Parameter Estimation of LFM Signals Based on Scaled Ambiguity Function

Abstract

A method based on the scaled ambiguity function for parameter estimation of LFM signals under noncooperative conditions is presented. In the proposed algorithm, the scaling principle is employed to remove the linear frequency migration brought by the coupling between the time variable and the lag variable. Afterward, the Fourier transform is performed for feature extraction of LFM signals on the centroid frequency versus chirp rate plane. The method avoids centroid frequency information loss, which is almost inevitable in the Radon-ambiguity transform. Furthermore, fractional lower-order statistics and the scaled ambiguity transform are combined to improve the performance in practical impulsive noise environments. Simulation results show that the fractional lower-order scaled ambiguity transform is robust for both Gaussian and impulsive noise, and it achieves significant performance improvement in a heavy noise environment.

Appendix

In the SAF, each signal component generates an auto term, and each pair of signal components generates a cross term. Cross terms in the SAF can be written as

$$\begin{aligned} \varGamma (R_{s_i s_j } )+\varGamma (R_{s_j s_i } )=2A_i A_j R_{s_i s_j }^{c1} \left( {t_n ,\tau } \right) \mathfrak {R}\left[ {R_{s_j s_i }^{c2} \left( {t_n ,\tau } \right) } \right] , \end{aligned}$$

(21)

with

$$\begin{aligned} R_{s_i s_j }^{c1} \left( {t_n ,\tau } \right) =\exp \left( {j\pi \left( {f_i +f_j } \right) \left( {\tau +1} \right) +j\pi \left( {r_i +r_j } \right) t_n } \right) , \end{aligned}$$

(22)

and

$$\begin{aligned}&R_{s_i s_j }^{c2} \left( {t_n ,\tau } \right) \nonumber \\&\quad =\exp \left( {j2\pi \left( {f_i -f_j } \right) \frac{t_n }{\tau +1}+j\pi \left( {r_i -r_j } \right) \left( {\frac{t_n }{\tau +1}} \right) ^{2}+j\pi \left( {r_i -r_j } \right) \left( {\frac{\tau +1}{2}} \right) ^{2}} \right) ,\nonumber \\ \end{aligned}$$

(23)

where $\mathfrak {R}\left( \cdot \right)$ denotes the operation of taking real part, $R_{s_i s_j }^{c1} \left( {t_n ,\tau } \right)$ and $\mathfrak {R}\left[ {R_{s_j s_i }^{c2} \left( {t_n ,\tau } \right) } \right]$ represent the complex exponent and envelope, respectively. Let $\Delta f_{ij} =\frac{f_i -f_j }{2}$, $\nabla f_{ij} =\frac{f_i +f_j }{2}$, $\widetilde{f}=f-\nabla f_{ij}$, $\Delta r_{ij} =\frac{r_i -r_j }{2}$, $\nabla r_{ij} =\frac{r_i +r_j }{2}$ and $\widetilde{r}=r-\nabla r_{ij}$. Each cross term can be expressed as:

Case 1 For $r_i =r_j \ne r$,

$$\begin{aligned} \Delta r_{ij} =0, \end{aligned}$$

(24)

then we obtain

$$\begin{aligned} \mathfrak {R}\left[ {R_{s_j s_i }^{c2} \left( {t_n ,\tau } \right) } \right] =\cos \left( {2\pi \left( {f_i -f_j } \right) \frac{t_n }{\tau +1}} \right) =\cos \left( {4\pi \Delta f_{ij} \frac{t_n }{\tau +1}} \right) . \end{aligned}$$

(25)

Equation (21) can be simplified as

$$\begin{aligned} \varGamma (R_{s_i s_j } )+\varGamma (R_{s_j s_i } )=2A_i A_j R_{s_i s_j }^{c1} \left( {t_n ,\tau } \right) \cos \left( {4\pi \Delta f_{ij} \frac{t_n }{\tau +1}} \right) . \end{aligned}$$

(26)

