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Communication Using Ground-based Cables and Satellites in SpacePowerPoint Presentation

Communication Using Ground-based Cables and Satellites in Space

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Now for the calculation: not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Now for the calculation: not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Speed of pulse not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Speed of pulse not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Speed of pulse not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Speed of pulse not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

What holds satellites in place? Nothing! rather than going via the optical fibre link we have assumed. Why does this make a difference?

This condition means that communications satellites need to be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

This condition means that communications satellites need to be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

This condition means that communications satellites need to be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

Equating these two expressions for v be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.2 we get

Equating these two expressions for v be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.2 we get

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

After all that hard work, what does it tell us about the time needed for signals to get from Los Angeles to London via satellite?

After all that hard work, what does it tell us about the time needed for signals to get from Los Angeles to London via satellite?

CommunicationUsing Ground-based Cablesand Satellites in Space

We send information down cables as electric currents.

What are these currents and how fast do these electrical signals travel?

We send information down cables as electric currents.

What are these currents and how fast do these electrical signals travel?

Electric currents consist of moving electrons (tiny, very light, negative particles).

We send information down cables as electric currents.

What are these currents and how fast do these electrical signals travel?

Electric currents consist of moving electrons (tiny, very light, negative particles).

These electrons travel quite slowly – about 6mm (¼inch) per second in a normal wire to a bedside lamp,

We send information down cables as electric currents.

What are these currents and how fast do these electrical signals travel?

Electric currents consist of moving electrons (tiny, very light, negative particles).

These electrons travel quite slowly – about 6mm (¼inch) per second in a normal wire to a bedside lamp, but the pulse, which tells each electron to move, travels at the speed of light. This is 300,000km/sec (186,000miles/sec) in a vacuum.

If the wire is not surrounded by a vacuum but by a plastic insulator then the pulse travels at the speed of light in that insulator, somewhat less than the speed in a vacuum.

If the wire is not surrounded by a vacuum but by a plastic insulator then the pulse travels at the speed of light in that insulator, somewhat less than the speed in a vacuum.

We can measure this speed in a piece of co-axial cable, the sort you use to send the signal to a TV set.

If the wire is not surrounded by a vacuum but by a plastic insulator then the pulse travels at the speed of light in that insulator, somewhat less than the speed in a vacuum.

We can measure this speed in a piece of co-axial cable, the sort you use to send the signal to a TV set.

To do this we need to measure two things – the length of the cable and the time taken for the pulse to travel that distance – since speed = distance/time.

Let’s decide how much cable and what sort of clock (timer) we might need.

Let’s decide how much cable and what sort of clock (timer) we might need.

Cables commonly come in 200m rolls. Will this do?

Let’s decide how much cable and what sort of clock (timer) we might need.

Cables commonly come in 200m rolls. Will this do?

If the speed is 300million metres/second then a pulse would take 2/3 of a millionth of a second to go the whole length.

Let’s decide how much cable and what sort of clock (timer) we might need.

Cables commonly come in 200m rolls. Will this do?

If the speed is 300million metres/second then a pulse would take 2/3 of a millionth of a second to go the whole length. The question then is, “Do we have a fast enough timer?”.

Let’s decide how much cable and what sort of clock (timer) we might need.

Cables commonly come in 200m rolls. Will this do?

If the speed is 300million metres/second then a pulse would take 2/3 of a millionth of a second to go the whole length. The question then is, “Do we have a fast enough timer?”.

Luckily, common laboratory devices called oscilloscopes will time this accurately.

This is what we’ll do: we might need.

This is what we’ll do: we might need.

The pulse generator works a bit like an electronic keyboard and produces a stream of short pulses at a rate of 250 thousand per second (250kHz).

The pulse generator works a bit like an electronic keyboard and produces a stream of short pulses at a rate of 250 thousand per second (250kHz). For sound, this would be about 4 octaves above the top of a normal piano keyboard.

The pulse generator works a bit like an electronic keyboard and produces a stream of short pulses at a rate of 250 thousand per second (250kHz). For sound, this would be about 4 octaves above the top of a normal piano keyboard.

These pulses occur with a 1/250,000 second gap between them.

The pulse generator works a bit like an electronic keyboard and produces a stream of short pulses at a rate of 250 thousand per second (250kHz). For sound, this would be about 4 octaves above the top of a normal piano keyboard.

