Analysing the Scattering of Electromagnetic Ultra-wideband Pulses from Large-Scale Objects by the Use of Wavelets

Abstract

The aim of this chapter is the study of the signal received by an antenna (RX) when a transmit antenna (TX) sends a short pulse in a large-scale space, for instance, in an urban environment. Integral equations are established, which link the densities of charges and currents inside the environment objects with the incident field created by the TX antenna. From these equations, we define an integral operator K. The densities can be obtained by inverting 1 − K. The introduction of Daubechies wavelets allows us to obtain sparse matrices for KK^∗, which is computationally convenient to get (1 − K)⁻¹.

Appendices

Appendix 1: Antennas Diagrams

For a small dipole antenna of length l, if the wire is along the z^′ direction,

$$\displaystyle \begin{aligned} E_{r^{\prime}}(\mathbf{u},\omega)&\simeq 0 \\ E_{\theta^{\prime}}(\mathbf{u},\omega)&\simeq j\frac{ e^{-j\omega|\mathbf{u}-\mathbf{a}|/c}} {|\mathbf{u}-\mathbf{a}|} \frac{ \eta \hat{i}_0(\omega) l \omega}{4\pi c} \sin\theta^{\prime} e^{j\omega t} \\ E_{\phi^{\prime}}(\mathbf{u},\omega)&\simeq 0, \end{aligned} $$

where η = 120π ohms is the intrinsic wave impedance and where |u −a|, θ^′, ϕ^′ are the spherical coordinates for u with a as origin.

If the applied current is $i(t)=\frac {1}{\sqrt {2 \pi }\omega _0}\int d\omega \hat {i}_0(\omega )e^{j\omega t}$ (we modify the usual Fourier transform introducing ω₀ that can be the central angular frequency of the signal, just to give to $\hat {i}_0(\omega )$ and i(t) the same units), the field becomes

$E_{\theta ^{\prime }}(\mathbf {u},t)\simeq \frac { 1} {|\mathbf {u}-\mathbf {a}|} \frac { l }{2\sqrt {2\pi } c} \sin \theta ^{\prime } \frac {dv}{dt}(t-|\mathbf {u}-\mathbf {a}|/c)$ with v(t) = η i(t).

In the case the antenna is a rectangular patch antenna at position a, if the x^′ axis is perpendicular to the patch, the y^′ axis is along the patch length L, and the z^′ axis is along the patch width W (see figure 14.16 in [1]) and if the applied voltage is $\hat {v}_0(\omega ) e^{j\omega t}$, the far-field components at point u are

$$\displaystyle \begin{aligned} E_{r^{\prime}}(\mathbf{u},\omega)&\simeq 0 \\ E_{\theta^{\prime}}(\mathbf{u},\omega)&\simeq 0 \\ E_{\phi^{\prime}}(\mathbf{u},\omega)&\simeq j \frac{ 2 \hat{v}_0(\omega)e^{-j\omega|\mathbf{u}-\mathbf{a}|/c}} {\pi|\mathbf{u}-\mathbf{a}|} \tan\theta^{\prime} \sin\Big(\frac{\omega W}{2c} \cos \theta^{\prime}\Big) \cos\Big(\frac{\omega L_e}{2c} \sin \theta^{\prime}\sin \phi^{\prime}\Big)e^{j\omega t}. \end{aligned} $$

L_e is the effective length ( see 14.2 and 14.3 formulas in [1] for the link in between L and L_e).

Notice that in the patch case $f_{\phi ^{\prime }}$ the variation with respect to ω is also approximately linear if W and L_e are small. So in this case we will write that the ϕ^′ field component is approximatively in the range of the frequencies considered of the form

$$\displaystyle \begin{aligned} E_{\phi^{\prime}}(\mathbf{u},t)\simeq \frac{1}{|\mathbf{u}-\mathbf{a}|}\frac{2 W}{\sqrt{2\pi} c} \sin\theta^{\prime} \frac{dv}{dt}(t-|\mathbf{u}-\mathbf{a}|/c). \end{aligned} $$

In the two cases, the field will be expressed as

$$\displaystyle \begin{aligned} \frac{1}{|\mathbf{u}-\mathbf{a}|}\mathbf{f}(\phi^{\prime},\theta^{\prime})\frac{dv}{dt}(t-|\mathbf{u}-\mathbf{a}|/c). \end{aligned} $$

Let us notice that we have been speaking about the total electrical field in the absence of the RX antenna. In [2], we discussed the conditions that have to be satisfied in order, and when we study the link between the received voltage at the RX antenna and the voltage sent at the TX antenna, we can separate into three steps the problem. First considering that the TX antenna is alone and emits a well-known field, then considering that the scatterers produce fields, and finally considering that the RX antenna receives a field that is the sum of the field emitted by the TX antenna and the field due to the charges inside the scatterers, this total field being considered as incident, to produce the final RX voltage.

