Conventional SAR imaging

Abstract

In this chapter, we explain the fundamental principles of SAR data collection and image formation, i.e., inversion of the received data. Synthetic aperture radar uses microwaves for imaging the surface of the Earth from airplanes or satellites. Unlike photography which generates the picture by essentially recoding the intensity of the light reflected off the different parts of the target, SAR imaging exploits the phase information of the interrogating signals and as such can be categorized as a coherent imaging technology.

Appendix 2.A Choosing the matched filter

Let the received field be given by the integral [cf. formula (2.14”)]

$$\displaystyle{ u(t,\boldsymbol{x}) =\int \nu (\boldsymbol{z})P\left (t - 2R_{\boldsymbol{z}}/c\right )d\boldsymbol{z}, }$$

(2.98)

where $\nu (\boldsymbol{z})$ incorporates both the ground reflectivity and the propagation attenuation, see (2.15’), and let the image be defined with the help of the function $K = K(t,\boldsymbol{y})$ [cf. formula (2.24)]:

$$\displaystyle{ I_{\boldsymbol{x}}(\boldsymbol{y}) =\int _{ -\infty }^{\infty }K(t,\boldsymbol{y})u(t,\boldsymbol{x})dt. }$$

(2.99)

Substituting (2.98) into (2.99) and changing the order of integration, we arrive at a mapping $\nu (\boldsymbol{z})\mapsto I_{\boldsymbol{x}}(\boldsymbol{y})$, which is known as the imaging operator:

$$\displaystyle{ I_{\boldsymbol{x}}(\boldsymbol{y}) =\int \nu (\boldsymbol{z})\int _{-\infty }^{\infty }K(t,\boldsymbol{y})P\left (t - 2R_{\boldsymbol{ z}}/c\right )dtd\boldsymbol{z}. }$$

(2.100)

The function $K(t,\boldsymbol{y})$, which is at our disposal, can be chosen to achieve some desirable properties of the imaging operator (2.100).

One of those properties may be to maximize the return from isolated point scatterers. Namely, let $\nu (\boldsymbol{z}) =\nu _{\boldsymbol{z}_{0}}\delta (\boldsymbol{z} -\boldsymbol{z}_{0})$, where $\nu _{\boldsymbol{z}_{0}}$ is a constant factor and $\boldsymbol{z}_{0}$ is given. Then, formula (2.100) yields:

$$\displaystyle{ I_{\boldsymbol{x}}(\boldsymbol{y}) =\nu _{\boldsymbol{z}_{0}}\int _{-\infty }^{\infty }K(t,\boldsymbol{y})P\left (t - 2R_{\boldsymbol{ z}_{0}}/c\right )dt. }$$

(2.101)

Assume that the function $K(t,\boldsymbol{y})$ is absolutely square integrable with respect to t uniformly in $\boldsymbol{y}$, i.e., $\forall \boldsymbol{y}:\ K(t,\boldsymbol{y}) \in L_{2}(-\infty,\infty )$, and the estimate

$$\displaystyle{ \int _{-\infty }^{\infty }\vert K(t,\boldsymbol{y})\vert ^{2}\,dt\leqslant E_{ K} }$$

(2.102)

holds with one and the same constant E _K for all $\boldsymbol{y}$. Consider the image (2.101) at $\boldsymbol{y} = \boldsymbol{z}_{0}$. We will seek $K(t,\boldsymbol{y})$ that maximizes the return $I_{\boldsymbol{x}}(\boldsymbol{z}_{0})$ relative to the input $\nu _{\boldsymbol{z}_{0}}$, i.e., maximizes the ratio $\frac{\vert I_{\boldsymbol{x}}(\boldsymbol{z}_{0})\vert } {\vert \nu _{\boldsymbol{z}_{0}}\vert }$, subject to constraint (2.102).

