SM 5 BSZ - What is Polarisation ? Signals v.s. Noise
(April 25 1997)

#### A radio signal is always polarised.

This statement may seem controversial, because black body radiation, galactic noise, is unpolarised. However, if black body radiation is received with two orthogonal antennas with identical radiation pattern, a cross yagi for example, the voltages that can be measured at the antenna terminals are completely uncorrelated. Therefore the black body radiation can be regarded as a pair of polarised signals.

It may seem uninteresting to discuss black body radiation, but for a radio amateur trying to hear an EME signal at 144MHz, the problem is that the black body radiation from our galaxy is much stronger than the signal. Only by use of very selective filters (a trained CW operator has a bandwidth below 50Hz), it is possible to receive these signals.

What we want to do is to maximise S/N, the signal to noise ratio, and it is may be helpful to understand how both signal and noise are affected by various things we do.

#### Orthogonal Antennas

Assume we have two similar antennas with orthogonal polarisation. Denote the voltage at their feed points H(t) and V(t), two functions of time (t). These antennas, with the feed points H and V, could be the horizontal and the vertical parts of a cross yagi array, but they could equally well be the left and right circular polarisation ports of a feed horn to a large dish. Orthogonal polarisation does not necessarily mean 90 degrees angle between something. Orthogonality is a mathematical concept. I am not going to be very mathematical, I try to describe these things by physical concepts.

Two antennas, occupying the same space, and having similar radiation pattern are orthogonal to each other, if V(t) remains more or less zero when H(t) is made large by connection of a transmitter (or vice versa). Orthogonality means that power (energy) is not transferred between the two ports H and V.

If the two feed points H and V are not orthogonal, it is possible to orthogonalise them by the use of an appropriate balancing network connected between the two feed points. Not being orthogonal means that feeding power into one produces an attenuated signal in the other. Balancing simply means feed an equal amount of signal of the opposite phase between the two ports for example by a tuning capacitor at an appropriate point along the two feed lines.

If H and V are not orthogonal, electronic polarisation control becomes impractical because the feed point impedance then becomes different for different polarisations.

#### Polarisation Transformation by Linear Combination of Signals

So we have H and V, and they are orthogonal. If the signal source is galactic noise, H(t) and V(t) are orthogonal to each other, which means that the average over time of H(t)*V(t) = 0. One interesting thing about H(t) and V(t) is that if we add them, for instance by the use of a cable network, to produce a new signal A(t)=c1*H(t) + c2*V(t). We will also be able to extract a second signal B(t)=c3*H(t) + c4*V(t) with interesting properties. The coefficients c1 to c4 are complex numbers, reflecting the amplitude and phase of the signals at the summation points. If the signal source is galactic noise only, A(t) and B(t) become orthogonal to each other if the two vectors (c1,c2) and (c3,c4) are orthogonal, which means c1*c3 + c2*c4 = 0. The signals from two orthogonal antennas can be linear combined to produce a perfect match to the polarisation of any incoming polarised signal. As a by product, the orthogonal polarisation is obtained, and it contains no signal, only noise. There is always one linear combination of the signals from an orthogonal pair that will produce exactly the signal that would have come from an antenna designed to match any polarisation, linear, circular or elliptic.

When we feed the A(t) signal through the receiver, through all filters and the product detector (which is not really a detector - it is a mixer) and into the head phones, all things we do are linear processes. This means that we will get exactly the same result even if the things are done in a different order. By the use of two pre amplifiers, H(t) and V(t) can be separately amplified. It is then trivial to add the signals by the use of some lossy device that can change amplitudes and phases to produce c1 and c2 before the actual summation. If H(t) and V(t) are also frequency shifted by the use of two frequency mixers driven from the same oscillator, and then filtered through similar filters, amplified, frequency shifted with a common oscillator again, and so on all the way to audio frequency, we can still produce the same signals A(t) and B(t) by adding the two audio signals H(t) and V(t) with amplitudes and phases as given by c1 to c4.

To be as concrete as possible, I now put numbers to everything. These numbers are typical for what I do myself on 144 MHz working EME, but the underlying arguments are independent of the actual example. The problem everything is about, is to decide: is the morse key at the other end up or down - and this should be decided as a function of time. The two signals H(t) and V(t) are amplified, filtered and frequency shifted to 300Hz. The filter bandwidth is 20Hz, so the filter is more or less a matched filter. Gain, bandwidth and phase shift is identical for H and V.

