I don’t have a problem with ISPs blocking outbound port 25 other than their own mail servers – that is a sensible anti-spam precaution. However blocking inbound connections can only be used to prevent the running of mail servers and the only reason I can think of for them preventing this is to force people to use their MPLS based Internet leased line which cost over USD 1500/month for 512K last time I looked at the price.

Using nmap on a server in Amsterdam I quickly discovered it was only port 25 which was blocked. Other ports remained unaffected.

So my problem was to find a way of getting mail into my server using a port other than 25 (SMTP). Internet mail requires access to port 25 so I had to find another machine to forward mail into my ADSL line in Qatar. Fortunately I have access to such machines running Debian and postfix in Amsterdam.

The solution turned out to be surprisingly easy:

1. Arrange for the ADSL link gateway to port forward traffic from a free port (I chose 200) to port 25 on the internal mail server. I used IPTABLES to do this but the same can be done with almost any ADSL gateway – look for “Application Sharing”. This can be tested by using netcat (nc) to the gateway from a machine in the outside world (some names obscured):

ams1:~# nc www.mydomain.com 200
220 www.mydomain.com ESMTP Postfix (Debian/GNU)
helo www.x31.com
250 www.mydomain.com
quit
221 2.0.0 Bye

2. Configure the relay machine (mine is in Amsterdam) to accept mail for the ADSL connected system (my system in Qatar). To do this edit /etc/postfix/main.cf and add the domain to the end of the line (note the use of the comma ‘,’):

relay_domains = $mydestination, mydomain.com

3. Setup a special delivery transport for mydomain.com. To do this edit /etc/postfix/transport and add the following lines (myddnshost.dyndns.org is the dynamic dns hostname for my system in Qatar and the 200 is the port number I am using):

mydomain.com smtp:[myddnshost.dyndns.org]:200
.mydomain.com smtp:[myddnshost.dyndns.org]:200


Now rebuild the map file:
$ postmap /etc/postfix/transport


and edit /etc/postfix/main.cf again. Ensure the following line is present:
transport_maps = hash:/etc/postfix/transport

Finally restart postfix:
$ /etc/init.d/postfix restart

4. Now update your DNS and point the MX record at the relay machine. Wait for things to settle down (can take a few hours) and your mail will start to be delivered into your ADSL connected server.

I run Debian Etch and Lenny systems with a mostly standard postfix configuration which uses hash tables. If your system is different then you may need to specify a different format. The /etc/postfix/transport file is extremely powerful and I recommend you look at “man transport” before making changes.

For communications systems to operate reliably a good signal needs to be received. Too weak a signal will cause loss of data and too strong a signal can result in overloading of the receiver or interference to other communications systems. Environmental factors will (normally) reduce the received signal below the level resulting from a simple line of sight computation. Note that this posting only considers propagation in the far field – antennas more than a few wavelengths apart and in an uncluttered environment.

The transmitting isotropic antenna

This is an antenna which radiates equally in all directions, in effect distributing the power over the surface of a sphere with the antenna at the centre. This is an important definition because all antennas are characterised in comparison with an isotropic antenna. A high gain antenna concentrates the power in a preferred direction at the expense of power being radiated in other directions. A useful analogy is squeezing a balloon where it shrinks where it is squeezed and pops out further in other areas.

Antenna gains are usually expressed in dB referenced to an isotropic antenna and these figures are denoted as dBi (dB relative to an isotropic antenna). An example of this is a dipole antenna which shows a gain of 2.2dBi in its preferred radiation directions.

An isotropic antenna is defined as being 100% efficient. This means all the power which is put into the antenna is transmitted onto its coverage area. This is known as radiation efficiency and all antennas are less than 100% efficient. Factors which can impact radiation efficiency include the impedance matching to the signal source (mismatches result in power being reflected to the source) and resistive losses of the surfaces of the antenna.

As the surface area of a sphere is equal to

[pmath]Area = 4 pi r^2[/pmath]

the power density at a distance r from the antenna can be computed as

[pmath]Pd = Pt / {4 pi r^2  }[/pmath] where Pt is the transmitted power.

From this it can be seen that the power density reduces by the square of the distance.

The receiving isotropic antenna.

Antennas receive as well as transmit and the amount of power they receive is a function of the area of the signal they intercept. The effective area of an isotropic antenna is defined as

[pmath]Ea = Lambda^2 / {4 pi}[/pmath] where Λ is the wavelength of the signal being received.

Isotropic antennas receive equally well in all directions and are defined as being 100% efficient. Real world antennas have efficiency losses and can increase their sensitivity in some directions at the expense of others. This figure is exactly the same as the transmit case as antennas are reciprocal in nature.

Path loss

Knowing the power density of a transmitting isotropic antenna and the effective area of a receiving antenna it is possible to work out the power transferred by two antennas.

[pmath]Pr = Pt (Lambda^2 / {4 pi} ) ( 1 / { 4 pi r^2 }) = Lambda^2 / { 16 pi^2 r^2 }[/pmath]

and the path loss is

[pmath]Path Loss = Pt / Pr = {16 pi^2 r^2 } / Lambda^2[/pmath]

As wavelength isn’t normally used we convert to frequency

[pmath]Path Loss = {16 pi^2 r^2 f^2} / {c^2}[/pmath] where c is the speed of light (300,000,000 m/s)

[pmath]Path Loss = ({{4 pi} / c} r f )^2[/pmath]

Converting everything to dB and separating the terms we end up with

[pmath]Path Loss dB = 20 log10({{4 pi} / c} r f )[/pmath]

[pmath]Path Loss dB = 20 log10({{4 pi} / c}) + 20 log10(r) + 20 log10(f)[/pmath]

[pmath]Path Loss dB = -138.02 + 20 log10(r) + 20 log10(f)[/pmath]

As antenna gains are expressed in dBi and we can easily add the effect of them to the pathloss computation:

[pmath]Path Loss dB = – Gtx – Grx -138.02 + 20 log10(r) + 20 log10(f)[/pmath]

This is a useful result as it allows us to quickly work out the loss between two antennas under optimal conditions. If you only have the natural logs rather than base 10 then you can convert the base using the substitution below:

[pmath]log10(x) = {loge(x)} / {loge(10)}[/pmath]

Because the antennas are receiving signals on the same frequency from two sources there exists a possibility of signal cancellation at certain frequencies. This would result in antennas failing to transmit (and receive) in certain areas as the signals cancelled at specific frequencies. The cancellation would depend on the relative distances from the amplifiers and the velocity factors of the intervening feeders.  The phasing of the signals would change with time and temperature making it near to impossible to predict the behaviour.

The solution was simple. Ensure that the signals at the antennas coming from the two sources are at different levels and therefore cannot completely cancel. This limits the variation in antenna gain across the band to a few dB. In this case a minimum distribution network design difference in power level of 6dB was adopted which results in a gain variation of 2.5dB across the band.

If the RF over fibre amplifiers are matched to within 2dB then this ripple could be about 4dB if the distribution network is set to 6dB difference. Obviously the coverage contour from the antenna would have to be at least at least 4dB above that level required for satisfactory performance to ensure that there are no dead spots.

The table below shows the worst case ripple resulting from the antenna being driven from two sources.