Minimum Discernable Signal
What does "minimum" really mean?
Notes by WA6PY

This "rule of thumb" MDS formula is widely used by system engineers, but it does not accurately calculate the actual value for the"minimum" discernable signal level, especially for very sensitive systems.

MDS = [-174 + NFs + 10 log (BWHz)] dBm

where:

    BWHz = system (generally IF) bandwidth in HZ

Remember, that noise figures for the Base Stations are > 5 dB ( T_noise >627 K ) and noise temperature of the antennas are close to the ground radiation temperature as main lobe is almost entirely covering ground. Then this simplification doesn't work badly.

Noise floor for sensitive system can be much lower then ambient 290 deg K or what we read on the thermometer. This is very often not considered by system engineers. The calculation is for an assumed "room temperature" of 290 or 293 deg K thermal noise floor. This is repeated in many text books. During my career, I have found it necessary to argue and explain this over and over again. Radiotelescopes are mounted in let say 290 deg K environments, but noise temperature of the entire RX system is often only 30 deg K.

EXAMPLE #1 - a good EME station on 1296:

    T_ant 50 deg K
    T_sky 16 deg K
    NF_rx = 0.3 dB = T_RX 21 deg K

    T_noise_RX = 50 + 16 + 21 = 87 deg K, or RX NF = 1.04 dB

    which, in turn, yields a noise floor = -179.2 dBm/Hz

This shows how the "-174" factor typically used in the formula for MDS can be erroneous. Assuming that we can hear (brain's signal processing) signals 10 dB below the noise floor, in BW = 10 Hz we can receive signals at:

    -179.2 + 10 log (10Hz) - 10(S/N) = -179.2 dBm

To illustrate further, the afore-mentioned typical assumption that noise temperature of the RX system is determined by 290K + NF means:

-174 dBm/Hz + NF which corresponds to NF = 3 + NF (dBm)

Our amateur radio EME systems are much better than this even at 24 GHz.

As in example #1, above, we have to calculate the Noise Temperature Density ( per Hz ) of the RX system, including Antenna and Sky temperature. Then we can multiply by the relative bandwidth.

T_rx_system = T_sky + T_ant + T_rx

where:

    T_rx = 290*( [10(NF/10)] -1 )

EXAMPLE #2:
Assume the following:

    T_sky 20K
    T_ant 60K
    NF_rx 0.4 dB ( cascaded entire RX system ) = T_rx 28K

    T_rx_system = 20 + 60 + 28 = 108 K

Then, recalculating to the corresponding NF:
    NF_rx_system = 1 - LOG( 1 + T_rx_system/290) = 1.37 dB

    Noise Pwr Density (per Hz) = kT = 1.38*10-23 [Wsec/K]*108 = 1.49*10-21 [Wsec]

    Expressed in dBm:

    Pn_dBm = 10 log ( 1.49*10-23) + 30 = -178.27 [dBm/Hz]

If the signal is equal to the noise, the total power at the RX output will increase by 3 dB. On EME we are often receiving signal much lower then the noise floor with good copy.

VY 73 Paul WA6PY

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