Target Image Enhancement in Radar Imaging Using Fractional Fourier Transform

Abstract

This paper presents a new Range-Doppler Algorithm based on Fractional Fourier Transform (RDA-FrFT) to obtain High-Resolution (HR) images for targets in radar imaging. The performance of the proposed RDA-FrFT is compared with the classical RDA algorithm, which is based on the Fast Fourier Transform (FFT). A closed-form expression for the range and azimuth compression of the proposed RDA-FrFT is mathematically derived and analyzed from the HR Synthetic Aperture Radar (SAR) imaging point of view. The proposed RDA-FrFT takes its advantage of the property of the FrFT to resolve chirp signals with high precision. Results show that the proposed RDA-FrFT gives low Peak Side-Lobe (PSL) and Integrated Side-Lobe (ISL) levels in range and azimuth directions for detected targets. HR images are obtained using the proposed RDA-FrFT algorithm.

Appendices

Appendix 1

1.1 Range FrFT

$$ S_{1} (f_{t} ,\eta ) = \int {A_{1} W_{r} \cdot W_{a} \cdot \exp [j\phi_{1} (t,\eta ,\,f_{t} )]} $$

(25)

where

$$ \phi_{1} (t,\eta ,\,f_{t} ) = \frac{{ - 4\pi f_{o} R(\eta )}}{c} + \pi K_{r} \left( {t - \frac{2R(\eta )}{c}} \right)^{2} + \pi \left( {\left( {f_{t}^{2} u_{o}^{\prime 2} + t^{2} u_{o}^{2} } \right)\cot \alpha - 2tf_{t} \csc \alpha } \right) $$

(26)

The derivative of ϕ ₁(t, η, f _t ) with respect to t is given by

$$ \frac{d\phi (t)}{dt} = 2\pi K_{r} \left( {t - \frac{2R(\eta )}{c}} \right) + 2\pi tu_{o}^{2} \cot \alpha - 2\pi f_{t} \csc \alpha $$

(27)

To use the PSP, we should find the range time t^* when $\frac{d\phi_{1} (t)}{dt} = 0$

$$ 2\pi K_{r} \left( {t - \frac{2R(\eta )}{c}} \right) + 2\pi tu_{o}^{2} \cot \alpha - 2\pi f_{t} \csc \alpha = 0 $$

(28)

$$ t\left[ {K_{r} + u_{o}^{2} \cot \alpha } \right] = f_{t} \csc \alpha + \frac{{2K_{r} R(\eta )}}{c} $$

(29)

$$ t^{*} = \frac{{\frac{{2K_{r} R(\eta )}}{c} + f_{t} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }} $$

(30)

The solution of the integral of S ₁(f _t , η) can be written as

$$ S_{1} (f_{t} ,\eta ) = A_{1} \cdot W_{r} \cdot W_{a} \cdot \exp \left[ {j\phi_{1} \left( {t^{*} ,\eta ,\,f_{t} } \right)} \right], $$

(31)

where

$$ \begin{aligned} \phi_{1} \left( {t^{*} ,\eta ,\,f_{t} } \right) & = \frac{{ - 4\pi f_{o} R(\eta )}}{c} + \left( {t^{*} - \frac{2R(\eta )}{c}} \right)^{2} + \pi f_{t}^{2} u_{o}^{\prime 2} \cot \alpha + \pi t^{*2} u_{o}^{2} \cot \alpha - 2\pi t^{*} f_{t} \csc \alpha \\ & = \frac{ - 4\pi R(\eta )}{c}\left( {f_{o} + \frac{{K_{r} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}f_{t} } \right) + \pi f_{t}^{2} \left( {u_{o}^{\prime 2} \cot \alpha - \frac{{2\csc^{2} \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) + \frac{{4\pi K_{r}^{2} R^{2} (\eta )}}{{c^{2} }}\left( {\frac{{u_{o}^{2} \cot \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) \\ \end{aligned} $$

(32)

The last term in Eq. 32 is the Residual Video Phase (RVP), which is a consequence of the range FrFT of the received chirp signal that should be removed to avoid defocusing of the constructed SAR image, hence

$$ \phi_{1} \left( {t^{*} ,\eta ,\,f_{t} } \right) \approx \frac{ - 4\pi R(\eta )}{c}\left( {f_{o} + \frac{{K_{r} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}f_{t} } \right) + \pi f_{t}^{2} \left( {u_{o}^{\prime 2} \cot \alpha - \frac{{2\csc^{2} \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right). $$

(33)

1.2 Range reference FrFT

$$ S_{R} (f_{t} ) = \int {A_{o} \cdot \exp \left( {j\phi_{R} (t,\, f_{t} )} \right)} dt , $$