Appendix 2: Conductivity and Polarisability

The total field polarizes the materials. At some given point, the polarisation vector P(u, t) is related to the total field. We can write this relation in the form P = 𝜖₀χ(E) where χ is an operator.

Denoting

$$\displaystyle \begin{aligned} (G_0 \rho_p)({\mathbf{u}}_2,t)&:=- \frac{1}{4 \pi \epsilon_0}\nabla \int_{V} d{\mathbf{u}}_1 \frac{\rho_p({\mathbf{u}}_1,t-|{\mathbf{u}}_2-{\mathbf{u}}_1|/c)}{|{\mathbf{u}}_2-{\mathbf{u}}_1|} \\ (G{\mathbf{J}}_p)({\mathbf{u}}_2,t)&:=-\frac{\mu_0}{4 \pi }\frac{\partial}{\partial t}\int_{V} d{\mathbf{u}}_1\frac{{\mathbf{J}}_p({\mathbf{u}}_1,t-|{\mathbf{u}}_2-{\mathbf{u}}_1|/c)}{|{\mathbf{u}}_2-{\mathbf{u}}_1|} \end{aligned} $$

and using ρ_p = −∇.P = −𝜖₀∇.χ(E) and ${\mathbf {J}}_{p}=\epsilon _0\frac {\partial }{\partial t}\boldsymbol {\chi } (\mathbf {E})$, if the materials are perfect insulators, the self-consistent equations for ρ_p(u, t) and J_p(u, t) can be written as

$$\displaystyle \begin{aligned} \rho_p&= - \epsilon_0\nabla.\boldsymbol{\chi} \Big({\mathbf{E}}_{\mathbf{in}}-\frac{1}{4 \pi \epsilon_0}\nabla (G_0 \rho_p) -\frac{\mu_0}{4 \pi }\frac{\partial}{\partial t}G{\mathbf{J}}_p\Big) \\ {\mathbf{J}}_p&=\epsilon_0\frac{\partial}{\partial t}\boldsymbol{\chi} \Big({\mathbf{E}}_{\mathbf{in}}-\frac{1}{4 \pi \epsilon_0}\nabla (G_0 \rho_p) -\frac{\mu_0}{4 \pi }\frac{\partial}{\partial t}G{\mathbf{J}}_p\Big). \end{aligned} $$

The materials used for constructions, such as concrete, can be neither metals nor perfect insulators. Some of their valence electrons are allowed to move freely in the whole material and generate electrical currents under the influence of the electric field.

$$\displaystyle \begin{aligned} {\mathbf{J}}_{f}= \boldsymbol{\sigma}^e(\mathbf{E}). \end{aligned} $$

In the literature appear some hypothesis on the relations P = 𝜖₀χ(E) and J_f = σ^e(E).

If the electric field is not large, it is considered that the operators χ are linear. In this case, the previous equations become

$$\displaystyle \begin{aligned} \rho_p&= - \epsilon_0\nabla.\boldsymbol{\chi} ({\mathbf{E}}_{\mathbf{in}})+\frac{1}{4 \pi } \nabla.\boldsymbol{\chi}( \nabla (G_0 \rho_p)) -\frac{\epsilon_0\mu_0}{4 \pi }\frac{\partial}{\partial t} \nabla.\boldsymbol{\chi} (G{\mathbf{J}}_p) \\ {\mathbf{J}}_p&=\epsilon_0\frac{\partial}{\partial t}\boldsymbol{\chi} ({\mathbf{E}}_{\mathbf{in}})-\frac{1}{4 \pi }\frac{\partial}{\partial t}\boldsymbol{\chi}(\nabla G_0 \rho_p) -\frac{\epsilon_0\mu_0}{4 \pi }\frac{\partial}{\partial t}\boldsymbol{\chi}(\frac{\partial}{\partial t}G{\mathbf{J}}_p). \end{aligned} $$

