mass in kg of target S: surface area in m 2 of contact between target and soil under (over) target
1D Model: Additional Assumptions and Conditions Assume zero displacement and zero velocity initially: u(z, 0) = 0 u(z, 0) = 0 z At the surface, the normal internal stress is balanced with the applied input force: ( ) κ u(z,t) z + η 2 u(z,t) t z = f(t), z=zp0 Computational assumption: the column is finite: u(z 00, t) = 0
1 Math Modeling and Differential Equations 2 1D Model Formulation 3 Simulation Setup and Results 4 Conclusion
Simulation Setup Surface: z p0 = 0m Observation point/target: z 10 = 0.3048m Far boundary: z 00 = 50m
Simulation Setup f(t) (units:n/m 2 ) 3 2 1 0 1 2 3 4 x 105 Forcing Function 4 0 0.002 0.004 0.006 0.008 0.01 t (units: s)
Questions to Ask If soil density ρ changes, what happens to the wave form and speed? What happens if the elastic modulus κ changes? What happens when the wave impacts the rigid body target? This is the Model Analysis part of the modeling cycle.
Results, Soil Only Changes in Soil: Density u(0.3048,t) (units: m) 10 x 10 4 8 6 4 2 0 Displacement around z=0.3048m (z=1 ft) κ=204000000 ρ=1440 κ=204000000 ρ=1800 κ=204000000 ρ=2250 2 0 0.002 0.004 0.006 0.008 0.01 t (units: s) Wave speed heuristic (no damping): v κ/ρ
Results, Soil Only Changes in Soil: Elastic Modulus u(0.3048,t) (units: m) 12 x 10 4 10 8 6 4 2 Displacement around z=0.3048m (z=1 ft) κ=102000000 ρ=1800 κ=204000000 ρ=1800 κ=408000000 ρ=1800 0 2 0 0.002 0.004 0.006 0.008 0.01 t (units: s) Heuristic: v κ/ρ
Results, Soil Only Changes in Soil: Density and Elastic Modulus u(0.3048,t) (units: m) 14 x Displacement around z=0.3048m (z=1 ft) 10 4 12 10 8 6 4 2 0 κ=102000000 ρ=1440 κ=204000000 ρ=1800 κ=408000000 ρ=2250 2 0 0.002 0.004 0.006 0.008 0.01 t (units: s) Heuristic: v κ/ρ
Results, Soil Only u(0.3048,t) (units: m) 12 x 10 4 10 8 6 4 2 0 Displacement around z=0.3048m (z=1 ft) κ=163200000 ρ=1440 κ=204000000 ρ=1800 κ=255000000 ρ=2250 2 0 0.002 0.004 0.006 0.008 0.01 t (units: s) Heuristic: v κ/ρ
Results, With Target: Wave Form Passing by Target 12 Wave form in z domain, at time t=0.00067831 14 x 10 5 Displacement, units: m 10 8 6 4 2 0 2 0 0.5 1 1.5 2 Distance under ground, units: m Dashed line represents location of target in column
Results, With Target: Wave Form Passing by Target Wave form in z domain, at time t=0.0020486 x 10 4 8 Displacement, units: m 6 4 2 0 0 0.5 1 1.5 2 Distance under ground, units: m
Results, With Target: Wave Form Passing by Target Wave form in z domain, at time t=0.0054744 x 10 4 8 Displacement, units: m 6 4 2 0 0 0.5 1 1.5 2 Distance under ground, units: m
Results, With Target: Wave Form Passing by Target Wave form in z domain, at time t=0.0068447 8 x 10 4 7 Displacement, units: m 6 5 4 3 2 1 0 0 0.5 1 1.5 2 Distance under ground, units: m
1 Math Modeling and Differential Equations 2 1D Model Formulation 3 Simulation Setup and Results 4 Conclusion
Discussion Model successes: The 1D model can capture basic wave dynamics in the soil Models results of changes in soil properties the way we would expect Model shows what we would expect when wave impacts a rigid body Modeled effects that were seen in field tests
Discussion Model drawbacks: Missing any dynamics outside soil column Assumes everything is uniform, clearly not true in practice Only models one wave type - in reality, multiple types result from a single impact Doesn t model effects of waves hitting target at an angle Here target is modeled as a point, real target is 3D
Future Work More complicated equations - need more spatial independent variables since real life is three-dimentional Further studies on how well we can estimate model parameters like density using real life data Couple with electromagnetic detection portion Discrimanants for target detection
Acknowledgements Project supported by Air Force Office of Scientific Research, grant FA9550-09-1-0226 My efforts were supported by a Department of Education GAANN Fellowship
Questions? For further details on elasticity theory and this computational example, see: "A Brief Review of Elasticity and Viscoelasticty", CRSC Technical Report CRSC-TR10-08, May 2010, www.ncsu.edu/crsc/reports/reports10.html or "A Brief Review of Elasticity and Viscoelasticity for Solids," Advances in Applied Mathematics, to appear