First assume we listen to a fairly strong continuous wave signal. It will show up in H(t) or V(t), or more likely in both of them. We may look at these functions of time by use of a two channel oscilloscope which is triggered by a 300 Hz local source. What we can see on the screen is two sine waves. These sine waves will move slowly over the screen, and the speed they move at is the frequency error. If the receiver is tuned for the signal to be exactly centred in the filter at 300 Hz, the two sine waves will stand still. If they move, both of them will move at the same speed - they have a fixed phase relation. Mathematically we describe H(t) and V(t) as 300Hz sine waves with constant amplitudes, and a phase that varies linearly with time while the phase difference is constant.

If the continuous carrier is removed, (and the gain is increased) we will ultimately hear galactic noise. If we look at that, with the same set-up - a two channel oscilloscope triggered with a local 300Hz source, two important differences show up:

1. The two sine waves do not have constant amplitude, and they move randomly backwards and forwards - but still slowly - over the screen.
2. The upper and lower track, although similar, are completely uncorrelated to each other.

As stated above, H(t) and V(t) originating in galactic noise, are orthogonal if the antennas are orthogonal. The orthogonality, by definition, means that the average value over time of H(t)*V(t) = 0. This is another way of saying that the two signals H(t) and V(t) are uncorrelated. They are two different, completely independent signals. The interesting thing is that, if we take H(t) and V(t) and for example by use of op-amps as summation amplifiers produce A(t)=c1*H(t)+c2*V(t), and B(t)=c3*H(t)+c*V(t), these new signals A(t) and B(t) will be orthogonal to each other if the vectors (c1,c2) and (c3,c4) are orthogonal. These A(t) and B(t) signals will be identical to the H(t) and V(t) signals that would have came out from a different pair of orthogonal antennas.

The summation operation with c1 to c4 is equivalent - in all details - to a physical rotation of the antenna if H and V come from a cross yagi, and c1 to c4 are real coefficients. If the coefficients are complex numbers - which is physically realised as a phase shift before summation, A and B is any possible orthogonal pair of polarisations. Linear, circular or elliptic.

If the scalar product (c1,c2)*(c3,c4)=0 the vectors (c1,c2) and (c3,c4) are orthogonal and the pair A(t),B(t) is an orthogonal transformation of the pair H(t),V(t). If also (c1,c2)*(c1,c2) = (c3,c4)*(c3*c4) = 1, A(t) and B(t) is an orthonormal transformation of H(t) and V(t). Keeping the transformation orthonormal means that the gain of our system is unchanged, which is the same as to say that the average over time of H(t)*H(t)+V(t)*V(t) is the same as the average value over time of A(t)*A(t)+B(t)*B(t).

If the input signal is galactic noise, there is no way to decide if your oscilloscope displays the pair H(t) and V(t) or if it displays A(t) and B(t), regardless of what values are chosen for the coefficients of an orthonormal transformation. An antenna picks up one polarisation and completely ignores another.

If the input signal is a continuous carrier, one particular orthogonal transformation will make B(t)=0 for all t, and consequently produce maximum for the average of A(t)*A(t). (remember, the average of A(t)*A(t)+B(t)*B(t) is a constant for any particular time interval). A(t) squared is the received power, so received signal energy in the A channel is maximised.

#### Practical Considerations

If the audio signals H(t) and V(t) contain a weak CW signal plus galactic noise, It is trivial to find out how to combine the signals in order to maximise the signal to noise ratio, S/N. This is described elsewhere in detail. Adaptive Polarisation and this is the way I normally run EME contacts.

If H and V are horizontal and vertical yagis, linearly polarised, and the signal is from VE3ONT, circularly polarised, The two tracks on the oscilloscope as described above, will be sine waves, of equal amplitudes, phase shifted by 90 degrees. If we use the transformation:

c1=0.707
c2=j0.707
c3=0.707
c4=-j0.7.7

Which would be physically realised as a phase shift by 90 degrees for one of the signals, followed by a 0.707 voltage divider for each signal and a summation amplifier to produce A(t), having 1.414 times the amplitude of the original H(t) and V(t) signals. A differential amplifier connected to the two voltage dividers will produce B(t), which is zero.

The power level in A(t) is 1.414 squared, = 2 times higher than the power levels originally available in H(t) and V(t), while there is no power at all in B(t).