These pulses occur with a 1/250,000 second gap between them.

(1/250,000 second is four millionths of a second, 4 microseconds, 4μs)

We can set the oscilloscope so that it just displays two pulses from the pulse generator,

We can set the oscilloscope so that it just displays two pulses from the pulse generator, with a time delay between them of 4millionths of a second.

If pulses from the pulse generator, with a time delay between them of 4millionths of a second.we now connect the 200m of cable, these pulses will travel to the far, joined end, get reflected, and return back to be shown on the oscilloscope before the next pulse starts.

If pulses from the pulse generator, with a time delay between them of 4millionths of a second.we now connect the 200m of cable, these pulses will travel to the far, joined end, get reflected, and return back to be shown on the oscilloscope before the next pulse starts.

As you can see, they arrive a bit distorted.

If pulses from the pulse generator, with a time delay between them of 4millionths of a second.we now connect the 200m of cable, these pulses will travel to the far, joined end, get reflected, and return back to be shown on the oscilloscope before the next pulse starts.

As you can see, they arrive a bit distorted.

How do we know there’s a reflection? pulses from the pulse generator, with a time delay between them of 4millionths of a second.

How do we know there’s a reflection pulses from the pulse generator, with a time delay between them of 4millionths of a second.? Undoing the ends of the cable produces a different type of reflection – the wrong way up!

We can even fool the pulse into thinking that the cable does not end by adding a suitable resistor

We can even fool the pulse into thinking that the cable does not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Now for the calculation: not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Now for the calculation: not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Distance travelled by pulse on return journey down the cable

Now for the calculation: not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

Distance travelled by pulse on return journey down the cable

= 2 x length of cable = 2 x 200m = 400m

Distance travelled by pulse on return journey down the cable

= 2 x length of cable = 2 x 200m = 400m

Time from start of pulse to start of reflected pulse

Distance travelled by pulse on return journey down the cable

= 2 x length of cable = 2 x 200m = 400m

Time from start of pulse to start of reflected pulse

= 1½microseconds (1.5μs)

Speed of pulse not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

= distance/time

Speed of pulse not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

= distance/time = 400m/1½x 10-6sec

Speed of pulse not end by adding a suitable resistor – we have now created an effective radio antenna – and the pulse reflection disappears.

= distance/time = 400m/1½x 10-6sec

= 270million metres/sec

= distance/time = 400m/1½x 10-6sec

= 270million metres/sec

Comparing this with the speed of light in a vacuum we get

= distance/time = 400m/1½x 10-6sec

= 270million metres/sec

Comparing this with the speed of light in a vacuum we get

speed of light in a vacuum = 300million m/s

speed of pulse along wire 270million m/s

= distance/time = 400m/1½x 10-6sec

= 270million metres/sec

Comparing this with the speed of light in a vacuum we get

speed of light in a vacuum = 300million m/s

speed of pulse along wire 270million m/s

= 1.12

= distance/time = 400m/1½x 10-6sec

= 270million metres/sec

Comparing this with the speed of light in a vacuum we get

speed of light in a vacuum = 300million m/s

speed of pulse along wire 270million m/s

= 1.12

= Refractive Index of the plastic insulation around the wire.

If we measure the speed of light pulses along an optical fibre, we get a very similar result.

If we measure the speed of light pulses along an optical fibre, we get a very similar result.

In order to telephone by cable from Los Angeles to London, a distance of about 8750km,

If we measure the speed of light pulses along an optical fibre, we get a very similar result.

In order to telephone by cable from Los Angeles to London, a distance of about 8750km, the signal takes 8.75 x 106m/2.7x108m/s, which is about 3 hundredths of a second.

This small time delay is not noticed as we hold a conversation, so why is there sometimes a time delay on this type of telephone call or on a TV signal?

Both TV and phone signals may travel by a satellite link, rather than going via the optical fibre link we have assumed. Why does this make a difference?

Both TV and phone signals may travel by a satellite link, rather than going via the optical fibre link we have assumed. Why does this make a difference?

To answer that we have to know where communications satellites are, and that requires some different Physics.

What holds satellites in place? rather than going via the optical fibre link we have assumed. Why does this make a difference?

What holds satellites in place? Nothing! rather than going via the optical fibre link we have assumed. Why does this make a difference?

What holds satellites in place? Nothing! rather than going via the optical fibre link we have assumed. Why does this make a difference?

