Telecommunications

AERSP 401B

Communication System Designers’ Goal

• Maximize information transfer

• Minimize errors/interference

• Minimize required power

• Minimize required system bandwidth

• Maximize system utilization

• Minimize cost

Useful Relationships

• Decibels– A logarithmic unit originally devised to express

power ratios but used today to express a variety of other ratios as well

where P1 and P2 are the two power levels being compared

210

1

10logP

Power ratio in dBP

Examples

• Loss1,000 watts (P1) 10 watts (P2)

• Gain10 watts (P1) 1,000 watts (P2)

Telephone cable line

10 1010

10log 10log 201,000

dB dB

10 101,000

10log 10log 2010

dB dB

20 dB means 100 times more

P2 P1

1 mile

Power inPower out

The unit decibel was named after Alexander Graham Bell. The unit originated as a measure of power loss in one mile of telephone cable. Also, hearing is based on decibel levels.

Derived Decibel Units

• The dBm:

• Example: 20W is what in dBm?

• The dBW:• Example conversions

10( )

( ) 10log1

Power mWPower dBm

mW

3

10 1020 20 10

( ) 10log 10log 431 1

W x mWPower dBm dBm

mW mW

10( )

( ) 10log1

Power WPower dBW

W

dBm dBW Watts Milliwatts (mW)

+50 +20 100 100,000

+30 0 1 1,000

+10 -20 -0.01 10

Voltages (examples)2

2

2

10 1010log 20log

1 0

10 20

100 40

: '

ref ref

out out out out

ref ref ref ref

Power V

P V

P V P V

P V P V

Examples

V dB

V dB

V dB

Note dB s are NOT absolute but RELATIVE measures

Gains and Losses

• Power is gained via amplification and lost via absorption or resistance

• Gains and losses are expressed in dB (usually the W or m are dropped)

Communications Example

10

10

10

10log

:

1 110log

10log

out

in

received

transmit uplink

out

in uplink

PA adB A

P

uplink

Px dB

P X X

PB bdB B

P

10

10

:

1 110log

10log

transmit

received downlink

out

in downlink

downlink

Py dB

P Y Y

PC cdB C

P

Pin

Attenuation: x dBAttenuation: y dB

Gain: a dB Gain: c dB

Gain: b dB

( )ABC

Total ratio Total gain a x b y c dBXY

Special Values

13 ( 3 )

2

2 3 ( 3 )

110 ( 10 )

10

out

in

out

in

out

in

PIf dB gain or dB loss

P

Por if dB gain or dB loss

P

PIf dB gain or dB loss

P

Other Examples

• Sound levels:– If Pref is the sound power resulting in a barely

audible sound,

60

90out

ref

dB normal conversationPin dB

dB jet aircraft on runwayP

Radio Frequency Radiation

• RF signals travel at the speed of light in air (atmosphere) and space (vacuum)

• c = speed of light in vacuum = 2.998x108 m/sec (186,200 miles/sec)

• Wavelength, =c/f – f – frequency

• Beam width: (rad) /D – D = aperture width or diameter– Defines how “spread out” the beam is

Half Power

• A 3 dB drop in power represents the half-power point

Isotropic Radiation

• Aperture – area of a receiving or transmitting antenna through which all signal is assumed to pass.

• If transmitting antenna radiates equally in all directions, it is called isotropic

• The fraction of power received from an isotropic radiator at a distance, d, is:

– where Ar is the aperture area of the receiving antenna

24received r

transmitted

P A

P d

Isotropic Radiation (cont’d)

• Receiver is not 100% efficient, so including efficiency factor, z,

– Z 0.55

• Transmitting antenna designed to focus radiation (i.e. not isotropic)

– Can also be expressed in dB

24received r

transmitted

P zA

P d

power received from the antennaGain of transmitting antenna

power received if antenna were isotropic

Typical Antenna Patterns

• Dipole

G<10 dBi

• Horn

G=10-20 dBi

• Slot

G< 10 dBi

Parabolic Reflector Antenna

parallel beams

focal point

2

1010log iD

G z dB

D – diameter

- wavelength

z - efficiency

Lobes

• Backlobes

• Sidelobes

Cassegrain Reflector Antenna

Modulation

• Definition– Altering a signal to make it

convey information (either analog or digital)

• AM (Amplitude Modulation)– Changes amplitude

(frequency constant)• FM (Frequency

Modulation)– Changes frequency

(amplitude constant)

Frequency modulation

Modulation (cont’d)

• Changing the phase of the signal

• For digital data, these methods are also called– ASK – amplitude shift keying– FSK – frequency shift keying– PSK – phase shift keying