(34)

where

$$ \phi_{R} (t,\,f_{t} ) = \pi K_{r} (t - t_{d} )^{2} + \pi \left( {f_{t}^{2} u_{o}^{\prime 2} + t^{2} u_{o}^{2} } \right)\cot \alpha - 2\pi tf_{t} \csc \alpha $$

(35)

To use the PSP, we should find the range time t^* when $\frac{d\phi_R (t)}{dt} = 0$

$$ \frac{d\phi_R (t)}{dt} = 2\pi K_{r} (t - t_{d} ) + 2\pi tu_{o}^{2} \cot \alpha - 2\pi f_{t} \csc \alpha = 0 $$

(36)

$$ t^{*} = \frac{{f_{t} \csc \alpha + K_{r} t_{d} }}{{K_{r} + u_{o}^{2} \cot \alpha }} $$

(37)

The solution of the integral of S _R (f _t ) can be written as

$$ S_{R} (f_{t} ) = A_{o} \cdot \exp \left( {j\phi_{R} \left( {t^{*} ,\,f_{t} } \right)} \right) $$

(38)

$$ \begin{aligned} \phi_{R} \left( {t^{*} ,\,f_{t} } \right) & = \pi K_{r} \left( {\frac{{f_{t} \csc \alpha + K_{r} t_{d} }}{{K_{r} + u_{o}^{2} \cot \alpha }} - t_{d} } \right)^{2} + \pi f_{t}^{2} u_{o}^{\prime 2} \cot \alpha + \pi u_{o}^{2} \cot \alpha \left( {\frac{{f_{t} \csc \alpha + K_{r} t_{d} }}{{K{}_{r} + u_{o}^{2} \cot \alpha }}} \right)^{2} \\ & \quad - 2\pi f_{t} \csc \alpha \left( {\frac{{f_{t} \csc \alpha + K_{r} t_{d} }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) \\ & = \pi f_{t}^{2} \left( {u_{o}^{\prime 2} \cot \alpha - \frac{{2\csc^{2} \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) - \frac{{2\pi f_{t} K_{r} t_{d} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }} + \pi K_{r}^{2} t_{d}^{2} \left( {\frac{{u_{o}^{2} \cot \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right). \\ \end{aligned} $$

(39)

1.3 Range-IFrFT

To find the range frequency $f_{t}^{*}$, the derivative $\frac{{d\phi_3 (f_{t} )}}{{df_{t} }}$ must be equal zero,

$$ \begin{aligned} \frac{{d\phi_{3} (f_{t} )}}{{df_{t} }} & = \frac{ - 4\pi R(\eta )}{c}\left( {\frac{{K_{r} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) - \frac{{2\pi K_{r} t_{d} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }} + 4\pi f_{t} \left( {u_{o}^{\prime 2} \cot \alpha - \frac{{2\csc^{2} \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) \\ & \quad + 2\pi t\csc \alpha - 2\pi f_{t} u_{o}^{\prime 2} \cot \alpha = 0 \\ \end{aligned} $$

(40)

Hence,

$$ f_{t}^{*} = M_{1} \cdot \left( {\frac{2R(\eta )}{c} + M_{2} (t)} \right) $$

(41)

where $M_{1} = {{K_{r} } \mathord{\left/ {\vphantom {{K_{r} } {\left[ {\left( {K_{r} + u_{o}^{2} \cot \alpha } \right) \cdot \left( {{{u_{o}^{\prime 2} \cos \alpha - 4\csc \alpha } \mathord{\left/ {\vphantom {{u_{o}^{\prime 2} \cos \alpha - 4\csc \alpha } {\left( {K_{r} + u_{o}^{2} \cot \alpha } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {K_{r} + u_{o}^{2} \cot \alpha } \right)}}} \right)} \right]}}} \right. \kern-\nulldelimiterspace} {\left[ {\left( {K_{r} + u_{o}^{2} \cot \alpha } \right) \cdot \left( {{{u_{o}^{\prime 2} \cos \alpha - 4\csc \alpha } \mathord{\left/ {\vphantom {{u_{o}^{\prime 2} \cos \alpha - 4\csc \alpha } {\left( {K_{r} + u_{o}^{2} \cot \alpha } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {K_{r} + u_{o}^{2} \cot \alpha } \right)}}} \right)} \right]}}$, is constant. $M_{2} (t) = t_{d} - t - tu_{o}^{2} \cot \alpha /K_{r}$, which is constant for azimuth FrFT.

