Satellite Links

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Satellite Links Satellite links are used to provide communications over very large distances (global coverage). This is achieved by using the satellite as a repeater. A ground station relays a signal up to the satellite at a frequency known as the uplink frequency. The satellite receives this signal and re-broadcasts it on a downlink frequency to another ground station. If digital communications signalling is used, the signal may be regenerated before it is re-transmitted to Earth. The analysis and design of satellite links is not unlike other links we have already studied in the course. There are particular details in evaluating certain components of the link budget that deserve particular attention: the characteristics of the satellite transponder, the fact that two frequencies (and hence two link budgets) are used, and the need to factor in the orbital geometry of the satellite to compute the free space loss.

1

Transponder Characteristics

A satellite receives a radio signal on its uplink receiver. It then downconverts this signal to the downlink frequency, and amplifies it before sending the downlink signal to the transmit antenna. These functions are carried out by the satellite transponder. Ideally, aside from the frequency shift, you can imagine this process as a 2-hop link (discussed earlier in the course) with some amplification between the links. At the satellite, apart from noise considerations, you would expect the downlink signal to be G dB higher than the received uplink signal, where G is the transponder gain. In practise, it is not quite that simple. The transponder amplification stage is realized from a solid-state amplifier or travelling-wave tube (TWT) which can only supply a maximum power level due to device constraints. Hence, the transponder only behaves as a linear device over a range of input power levels. This is demonstrated graphically in Figure 1. The power of the uplink signal is shown on the x-axis. In this example, the transponder has a nominal gain of 30 dB, so in theory, the output signal power should be increased by 30 dB. We can see that over the majority of the characteristic, this is true. However, at a certain point, the output becomes saturated as the amplifier is no longer capable of delivering additional power (it is essentially clipping the input signal). In this example, the saturated output power is 35 dBW (about 3.2 kW). Well below this point, the curve departs from the ideal linear characteristic (shown as a dotted line) and the amplifier is said to be in compression or saturation. Operating beyond the compression point of the amplifier causes additional problems. The amplifier is no longer linear. A nonlinear device is capable of generating new frequencies from the frequencies present in the input signal; in this case, the input communication signal will be distorted to produce intermodulation products which extend beyond the bandwidth of the input signal. This can create interference problems: if you imagine a single channel in the input signal as occupying a specific bandwidth, then the spillover will produce unwanted signals in the adjacent channels, or interference. This is essentially noise, so in effect we are increasing the noise in the re-broadcast signal when the amplifier is operated in compression. Therefore, it is important to limit the input power to the transponder, which equates to managing Prof. Sean Victor Hum

ECE422: Radio and Microwave Wireless Systems

Satellite Links

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Figure 1: Satellite transponder input/output power characteristic the EIRP of the ground station carefully. The goal is to obtain as much output power available from the satellite while minimizing the signal distortion. This usually means reducing or backing off the input power, so that the amplifier is operated just below compression. This point gives decent output power while producing manageable levels of distortion/interference. The saturated input power density may be specified for a satellite, as opposed to the saturated input power. In this example, the output begins to compress at an input power of about 5 dBW. Increasing the input power beyond this point does not yield a linear increase in output power: for example, at an input power of 10 dBW, we would expect 40 dBW of output power and yet only are able to achieve a power of 35 dBW. The output is fully saturated at this point. A suitable input power level could be right before the saturated region is entered, along with some margin. If the input power level was chosen to be 4 dBW, we would say the the input power back-off (IPBO) is 6 dB, since the power has been reduced 6 dB from the saturated power point. Correspondingly there is an output power back-off (OPBO) which in this case is 2.5 dB; that is, the output is 2.5 dB from being fully saturated. A common empirical formula that is used to model the input/output characteristic described

Prof. Sean Victor Hum

ECE422: Radio and Microwave Wireless Systems

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above is

4Pin /Psat Wout = , (1) Wsat [1 + (Pin /Psat )]2 where Wsat is the saturated output power from the transponder, Wout is the actual power power produced at the output of the transponder, Pin is the input power density to the transponder’s receiving antenna, and Psat is the power density at the transponders’ receiving antenna that is known to saturate the output of the transponder.

