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 below a 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:
4′
90° − tan−1 4′10" = 50.9°
(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 subtracted into 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.
This is the quiet zone reflectivity for the chamber being analyzed.
(5)
Chamber Absorber Distance Angle of
Size
Height
Tx-Rx
Incidence
59°
4x4x8
flat
6'8"
4x4x8
4x4x8
6x6x12
6x6x12
8x8x12
8x8x12
Angle
from Tx
Normal
31°
Absorber
Coefficient
Absorber
at
Effective
Reflectivity incidence Reflectivity
-15
.31
-4.65
Quiet
Zone
8”
6'2”
57º
33º
-35
.7
-24.5
12”
5'10”
55.6°
34.4º
-40
.8
-32
8”
10'2”
59.5º
30.5º
-35
.66
-23.1
12”
9'10”
58.6º
31.4º
-40
.705
-28.2
12”
9’10”
50.9º
39.1º
-40
.885
-35.4
-33.4
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.)