After applying the two-dimensional Fourier transform to $2A_i A_j R_{s_i s_j }^{c1} \left( {t_n ,\tau } \right)$, we have

$$\begin{aligned} F_\tau \left[ {F_{t_n } \left( {2A_i A_j R_{s_i s_j }^{c1} \left( {t_n ,\tau } \right) } \right) } \right] =2A_i A_j \exp ( {j2\pi \nabla f_{ij} } )\delta ( {r-\nabla r_{ij} })\delta ( {f-\nabla f_{ij} }).\nonumber \\ \end{aligned}$$

(27)

Similarly, applying the two-dimensional Fourier transform to (25), we obtain

$$\begin{aligned}&F_\tau \left[ {F_{t_n } \left( {\cos \left( {4\pi \Delta f_{ij} \frac{t_n }{\tau +1}} \right) } \right) } \right] \nonumber \\&\quad =F_\tau \left[ {\frac{1}{2}\left( {\delta \left( {r+2\frac{\Delta f_{ij} }{\tau +1}} \right) +\delta \left( {r-2\frac{\Delta f_{ij} }{\tau +1}} \right) } \right) } \right] \nonumber \\&\quad =\left\{ {{\begin{array}{ll} {F_\tau \left[ {\frac{\Delta f_{ij} }{\widetilde{r}^{2}}\left( {\delta \left( {\tau +1+\frac{2\Delta f_{ij} }{\widetilde{r}}} \right) +\delta \left( {\tau +1-\frac{2\Delta f_{ij} }{\widetilde{r}}} \right) } \right) } \right] ,} &{}\quad {\hbox {for}\;\Delta f_{ij} \ne 0.} \\ {\delta \left( {\widetilde{f}} \right) \delta \left( {\widetilde{r}} \right) ,} &{}\quad {\hbox {for}\;\Delta f_{ij} =0}\\ \end{array} }} \right. \nonumber \\&\quad =\left\{ {{\begin{array}{ll} {\frac{2\Delta f_{ij} }{\widetilde{r}^{2}}\exp \left( {j2\pi \nabla f_{ij} } \right) \cos \left( {\frac{4\pi \Delta f_{ij} }{\widetilde{r}}\widetilde{f}} \right) ,} &{}\quad {\hbox {for}\;\Delta f_{ij} \ne 0} \\ {\delta \left( {\widetilde{f}} \right) \delta \left( {\widetilde{r}} \right) ,} &{}\quad {\hbox {for}\;\Delta f_{ij} =0} \\ \end{array} }} \right. \end{aligned}$$

(28)

Combining (27) with (28) and exploiting the two-dimensional convolution theorem, we can obtain that for $r_i =r_j \ne r$ , $\hbox {SAT}_{s_i s_j }$ in (9) is given by $\hbox {SAT}_{s_i s_j } =4A_i A_j \frac{\Delta f_{ij} }{\widetilde{r}^{2}}\exp \left( {j2\pi \nabla f_{ij} } \right) \cos \left( {\frac{4\pi \Delta f_{ij} }{\widetilde{r}}\widetilde{f}} \right)$. And for $r_i =r_j =r$, $\hbox {SAT}_{s_i s_j }$ is simplified to meet $\hbox {SAT}_{s_i s_j } =0$, that is, when $\widetilde{r}=0$, the corresponding cross term turns out to be zero.

Case 2 $r_i \ne r_j$ .

Since the Fourier transform of $\mathfrak {R}\left[ {R_{s_j s_i }^{c^{2}} (t_n , \tau )} \right]$ with respect to $t_n$ or $\tau$ is an oscillatory integral and the amplitude of integrand is slowly varying as compared to the oscillations controlled by the phase terms, the principle of stationary phase in [16, Ch.2] can be used. Then, we have

$$\begin{aligned} t_n ^{{*}}=\left( {r-\frac{2\Delta f_{ij} }{\tau +1}} \right) \frac{\left( {\tau +1} \right) ^{2}}{2\Delta r_{ij} }, \end{aligned}$$