It is usually supposed that the linear operator χ is a 3 × 3 matrix whose elements are integral operators with kernels χ_ij(x, x^′, t − t^′), and each of them, supposing that the polarisation at point x depends only on the values of the electric field at this point, is rewritten as χ_ij(x, t − t^′)δ(x −x^′). Another assumption consists in supposing that $\chi _{ij}(\mathbf {x}, t-t^{\prime })=\chi ^{(1)}_{ij}(\mathbf {x})\chi ^{(2)}_{ij}( t-t^{\prime })$. The temporal dependence is generally described giving us the frequency dependence of the polarisability, so $\boldsymbol {\chi }_{ij}^{(2)}( t-t^{\prime })=\int d\omega e^{-i \omega (t-t^{\prime })}\hat {\chi }_{ij}(\omega )$. This means that the polarisation at a time does not only depend on the value of the field at this time but also on the previous values of the field. There is a delay in the response. In the case the material is isotropic, the matrix is diagonal, and the integral operators on the diagonal are the same.

Notice that the current density in the material is the sum of two terms whose origin is different.

It appears that to calculate the charge and current densities, it is necessary to know for each material the dependence of the polarisability and the conductivity at least on the frequency range corresponding to the voltage excitation of the antenna.

Finally, if the polarisability and the conductivity are almost flat in the considered frequency range, we can consider that the operator χ acts simply as a multiplication by χ₀(x) and that σ acts simply as a multiplication by σ₀(x) where χ₀(x) and σ₀(x) are, respectively, the values of the polarisability and the conductivity at x, at the central frequency f₀.

Appendix 3: Daubechies Wavelets and $W_{jj^{\prime }}$ Function

To calculate numerically the matrix elements $(w_{{\mathbf {m}}^{\prime }{\mathbf {n}}^{\prime }}w_{j^{\prime }k^{\prime }}K_{11}^*K_{11} w_{\mathbf {m}\mathbf {n}}w_{jk})$, we have to calculate the functions $W_{jj^{\prime }}(t_0)=\int _{-\infty }^{\infty }dt \frac {d^2w_{j^{\prime }0}}{dt^2}(t) \frac {d^2w_{j0}}{dt^2}(t-t_0)$ from the Daubechies wavelets. We are using Mathematica. If we directly calculate the second derivatives of the wavelets and use NIntegrate, the numerical integration, we get the indication that “it failed to converge to prescribed accuracy after 9 recursive bisections” or “is converging too slowly; suspect one of the following: highly oscillatory integrand, or WorkingPrecision too small”.

Then, instead, the Fourier images of the wavelets $\hat {w}(\omega )$ are introduced, and we get the alternative form $W_{jj^{\prime }}(t_0)= 2^{-j/2}2^{-j^{\prime }/2}\int _{-\infty }^{\infty }d\omega e^{i \omega t_0} \omega ^4 \hat {w}(2^{-j^{\prime }}\omega )\hat {w}(-2^{-j}\omega )$. The Fourier images are calculated from the wavelet filter coefficients.

Here we can see in Fig 2, the plot of the Daubechies-10 wavelet w(t) calculated from the filter coefficients (Fig. 2), from the filter coefficients [18, 20], and the plot

Fig. 3

The W₀₀ function

for W₀₀(t₀) calculated from $W_{00}(t_0)= \int _{-\infty }^{\infty }d\omega e^{i \omega t_0} \omega ^4 \hat {w}(\omega )\hat {w}(-\omega )$ through the Fourier image, $\hat {w}(\omega )$, of the Daubechies-10 wavelet (Fig. 3).

Appendix 4: Singularities for the Parallelipedic Case

In this appendix, we discuss about the Hypothesis R_s in the case the volume is a parallelepiped with constant permittivity χ₀.

We look at the singularities of $f(\sigma ,\mathbf {x},\mathbf {y})= \int _{1}^{\infty }d\tau \int _{0}^{2\pi }d\varphi |\chi (\sigma ,\tau ,\varphi ,\mathbf {x},\mathbf {y})|{ }^2$.

