## 3 Digital Pulse Compression
### Linear Phase Sampling and Accumulating Method for Fast Pulse Compression
\[A_{2}=\frac{1}{2}\sum_{n=1}^{N}\sum_{i=1}^{m}a_{i}\cos(2\pi(f_{bpfstop}(t_{j}+t_{n}^ {2}-t_{i})-\frac{1}{2}k\Big{(}t_{j}+t_{n}^{2}-t_{i}\Big{)}^{2})-\varphi_{ec0}+ \varphi_{i}) \tag{21}\]  
where \(t_{n}^{1}\) and \(t_{n}^{2}\) satisfy the following conditions:  
\[t_{n}^{1},t_{n}^{2}\in(t_{i}-t_{j},t_{i}-t_{j}+\frac{B_{bpf}}{k})\cap\left(0, \frac{B_{bpf}}{k}\right) \tag{22}\]  
If \(\Delta t_{ji}=t_{i}-t_{j}\), then (22) can be rewritten as:  
\[t_{n}^{1},t_{n}^{2}\in(\Delta t_{ji},\frac{B_{bpf}}{k})\ \ \Delta t_{ji}\geq 0 \tag{23}\]  
\[t_{n}^{1},t_{n}^{2}\in(0,\Delta t_{ji}+\frac{B_{bpf}}{k})\ \ \Delta t_{ji}<0 \tag{24}\]  
Equations (23) and (24) indicate that only the common parts between the i-th and j-th IF chirp components will be effective in the data accumulation. If there is only the j-th IF chirp component, which means \(i=j\), (20) and (21) turn into:  
\[2\pi\bigg{(}f_{bpfstop}t_{n}^{2}-\frac{1}{2}k\Big{(}t_{n}^{2}\Big{)}^{2}\bigg{)} =\varphi_{r}+\frac{\pi}{2}+2n\pi\ \ n=1,2,\ldots,N \tag{17}\]0  
\[2\pi\bigg{(}f_{bpfstop}t_{n}^{2}-\frac{1}{2}k\Big{(}t_{n}^{2}\Big{)}^{2}\bigg{)} =\varphi_{r}+\frac{\pi}{2}+2n\pi\ \ n=1,2,\ldots,N \tag{17}\]1  
Thus, the amplitude of the input signal \(a_{j}\) can be obtained:  
\[2\pi\bigg{(}f_{bpfstop}t_{n}^{2}-\frac{1}{2}k\Big{(}t_{n}^{2}\Big{)}^{2}\bigg{)} =\varphi_{r}+\frac{\pi}{2}+2n\pi\ \ n=1,2,\ldots,N \tag{17}\]2  
For any other IF chirp components (\(i\
eq j\)), (20) can be rewritten as:  
\[2\pi\bigg{(}f_{bpfstop}t_{n}^{2}-\frac{1}{2}k\Big{(}t_{n}^{2}\Big{)}^{2}\bigg{)} =\varphi_{r}+\frac{\pi}{2}+2n\pi\ \ n=1,2,\ldots,N \tag{17}\]3  
If \(\varphi_{ji}=2\pi(-f_{bpfstop}\Delta t_{ji}-\frac{1}{2}k(\Delta t_{ji})^{2})\) and \(\Delta f_{ji}=k\Delta t_{ji}\), then (28) can be further simplified:  
\[2\pi\bigg{(}f_{bpfstop}t_{n}^{2}-\frac{1}{2}k\Big{(}t_{n}^{2}\Big{)}^{2}\bigg{)} =\varphi_{r}+\frac{\pi}{2}+2n\pi\ \ n=1,2,\ldots,N \tag{17}\]4  
Similarly, (21) can be simplified:  
\[2\pi\bigg{(}f_{bpfstop}t_{n}^{2}-\frac{1}{2}k\Big{(}t_{n}^{2}\Big{)}^{2}\bigg{)} =\varphi_{r}+\frac{\pi}{2}+2n\pi\ \ n=1,2,\ldots,N \tag{17}\]5  
where \(t_{n}^{1}\) and \(t_{n}^{2}\) satisfy (23) and (24). Equation (29) indicates that the accumulation result is actually equal to the amplitudes accumulation of a cosine signal, the frequency of which is equal to the frequency interval of the i-th and j-th measured signals. For a fixed compression time \(T_{c}\), comparing (25) and (29), the relationship between the compression time and the frequency resolution (\(f_{r}\)) can be evaluated: \(f_{r}=\frac{\sqrt{2}}{\pi T_{c}}+\frac{\sqrt{2}T_{c}}{\pi T_{c}^{2}+\frac{ \sqrt{2}}{\pi T_{c}}}\), where \(\frac{\sqrt{2}}{\pi T_{c}}\) and \(\frac{\sqrt{2}T_{c}}{\pi T_{c}^{2}+\frac{\sqrt{2}}{\pi}}\) correspond to the frequency resolution for the situation of \(\Delta t_{ji}<0\) and \(\Delta t_{ji}\geq 0\), individually. It can be seen that the frequency resolution is related to the compression time and the chirp rate. However, if \({T_{c}}^{2}\gg\frac{1}{k}\), the influence of the chirp rate on the frequency resolution can be ignored. In other words, the chirp rate k only matters for low frequency resolution. When the chirp rate is predetermined, the frequency resolution is only relative to the compression time. Thisallows us to obtain different frequency resolutions by simply changing the compression time. It should be noted that the sampling rate in (16) and (17) is changes linearly, which is different from uniformly-spaced sampling. So, the spectral periodic extension due to the sampling in the time domain will also be different. Assuming that the sampling rate in the algorithm ranges from \(f_{s1}\) to \(f_{s2}\) with a bandwidth of \(B_{s}=f_{s2}-f_{s1}\), the extended spectrum will be distributed at \(M\):\((f_{s1},f_{s2})\)(M is an arbitrary positive integer), and the amplitude at the frequency point \(f_{x}(f_{x}\in M\cdot(f_{s1},f_{s2}))\) will have a weight factor of \(\frac{f_{x}}{\sqrt{M\cdot B_{s}\cdot T_{c}}}\frac{\frac{\delta\mu+1/2}{2}}\) compared to the major signal.