From the Cauchy-Schwarz inequality we get the following estimate:

$$\displaystyle{ \vert I_{\boldsymbol{x}}(\boldsymbol{y})\vert = \left \vert \nu _{\boldsymbol{z}_{0}}\int _{-\infty }^{\infty }K(t,\boldsymbol{y})P\left (t - 2R_{\boldsymbol{ z}_{0}}/c\right )\,dt\right \vert \leqslant \vert \nu _{z_{0}}\vert \sqrt{E_{K } E_{P}}, }$$

(2.103)

where

$$\displaystyle{E_{P} =\int _{ -\infty }^{\infty }\vert P(t)\vert ^{2}\,dt <\infty.}$$

For the chirp (2.10), (2.11), E _P  = τ. The equality in (2.103) is reached only for

$$\displaystyle{ K(t,\boldsymbol{y}) = \sqrt{\frac{E_{K } } {E_{P}}} \overline{P\left (t - 2R_{\boldsymbol{y}}/c\right )}. }$$

(2.104)

Formula (2.104) defines the matched filter: the kernel in the integral operator (2.99) is a scaled delayed complex conjugate replica of the original signal. Moreover, it is easy to see that the resulting maximum value of $\frac{\vert I_{\boldsymbol{x}}(\boldsymbol{z}_{0})\vert } {\vert \nu _{\boldsymbol{z}_{0}}\vert }$ appears independent of $\boldsymbol{z}_{0}$. If we take E _K  = τ, then formula (2.104) reduces to the expression (2.23) for the matched filter that we introduced in Section 2.3.1.

The consideration based solely on point scatterers is deficient though in that the real radar targets may have a different composition. As such, instead of maximizing the return from point scatterers we may require that the kernel of the imaging operator (2.100), i.e., the PSF

$$\displaystyle{ W_{\boldsymbol{x}}(\boldsymbol{y},\boldsymbol{z}) =\int _{ -\infty }^{\infty }K(t,\boldsymbol{y})P\left (t - 2R_{\boldsymbol{ z}}/c\right )\,dt, }$$

(2.105)

be close to the delta-function $\delta (\boldsymbol{y} -\boldsymbol{ z})$. This requirement is more general than the previous one because in the case of a true equality: $W_{\boldsymbol{x}}(\boldsymbol{y},\boldsymbol{z}) =\delta (\boldsymbol{y} -\boldsymbol{ z})$, the image $I(\boldsymbol{y})$ on the left-hand side of (2.100) exactly reconstructs the unknown function $\nu (\boldsymbol{z})$ regardless of its actual form.

Yet the question of how one shall understand the “closeness” between $W_{\boldsymbol{x}}(\boldsymbol{y},z)$ and $\delta (\boldsymbol{y} -\boldsymbol{ z})$ requires attention, because spaces of distributions are not equipped with norms. We will use the spectral interpretation, i.e., employ the Fourier transform. Namely, let $\tilde{K}$ and $\tilde{P}$ denote the Fourier transforms of K and P, respectively, in time:

$$\displaystyle{ \tilde{K}(\omega,\boldsymbol{y}) =\int _{ \infty }^{\infty }K(t,\boldsymbol{y})e^{i\omega t}dt,\quad \tilde{P}(\omega ) =\int _{ \infty }^{\infty }P(t)e^{i\omega t}dt. }$$

Then, taking into account that the Fourier transform of a product is the convolution of Fourier transforms, for the PSF (2.105) we can write:

$$\displaystyle{ W_{\boldsymbol{x}}(\boldsymbol{y},\boldsymbol{z}) = \frac{1} {2\pi }\int _{-\infty }^{\infty }e^{i\omega 2R_{\boldsymbol{z}}/c}\tilde{P}(\omega )\tilde{K}(-\omega,\boldsymbol{y})\,d\omega. }$$

(2.106)