The solution of the integral of $S_{3} (t,\eta )$can be written as

$$ S_{3} (t,\eta ) = A_{o}^{2} \cdot A_{1} \cdot W_{r} \cdot W_{a} \cdot \exp \left( {j\phi_{3} (t,\, f_{t}^{*} )} \right) $$

(42)

where

$$ \begin{aligned} \phi_{3} \left( {t,\, f_{t}^{*} } \right) & = \frac{ - 4\pi R(\eta )}{c}\left( {f_{o} + M_{1} \left( {\frac{{K_{r} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right)\left( {\frac{2R(\eta )}{c} + M_{2} (t)} \right)} \right) \\ & \quad - \frac{{2\pi K_{r} t_{d} \csc \alpha \cdot M_{1} }}{{K_{r} + u_{o}^{2} \cot \alpha }}\left( {\frac{2R(\eta )}{c} + M_{2} (t)} \right) + 2\pi M_{1}^{2} \left( {u_{o}^{\prime 2} \cot \alpha - \frac{{2\csc^{2} \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right)\left( {\frac{2R(\eta )}{c} + M_{2} (t)} \right)^{2} \\ & \quad + 2\pi t\csc \alpha \cdot M_{1} \left( {\frac{2R(\eta )}{c} + M_{2} (t)} \right) - \pi u_{o}^{\prime 2} \cot \alpha \cdot M_{1}^{2} \left( {\frac{2R(\eta )}{c} + M_{2} (t)} \right)^{2} - \pi t^{2} u_{o}^{2} \cot \alpha \\ \end{aligned} $$

(43)

The above expression contains the RVP. We can remove it from our derivation, since it only adds additional complexity to the mathematical calculations.

$$ \phi_{3} \left( {t,\, f_{t}^{*} } \right) \approx \frac{2\pi R(\eta )}{c}\left\{ \begin{gathered} - 2f_{o} - 2M_{1} \cdot M_{2} (t)\left( {\frac{{K_{r} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) - \frac{{2K_{r} t_{d} M_{1} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }} + 4M_{1}^{2} M_{4} M_{2} (t) \hfill \\ + t \cdot M_{1} \csc \alpha - 2M_{1}^{2} M_{2} (t)u_{o}^{\prime 2} \cot \alpha \hfill \\ \end{gathered} \right\} + M_{3} (t) $$

where

$$ M_{4} = u_{o}^{\prime 2} \cot \alpha - \left( {\frac{{2\csc^{2} \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right)\quad {\text{is}}\;{\text{constant}}. $$

(44)

Appendix 2

2.1 Azimuth FrFT

$$ S_{4} (t,\,f_{\eta } ) = \int {S_{3} (t,\eta ) \cdot K_{b} (\eta ,f_{\eta } )d\eta } = \int {A_{o}^{3} \cdot A_{1} \cdot W_{r} \cdot W_{a} \cdot \exp \left( {j\phi_{4} (t,\,f_{\eta } ,\eta )} \right)d\eta } $$

(45)

where

$$ K_{b} (\eta ,f_{\eta } ) = A_{o} \cdot \exp \left( {j\pi \left( {f_{\eta }^{2} u_{o}^{\prime 2} + \eta^{2} u_{o}^{2} } \right)\cot \beta - 2\eta f_{\eta } \csc \beta } \right),\quad \beta = b\pi /2. $$

(46)

After applying the PSP, the final expression is

$$ S_{4} (t,\,f_{\eta } ) = A_{o}^{3} \cdot A_{1} \cdot W_{r} \cdot W_{a} \cdot \exp \left( {j\phi_{4} \left( {t,\,f_{\eta } ,\eta^{*} } \right)} \right) $$

(47)

where

$$ M_{5} (t) = \frac{1}{c}\left\{ \begin{gathered} - 2f_{o} - 2M_{1} \cdot M_{2} (t)\left( {\frac{{K_{r} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }}} \right) - \frac{{2K_{r} t_{d} M_{1} \csc \alpha }}{{K_{r} + u_{o}^{2} \cot \alpha }} + 4M_{1}^{2} M_{4} M_{2} (t) \hfill \\ + t \cdot M_{1} \csc \alpha - 2M_{1}^{2} M_{2} (t)u_{o}^{\prime 2} \cot \alpha \hfill \\ \end{gathered} \right\} $$

(48)

$$ \begin{aligned} \phi_{4} \left( {t, f_{\eta } ,\eta^{*} } \right) & = 2\pi \cdot M_{5} (t) \cdot \left( {R_{o} + \frac{{V^{2} }}{{2R_{o} }} \cdot \left( {\frac{{f_{\eta } \csc \beta }}{{u_{o}^{2} \cot \beta + V^{2} M_{5} (t)/R_{o} }}} \right)^{2} } \right) + M_{3} (t) + \pi \cdot f_{\eta }^{2} u_{o}^{\prime 2} \cot \beta \\ & \quad + \pi \cdot u_{o}^{2} \cot \beta \cdot \left( {\frac{{f_{\eta } \csc \beta }}{{u_{o}^{2} \cot \beta + V^{2} M_{5} (t)/R_{o} }}} \right)^{2} - 2\pi \cdot f_{\eta } \csc \beta \cdot \left( {\frac{{f_{\eta } \csc \beta }}{{u_{o}^{2} \cot \beta + V^{2} M_{5} (t)/R_{o} }}} \right). \\ \end{aligned} $$