(29)

and

$$\begin{aligned} \frac{\mathrm{d}^{2}}{\mathrm{d}t_n ^{*2}}\left( {\hbox {phase}\left( {R_{{s_j} {s_i}}^{c2} \left( {t_n ^{{*}},\tau } \right) } \right) -2\pi rt_n ^{{*}}} \right) =\frac{4\pi \Delta r_{ij} }{\left( {\tau +1} \right) ^{2}}, \end{aligned}$$

(30)

where $t_n ^{{*}}$ denotes the stationary point. Using (29) and (30), we obtain

$$\begin{aligned} \hbox {SAF}_{s_i s_j } \left( {\tau ,r} \right)= & {} F_{t_n } \left( {R_{s_j s_i }^{c2} \left( {t_n ,\tau } \right) } \right) \nonumber \\= & {} \frac{\left| {\tau +1} \right| }{\sqrt{2}\left| {\Delta r_{ij} } \right| ^{1/2}}\exp \left( {j\hbox {sgn}\left( {\Delta r_{ij} } \right) \frac{\pi }{4}} \right) \nonumber \\&\times \exp \left\{ {j\pi \left( {2\Delta f_{ij} \frac{\tau +1}{\Delta r_{ij} }r-\frac{\left( {\tau +1} \right) ^{2}}{2\Delta r_{ij} }r^{2}-\frac{2\Delta f_{ij} ^{2}}{\Delta r_{ij} }+\Delta r_{ij} \frac{\left( {\tau +1} \right) ^{2}}{2}} \right) } \right\} ,\nonumber \\ \end{aligned}$$

(31)

where $\hbox {sgn}\left( \cdot \right)$ denotes the sign function.

Case 2-1 $r=\pm \left| {\Delta r_{ij} } \right|$. Letting ${\tau }'=\tau +1$ and taking Fourier transform of $\hbox {SAF}_{s_i s_j } \left( {{\tau }',r} \right)$ with respect to ${\tau }'$, we have

$$\begin{aligned} {S}'\left( {f,r} \right)= & {} F_{{\tau }'} \left( {\hbox {SAF}_{s_i s_j } \left( {{\tau }',r} \right) } \right) \nonumber \\= & {} -\frac{1}{\left| {\Delta r_{ij} } \right| ^{1/2}}\exp \left( {-j2\pi \frac{\Delta f_{ij} ^{2}}{\Delta r_{ij} }} \right) \exp \left( {j\hbox {sgn}\left( {\Delta r_{ij} } \right) \frac{\pi }{4}} \right) \frac{1}{2\sqrt{2}\pi ^{2}\left( {f-\frac{\Delta f_{ij} }{\Delta r_{ij} }r} \right) ^{2}}.\nonumber \\ \end{aligned}$$

(32)

According to the conjugation property of the Fourier transform, we obtain

$$\begin{aligned} F_\tau \left[ {F_{t_n } \left( {\mathfrak {R}\left[ {R_{s_j s_i }^{c2} \left( {t_n ,\tau } \right) } \right] } \right) } \right] =\frac{1}{2}\left( {{S}'\left( {f,r} \right) +{S}'^{{*}}\left( {-f,-r} \right) } \right) , \end{aligned}$$

(33)

where ${*}$ Denotes complex conjugation as in (2). Combining (33) with (27), we have

$$\begin{aligned}&F_\tau \left[ {F_{t_n } \left( {2R_{s_i s_j }^{c1} \left( {t_n ,\tau } \right) \mathfrak {R}\left[ {R_{s_j s_i }^{c2} \left( {t_n ,\tau } \right) } \right] } \right) } \right] \nonumber \\&\qquad ={S}'\left( {f-\nabla f_{ij} ,r-\nabla r_{ij} } \right) +{S}'^{{*}}\left( {f-\nabla f_{ij} ,r-\nabla r_{ij} } \right) \end{aligned}$$

(34)

