**2. Raw echo algorithm of squint airborne strip SAR with trajectory offset error**

In this section, an algorithm of raw echo with trajectory error in squint is proposed. The approach adopts the Acquisition Doppler (AD) [24] geometry instead of the standard cylindrical reference system (see **Figure 1**).

## **2.1 Raw echo algorithm of squint airborne strip SAR with trajectory offset error**

The echo generation algorithm with trajectory error in squint is deduced by rotating the trajectory error along the line of sight. The motion error in the cone coordinate system is shown in **Figure 2**. The transformation relationship between a cylindrical coordinate system and a conical coordinate system is as follows:

$$r r\_0 = r \cos \phi, \varkappa\_0 = \varkappa + r \sin \phi\_d \tag{1}$$

**Figure 1.** *Geometric diagram of SAR with squint and trajectory offset.*

*Efficient Simulation of Airborne SAR Raw Data in Case of Motion Errors DOI: http://dx.doi.org/10.5772/intechopen.99378*

#### **Figure 2.** *AD processing geometry.*

where *x* are *r* the coordinates of the point target in the cone coordinate system, *x*<sup>0</sup> and *r*<sup>0</sup> are the coordinates of the point target in the cylinder coordinate system. *d* is the trajectory error, *β* is related to the horizontal and vertical components of the trajectory error. *ϕ<sup>d</sup>* is the squint angle. *R*<sup>n</sup> and *R* are the ideal distance and the distance from the carrier to the target, respectively. In the cone coordinate system, *R*<sup>n</sup> and *R* are transformed into:

$$R\_n = \sqrt{r\_0^2 + \left(\mathbf{x'} - \mathbf{x}\_0\right)^2} = \sqrt{\left(r\cos\phi\right)^2 + \left(\mathbf{x'} - \mathbf{x} - r\sin\phi\_d\right)^2} \tag{2}$$

$$R = \sqrt{\left(r\cos\phi\_d + \delta\dot{r}\_{xr}\right)^2 + \left(\varkappa' - \varkappa - r\sin\phi\_d\right)^2} \tag{3}$$

where [25].

$$\begin{aligned} \delta\hat{r}(\mathbf{x}',\mathbf{x},r,\phi\_d) &\approx d(\mathbf{x}') \Big\{ \sin\theta(\mathbf{x}')\cos\phi\_d \cos\theta(\mathbf{x},r) - \cos\theta(\mathbf{x}')\sin\theta(\mathbf{x},r) \\ &+ \frac{1}{2} (\sin\phi\_d)^2 [\cos\theta(\mathbf{x},r)]^2 \sin\theta(\mathbf{x}')\cos\phi\_d \cos\theta(\mathbf{x},r) \\ &+ \frac{1}{2} \cos\theta(\mathbf{x}')\sin\theta(\mathbf{x},r) (\sin\phi\_d)^2 \big[\cos\theta(\mathbf{x},r)\big]^2 \end{aligned} \tag{4}$$

Then the trajectory deviations *δR x*<sup>0</sup> , *x*,*r*, *ϕ<sup>d</sup>* ð Þ can be decomposed into the following terms:

$$
\delta \mathcal{R}(\mathbf{x}', \mathbf{x}, r, \phi\_d) \approx \delta \bar{r}(\mathbf{x}', \phi\_d) + \bar{\psi}(\mathbf{x}', r, \phi\_d) + \bar{\phi}(\mathbf{x}', \mathbf{x}, r, \phi\_d) \tag{5}
$$

where *δ*~*r x*<sup>0</sup> , *ϕ<sup>d</sup>* ð Þ is the track deviation caused by the deviation of radar track from the ideal track when the radar beam points to the center of the scene, which is only related to the azimuth position of radars and is the main component of track deviation, *ψ*~ *x*<sup>0</sup> ,*r*, *ϕ<sup>d</sup>* ð Þ is the range to space variability of track deviation, and *φ*~ *x*<sup>0</sup> , *x*,*r*, *ϕ<sup>d</sup>* ð Þ accounts for the influence of radar platform position, azimuth and range variations.

