## 2 ISAR Imaging Model of Maneuvering Targets  
As shown in Figure 1, we established a three-dimensional ISAR imaging geometric model [23] of a maneuvering target with the center of the target turntable as the coordinate origin (O), where \(\mathbf{r}_{p}\) is the position vector of any scattering point \(P(x_{p},y_{p},z_{p})\) on the target. The ISAR imaging projection plane \(\Gamma\) is determined by the three-dimensional rotation vector \(\mathbf{\omega}\) and the range vector \(\mathbf{R}\) of the radar to the target origin. \(\mathbf{\omega}\) consists of the radial component \(\mathbf{\omega}_{r}\) in the \(\Gamma\) plane and the normal component \(\mathbf{\omega}_{e}\) perpendicular to the \(\Gamma\) plane. \(\mathbf{\omega}_{e}\) moves the scattering point, resulting in a change in the echo phase and producing a Doppler shift related to the target azimuth that can result in high-resolution azimuth imaging of the target, which is called the effective rotation component. Conversely, \(\mathbf{\omega}_{r}\) does not cause radial motion of the scattering point, which is called the invalid rotation component. Similarly, the three-dimensional velocity \(\mathbf{v}\) of the target can be decomposed into a radial component \(\mathbf{v}_{r}\) along the line-of-sight (LOS) direction and a normal component \(\mathbf{v}_{e}\) perpendicular to the LOS. Notably, \(\mathbf{v}_{e}\) is not necessarily in the same direction as \(\mathbf{\omega}_{e}\). \(\mathbf{v}_{r}\) causes the target to move in the radial direction, causing the range to move, which needs to be compensated for, whereas \(\mathbf{v}_{e}\) has no effect on the target's radial motion. ISAR imaging projects the three-dimensional effective component onto a two-dimensional imaging plane for turntable imaging, as shown in Figure 1b, where \(\mathbf{\omega}_{e}\) and \(\mathbf{v}_{r}\) are the effective projection component of angular velocity and translational velocity, respectively. As can be seen from Figure 1a, the scattering point \(P\) is rotated by a linear velocity of \(\mathbf{\omega}_{\text{e}}\times r_{p}\); its corresponding radial component is \((\mathbf{\omega}_{\text{e}}\times\mathbf{r}_{p})\cdot\mathbf{R}\). Therefore, the instantaneous Doppler frequency of the scattering point \(P\) is:  
\[f_{d}=\frac{2}{\lambda}\big{[}\mathbf{v}_{r}+(\mathbf{\omega}_{\text{e}}\times\mathbf{r}_{ p})\cdot\mathbf{R}\big{]} \tag{1}\]  
where \(\lambda\) is the wavelength of the radar signal, and \(\times\) and \(\cdot\) represent the outer product and inner product of the vector, respectively. Since the ISAR imaging time is short, we assumed that the target imaging projection plane is approximately constant during the short imaging time, so the Doppler frequency of the scattering point \(P\) can be further expanded to:  
\[f_{d}=\frac{2}{\lambda}\big{[}\mathbf{v}_{r}+\omega_{ex}\big{(}r_{py}R_{z}-r_{py}R _{z}\big{)}+\omega_{ey}\big{(}r_{pz}R_{x}-r_{px}R_{z}\big{)}+\omega_{ez}\big{(} r_{px}R_{y}-r_{py}R_{x}\big{)}\big{]} \tag{2}\]  
where \(\big{(}\omega_{ex},\omega_{ey},\omega_{ez}\big{)}\), \(\big{(}r_{py},r_{pz},r_{px}\big{)}\), and \(\big{(}R_{x},R_{y},R_{z}\big{)}\) are projected components of \(\mathbf{\omega}_{\text{e}}\), \(\mathbf{r}_{p}\), and \(\mathbf{R}\) on the three axes \((x,y,z)\), respectively. When a target is maneuvering, \(\big{(}v_{r},\omega_{ex},\omega_{ey},\omega_{ez}\big{)}\) generally changes with time. Because the imaging time is short, the motion of the target can approximately change linearly with time [7], ignoring the influence of high-order components on the imaging, so we can obtain:  
\[\left\{ \begin{array}{l}v_{r}=v_{0}+at_{m}\\ \omega_{ex}=\omega_{x}+\Omega_{x}t_{m}\\ \omega_{ey}=\omega_{y}+\Omega_{y}t_{m}\\ \omega_{ez}=\omega_{z}+\Omega_{z}t_{m}\end{array} \right.