2

Link Budget

As discussed, the link budgets on the uplink and downlink are treated separately, since different transmission distances may be involved, and there is a significant different in frequency/wavelength. On a per-Hz basis, the link budget equations are:   GS C = EIRPG − F SLU − LU + + 228.6 (units: dB-Hz) (2) N0 U TS   C GG + 228.6 (units: dB-Hz) (3) = EIRPS − F SLD − LD + N0 D TG where the subscripts U and D refer to the uplink and downlink, respectively, and the subscripts S and G refer to the satellite and ground station respectively. Note that the EIRP of the ground station is selected so that the transponder is suitably backed-off from saturation. The terms LU and LD refer to other losses on the uplink and downlink, such as atmospheric absorption, rain attenuation, etc. Free space loss forms the bulk of the loss in the link budget. This is especially true for satellite systems due to the huge distances involved which are typically tens of thousands of kilometres from station to station. Hence, it is important to understand the orbit geometry of the satellite so that the link distance and hence free space loss can be estimated accurately. The calculation of the link distance requires an understanding of orbital mechanics, which is discussed later in Section 3. Knowing the link distance the free-space loss can be calculated on the link being considered (uplink/downlink). The link losses on the uplink and downlink are not the same, even if the distance r is the same, because the uplink and downlink frequencies are not the same. In fact, usually the downlink is at lower frequencies. This is done for multiple reasons: • Atmospheric attenuation is less at lower frequencies. • The ground station is not power-limited like the satellite, and hence is better equipped to deal with attenuation on the uplink. The maximum EIRP of the ground station is ultimately limited by the satellite’s transponder (see the next section) • The G/T on the uplink is very poor, since the satellite “sees” a warm earth. Hence, there is little point in using a lower frequency with less attenuation since the SNR is going to be relatively poor, anyway. • Improvements are thus best reserved for the downlink. Prof. Sean Victor Hum

ECE422: Radio and Microwave Wireless Systems

Satellite Links

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3

Introduction to Orbital Mechanics

We wish to work out the specifics of the orbital geometry of satellites. We begin by employing Newton’s laws of motion to determine the orbital period of a satellite. The first equation of motion is F = ma

(4)

where m is mass, in kg, and a is acceleration, in m/s2 . The Earth produces a gravitational field equal to g(r) = −

GmE rˆ r2

(5)

where mE = 5.972 × 1024 kg is the mass of the Earth and G = 6.674 × 10−11 N · m2 /kg2 is the gravitational constant. The gravitational force acting on the satellite is then equal to F in = −

GmE mr GmE mˆ r =− . 2 r r3

(6)

This force is inward (towards the Earth), and as such is defined as a centripetal force acting on the satellite. Since the product of mE and G is a constant, we can define µ = mE G = 3.986 × 1014 N · m2 /kg which is known as Kepler’s constant. Then, µmr (7) F in = − 3 . r This force is illustrated in Figure 2(a). An equal and opposite force acts on the satellite called the centrifugal force, also shown. This force keeps the satellite moving in a circular path with linear speed, away from the axis of rotation. Hence we can define a centrifugal acceleration as the change in velocity produced by the satellite moving in a circular path with respect to time, which keeps the satellite moving in a circular path without falling into the centre. v

v2

v1

F out B

r

r

F in O

(a) Forces on satellite

A

v1 ∆v v2

O

(b) Velocity vectors at two times

(c) Velocity detail

vector

Figure 2: Illustration of motion of satellite in its orbit a distance r from a centre of mass O The satellite moves with constant angular velocity ω, which is defined by the angle θ covered by the satellite in time t. But, rθ = ` (8) Prof. Sean Victor Hum

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is the arc length traversed by the satellite, therefore, ω=