Suppose that the radar transmits the following Linear frequency modulation (LFM) signal. After heterodyne, the echo signal obtained is

$$\begin{split} h(\mathbf{x}',r') &= \iint d\mathbf{x} dr \chi(\mathbf{x},r) \text{rect}\left[\frac{r'-R}{D}\right] \alpha^2 \left[\mathbf{x}' - \mathbf{x} - r \frac{\sin\left(\phi - \phi\_d\right)}{\cos\phi\_d}\right] \\ &\times \exp\left[-j\frac{4\pi}{\lambda}R - ja\_r(r'-R)^2\right] \end{split} \tag{6}$$

where *r*<sup>0</sup> is the sampling coordinate of the range signal, *λ* is the wavelength, *γ*ð Þ *x*,*r* is the reflection function, *rect*½ �� represents the rectangular envelope of the pulse. After applying the range FT of *h x*<sup>0</sup> ,*r*<sup>0</sup> ð Þ, we obtain

$$\begin{split} H(\mathbf{x}',\eta) &= \text{rect}\left[\frac{\eta}{2a\_rD}\right] \cdot \exp\left[j\frac{\eta^2}{4a\_r}\right] \times \iint d\mathbf{x} d\mathbf{y} \langle \mathbf{x}, r \rangle \\ &\times \exp\left[-j\left(\eta + \frac{4\pi}{\lambda}\right) \cdot (r + \Delta R + \delta R)\right] \cdot \alpha^2 \left(\mathbf{x}' - \mathbf{x} - r \frac{\sin\left(\phi - \phi\_d\right)}{\cos\phi\_d}\right) \end{split} \tag{7}$$

Furthermore, by separating the factor exp �*<sup>j</sup> <sup>η</sup>* <sup>þ</sup> <sup>4</sup>*<sup>π</sup> λ* � �*δr x*<sup>0</sup> , *<sup>ϕ</sup><sup>d</sup>* ð Þ � �, we obtain

$$H(\mathbf{x}',\eta) = \exp\left[-j\left(\eta + \frac{4\pi}{\lambda}\right)\delta r(\mathbf{x}',\phi\_d)\right] \tilde{H}(\mathbf{x}',\eta) \tag{8}$$

where

$$\begin{split} \dot{H}(\mathbf{x}',\eta) &= \text{rect}\left[\frac{\eta}{2a\_rD}\right] \cdot \exp\left[j\frac{\eta^2}{4a\_r}\right] \times \iint d\mathbf{x}d\mathbf{r} \gamma(\mathbf{x},r) \\ &\times \exp\left[-j\left(\eta + \frac{4\pi}{\lambda}\right) \cdot (r + \Delta R(\mathbf{x}',\mathbf{x},r,\phi\_d) + \bar{\eta}(\mathbf{x}',r,\phi\_d) + \bar{\eta}(\mathbf{x}',\mathbf{x},r,\phi\_d))\right] \\ &\times o^2\left(\mathbf{x}' - \mathbf{x} - r\frac{\sin\left(\phi - \phi\_d\right)}{\cos\phi\_d}\right) \end{split} \tag{9}$$

If the following conditions are satisfied:

$$|\ddot{\rho}(\mathbf{x}', \mathbf{x}, r, \phi\_d)| \ll \frac{\lambda}{4\pi} \tag{10}$$

$$|\ddot{\Psi}(\mathbf{x}', \mathbf{x}, r, \phi\_d)| \ll \frac{f}{\Delta f} \frac{\lambda}{2\pi} \tag{11}$$

we can obtain the azimuth FT of *H x* <sup>~</sup> <sup>0</sup> ð Þ , *<sup>η</sup>* as follows:

$$\begin{split} \tilde{H}(\xi,\eta) &= \int dr \, \exp\left[-j\overline{\eta}r\right] \left[dr G(\xi-l,\eta,r)\Gamma(\xi-l,r)Q\_{\eta}(l,\eta,r) \\ &= \int dr \, \exp\left[-j\overline{\eta}r\right] \left\{Q\_{\eta}(\xi,\eta,r)\otimes\_{\xi}[G(\xi,\eta,r)\Gamma(\xi,r)]\right\} \end{split} \tag{12}$$

where

$$Q\_{\eta}(l,\eta,r) = FT\_{\mathbf{x'}}\{\exp\left[-j\overline{\eta}\tilde{\boldsymbol{\eta}}(\mathbf{x'},r,\phi\_d)\right]\}\tag{13}$$

with *<sup>η</sup>* <sup>¼</sup> *<sup>η</sup>* <sup>þ</sup> <sup>4</sup>*<sup>π</sup> <sup>λ</sup>* , After evaluating *<sup>H</sup>*<sup>~</sup> ð Þ *<sup>ξ</sup>*, *<sup>η</sup>* , it can be direct get *H x* <sup>~</sup> <sup>0</sup> ð Þ , *<sup>η</sup>* by using the inverse FT of *<sup>H</sup>*<sup>~</sup> ð Þ *<sup>ξ</sup>*, *<sup>η</sup>* along the azimuth direction. Then, using Eq. (8) and the range inverse FT, the original echo signal can be computed. The flowchart of the algorithm is shown in **Figure 3**. The validity of the range needs to satisfy Eqs. (10) and (11).

