Need to determine/analyse link performance (e.g Bit Error Rate (BER) or Packet Loss Rate) in terms of:

• Transmit power
  

• Antenna parameters (e.g. gain)
  

• Received system noise levels (usually specified as noise temperature)
  

• Other factors (e.g. propagation losses, interference, intermodulation)
  

Consider an isotropic antenna - radiates in all directions. Imagine antenna at centre of hollow sphere.

Power flux density at radius d is:

PFD = P_T / (4πd²) W/m²

'Inverse Square Law'

 

 

 

 

Antennas have gain to concentrate power in a wanted direction.

Gain is relative to Isotropic with units of dBi

Trivial Example:

• What is the gain of an isotropic antenna?
  Unity. So G = 0 dBi.

Transmitter and receiver spaced distance d.

•  PFD in wanted direction is increased by a factor of G_T
  

• Then at receiver it looks as though there is an isotropic transmitter with power P_TG_T
  

• P_TG_T = Effective Isotropic Radiated Power (EIRP).
  

• In dB: P_T + G_T = EIRP (dBW)

Examples:

• A transmit antenna has a gain of 50 dB and is fed with a power of 20 watts. What is its EIRP?
  P_T = 13 dBW,  G_T = 50 dB, so EIRP = 13 + 50 = 63 dBW
  

• A transmit antenna has a gain of 27 dB and is fed with a power of 2 mW. What is its EIRP?
  P_T = -27 dBW,  G_T = 27 dB, so EIRP = 0 dBW.

Incident Power Flux Density

P_R = PFD x A = (G_TP_T / 4πd²) x A

A is the aperture. Use effective aperture: A_e = ηA_actual (η = Efficiency, take as 70% if in doubt)

Can relate antenna gain and aperture:

G = (4πA_e) / λ²_, so A_e = (Gλ²) / 4π

⁽ λ = c / f)

 

P_R = PFD x A = (G_TP_T / 4πd²) x (G_Rλ² / 4π) = G_RG_TP_T(λ/4πd)²

Express in dB:

P_R = G_R + G_T + P_T + 10Log((λ/4πd)²) = G_R + G_T + P_T + 20Log(λ/4πd)

Now, G_T + P_T is EIRP and 20Log(λ/4πd) is Free Space Path Loss (FSPL).

So, P_R = EIRP  - FSPL + G_R

• A dimensionless quantity in dB
  
• An artefact to make equations work
  
• Represents loss between two isotropic antennas
  
• Function of range and wavelength
  

Example:

• What is the Free Space Path Loss to a geostationary satellite at a range of 40,000 km, operating at 15 GHz?
  λ = c / f = 3 x 10⁸ / 15 x 10⁹ = 0.02 m.
  d = 4 x 10⁷ m.
  So FSPL = 20log(0.02/(4π x 4 x 10⁷) = 20log(3.98 x 10^-11) = -208 dB

(P_R / P_T) is approximately (Dish Aperture / Coverage Area)

This is a frequency independent model.

A larger dish (or smaller coverage region) means:

• more received power
  
• higher data rate

• Inevitable
  
• In Satcom usually Additive White Gaussian Noise (AWGN)
  
• Flat power spectral density

N_o = kT_N

T_N = Noise Temperature
k = The Boltzmann constant (1.38 x 10^-23 J/K)

k can have units dBJ/K or dBW/Hz/K since 1 joule is a unit of energy  = 1 watt second, and seconds = 1/Hz.

Example:

• Determine k in dB units
  If k were simply 1 x 10^-23 then it would be -230 dBW/Hz/K. The 1.38 factor increases this a little.
  Actual value is -228.6 dBW/Hz/K (-229 is OK for many calculations)

N_o = kT_N and N = kT_NB.

Example:

• An amplifier has a gain of 60 dB, a noise temperature of 1000K, and a bandwidth of 1 MHz. What is the noise power at its output?
  N = GkT_NB = 60 - 229 + 30 + 60 = -79 dBW

T_N = system receive noise temperature

All noise sources 'referred' to a common point, usually the antenna input.

C / N_o = P_R / (kT_N) = P_T + G_T - FSPL + G_R - k - T_N dB-Hz

Note: The -k term gives a double negative (- - 229 dBW/Hz/K), so is + 229 dBW/Hz/K. This large number opposes the large FSPL subtraction, and the two may very roughly cancel each other out.

Simple Example:

Transmit Power = 40W = 16 dBW
Transmit Antenna Gain = 30dB
So, EIRP = 30 + 16 =	46dBW
Geostationary satellite at 37000 km range Downlink (Space to Earth) at 12GHz So, FSPL = 20log(4πd/λ) =	206 dB
Receive terminal antenna gain	35 dB
Resultant received power (= 46 - 206 + 35)	-125 dBW
Receive noise temperature = 200K = 23 dBK k = -229 dBW/Hz/K, so No = kT =	-206 dBW/Hz
Resultant Downlink C/No = (-125 - - 206)	82 dB-Hz

• C/N_o is Carrier-to-Noise Density Ratio
  
• Units of dB-Hz
  
• Reflects situation, viz: P_R and N_o
  
• Does not require knowledge of Receive bandwidth
  
• Could regard as C/N (or Signal-to-Noise Ratio) in a 1Hz bandwidth

• Could alternatively do all calculations within a fixed bandwidth. Then this would yield C/N (dB)
  
• But relevant bandwidth often not known precisely.
  
• (Really needs to be the Noise Bandwidth of the modem)

Modems are usually specified in terms of C/N_o. We can estimate the data rate achievable, since:

Signal Power = Bit Energy x Number of Bits  i.e. C = E_bR

E_b/N_o requirement is fixed for a given modulation and coding scheme.

So, C / N_o = R(E_b / N_o), or in dB units, (C / N_o) = R + (E_b / N_o)

C/N Example:

• A link yields a C/No of 53 dB-Hz. Assuming it uses coherent BPSK with a desired BER of 10-6, what bit rate might be achieved?

From standard results for BER vs. Eb/No, we see the required Eb/No is approximately 10 dB.

Hence the achievable data rate R is about 53 - 10 = 43 dB-Hz, or approximately 20 kbit/s.

(C / N_o)^-1 = (C/N_up)^-1 + (C/N_down)^-1

• A link has uplink C/No of 60 dB-Hz and a downlink C/No of 60 dB-Hz. What is the approximate overall C/No?

Reciprocating (10^-6) adding and de-reciprocating will yield 57 dB-Hz.
Alternatively, note that the noise is simply doubled so the result is 3 dB worse.

• A link has uplink C/No of 63 dB-Hz and a downlink C/No of 43 dB-Hz. What is the approximate overall C/No?

The lower figure will dominate, and the higher figure add only about 1% additional noise (negligible in dB terms). So result is approx. 43 dB-Hz still.