• Reflectivity density function: $g(x, y)$. The goal of SAR imaging is to estimate $|g(x, y)|$.
• 1-D simple case:
  • $u$ is the slant-range position
  • Round-trip propagation time: $\tau_0 + \tau(u)$
  • $\tau_0 = \frac{2u_0}{c}$
  • $\tau(u) = \frac{2u}{c}$
  • $r = u_0 + u = u_0 + y cos\psi$
  • $\psi$: Grazing angle
• Projection Slice Theorem
  • Linear trace: $P_1(X) = \int p_1(x) exp(-jxX)dx = G(X, 0, 0)$
  • Planar trace: $P_2(X, Y) = \int \int p_2(x, y) exp(-j(xX+yY))dxdy = G(X, Y, 0)$
• After using the Projection Slice Theorem, we can derive the radar return received at position $(\theta, \psi)$ to be: $\bar r_{\theta, \psi}(t) = \frac{A}{2}P_{\theta, \psi}[\frac{2}{c}(\omega_0 + 2 \alpha (t - \tau_0) )]$
  • P is Fourier transform of the projection function p

The second part of this article is to derive the bandwidths and resolutions.

• $\phi$ (phase) $= - \frac{2 \omega R}{c}$
  • $\omega$: angular frequency (rad/s)
• $U$ (spatial frequency) $=\frac{\partial \phi}{\partial \R} = - \frac{2 \omega}{c}$
• $\Delta Y'$ (slant plane range bandwidth) $=\frac{2}{c}\Delta \omega, \Delta \omega = (\omega_0 + \pi B_c) - (\omega_0 - \pi B_c) = 2 \pi B_c$
  • $B_c$: Hz. The angular frequency of $B_c$ is $\omega = 2 \pi B_c$
  • So $\Delta Y’ = \frac{2}{c}(2 \pi B_c)$
• Radius of the annulus is $U_0 = \frac{2 \omega_0}{c} = \frac{2}{c} 2 \pi f_0 = \frac{2}{c}2 \pi \frac{c}{\lambda}=\frac{4 \pi}{\lambda}$
• $\Delta X’=2(\frac{4 \pi}{\lambda}) sin(\frac{\Delta \theta}{2}) \approx \frac{4 \pi}{\lambda} \Delta \theta$
• $\rho_y’$ (slant plane range resolution) $=\frac{2 \pi}{\Delta Y'} = \frac{c}{2 B_c}$
  • $\Delta k$ in frequency domain is $\frac{2 \pi}{ \Delta k}$ in time domain.
• $\rho_x’$ (slant plane cross range resolution) $= \frac{2 \pi}{ \Delta X’} = \frac{\lambda}{2 \Delta \theta}$
• Sampling rate: the minimum X’ sampling interval in the phase-history domain required to reconstruct alias-free of diameter 2L is: $\delta X’ = \frac{2 \pi}{ 2L}$
• $\delta \theta$ (angular sampling interval) $=\delta X’ \frac{\lambda}{4 \pi} = \frac{\lambda}{ 4L}$
• $\delta A$ (along-track sampling interval) $=R \delta \theta = \frac{\lambda R}{ 4L} \approx \frac{D}{2}$
  • $\frac{\lambda}{D} \approx \frac{2L}{R}$