Satellites are in free-fall towards the Earth

What holds satellites in place? Nothing! rather than going via the optical fibre link we have assumed. Why does this make a difference?

Satellites are in free-fall towards the Earth and also moving around the Earth with a speed which allows them to follow the curvature of their orbit.

Satellites are in free-fall towards the Earth and also moving around the Earth with a speed which allows them to follow the curvature of their orbit.

As a shell is fired

from a gun, it will

travel further as

its launching speed

is increased

In all these trajectories (neglecting any air-resistance) the only force after launch will be gravity.

In all these trajectories (neglecting any air-resistance) the only force after launch will be gravity.

If the shell makes a circular orbit then we need to consider that everything moving in a circle has to have a force on it of mv2/r towards the centre of the circle.

In all these trajectories (neglecting any air-resistance) the only force after launch will be gravity.

If the shell makes a circular orbit then we need to consider that everything moving in a circle has to have a force on it of mv2/r towards the centre of the circle.

Going round a curve in a car is more exciting if the speed (v) is greater or the corner is sharper (its radius (r) is smaller).

In any circular orbit where gravity is the only force, we have:

Gravitational force = centripetal force

In any circular orbit where gravity is the only force, we have:

Gravitational force = centripetal force

mg = mv2/r

In any circular orbit where gravity is the only force, we have:

Gravitational force = centripetal force

mg = mv2/r

or g = v2/r

In any circular orbit where gravity is the only force, we have:

Gravitational force = centripetal force

mg = mv2/r

or g = v2/r

For a very low Earth orbit, the speed needed (often called the escape velocity) is given by

v = √gr

In any circular orbit where gravity is the only force, we have:

Gravitational force = centripetal force

mg = mv2/r

or g = v2/r

For a very low Earth orbit, the speed needed (often called the escape velocity) is given by

v = √gr

where g = acceleration due to gravity at the Earth’s surface (10m/s2) and r = radius of orbit (radius of Earth) = 6400km

v have:= √(10m/s2 x 6.4 x 106)m

v have:= √(10m/s2 x 6.4 x 106)m

= √(64 x 106)m2/s2

v have:= √(10m/s2 x 6.4 x 106)m

= √(64 x 106)m2/s2

= 8000m/s (18,000mph)

We can’t have such low orbiting satellites because of air resistance,

v have:= √(10m/s2 x 6.4 x 106)m

= √(64 x 106)m2/s2

= 8000m/s (18,000mph)

We can’t have such low orbiting satellites because of air resistance, but at 300km altitude (giving a radius of 6700km) we can.

v have:= √(10m/s2 x 6.4 x 106)m

= √(64 x 106)m2/s2

= 8000m/s (18,000mph)

We can’t have such low orbiting satellites because of air resistance, but at 300km altitude (giving a radius of 6700km) we can.

This gives a speed of 8185m/s (18420mph) and this means that orbiting takes a time

t = circumference of orbit / speed

t = circumference of orbit / speed = 2 have:πro/v

t = circumference of orbit / speed = 2 have:πro/v

= 2πx 6.7 x 106m / 8185m/s

t = circumference of orbit / speed = 2 have:πro/v

= 2πx 6.7 x 106m / 8185m/s

= 5143sec = 1hr 26min, which is the minimum time for an orbiting Earth satellite.

t = circumference of orbit / speed = 2 have:πro/v

= 2πx 6.7 x 106m / 8185m/s

= 5143sec = 1hr 26min, which is the minimum time for an orbiting Earth satellite.

When we consider higher altitude satellites, we also have to consider that the gravitational force gets less according to Newton’s relationship g α1/r2.

t = circumference of orbit / speed = 2 have:πro/v

= 2πx 6.7 x 106m / 8185m/s

= 5143sec = 1hr 26min, which is the minimum time for an orbiting Earth satellite.

When we consider higher altitude satellites, we also have to consider that the gravitational force gets less according to Newton’s relationship g α1/r2. What we want for a communications satellite is that it appears to hover over the same spot on the Earth’s surface, so that our antennas can always point to the same place.