Link Budget

• Allocation of various losses and gains in the communication link between Earth and the spacecraft

• Similar to signal-to-noise ratio, but Eb/No pertains to digital data

b l t s a r i

o sestimated

E PL G L L G L received energy per bit

N kT R noise density

Link Requirements

• For data• (Eb/No)estimated – (Eb/No)required 3 dB

• For commands• (Eb/No)estimated – (Eb/No)required 20 dB

• This difference is known as the link margin

Terms

• P – transmitter power

• Ll – line loss (between transmitter and antenna)

• Gt – transmitter antenna gain

• Ls – space loss (inverse square in distance)

• k – Boltzmann’s constant

• La – transmission path loss (atmosphere and rain absorption)

• Gr – receive antenna gain

• Ts – system noise temperature

• R – data rate

• Li – implementation loss (-2 dB)

,

228.60b rl t s a i

o sdB est dB

E GP L G L L R L

N T

More Details

10( ) 10log

10log10(100) 20

1 ( )

indBW transmitter power in Watts

Example: transmitter generates 100 Watts output power,

then

line loss typical value

= peak transmitter antenna gain+ trans

l

t pt

P

P dBW

L dB

G G L

2

3

3

10 10 10

12

21

159.59 20log 20log 10log

loss due to pointing errror

pointing error (deg)

transmitter frequency (GHz),

antenna diameter (meters)

efficiency (

transdB

dB GHzGHz

pt Hz

eL e

ff D

D

G D f

=0.55, typical value)

More Details2

10

10 10

10log4

147.55 20log 20log

space loss

speed of light

propagation distance (meters)

path loss (calculate using Figs. 13-10 and 13-11 in SMAD)

(note that these figs. show att

sHz

Hz

a

cL

Sf

c

S

S f

L

15

o

enuation, so convert the values

to negative numbers)

Example: at elevation angle of 20 (99.5% availability) with

a frequency of 40 GHz will have a path loss of

bit rate (bits per second)

example

aL dB

R

1010log 50 - data downloaded at 100,000 bps has dBR R dB

More Details

1614 10log

antenna gain to noise-temperature ratio (Use Table 13-10 to get )

Example: Receiving antenna on spacecraft is receiving uplink.

If the frequency is in the range of 0.2-20 GHz, then

rs

s

s

GT

T

T K

0 (614) 27.8

calculated the same way as was done for the transmitting antenna but uses receiver antenna paramters

Example: Receiver antenna has diameter, = 0.5 m, transmitter frequen

recr pr

dB

G G L

D

3

2

910 10

2117.5deg

2.4 0.5

1deg12 0.5

17.5deg

159.59 20log (0.5 ) 20log (2.4 10

cy of 2.4 GHz,

If the attitude control on the spacecraft can maintain pointing accuracy of 1 deg,

then, rec

dB

pr

L dB

G m

10) 10log 0.55 19.4

19.4 ( 0.5 ) 18.9

18.9 27.8 9

r

r

s dB

Hz dB

G dB dB dB

GdB dB dB

T

More Details

• Calculate Link Margin = (Eb/No)est – (Eb/No)req

Fig. 13-9,

SMAD

Acceptable BER

Example

• If acceptable BER (bit error rate) is one bit error in every 100,000 bits, then BER=10-5

• Using BPSK modulation with Reed/Solomon coding, this requires an Eb/No=2.5dB

• If BPSK is used without coding, Eb/No=9.5dB– Increase transmitter power by 7 dB

• Multiplicative factor of 100.7=5

– Increasing the transmitter and receiver antenna gains by 7dB (combined)

• Antennas then more sensitive to pointing errors

Data Rates

• For each sensor, data rate

• Sample size is determined based upon required level of accuracy– Example – temperature sensor needed to monitor

propellant tank temperature in range -10C to +80C– Amplitude range=80C-(-10C)= 90C

ibits samples

R sample size sampling ratesample second

Prop. tank

sensor

A/D Converter Microprocessor C&DH

Data Rates (cont’d)

• Sensor generates voltage proportional to temperature

• A/D converter generates a digitized representation of this temperature – an n-bit word

• Number of quantized levels that are represented = 2n

• Quantization step here=90

2 2n n

C amplitude

80C

-10C

Quantized steps

Data Rates (cont’d)

• Example continued

12 step quantized

2

1

erroron quantizati max.

n

q

amplitude

E

So, if n=8, then quantized step = 0.35oC and Eq = 0.175oC

Typically, one needs to find the required value of n. Using same example, if required Eq 0.05oC, then quantized step = 0.1oC and

10 toup round81.91

2ln

ln

n

Eamplituden q

Sample Rate

• Determined based upon estimated rate of change of quantity being measured– Examples