The A(t) signal is identical to the signal that would come from a single channel receiver if the two yagis were connected through a power divider, and one of them was phase shifted 90 degrees by means of cable lengths, producing circular polarisation in the conventional way. S/N will be identical.

#### Unpolarised Waves - Polarised Signals

There is something funny with black body radiation. The electromagnetic wave coming from outer space does not have any polarisation. The polarisation is created by the antenna, when the field is converted to a voltage across a gap in a conductor. What really happens can be described in confusing ways by quantum mechanics, but a quite simple argument will give the essentials: An electromagnetic wave travels with the speed of light. Travels means direction of movement. Let us call the direction along which the wave travels the Z direction. Space has 3 dimensions, so there are two orthogonal dimensions X and Y left for the electric field that constitutes the wave to oscillate in. To describe the wave you need two independent variables in order to describe the random oscillation of the electric field vector in the XY plane. The voltage across a single feed point, being just a single variable, can therefore only describe certain aspects of the wave. To fully describe the wave, you need two simultaneous voltages, one for each dimension.

Of course coordinate transformations will make no changes - if you prefer polar coordinates or anything else, that is OK, but still two independent variables are needed to describe the electric field vector in a plane. And consequently, regardless what they mean, you need two voltages to receive an electromagnetic wave if you do not happen to know the polarisation in advance. But what the voltages represent - a pair of linear, or a pair of circular, or generally a pair of elliptic polarisations does not matter. They do not even have to be orthogonal, it is always possible to orthogonalise afterwards by use of a non orthogonal transformation. (Although noise figure or ohmic losses may suffer)

#### Detecting weak signals

What we do in our brains, when we detect a weak CW signal may be difficult to understand. To get some idea, we may take a look at what happens when the signal has passed through a matched filter. The matched filter will make random noise sound like the signal, it is impossible to distinguish signal from noise except by the amplitude. Many amateurs find such filters useless, "they create too much ringing". Personally, I prefer to use them, and I think they improve my ability to receive weak signals compared to all other possible choices. If you like to listen to signals filtered through nearly matched filters, some audio files can be found here: Audio Demonstration and Spectrum Images of Weak EME signals.

After passage through a matched filter, a morse code dot will not differ from a situation when the noise, just by random, is a little higher at a particular moment. If we define a threshold in such a way that the pure noise is above it, say 5% of the time, signal plus noise will be above that threshold more frequently than noise alone. BUT, at any given moment, we have no certain information. A level above the threshold may be noise only, AND a level below threshold may be a signal present together with some noise having the opposite phase. The noise may equally well cancel out signal as add to the signal by having the same phase. Since the signal with equal probability as 0 and 180 may be +/- 90 degrees, the signal and noise amplitudes will add as if they were 90 degrees phase shifted on the average. The amplitudes add according to the law of Pythagoras. This is the same as to say that the signal is orthogonal to the noise, or that the power of the signal adds to the power of the noise.

Although it is very difficult to receive morse coded messages at low S/N (one has to rely on repetitions of the message), it is very easy to notice the presence of the signal as an increase of the total amount of energy received. Together with audio files, this is demonstrated here: Audio Demonstration and Spectrum Images of Weak EME signals.

It should be obvious, that if a morse coded signal is received in both signals H(t) and V(t) from an orthogonal pair of antennas, the signal amplitudes can be determined quite accurately just by looking at the total amount of energy received during a long time - say one or two seconds. By looking at the correlation of H(t) and V(t), which becomes nonzero due to the presence of the signal, one can determine the phase relation between H(t) and V(t). The implication is that the polarisation can be determined accurately for a signal that is weaker than the noise, although it takes some time.

By use of a pair of orthogonal antennas, it is possible to receive very weak signals also when the polarisation is unknown, as is the case in EME, because by use of both signals it is possible to adapt the polarisation to exactly fit the polarisation of the incoming wave. This is described in more detail here: Adaptive Polarisation

In practical EME, even if one has the ability to rotate the polarisation, it is very difficult to turn the polarisation plane correctly for a very weak signal if a single channel receiver is used. The amplitude of the signal varies a lot due to the characteristics of the EME channel, and it is very difficult to decide if the amplitude change that follows an adjustment of the polarisation is due to the polarisation change or due to the normal amplitude variations. One has to try many times before the optimum setting is found, and by then a lot of time is lost, for me, when I used this method sometimes several minutes - and the weak signal may have disappeared because the station at the other end decided I do not hear him well enough!!!