Anechoic Chamber Quiet Zone Calculation
(Rectangular Chamber)
The Quiet zone of an anechoic chamber describes a rectangular volume where electromagnetic waves reflected from the walls, floor and ceiling are stated to be belowa certain specified minimum. There are two main methods to calculate the quiet zone for a given chamber geometry. The first is a detailed mathematical model, accounting for a volume of reflections converging in the quiet zone, with a power gradient across them for how far inside and outside the HPBW of the antenna the reflection is, and how many times it reflects. The second method is to take the largest factor in this calculation, and estimate based off of it. For the case of a rectangular chamber, the largest factor in the detailed calculation is the wave that only reflects once to reach the receiving antenna. There will be four such waves, two from the sides, one from the ceiling, and one from the floor.
To calculate the quiet zone, first the chamber size must be decided. With the chamber size selected, the size of the absorber cones must be decided. To get a rough estimate of what to select a table for the specific material used must be consulted. If a specific quiet zone is expected, but it requires large cones, the large cones can be placed in key areas (such as the walls where the bounce will take place, and the rear wall behind the DUT). Table 1 was used for the case of the 8x8x12 chamber that will serve as an example for the duration of this paper. For the example chamber the SFC-12 cones will be used. The number denotes the height of the cones. The example chamber will be used at 2-3 GHz, and the reflectivity for this frequency is -40 dB.
Next the angle of incidence from the normal, that the wave will impact the wall must be determined. For a rectangular chamber this is a simple calculation. First the distance between the Tx and Rx antenna must be determined. This is done by determining how far from the ends each antenna will be. For the Tx 10” is selected since this is the length of the example antenna that will be used. For the Rx, 16” is selected, since the cones at the back wall are 12” and the platform requires 4” space to rotate a microstrip DUT.
This leaves 9’10” between Tx and Rx:
12’-1’4”-10”=9’10”(1)
Once this distance is determined, the height off the chamber floor of the antennas must be determined. For the example chamber, half way up was selected (4’). Once these two distances are determined, a rectangle can be drawn, for the example chamber this rectangle is 4’x9’10”. The rectangle is then divided by two lines that meet in the middle of the rectangle at the top. These represent the wave reflecting off the chamber wall then hitting the receiving antenna. The angle that is needed is the angle this wave reflection makes with the normal of the chamber wall. It is found by taking the arctangent of the height of the rectangle divided by ½ the length of the rectangle. The result is then subtracted from 90°. For the example chamber the calculation was:
(2)
With the angle of incidence calculated, Table 2 can be used to look up the multiplier used to calculate the off incidence reflectivity of the material. For the example chamber 12” cones are 3 wavelengths tall, so the 2 wavelength coefficient .82 and the 4 wavelength coefficient .95 are averaged together for a coefficient of .885. This coefficient is then multiplied with the normal incidence reflectivity of the absorber selected. For the example chamber this works out to:
-40 dB * .885 = -35.4 dB(3)
(3) is the adjusted reflectivity.
Next, the transmitting antenna’s radiation pattern is analyzed. 90° minus the angle of incidence calculated previously is the angle from the normal of the Tx antenna that the wave leaves from. Looking at the radiation pattern from the Tx antenna, the angle from Tx’ing normal is located, then the power at that angle is read out. This power is subtracted from the maximum power of the antenna. This value (in dB) is the how much less powerful the wave is when it reflects off the wall. This number (positive) is subtractedinto the adjusted reflectivity (negative) to quantify the effect of a less powerful wave interfering at the Rx DUT. For the example chamber and Tx antenna, the power difference at the angle from transmitting normal was found to be -4 dB so the new adjusted reflectivity is:
-35.4 dB – 4 dB= -39.4 dB (4)
The last value needed to calculate the quiet zone is a random phase correction. The waves reflecting off the chamber walls will be converging at the Rx DUT with random phases. This will cause some random cancellation of the main transmitted beam in the quiet zone. Therefore an industry standard value of -6 dB is subtracted from the adjusted reflectivity (negative) thus decreasing the magnitude of the reflectivity. In the example chamber this works out to be:
-39.4 dB – (-6 dB) = -33.4 dB.(5)
This is the quiet zone reflectivity for the chamber being analyzed.
Chamber Size / Absorber Height / Distance Tx-Rx / Angle of Incidence / Angle from Tx Normal / Absorber Reflectivity / Absorber Coefficient at incidence / Effective Reflectivity / Quiet Zone4x4x8 / flat / 6'8" / 59° / 31° / -15 / .31 / -4.65
4x4x8 / 8” / 6'2” / 57º / 33º / -35 / .7 / -24.5
4x4x8 / 12” / 5'10” / 55.6° / 34.4º / -40 / .8 / -32
6x6x12 / 8” / 10'2” / 59.5º / 30.5º / -35 / .66 / -23.1
6x6x12 / 12” / 9'10” / 58.6º / 31.4º / -40 / .705 / -28.2
8x8x12 / 12” / 9’10” / 50.9º / 39.1º / -40 / .885 / -35.4 / -33.4
8x8x12 / 24” / 8'10” / 47.8 / 42.2° / -50 / .999 / -50 / -47
Table 3: Sample calculations for various chamber sizes
(no Tx antenna selected so calculations could not be completed.)