Transcript Telecommunications AERSP 401B

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
Power ratio in dB  10log10
P2
P1
where P1 and P2 are the two power levels being
compared
Examples
• Loss
Telephone cable line
1,000 watts (P1)
10 watts (P2)
dB  10log10  10log10
10
 20 dB
1,000
Power out
• Gain
10 watts (P1)
dB  10 log10
P1
P2
1,000 watts (P2)
1, 000
 10 log10
 20 dB
10
20 dB means 100 times more
Power in
1 mile
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: Power (dBm)  10log10
Power (mW )
1 mW
• Example: 20W is what in dBm?
20W
20 x103 mW
Power (dBm)  10log10
 10log10
 43 dBm
1mW
1mW
• The dBW: Power (dBW )  10log10
• Example conversions
Power (W )
1W
dBm
dBW
Watts
Milliwatts (mW)
+50
+20
100
100,000
+30
0
1
1,000
+10
-20
-0.01
10
Voltages (examples)
Power  V 2
2
Pref  Vref
Pout  Vout

 Vref
Pref

2

 Pout
  10 log10 

 Pref



 Vout
  20 log10 

 Vref






Examples
1V  0dB
10V  20dB
100V  40dB
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
Gain: b dB
Pin
Gain: c dB
Gain: a dB
Pout
 A  a dB  10 log10 A
Pin
downlink :
uplink :
 Preceived 
1
 1 



x
dB

10
log


10 

P
X
X


 transmit uplink
 Pout 
 B  b dB  10 log10 B


 Pin uplink
ABC
Total ratio 
XY
 Ptransmit 
1
1
   y dB  10 log10  


Y 
 Preceived downlink Y
 Pout 
 C  c dB  10 log10 C


 Pin downlink
Total gain  (a  x  b  y  c) dB
Special Values
Pout 1
If
  3 dB gain (or 3 dB loss)
Pin
2
Pout
or if
 2  3 dB gain (or  3 dB loss)
Pin
Pout
1
If

 10 dB gain (or 10 dB loss)
Pin 10
Other Examples
• Sound levels:
– If Pref is the sound power resulting in a barely
audible sound,
60 dB normal conversation
Pout
in dB 
Pref
 90 dB jet aircraft on runway
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:
Preceived
Ar

Ptransmitted 4 d 2
– where Ar is the aperture area of the receiving
antenna
Isotropic Radiation (cont’d)
• Receiver is not 100% efficient, so including
efficiency factor, z,
Preceived
zAr

Ptransmitted 4 d 2
– Z  0.55
• Transmitting antenna designed to focus
radiation (i.e. not isotropic)
Gain of transmitting antenna 
power received from the antenna
power received if antenna were isotropic
– Can also be expressed in dB
Typical Antenna Patterns
• Dipole
G<10 dBi
• Horn
G=10-20 dBi
• Slot
G< 10 dBi
Parabolic Reflector Antenna
parallel beams
focal point
   D 2 
G  10 log10  z 
 dBi

    
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
 Eb 
PLl Gt Ls La Gr Li received energy per bit




kTs R
noise density
 N o estimated
• Similar to signal-to-noise ratio, but Eb/No
pertains to digital data
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)

 Eb 
 Gr
 228.60   P  Ll  Gt  Ls  La  



 No dB,est
 Ts


  R  Li 
 dB

More Details
P (in dBW )  10 log10  transmitter power in Watts 
Example: transmitter generates 100 Watts output power,
then P  10 log10(100)  20 dBW
Ll  line loss   1dB (typical value)
Gt  G pt  Ltrans = peak transmitter antenna gain+ loss due to pointing errror
2
 e 
Ltrans  12 
 e  pointing error (deg)
 3dB 
21
3dB 
fGHz  transmitter frequency (GHz),
fGHz D
G pt
D  antenna diameter (meters)
 159.59  20 log10 D  20 log10 f Hz  10 log10 
  efficiency (=0.55, typical value)
More Details
2
Ls
 c 
 space loss  10 log10 

4

Sf
Hz 

c  speed of light
S  propagation distance (meters)
 147.55  20 log10 S  20 log10 f Hz
La  path loss (calculate using Figs. 13-10 and 13-11 in SMAD)
(note that these figs. show attenuation, so convert the values
to negative numbers)
Example: at elevation angle of 20o (99.5% availability) with
a frequency of 40 GHz will have a path loss of La  15 dB
R  bit rate (bits per second)
example - data downloaded at 100,000 bps has RdB  10 log10 R  50dB
More Details
Gr
 antenna gain to noise-temperature ratio (Use Table 13-10 to get Ts )
Ts
Example: Receiving antenna on spacecraft is receiving uplink.
If the frequency is in the range of 0.2-20 GHz, then
Ts  614 K  10 log10 (614)  27.8dB
Gr 
G pr  Lrec
calculated the same way as was done for the transmitting
antenna but uses receiver antenna paramters
Example: Receiver antenna has diameter, D = 0.5 m, transmitter frequency of 2.4 GHz,
3dB 
21
 17.5deg
 2.4  0.5
If the attitude control on the spacecraft can maintain pointing accuracy of 1 deg,
2
then, Lrec
 1deg 
 12 
  0.5dB
17.5deg


G pr  159.59  20 log10 (0.5m)  20 log10 (2.4 109 Hz )  10 log10  0.55   19.4dB
Gr  19.4dB  (0.5dB)  18.9dB
 Gr 

  18.9dB  27.8dB  9dB
 Ts dB
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
bits
samples
Ri 

 sample size  sampling rate
sample second
• Sample size is determined based upon required
level of accuracy
– Example – temperature sensor needed to monitor
propellant tank temperature in range -10C to +80C
– Amplitude range=80C-(-10C)= 90C
Prop. tank
A/D Converter
sensor
Microprocessor
C&DH
Data Rates (cont’d)
• Sensor generates voltage
proportional to temperature
80C
• A/D converter generates a
digitized representation of this
temperature – an n-bit word
-10C
• Number of quantized levels
that are represented = 2n
• Quantization step here=
90C
2n
Quantized
steps

amplitude
2n
Data Rates (cont’d)
• Example continued
Eq  max. quantizati on error
1
amplitude
 quantized step 
2
2n 1
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
n
ln amplitude Eq 
ln 2
 1  9.81  round up to n  10
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
Microcomputer
Sensor B
Microcomputer
Sensor C
Microcomputer
Main
Computer
Modulator
Transmitter
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,
0
0
H 
0

1
0 0 0 1 1 1 1
0 1 1 0 0 1 1
1 0 1 0 1 0 1

1 1 1 1 1 1 1
to get S=HMT (syndrome vector)
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)
p4  b  c  d  0
p3  a  c  d  0



 mod 2
p2  a  b  d  0

p1  p2  p3  a  p4  b  c  d  0 
(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
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]
1   s1 
0   s 
2



S

1   s3 
   
1   s4 
• 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
Hamming Example (cont’d)
• M= [b0 b1,b2 b3,b4 b5,b6 ,b7]
• M1=[0, 0, 1, 1, 1, 0, 0, 0 ]
Error
• 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]
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.