#### **2.2 Computational complexity of the algorithm**

Let us now consider the computational complexity of the algorithm in **Figure 3**. Suppose the data size is *Na* � *Nr*, with *Na* corresponding to the number of azimuth *Efficient Simulation of Airborne SAR Raw Data in Case of Motion Errors DOI: http://dx.doi.org/10.5772/intechopen.99378*

**Figure 3.** *Raw data simulation process of strip squint SAR with trajectory error.*

sampling points, and *Nr* to the number of range sampling points. It follows that the amount of computation required by the algorithm is

$$\begin{split} N &\approx N\_a N\_{r2} \left( 3 + \log\_2 N\_d \right) + N\_d N\_r \left( 1 + \log\_2 N\_d + \frac{1}{2} \log\_2 N\_r \right) \\ &\approx N\_a N\_r^2 \left( 3 + \log\_2 N\_d \right) \end{split} \tag{14}$$

Recalled that the required computation of time domain algorithm is as follows

$$
\tilde{N} \approx N\_a N\_r N\_{tp} N\_{sa} \tag{15}
$$

where *N*tp is the size of the emitted pulse and *N*sa is the synthetic aperture, *N*tp is usually equal to the size of *N*r. The ratio of Eqs. (14) and (15) takes the form

$$\frac{\bar{N}}{N} \approx \frac{N\_{sa}}{3 + \log\_2 N\_a} \tag{16}$$

It is evident that compared with the time domain simulation method, the proposed method has higher computational efficiency.

#### **2.3 Algorithm simulation verification**

In what follows, we use the SAR system parameters listed in **Table 1** to verify the SAR raw data simulation algorithm of non-ideal track under squint condition. **Figure 4** shows the track deviation between the actual track and the ideal track. For a point target located at the near range *<sup>r</sup>* <sup>¼</sup> <sup>4840</sup>*<sup>=</sup>* cos 4*<sup>o</sup>* ð Þ*m*, the phase difference between the simulated and the exact raw signal, along the range direction and the azimuth direction, is displayed in **Figure 5(a)** and **(b)**, respectively. We see that the


**Table 1.**

*Simulation parameters.*

**Figure 4.** *Sensor displacements.*

**Figure 5.** *(a) Range cut, (b) azimuth cut, the point target is located in the close range position (r* <sup>¼</sup> <sup>4840</sup>*<sup>=</sup>* cos 4*<sup>o</sup>* ð Þ *m) of the scene.*

phase error is much smaller than *π=*10, which in practice is negligible. By comparing the "depurated" phase error and the expected phase error 4ð Þ *π=λ φ*~ *x*<sup>0</sup> , *x*,*r*, *ϕ<sup>d</sup>* ð Þ (see **Figure 6**), we may conclude that the azimuth "depurated" phase error is given rise by the approximation of Eq. (10). Moreover, the point target impulse response for midrange at squint angle 15<sup>o</sup> after the motion compensation are shown in **Figure 7 (a)** and **(b)**. In addition, the figure of merits by point target analysis for midrange shown in **Figure 7(a)** and **(b)** between the proposed approach and the time-domain one is given in **Table 2**. The proposed raw data generator has high precision at least as that by the time domain algorithm.

*Efficient Simulation of Airborne SAR Raw Data in Case of Motion Errors DOI: http://dx.doi.org/10.5772/intechopen.99378*

#### **Figure 6.**

*(a) The phase error (smooth line) after removing the stationary phase influence. (b) the phase error along the azimuth direction (r* ¼ 4840*=* cos *ϕ<sup>d</sup>* ð Þ*m) according to the formula φ x*<sup>0</sup> , *<sup>x</sup>*,*r*, *<sup>ϕ</sup><sup>d</sup>* j j ð Þ < < <sup>4</sup>*<sup>π</sup> λ .*

#### **Figure 7.**

*Point target imaging results after motion error compensation. (a) Range cut and (b) azimuth cut.*


#### **Table 2.**

*Point target evaluation results.*

Finally, the extended scene is considered. The final results are given in **Figure 8**, where **Figure 8(a)** shows the image without motion compensation, **Figure 8(b)** displays the results by a two-step MOCO algorithm, and **Figure 8(c)** is the image after the motion compensation [25]. Concerning the computational efficiency, it took only about 2 min for the simulation run, as shown in **Figure 8(c)**, on an 8-GHz Intel Core i5 personal computer.

**Figure 8.**

*Comparison of imaging results: (a) No motion compensation, (b) second order motion error compensation, (c) the compensation method in ref. [25] is used.*
