SEDSAT-2 Communications Design Notes 20071124

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GHz communications feasability

This page summarises the results of my investigation into the feasibility of GHz communication systems for SEDSAT-2 . I'll give an overview of why we might (or might not) want a GHz communications system and then I'll move onto some of the maths behind my understanding of GHz communications.

--Steve

Background

TODO

Theory

Radio Propagation

Radio waves are transmitted as electromagnetic waves and the amount electromagnetic energy passing through a unit area (1 meter squared) of space is called the flux density. If the transmitter radiates electromagnetic energy evenly in all directions - in the shape of a sphere with the transmitter at the center - then the flux density at a receiver is equal to the power of the transmitter Pt divided by the area of a sphere \frac{1}{4 \pi R^{2}}. Hence:

flux\ density\ per\ unit\ area\ at\ receiver = \frac{P_{t}}{4 \pi R^{2}}

The above gives the amount of energy a receiver receives per unit area. So for a large receiver area, we will receive a lot of energy and for a small receiver area, we will receive a small amount of energy - we just scale the above equation by the size of the receiver.

A radio receiver receives the transmitted electromagnetic energy using an antenna and the amount of energy captured by the antenna is determined by the antennas "effective area", Ae. This "effective area" is just a number used to multiply the flux density per unit area, to get the received signal power, but the expression for the "effective area" doesn't directly depend on the physical size of the antenna. Instead, antennas can boost the power of the received signal - by an amount called the Gain - so that they intercept a larger amount of electromagnetic energy traveling through space than their physical size alone would indicate. The equation for the "effective area" is:

A_{e} = \frac{\lambda^{2} G}{4\pi}

where λ is the wavelength of the signal and G is the gain of the antenna.

If the above two equations are combined, then we can calculate the power at a receiver Pr for a given effective antenna as follows:

P_{r} = \frac{P_{t}}{4 \pi R^{2}} \times A_{e} = \frac{P_{t}A_{e}}{4 \pi R^{2}}

Now an example of how we can use this equation. Let's assume we have an antenna which gives 20dB (= 100x) of gain. The transmitter has a power of 1W and the transmitter and receiver are 700km apart. We've got a 12cm wavelength signal, so how much power do we have at the receiver? First, calculate the effective area of the antenna:

A_{e} = \frac{0.12^{2} \times 100}{4\pi} = 0.115m^{2}

Now calculate the received power:

P_{r} = \frac{1 \times 0.115 }{4 \pi (700\times10^{3})^{2}} = 1.86\times10^{-14} W = -107dBm

Received power and frequency

So how is all of this affected by frequency? This is where I'm a little confused, as everyone else seems to think that path loss increases with increasing frequency, but I think this is incorrect. My understanding is that path loss itself is independent of frequency and depends only on the area of a sphere (as above), and only if you let your antenna physical size scale down as you increase frequency (in order to keep a constant antenna gain), do you see an increased post-antenna received power loss with increasing frequency. But when will you ever do this? Surely you'll fix your antenna size (and not the antenna gain), and if you do that, your post-antenna received power isn't affected by frequency. Here is the semi-mathematical explanation:

If we re-write the equation for effective aperture in terms of frequency, we see that:

A_{e} = \frac{c^{2}}{f^{2}} \times \frac{G}{4 \pi}

The effect of frequency on received power depends on what happens to the other variables in the equation. If Ae is fixed then holding Ae fixed in the Pr equation above shows that the received power doesn't depend on frequency. If however, G is fixed, then as frequency increases, Ae decreases and so received power decreases. But this should be no surprise because as you go to higher frequencies, the antenna gets smaller for the same amount of gain because the wavelength is shorter. So while the gain the antenna provides stays the same, because the antenna is smaller, it captures less of the electromagnetic flux so there is less electromagnetic energy to be "gained up" in the first place - so the resulting received signal power is less.

With two antennas - one at the transmitter and one at the receiver - an extra f2 term is added to the above equations (among other things). If both antennas have fixed Ae, then received power actually goes up with frequency. If both antennas have fixed G, then received power goes down with frequency. And if one antenna is fixed in Ae and the other antenna is fixed in G, then the received power is again independent of frequency.

Implications for SEDSAT-2

My understanding is that we'll have some fixed amount of space for an antenna - and the bigger, the better, because we intercept more of the electromagnetic flux. Because we have a fixed antenna size, we fall into the first case above - fixed effective aperture - so frequency doesn't affect our received power.

Have I got something wrong here? If so, please tell me!

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