This condition means that communications satellites need to be above the equator

This condition means that communications satellites need to be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

This condition means that communications satellites need to be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

We have found that gorbit = v2/rorbitfor any orbit from the equation for circular motion

This condition means that communications satellites need to be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

We have found that gorbit = v2/rorbitfor any orbit from the equation for circular motion

and gorbit = gsurface rEarth2 / rorbit2 from Newton’s law of gravitation

We have found that gorbit = v2/rorbitfor any orbit from the equation for circular motion

and gorbit = gsurface rEarth2 / rorbit2 from Newton’s law of gravitation

and so gsrE2/ ro2 = v2/ro

We have found that gorbit = v2/rorbitfor any orbit from the equation for circular motion

and gorbit = gsurface rEarth2 / rorbit2 from Newton’s law of gravitation

and so gsrE2/ ro2 = v2/ro

- v2= gsrE2/ro

We have found that gorbit = v2/rorbitfor any orbit from the equation for circular motion

and gorbit = gsurface rEarth2 / rorbit2 from Newton’s law of gravitation

and so gsrE2/ ro2 = v2/ro

- v2= gsrE2/rowhich relates the velocity and radius of this orbit: at this point we know neither.

But, we do know that the time of orbit be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

t = 2πro/v = 24 hours

giving v2 = (2πro/t)2

which also relates v and ro

Equating these two expressions for v be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.2 we get

gsrE2/ ro = (2πro/t)2

Equating these two expressions for v be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.2 we get

gsrE2/ ro = (2πro/t)2 where now we know all the terms except ro.

Equating these two expressions for v be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.2 we get

gsrE2/ ro = (2πro/t)2 where now we know all the terms except ro.

Expanding the bracket gives gsrE2 = 4π2ro2

rot2

gsrE2/ ro = (2πro/t)2 where now we know all the terms except ro.

Expanding the bracket gives gsrE2 = 4π2ro2

rot2

and rearranging gives ro3 = gsrE2 t2/ 4π2

gsrE2/ ro = (2πro/t)2 where now we know all the terms except ro.

Expanding the bracket gives gsrE2 = 4π2ro2

rot2

and rearranging gives ro3 = gsrE2 t2/ 4π2

Now we can enter the numbers we know

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

= 10m/s2

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

= 10m/s2 (6.4 x 106m)2

r be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.o3 = gsrE2 t2/ 4π2

= 10m/s2 (6.4 x 106m)2 (24 x 3600s)2

= 10m/s2 (6.4 x 106m)2 (24 x 3600s)2 / 39.5

= 10m/s2 (6.4 x 106m)2 (24 x 3600s)2 / 39.5

= 7.745 x 1022

= 10m/s2 (6.4 x 106m)2 (24 x 3600s)2 / 39.5

= 7.745 x 1022

and so ro = 3√7.745 x 1022

= 10m/s2 (6.4 x 106m)2 (24 x 3600s)2 / 39.5

= 7.745 x 1022

and so ro = 3√7.745 x 1022

= 42,630,000m

= 10m/s2 (6.4 x 106m)2 (24 x 3600s)2 / 39.5

= 7.745 x 1022

and so ro = 3√7.745 x 1022

= 42,630,000m

= 42,630km ( = 26,600miles)

= 10m/s2 (6.4 x 106m)2 (24 x 3600s)2 / 39.5

= 7.745 x 1022

and so ro = 3√7.745 x 1022

= 42,630,000m

= 42,630km ( = 26,600miles)

which is an altitude of 42630-6400

=36,230km or 22,640miles

Earth and Moon be above the equator and to orbit the Earth in the same time as the Earth rotates, i.e. 24 hours.

After all that hard work, what does it tell us about the time needed for signals to get from Los Angeles to London via satellite?

After all that hard work, what does it tell us about the time needed for signals to get from Los Angeles to London via satellite?

After all that hard work, what does it tell us about the time needed for signals to get from Los Angeles to London via satellite?

To make this trip, we need to use two satellites, with the signal travelling up and down to each.

To make this trip, we need to use two satellites, with the signal travelling up and down to each. This means going at least 36,230km four times, i.e. about 145,000km.

To make this trip, we need to use two satellites, with the signal travelling up and down to each. This means going at least 36,230km four times, i.e. about 145,000km.

At the speed of light this takes just about half a second, a very noticeable delay when holding a conversation. We normally reply just before our speaking partner finishes.

This delay is often noticeable on news reports from across the world as they use satellites to link. If you are able to receive radio or TV signals via both cable or normal antenna and also by satellite dish, there is a noticeable time-lag on the satellite signal. This causes some problems when giving accurate times via the airwaves.

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