`1 v ` = = , rt tr r

(9)

where v is the linear velocity of the satellite. If we define an angular acceleration Ω, then Ω=

dω d2 θ = 2. dt dt

(10)

If the points A and B shown in Figure 2(b) are very close, then from Figure 2(c), vdt ∆v dv AB l ≡ ≈ . = = r r v v OA

(11)

The last equality results from the triangle showing the velocity vectors in Figure 2. Rearranging, dv v2 = = Ω. dt r

(12)

The centrifugal force on the satellite is then Fout = mΩ

(13)

or

mv 2 rˆ mv 2 r F out = = . r r2 The net force on the satellite must be zero, so the sum of (7) and (14) must be zero, µm mv 2 √ = ⇒ v = µr. 2 r r

(14)

(15)

A direct consequence of this equation is that the velocity of the satellite is inversely proportional to its orbital altitude r; the lower the orbit of the satellite, the faster it travels. The time T it takes the satellite to transit through one orbit is determined knowing the circumference of the orbit, 2πr 2πr3/2 S = √ . (16) T = =p v µ µ/r We can see that the higher the orbital altitude r of a satellite, the longer its orbital period. In fact, the orbit of a satellite is classified according to its altitude and corresponding orbital period, as shown in Table 1. Our assumption of a circular orbit is sufficient for calculating the orbital period, but in general satellites do not move in circular orbits around the earth. We now wish to determine the shape of the orbital path taken by a satellite. Let us say the position of the satellite is described by a position vector r pointing from the centre of the earth to the satellite, as shown in Figure 3. If the satellite is accelerated then the force on the satellite is described by d2 r . (17) dt2 ECE422: Radio and Microwave Wireless Systems

F =m Prof. Sean Victor Hum

Satellite Links

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Orbit Low earth orbit (LEO) Medium earth orbit (MEO) Geostationary earth orbit (GEO)

Orbital altitude (km) Orbital period T 160–2,000 87 – 127 min 2,000–35,786 127 min – 24 hr 35,786 23 hr 56 min 4.1 sec

Table 1: Orbit types However, there is also a centripetal force acting on the satellite given by (7). Therefore, −

µr d2 r = , r3 dt2

(18)

or

d2 r µr + 3 = 0. (19) dt2 r This is a second-order linear differential equation which we wish to solve for r. This is challenging because both r and its unit vector rˆ are functions of time. That is, r = r(t)ˆ r (t),

(20)

ˆ sin θ(t) cos φ(t) + y ˆ sin θ(t) sin φ(t) + z ˆ cos θ(t). rˆ (t) = x

(21)

where We need to use the product rule to find the derivatives dr/dt and d2 r/dt2 . For example, ˆ dr(t) dr(t) dr = rˆ (t) + r(t). dt dt dt

(22)

Figure 3: Initial satellite coordinate system It is better to express r in a coordinate system with a simpler dependence on time and angles. A good choice is a rotated coordinate system where the orbital plane of the satellite coincides with the xy plane. We will call this new plane the x0 − y0 plane, and the coordinate system is shown in Figure 4. Prof. Sean Victor Hum

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y0

r0 φ0 z0 (a) Rotated coordinate system

x0

(b) x0 − y0 system viewed from above

Figure 4: Orbital plane coordinate system We can convert these local coordinates to cylindrical form as r 0 = x0 xˆ0 + y0 yˆ0 = r0 rˆ0 rˆ0 = cos φ0 xˆ0 + sin φ0 yˆ0 ˆ0 = − sin φ0 xˆ0 + cos φ0 yˆ0 φ

(23) (24) (25)

d2 r 0 µr 0 + 3 = 0. dt2 r0

(26)

