Brief on Synthetic Aperture Radar Imaging Resolution

I assume readers already know the definitions of:

  • Frequency of a wave \(f\): number of periods per unit time
  • Wavelength \(\lambda\): distance between each period
  • Propagation velocity \(v=\frac{1}{\sqrt{\mu\epsilon}}\): wave traveling speed, here assuming the speed of light \(c=\lambda f\) in the air
  • Beam width \(\theta\): the angular spread of the radar beam, approximated by \(\theta=\frac{\lambda}{L}\) in radian, where \(L\) is the size of real aperture

Synthetic Aperture Radar (SAR)

By radar theory, the (azimuth) resolution \(R_a\) of a real aperture radar is proportional to the wavelength \(\lambda\), the distance to the object \(D\), and the inverse of real aperture size \(1/L_a\): (derivation 1)

\[R_a = \frac{D\lambda}{L_a}\]

Azimuth is the direction parallel to the moving path. So if the size of the real antenna is big enough, we need no additional algorithm to achieve a precise imaging results.

Screen Shot 2015-10-28 at 3.13.25 PM

But this is not the case usually. When distance is long and wavelength is large, the required antenna size can be over meters. So here we get the Synthetic Aperture Radar, or simply SAR. The basic idea is to leverage the movement to emulate a large antenna with a small one.

Depending on the way of SAR algorithm, the azimuth resolution is different. The first one is called spotlight SAR, also known as focused SAR.


The benefit of this type of SAR is that we can leverage the entire radar footprint width to be used as the synthetic aperture length. In this case, we can obtain an optimal azimuth resolution: (derivation 2)


From the equation above, the smaller the actual antenna, the finer the azimuth resolution. This might seem wrong and non-intuitive at all. The physical meaning of this is that a small antenna will have a wide lobe, and thus the target will be seen for a longer time. Thus assuming the synthetic aperture length large enough (larger than the footprint), we will have such an optimal resolution.

The resolution of Spotlight SAR is not very practical since

  1. The movement is not always straight as the Earth is not a flat surface;
  2. The beam steering has side beams that affects the final imaging results too.

Therefore, along with the movement of device (like plane), the radar footprint width is actually limited. So as the size of the antenna aperture.

The second is named stripmap SAR, which tends to have a more practical analysis.


Assuming allowing a maximum of \(45^\circ\) phase variation, the resolution of the Stripmap SAR is: (derivation 3)

\[R_a^{unfocused}=\sqrt{\frac{\lambda D}{2}}\]

Notice that this method does not account for the phase change while maintaining a small resolution comparing with the real aperture radar.


The assumption here is that the beam width is small enough to allow us to make the approximation \(\theta \approx \sin{\theta}\).

  1. Azimuth resolution \(R_a\) of real aperture radar
    It is in fact the arc length defined by radius/distance \(D\) and angular spread (or beam width) \(\theta_a\) (azimuth direction). We have


  2. Azimuth resolution \(R_a^{focused}\) of SAR
    When a device moves as the source of wave, it results in a frequency shift known as Doppler effect. Suppose a receiver moves at \(v\) away from the transmitter. Then the actual frequency receiver is \(f_{rec}=(\frac{1-v/c}{1+v/c})^{1/2}f\). Thus, a receding receiver sees lower frequency than the transmitted one, while approaching receiver sees higher frequency.
    Assuming \(v \ll c\), we can derive a simpler form \(f_{rec}=(1+\frac{v}{c})f\). So the frequency difference we will observe is \((f_{rec}-f)=(\frac{v}{c})f\). This is just one way. Since there is a return echo that has the same property, the final doppler frequency shift seen by the transceiver would be \(f_d=(\frac{2v}{c})f=\frac{2v}{\lambda}\), assuming constant throughout the observation (which is not true in general, see derivation 3).
    Screen Shot 2015-10-28 at 3.13.57 PM
    Suppose the target's relative azimuth speed is \(v_{rel}\). Then \(v\approx v_{rel}\frac{x}{D}\), where \(x\) is entire radar footprint length (and \(\delta x = R_a\)). So the resolution of Doppler frequency shift would be \(\delta f_d=\frac{2v_{rel}R_a}{\lambda D}\).
    From the equations above, the azimuth resolution is \(R_a=(\frac{\lambda D}{2v_{rel}})\delta f_d\).
    Because the resolution of Doppler shift is approximately equal to the inverse of time \(t\) while the point target is in the beam coverage\(\delta f_d\approx \frac{1}{t}=\frac{v_{rel}}{D\theta}=\frac{L_a v_{rel}}{D\lambda}\).
    Thus, the focused SAR can have a maximum azimuth resolution:


  3. Azimuth resolution \(R_a^{unfocused}\) of SAR
    Since the Earth is curved and can only be considered flat in a short distance, the Doppler shift is not a constant. In fact, it varies with \(D\).
    Simple geometry tells that \((\frac{x}{2})^2+D^2=(D+\delta d)^2\), where \(\delta d\) is the range difference.
    Screen Shot 2015-10-28 at 3.14.27 PM
    Assuming \(\delta d\ll D\), we get \(x\approx\sqrt{8D \delta d}\). Also, \(\delta d\) in terms of phase delay satisfies \(2\delta d=\frac{\phi\lambda}{2\pi}\), where \(\phi\) is the phase delay in radian and the \(2\) comes from the round trip of signal. Thus, if we allow a maximum \(45^\circ=\pi/4\) phase variation, \(x=\sqrt{8D\frac{\lambda\pi}{16\pi}}=\sqrt{\frac{\lambda}{2}}\).
    From above, the time \(t_{unfocused}=x/v_{rel}\). Therefore,

    \[R_a^{unfocused}=(\frac{\lambda D}{2v_{rel}})\cdot(\frac{v_{rel}}{\sqrt{\lambda/2}})=\sqrt{\frac{\lambda D}{2}}\]

Leave a Reply

Your email address will not be published. Required fields are marked *