Inserting (32) into (34) and applying the translation and scaling properties to (34), it can be obtained that $\hbox {SAT}_{s_i s_j }$ in (9) is given by

$$\begin{aligned} \hbox {SAT}_{s_i s_j }= & {} -\frac{A_i A_i }{\left| {\Delta r_{ij} } \right| ^{1/2}}\cos \left( {\hbox {sgn}\left( {\Delta r_{ij} } \right) \frac{\pi }{4}-2\pi \frac{\Delta f_{ij} ^{2}}{\Delta r_{ij} }} \right) \\&\times \frac{\exp \left( {j2\pi \nabla f_{ij} } \right) }{\sqrt{2}\pi ^{2}\left( {\widetilde{f}-\frac{\Delta f_{ij} }{\Delta r_{ij} }\left( {\pm \left| {\Delta r_{ij} } \right| -\nabla r_{ij} } \right) } \right) ^{2}}. \end{aligned}$$

Case 2-2 $r\ne \pm \left| {\Delta r_{ij} } \right|$. Similarly, the stationary point can be expressed as

$$\begin{aligned} {\tau }'^{{*}}=\frac{2\Delta r_{ij} f-2\Delta f_{ij} r}{\Delta r_{ij}^2 -r^{2}}, \end{aligned}$$

(35)

and

$$\begin{aligned} \frac{\mathrm{d}^{2}}{\mathrm{d}{\tau }'^{2}}( {\hbox {phase}( {\hbox {SAF}_{s_i s_j } ( {{\tau }',r} )} )-2\pi f{\tau }'})=\pi \Delta r_{ij} -\pi \frac{r^{2}}{\Delta r_{ij} }. \end{aligned}$$

(36)

Using (35) and (36), the Fourier transform of $\hbox {SAF}_{s_i s_j } \left( {{\tau }',r} \right)$ with respect to ${\tau }'$ can be written as

$$\begin{aligned}&{S}'\left( {f,r} \right) \nonumber \\&\quad =F_{{\tau }'} \left( {\hbox {SAF}_{s_i s_j } \left( {{\tau }',r} \right) } \right) \nonumber \\&\quad =\frac{2\left| {\Delta r_{ij} f-\Delta f_{ij} r} \right| }{\left| {r^{2}-\Delta r_{ij}^2 } \right| ^{3/2}}\exp \left( {j\frac{\pi }{4}\hbox {sgn}\left( {\Delta r_{ij} } \right) \left( {1+\hbox {sgn}\left( {\Delta r_{ij}^2 -r^{2}} \right) } \right) } \right) \nonumber \\&\qquad \exp \left\{ {-j\pi \frac{2\Delta r_{ij} }{\Delta r_{ij}^2 -r^{2}}\left( {f-\Delta f_{ij} \frac{r}{\Delta r_{ij} }} \right) ^{2}-j2\pi \frac{\Delta f_{ij}^2 }{\Delta r_{ij} }} \right\} . \end{aligned}$$

(37)

Substituting (37) in (34) and applying the translation and scaling principles to (34), we can obtain that $\hbox {SAT}_{s_i s_j }$ in (9) is given by

$$\begin{aligned} \hbox {SAT}_{s_i s_j }= & {} 4A_i A_j \exp \left( {j2\pi \nabla f_{ij} } \right) \nonumber \\&\times \frac{\left| {\Delta r_{ij} \widetilde{f}-\Delta f_{ij} \widetilde{r}} \right| }{\left| {\widetilde{r}^{2}-\Delta r_{ij} ^{2}} \right| ^{3/2}}\cos \left[ {\left( {\frac{\pi }{4}\hbox {sgn}\left( {\Delta r_{ij} } \right) \left( {1+\hbox {sgn}\left( {\Delta r_{ij}^2 -r^{2}} \right) } \right) } \right) } \right. \nonumber \\&\left. {-\frac{2\pi \Delta r_{ij} }{\Delta r_{ij}^2 -r^{2}}\left( {\widetilde{f}-\Delta f_{ij} \frac{\widetilde{r}}{\Delta r_{ij} }} \right) ^{2}-2\pi \frac{\Delta f_{ij}^2 }{\Delta r_{ij} }} \right] . \end{aligned}$$

(38)

As shown in (38), when $\widetilde{r}=\pm \left| {\Delta r_{ij} } \right|$, $\hbox {SAT}_{s_i s_j } =0$, that is, the cross term located at the center point of two auto terms is eliminated, which completes the proof.