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 = PT / (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 GT

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

• PTGT = Effective Isotropic Radiated Power (EIRP).

• In dB: PT + GT = 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?
PT = 13 dBW,  GT = 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?
PT = -27 dBW,  GT = 27 dB, so EIRP = 0 dBW.

Incident Power Flux Density

PR = PFD x A = (GTPT / 4πd2) x A

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

Can relate antenna gain and aperture:

G = (4πAe) / λ2, so Ae = (Gλ2) / 4π

( λ = c / f)

PR = PFD x A = (GTPT / 4πd2) x (GRλ2 / 4π) = GRGTPT(λ/4πd)2

Express in dB:

PR = GR + GT + PT + 10Log((λ/4πd)2) = GR + GT + PT + 20Log(λ/4πd)

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

So, PR = EIRP  - FSPL + GR

Free Space Path Loss
Free Space Path Loss (FSPL) is:
• 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
Step back: the ‘zeroth order’ model

(PR / PT) is approximately (Dish Aperture / Coverage Area)

This is a frequency independent model.

A larger dish (or smaller coverage region) means:

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

No = kTN

TN = 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)

No = kTN and N = kTNB.

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 = GkTNB = 60 - 229 + 30 + 60 = -79 dBW

TN = system receive noise temperature

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

C / No = PR / (kTN) = PT + GT - FSPL + GR - k - TN 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/No

• C/No is Carrier-to-Noise Density Ratio
• Units of dB-Hz
• Reflects situation, viz: PR and No
• 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/No. We can estimate the data rate achievable, since:

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

Eb/No requirement is fixed for a given modulation and coding scheme.

So, C / No = R(Eb / No), or in dB units, (C / No) = R + (Eb / No)

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 / No)-1 = (C/Nup)-1 + (C/Ndown)-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.