(49)

2.2 Azimuth Reference FrFT

The azimuth reference function used for the RDA is given by:

$$ S_{a} (\eta ) = \exp (j2\pi K_{a} \eta ) $$

(50)

Its FrFT is given by:

$$ S_{a} (f_{\eta } ) = \int {S_{a} (\eta ) \cdot K_{b} (\eta ,f_{\eta } )d\eta } = \int {A_{o} \cdot \exp \left( {j\phi_{a} (\eta ,f_{\eta } )} \right)d\eta } $$

(51)

then,

$$ S_{a} (f_{\eta } ) = A_{o} \cdot \exp \left( {j\phi_{a} \left( {\eta^{*} , f_{\eta } } \right)} \right) $$

(52)

The next step is multiplying $S_{4} (t,\,f_{\eta } )\,{\text {by}}\,S_{a} (f_{\eta } )\,{\text {to}\,{get}}$:

$$ S_{5} (t,\,f_{\eta } ) = A_{o}^{4} \cdot A_{1} \cdot W_{r} \cdot W_{a} \cdot \exp \left( {j\phi_{5} \left( {t, f_{\eta }, \eta^{*} } \right)} \right) $$

(53)

where

$$ \phi_{5} \left( {t,\,f_{\eta } ,\eta^{*} } \right) = \phi_{4} \left( {t,\,f_{\eta } ,\eta^{*} } \right) + \phi_{a} \left( {\eta^{*} , f_{\eta } } \right). $$

(54)

2.3 Azimuth IFrFT

$$ S_{6} (t,\eta ) = \int {S_{5} (t,\,f_{\eta } ) \cdot K_{ - b} (f_{\eta } ,\eta )df_{\eta } = \int {A_{o}^{5} \cdot A_{1} \cdot W_{r} \cdot W_{a} \cdot \exp \left( {j\phi_{6} \left( {t, f_{\eta }, \eta^{*} } \right)} \right)} df_{\eta } } $$

(55)

where

$$ \begin{aligned} \phi_{6} \left( {t, f_{\eta } ,\eta^{*} } \right) & = 2\pi \cdot M_{5} (t) \cdot \left( {R_{o} + \frac{{V^{2} M_{6}^{2} (t)}}{{2R_{o} }}f_{\eta }^{2} } \right) + M_{3} (t) + \pi \cdot u_{o}^{2} M_{6}^{2} (t) \cdot f_{\eta }^{2} \cot \beta - 2\pi \cdot M_{6} (t)f_{\eta } \csc \beta \\ & \quad + \frac{{2\pi \cdot K_{a} }}{{u_{o}^{2} \cot \beta }}\left( {f_{\eta } \csc \beta - K_{a} } \right) + \pi \cdot f_{\eta } u_{o}^{\prime 2} \cot \beta - \frac{2\pi \csc \beta }{{u_{o}^{2} \cot \beta }}\left( {f_{\eta }^{2} \csc \beta - K_{a} f_{\eta } } \right) - \pi \eta^{2} u_{o}^{2} \cot \beta + 2\pi \eta \cdot f_{\eta } \csc \beta \\ \end{aligned} $$

(56)

After applying the PSP, the azimuth frequency is

$$ f_{\eta }^{*} = \frac{{M_{6} (t)\csc \beta - 2K_{a} /u_{o}^{2} \cos \beta - u_{o}^{'2} \cot \beta /2 - \eta \csc \beta }}{{V^{2} M_{5} (t) \cdot M_{6}^{2} (t)/R_{o} + u_{o}^{2} M_{6}^{2} (t)\cot \beta - 2\csc \beta /u_{o}^{2} \cos \beta }} $$

(57)

and

$$ M_{6} (t) = \frac{\csc \beta }{{u_{o}^{2} \cot \beta + V^{2} M_{5} (t)/R_{o} }} $$

(58)

The solution to the integral of $S_{6} (t,\eta )$ can be written as

$$ S_{6} (t,\eta ) = A_{o}^{5} \cdot A_{1} \cdot W_{r} \cdot W_{a} \cdot \exp \left( {j\phi_{6} \left( {t,\,f_{\eta }^{*} ,\eta^{*} } \right)} \right) $$

(59)

where $\phi_{6} \left( {t,\,f_{\eta }^{*} ,\eta^{*} } \right)$ is given in equation (56) replacing f _η by $f_{\eta }^{*}$ in equation (57).