For the rest of this section, we will adopt a simplified one-dimensional setting, for which $\boldsymbol{y} = R_{\boldsymbol{y}} \equiv y$ and $\boldsymbol{z} = R_{\boldsymbol{z}} \equiv z$. Then, the following identity holds:

$$\displaystyle{\delta (y - z) = \frac{2} {c}\delta \left (2R_{\boldsymbol{y}}/c - 2R_{\boldsymbol{z}}/c\right ) = \frac{2} {c} \frac{1} {2\pi }\int _{-\infty }^{\infty }e^{-i\omega 2(R_{\boldsymbol{y}}-R_{\boldsymbol{z}})/c}d\omega.}$$

Matching the right-hand side of the previous equality against that of (2.106), we see that if

$$\displaystyle{ \tilde{P}(\omega )\tilde{K}(-\omega,y) = e^{-i\omega 2R_{\boldsymbol{y}}/c}, }$$

(2.107)

then the spectra of the two expressions differ only by the constant $\frac{2} {c}$, which is not essential. Consequently, if (2.107) could be satisfied for all ω, then $W_{\boldsymbol{x}}(y,z)$ will be proportional to δ(y − z), which is our goal.

However, choosing $\tilde{K}(\omega,y)$ to satisfy (2.107) for all $\omega \in \mathbb{R}$ is not possible if $\tilde{P}(\omega )$ is zero (or very small by absolute value) for a range of frequencies. This appears to be the case for the chirped signals (2.10), (2.11) that are designed to have their spectrum confined to a band of width B around the central carrier frequency ω ₀:

$$\displaystyle{ \begin{array}{rcl} \tilde{P}(\omega )& =& \int _{-\infty }^{\infty }\chi _{\tau }(t)e^{-i\alpha t^{2} }e^{-i\omega _{0}t}e^{i\omega t}\,dt \\ & \approx &\sqrt{\frac{\pi }{\alpha }}e^{-i\pi /4}e^{i(\omega -\omega _{0})^{2}/(4\alpha ) }\chi _{B}(\omega -\omega _{0}).\end{array} }$$

(2.108)

The integration in (2.108) has been performed using the method of stationary phase, and the factor

$$\displaystyle{ \chi _{B}(\omega -\omega _{0})\stackrel{\text{def}}{=}\left \{\begin{array}{@{}l@{\quad }l@{}} 1,\quad &\vert \omega -\omega _{0}\vert \leqslant B/2, \\ 0,\quad &\text{otherwise}, \end{array} \right. }$$

(2.109)

on the last line of (2.108) originates from the condition that the stationary point of the phase

$$\displaystyle{ t_{\text{st}} = \frac{\omega -\omega _{0}} {2\alpha } }$$

(2.110)

must belong to the interval [−τ∕2, τ∕2] defined by the indicator function χ _τ (t) of (2.11).⁹ It is only in the frequency band $\vert \omega -\omega _{0}\vert \leqslant B/2$ defined by (2.109) that $\tilde{K}(\omega,y)$ can be chosen to satisfy (2.107):

$$\displaystyle{ \tilde{K}(\omega,y) = \frac{1} {\tilde{P}(-\omega )}e^{i\omega 2R_{\boldsymbol{y}}/c} = \text{const} \cdot \tilde{\overline{P}}(\omega )e^{i\omega 2R_{\boldsymbol{y}}/c}, }$$

(2.111)

where $\tilde{\overline{P}}(\omega )$ denotes the Fourier transform of the complex conjugate of P(t), and the second equality in (2.111) is derived with the help of (2.108) and (2.109). Applying the inverse Fourier transform to (2.111), we obtain:

$$\displaystyle{ K(t,y) \propto \overline{P(t - 2R_{\boldsymbol{y}}/c)}, }$$

(2.112)

which coincides with the matched filter expression (2.104). Hence, for the chirped signals the matched filter formula (2.104) also satisfies (2.107) so that the spectrum of $W_{\boldsymbol{x}}(y,z) = W_{\boldsymbol{x}}(y - z)$ coincides with that of δ(y − z) for $\vert \omega -\omega _{0}\vert \leqslant B/2$. In this sense, $W_{\boldsymbol{x}}(y - z)$ can be considered an approximation to δ(y − z).