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SOLVE HOLOGRAPHY

Holography is powerful way of caracterizing an antenna. It builds on the fact that there exists a Fourier transform relationship between the far-field antenna beam (including its amplitude and phase) and the (complex) aperture field distribution. A modified Fourier relationship can be derived in the near-field case (far field is $ > 2d^2/\lambda$ , or $ ~130$ km for a 15-m antenna at a wavelength of 3.4mm). The interested readers will find in Baars et al. 2007 a lot more of details. CLIC can handle both near-field (Fresnel) and far-field (Fraunhofer) measurements.

In essence, a source is scanned with one or more antennas while one or more antennas are pointing at the source. The data are gridded and Fast Fourier Transformed to an aperture plane. The aperture plane is then analyzed in terms of amplitude and phase (left and right plots in Fig. 7)

Figure 7: Example of SOLVE HOLOGRAPHY.

The amplitude map is normalized and transformed in decibels (dB), so that it peaks at 0dB. In radio-astronomy, the aperture is often tapered to reduce the beam sidelobes. CLIC does a fit of this illumination with a 2-d gaussian (2-d parabola in the dB maps) to measure the actual taper and center of illumation. The values are given in the first line of the header.

The phase map is more interesting in that it gives the deviations w.r.t. a perfect surface and can be used to derive corrections to be applied to the individual panels of the parabola. The advantage is also that it can be measured, and panel corrections derived, at any antenna elevation.

A least squares fit to the phase map of the following terms is performed, where $ \left(\xi,\eta\right)$ are aperture coordinates, $ r=\sqrt{\xi^2+\eta^2}$ is the radius coordinate from the center of the aperture, $ f$ is the focal length of the antenna:

$\displaystyle \phi_1 (\xi,\eta)$ $\displaystyle =$ $\displaystyle a$ (25)
$\displaystyle \phi_2 (\xi,\eta)$ $\displaystyle =$ $\displaystyle b\xi$ (26)
$\displaystyle \phi_3 (\xi,\eta)$ $\displaystyle =$ $\displaystyle c\eta$ (27)
$\displaystyle \phi_4 (\xi,\eta)$ $\displaystyle =$ $\displaystyle \delta z \left(1-\frac{1-\frac{r^2}{4f^2}}{\sqrt{\frac{r^2}{f^2}+(1- \frac{r^2}{4f^2})^2}}\right)$ (28)
$\displaystyle \phi_5 (\xi,\eta)$ $\displaystyle =$ $\displaystyle \delta x\frac{\xi}{f}\left(1-\frac{1}{\sqrt{\frac{r^2}{f^2}
+\left(1-\frac{r^2}{4f^2}\right)^2}}\right)$ (29)
$\displaystyle \phi_6 (\xi,\eta)$ $\displaystyle =$ $\displaystyle \delta y\frac{\eta}{f}\left(1-\frac{1}{\sqrt{\frac{r^2}{f^2}
+\left(1-\frac{r^2}{4f^2}\right)^2}}\right)$ (30)
$\displaystyle \phi_7 (\xi,\eta)$ $\displaystyle =$ $\displaystyle a_{+}\frac{\xi^2-\eta^2}{2f^2}$ (31)
$\displaystyle \phi_8 (\xi,\eta)$ $\displaystyle =$ $\displaystyle a_{\times}\frac{2\xi \eta}{2f^2}$ (32)

$ a$ is a constant phase offset, $ b$ and $ c$ account for pointing offset that would be constant during the observations, $ \delta x$ , $ \delta y$ and $ \delta z$ correspond to focus offsets in the x (horizontal), y (vertical) or z (along optical axis) directions. $ a_{+}$ and $ a_{\times}$ are the antenna astigmatism along the horizontal or vertical axis, $ a_{\times}$ along directions rotated by 45$ ^\circ$ . The corresponding astigmatism angle $ \psi_a$ is:

$\displaystyle \psi_a = \frac{\arctan{(a_{\times},a_{+})}}{2}$ (33)

In addition, holographies provide an estimate of the actual antenna surface, one can compute the aperture and illumination efficiencies $ \eta_A$ and $ \eta_I$ :


$\displaystyle \eta_A$ $\displaystyle =$ $\displaystyle \frac{\lvert \int_A A(\xi,\eta)\exp(i \phi (\xi,\eta))d\xi d\eta
\rvert^2}{\pi r^2\int_A A(\xi,\eta)^2d\xi d\eta}$ (34)
       
$\displaystyle \eta_I$ $\displaystyle =$ $\displaystyle \frac{\lvert \int_A A(\xi,\eta)d\xi d\eta
\rvert^2}{\pi r^2\int_A A(\xi,\eta)^2d\xi d\eta}$ (35)

and the antenna efficiency:

$\displaystyle J=\frac{2k}{\eta_A \pi r^2}$ (36)

One can also fit panel displacements. According to the panel mode selected, translation only, tilt in both radial and tangential directions, torsion mode and a fit of the inner fifth support point can be fitted.

Finally, one can produce either a phase map or a surface error map. The latter is linked to the former through:

$\displaystyle \delta p = \delta \phi \times \frac{\lambda}{4\pi}\times\sqrt{1+\frac{r^2} {4f^2}}$ (37)

The term to the right corresponds to the projection factor between the incident or reflected ray and the normal to the surface.


next up previous contents index
Next: Storing Up: Solving Previous: SOLVE BASE ; RESIDUAL   Contents   Index
Gildas manager 2022-01-17