Equation (26) becomes

ˆ ) component. This equation has two components: a radial (ˆ r 0 ) component and an axial (φ 0 Taking the derivative on the left hand side, the radial component of this equation is  2 dφ0 µ d2 r0 − r = − (27) 0 dt2 dt r02 while the axial component is dr0 dφ0 d2 φ 0 + r0 2 = 0. (28) dt dt dt Let us begin by solving the second equation (28. Consider the quantity on the left hand side below, which we apply the product rule to to yield     2 1 d 1 dφ0 2 dφ0 2 d φ0 r = 2r + r0 2 (29) r0 dt dt r0 dt dt 2

This equation is the same equation as (28). Therefore, we conclude that   1 d 2 dφ0 r = 0, r0 dt dt which means that r2 Prof. Sean Victor Hum

(30)

dφ0 = constant ≡ h. (31) dt ECE422: Radio and Microwave Wireless Systems

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h is a quantity which we call angular momentum per unit mass. We now return to the first differential equation (27). The solution to this equation can be shown to be h2 , (32) r0 = µ + Ah2 cos(φ0 + θ0 ) where A is a constant. This equation can be rewritten as h2 /µ p r0 = , ≡ Ah2 1 + e cos(φ0 + θ0 ) 1 + µ cos(φ0 + θ0 )

(33)

which is recognized as the equation of an ellipse in polar form. The quantity p = h2 /µ is called the semilatus rectum of the ellipse, while e = h2 A/µ is the eccentricity of the ellipse. We can eliminate θ0 from this equation by aligning the x0 -axis of the coordinate system to be coincident with the major axis of the ellipse, so that r0 =

p . 1 + e cos φ0

(34)

The orbital path in this coordinate system is illustrated in Figure 5. The satellite moves in an elliptical path about the origin. The foci of the ellipse are located at points O and F ; the Earth is located at focal point O. This constitutes the first of Kepler’s Three Laws of Planetary Motion: the orbit of a smaller body about a larger body is always an ellipse, with the centre of mass of the larger body coinciding with on of the two foci of the ellipse. y0

r0 A

a

C

O

P

F

x0

ae

a(1 − e)

a(1 + e)

Figure 5: Orbital path of satellite in x0 y0 plane The length of the semi-major axis of the ellipse is a= Prof. Sean Victor Hum

p (35) 1 − e2 ECE422: Radio and Microwave Wireless Systems

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while the length of the semi-minor axis of the ellipse is b = a(1 − e2 )1/2 .

(36)

With the Earth at point O, we can see that as the satellite traverses its orbital path, it reaches points A and P , which are the points furthest and closest to the Earth, respectively. These points are called apogee and perigee. Kepler’s laws of planetary motion are: 1. The orbit of a smaller body about a larger body is always an ellipse, with the centre of mass of the larger body coinciding with on of the two foci of the ellipse. 2. The orbit of the smaller body sweeps out equal areas in equal time. This is graphically depicted in Figure 6, whereby the the areas A12 and A34 are equal if the time differences t2 − t1 and t4 − t3 are the same. 3. The square of the period of revolution T of the smaller body is equal to a constant multiplied by the 3rd power of the semi-major axis length a, i.e. T2 =

4π 2 a3 µ

(37)

Comparing this to (16), we see that if a = r, the expression derived earlier for circular orbits applies equally to elliptical orbits.

Figure 6: Illustration of Kepler’s second law

Point of interest: GEO orbit Geostationary orbit is a special case of a satellite orbit meeting the following conditions: 1. The orbit is perfectly circular: a = r0 or e = 0. Prof. Sean Victor Hum

ECE422: Radio and Microwave Wireless Systems

Satellite Links

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2. The altitude of the satellite is such that the orbital period of the satellite is 23 hours, 56 minutes, and 4.1 seconds, which is one sidereal day which is the time it takes for the Earth to rotate about its axis exactly once. 3. The satellite’s orbital plane must coincide with the the plane of the equator. Under these conditions, the satellite appears perfectly stationary with respect to an stationary station on the Earth’s surface. This makes this orbit highly coveted for telecommunications and broadcast applications, such as television. The geostationary orbital altitude is, from Table 1, H = e = 0.151 = 1/6.61. 35, 786 km. It follows knowing that Re = 6, 378.2 km that the ratio ReR+H Another way of remembering this is that the geosynchronous orbit altitude is approximately 6.6 Earth radii. There is another kind of orbit where θ 6= 0 or e 6= 0 whereby the orbital period is still the same as a sidereal day. This orbit is called a geosynchronous orbit, but the satellite will appear to oscillate about a mean look angle in this case and hence will not appear station with respect to the earth station.