Let us notice that if S_σ denotes the surface of the hyperboloid σ inside the parallelepiped, then f(σ, x, y) = |χ₀|²S_σ. It is clear that S_σ is very regular as long as the hyperboloid σ does not hit the parallelepiped corners. So there are 8 values, σ_i(x, y), i = 1, ...8, for which the derivative of f(σ, x, y) is discontinuous. The σ_i(x, y) are easily calculable.

An expression for S_σ can be obtained analytically. We have to find the intersection of the circle parametrized by τ and σ and the surface planes P₁, P₂, , , P₆ of the parallelepiped.

If the components of the normal to the plane P_i are (α_i, β_i, γ_i), the equation for plane P_i is $\alpha _{i} u^{\prime \prime }_{1}+\beta _{ni}u^{\prime \prime }_{2}+\gamma _{i} u^{\prime \prime }_{3}=\gamma _{i} s_i$, where $u^{\prime \prime }_{1},u^{\prime \prime }_{2},u^{\prime \prime }_{3}$ are the Cartesian coordinates of a point with respect to three orthogonal axes whose origin is (x + y)∕2, and the third axis direction is given by the direction of x −y. In spheroidal coordinates, the values for the ϕ_i(τ, σ) of the intersection points can be extracted from $\alpha _{i}\Big (a\sqrt {(\tau ^2-1)(1-\sigma ^2)}\cos \phi \Big ) + \beta _{i}\Big (a\sqrt {(\tau ^2-1)(1-\sigma ^2)}\sin \phi \Big )+ \gamma _{i}\Big ( a\tau \sigma \Big )=\gamma _{i} s$, where s is the Cartesian coordinate of the point at the intersection of the x −y axis with P_i in the Cartesian coordinates.

If we denote by t_i(x, y) the images of the σ_i(x, y) by the transform $\sigma \rightarrow \mathcal {T}(\sigma )=\sigma |\mathbf {x}-\mathbf {y}|/c+2^{-j}k-2^{-j^{\prime }}k^{\prime }$, it is possible to see if the t_i(x, y) belong or not to the interval (t_m, t_M). If they do not belong to (t_m, t_M), one concludes that the matrix elements $\Big (w_{{\mathbf {m}}^{\prime }{\mathbf {n}}^{\prime }}w_{j^{\prime }k^{\prime }},K^*K w_{\mathbf {m}\mathbf {n}}w_{jk}\Big )$, $\Big (w_{{\mathbf {m}}^{\prime }{\mathbf {n}}^{\prime }}w_{j^{\prime }k^{\prime }},KK^* w_{\mathbf {m}\mathbf {n}}w_{jk}\Big )$ are of the order $(2^{-j}+2^{-j^{\prime }})^{(p+1)}$. If one of the t_i(x, y) belongs to the interval (t_m, t_M), the matrix elements are of the order $(2^{-j}+2^{-j^{\prime }})$. When j and j^′ become big enough, in general the t_i(x, y) are outside the interval (t_m, t_M). This is not the case when x and y are at equal distance from a corner ( or at approximatively equal distance from a corner), i.e. if x is on the sphere of radius |y −c_i| or close to this sphere.

So once x_mn, j, k are fixed, if $|{\mathbf {x}}_{{\mathbf {m}}^{\prime }{\mathbf {n}}^{\prime }}-{\mathbf {x}}_{\mathbf {m}\mathbf {n}}| >d_{\mathbf {m}}+d_{{\mathbf {m}}^{\prime }}+c(2^{-j^{\prime }}k^{\prime }-2^{-j}k+(2^{-j}+2^{-j^{\prime }})p-2^{-j})$ and $2^{-j^{\prime }}k^{\prime }-2^{-j}k >(2^{-j}+2^{-j^{\prime }})p-2^{-j^{\prime }}$, the matrix elements are of order $C(2^{-j}+2^{-j^{\prime }})^{p}$ except those for which ${\mathbf {x}}_{{\mathbf {m}}^{\prime }{\mathbf {n}}^{\prime }}$ is close to one of the spheres centred at corners c_i of radius |y −c_i|, in which case they are of order $(2^{-j}+2^{-j^{\prime }})$.

In Fig. 1, the red scars correspond to the singularities of $\hat {f}$. The scars are in the vicinity of the spheres centred at the corners of the parallelepiped c_i and whose radius is |c_i −x_mn|. Notice that they disappeared in the lower figure as for large time the intersection in the outer region of the crown with the spheres centred at the corners is void.