• Thermal sensors typically sample at low rates (once per minute)

• Attitude sensors sample at high rates, especially during attitude maneuvers (1-5 samples/sec)

Sampling Oscillatory Phenomena

• Must sample at 2.2 times the highest frequency present– Human voice has frequency range of ~3.5 KHz

• Sample at 7.7 KHz (7,000 samples/second)

– Commercial audio (telephony) requires ~8 bits/sample

– Data rate = 7,700 samples/sec x 8 bits/sample

~62,000 bits/sec (bps)

Data Compression

• Compression/encoding allow lower data rates– Make use of repeated patterns in the data

and/or transmit only parts of data that changes since previous sample

– Voice data can be reduced to ~9.6 Kbps– Compressed video (videophone) ~28 Kbps

• Full video with color 256 Mbps (~40 Mbps with coding)

Telemetry

• Packet telemetry format– Each sensor forms packet of data– When packet complete, microcomputer interrupts main computer– Main computer formats main block– Main block transmitted

• Advantages– Flexible data rates for sensors

• Disadvantages– Spacecraft processing more complex– Ground station equipment more complex

Sensor A

Sensor B

Microcomputer

Sensor C

Main Computer

Modulator TransmitterMicrocomputer

Microcomputer

Error Detection and Correction

• Once our telemetry data is set to transmit, we must concern ourselves with possible induced errors in the transmission

• With digital data, there are several ways to check for errors– Parity check (with retransmission)– Error correction (without retransmission)

Ref: Spacecraft Attitude Determination and Control, J.R. Wertz (ed), Reidel Publishing Co., 1978

Parity Check

• Simplest method of detection• Example:

– M = [1,1,0,0] = original message– Add another parity bit to M– M now becomes [1,1,0,0,p]– Even parity scheme:

• m1+ m2 + m3+ m4 + p = even number p=0

– Odd parity scheme:• m1+ m2 + m3+ m4 + p = odd number p=1

– Receiving equipment then checks each message vector

Parity Check

• Suppose receiving equipment receives:– M = [1,1,0,0,1]– If both transmitter and receiver are employing

even parity scheme, then an error has occurred

• m1+ m2 + m3+ m4 + p = 3; not an even number

– Receiver requests retransmission

• What if two bits are flipped?– Parity scheme fails (much lower probability of

two bit flips than one bit flip)

Error Correction without Retransmission

• Example self-correcting developed by Hamming

• Extra set of bits equal in number to the original message bits added to message vector– Before: M = [a,b,c,d]

– After: M = [p1,p2,p3,a,p4,b,c,d]

Hamming (cont’d)

• Multiply MT by the Hamming matrix,

to get S=HMT (syndrome vector)

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

1 1 1 1 1 1 1 1

H

Hamming (cont’d)

• Need to determine the values of p1,p2,p3, and p4

• Set these such that S = [0,0,0,0]T (mod 2)

(Any even number = 0 mod 2)• Arrangement of parity bits in M so that only one new

parity bit is involved in each successive calculation of p1,p2,p3,p4

4

3

2

1 2 3 4

0

0mod 2

0

0

p b c d

p a c d

p a b d

p p p a p b c d

Hamming Example

• Intended message vector: Mo=[0,0,1,1,1,1,0,0]

• Received message vector: M1=[0,0,1,1,1,0,0,0]

• Correction scheme– If s4 = 0, then a, b, c, and d are correct

– If s4 = 1, then error occurred in message bit s1s2s3 (101)2=5

1

2

3

4

1

0

1

1

S

s

s

s

s

Hamming Example (cont’d)

• M= [b0 b1,b2 b3,b4 b5,b6 ,b7]

• M1=[0, 0, 1, 1, 1, 0, 0, 0 ]

• Correct M1 is M1=[0, 0, 1, 1, 1, 1, 0, 0 ]

• So the original message data is [1 1 0 0]

[a b c d]

Error

Probability of Errors – Simple Parity

• If probability of error in 1 bit is 1%, probability of at least one error in a 4-bit message is 4%

• Adding one parity bit increases error rate to 5%– Can detect, but not correct this error– Need to retransmit 5% of the data

• Probability of 0.25% that error occurs in 2 or more of the original 4 bits

Probability of Errors – Hamming Code

• Using the 8-bit Hamming code will increase probability of error to 8%– One bit error can be corrected

• Errors in 2 bits of M will occur in 0.64% of messages received– Two bit errors cannot be corrected

• Hamming will detect two errors, so retransmission can be requested

• Undetected errors in 3 or more bits will occur in ~0.051% of the messages received.