4

Orbital Geometry

We have seen that that satellites orbit the Earth along an elliptical path and that this ellipse is actually inclined relative to the Earth as shown in Figure 4. Satellites also orbit the Earth at different speeds, since the orbital periods depend on the orbital altitude of the satellite. Generally, to locate a satellite at an instant in time requires knowledge of the geometry of the ellipse relative the the Earth as well as times that the satellite passes reference points along the ellipse. The parameters are known as the Keplerian elements of the satellite and are precisely known by satellite operators. The relation of these elements to the elliptical geometry we have described here is beyond the scope of this document, but it suffices to say that in general the distance from a ground station to the satellite depends on time, and that the satellite may not always be visible (above the horizon). satellite nadir direction zenith direction C Earth

sub-satellite point

Figure 7: The sub-satellite point Prof. Sean Victor Hum

ECE422: Radio and Microwave Wireless Systems

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From the perspective of a ground station, what matters is the location of the sub-satellite point on the Earth’s surface which is a point on the ground in the nadir (downward) pointing direction of the satellite. This point is shown in Figure 7. This point can be described by its latitude and longitude. The Earth station also can be located by its own latitude and longitude. Combined, these parameters are: • Le , the Earth station latitude (the number of degrees the station is north of the equator); • `e , the Earth station longitude (the number of degrees the station is west of the prime meridian); • Ls , the latitude of the sub-satellite point; and • `s , the longitude of the sub-satellite point. Consider the link between a ground station and orbiting satellite, shown in Figure 8.

Figure 8: Orbit geometry The angle ψ is related to the coordinates of the Earth station and the sub-satellite point according to cos ψ = cos Le cos Ls cos(`s − `e ) + sin Le sin Ls . (38) The remaining angles shown in the diagram are easily related by applying the law of sines to the triangle formed by the orbit geometry. Hence, Re Re + H r = = . ◦ sin β sin(α + 90 ) sin ψ

(39)

Since sin(α + 90◦ ) = cos α, −1

α = cos Prof. Sean Victor Hum



 Re + H sin ψ . r

(40)

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α is the elevation angle (angle above the horizon) that the Earth station antenna needs to be pointed to to make contact with the satellite. Therefore, knowing the coordinates of the Earth station and the sub-satellite point, the appropriate elevation angle can be found. An additional angle, the azimuth angle, is also needed, but is not discussed here. Knowing the angle ψ we can find the total distance r the satellite is from the ground station. Using the law of cosines, r2 = (Re + H)2 + Re2 − 2Re (Re + H) cos ψ, or "



r = (Re + H) 1 +

2

Re Re + H

Re cos ψ −2 Re + H

(41) #1/2 .

(42)

The elevation angle can then be found knowing sin(α + 90◦ ) = cos α, so cos α =

(Re + H) sin ψ = r

sin ψ 1+



Re Re +H

2

e cos ψ − 2 ReR+H

1/2 .

(43)

If instead, we wish to determine ψ for a given elevation angle α in advance, we can use the fact that β = 90◦ − (α + ψ), sin β = cos(α + ψ) and   Re −1 cos α − α. (44) ψ = cos Re + H The satellite is only visible from an earth station is the elevation angle α to be above some minimum value, which is at least 0◦ . From Figure 8, this requires Re + H ≥

Re cos ψ

(45)

or

Re . Re + H For a nominal GEO orbit, ψ ≤ 81.3◦ for the satellite to be visible. ψ ≤ cos−1

Prof. Sean Victor Hum

(46)

ECE422: Radio and Microwave Wireless Systems