1. MITP 413: Wireless Technologies Week 5
Michael L. Honig
Department of ECE
Northwestern University
April 2005

2. Binary Frequency-Shift Keying (FSK)
Bits:

3. Amplitude Shift Keying (4-Level ASK)
Bits:
Baseband signal
symbol duration

4. Binary Phase Shift Keying (BPSK)
Bits:
Baseband signal

5. Quadrature Phase Shift Keying (QPSK)
Bits:

6. Baseband  RF Conversion
Passband (RF) signal
Baseband signal
sin 2fct
time X T
fc is the carrier frequency
Why not transmit the baseband signal?
Power
signal bandwidth is roughly 1/T
Power
frequency
frequency
 0 fc

7. Selection Criteria
How do we decide on which modulation technique to use?
Performance: probability of error Pe.
Probability that a 0 (1) is transmitted and the receiver decodes as a 1 (0).
Complexity: how difficult is it for the receiver to recover the bits (demodulate)?
FSK was used in early voiceband modems because it is simple to implement.
Bandwidth or spectral efficiency: bandwidth (B) needed to accommodate data rate R bps, i.e., R/B measured in bits per second per Hz.
Power efficiency: energy needed per bit to achieve a satisfactory Pe.
Performance in the presence of fading, multipath, and interference.
Which of these criteria are especially important for wireless channels (as opposed to wired channels)?

8. Probability of Error 4-ASK BPSK 7 dB (factor of 5)
Signal-to-Noise Ratio (dB)

9. How to Increase Bandwidth Efficiency?
Increase number of signal levels.
Use more bandwidth efficient modulation scheme (e.g., PSK).
Apply coding techniques: protect against errors by adding redundant bits.
Note that reducing T increases the symbol rate, but also increases the signal bandwidth.
There is a fundamental tradeoff between power efficiency and
bandwidth efficiency.

10. Shannon Capacity Channel capacity: C = B log(1+S/N) bits/second
noise
Information
Source
Encoder
Channel
Decoder
bits
input
x(t)
output
y(t)
Estimated
bits
Channel capacity: C = B log(1+S/N) bits/second
B= Bandwidth, S= Signal Power, N= Noise Power
No fading

11. Colored Balls 2
The width of the funnel tube is analogous to the available bandwidth. Now the narrower the funnel, the longer it takes to get the jellybeans through the funnel. So, the less bandwidth we have, the longer it takes to transmit our symbols.

12. Binary Phase Shift Keying (BPSK)
Bits:
Baseband signal

13. Minimum Bandwidth (Nyquist) Pulse Shape
This pulse has the minimum bandwidth for a given
symbol rate.
Given bandwidth B, the maximum symbol rate without
intersymbol interference (ISI) is B, the “Nyquist rate”.

14. Pulse Shape and Bandwidth
Rectangular pulse
Power
amplitude
Bandwidth B > 2/T
2/T T
frequency
time
Nyquist pulse
Power
bandwidth B = 1/T
time
frequency T

15. Shifted Nyquist Pulses
Bits:

16. Baseband Waveform (Nyquist Signaling). . .
Bits: . . .

17. Baseband  RF Conversion
Passband (RF) signal
Baseband signal
sin 2fct X T
time
fc is the carrier frequency
Why not transmit the baseband signal?
Power
signal bandwidth is roughly 1/T
Power
frequency
frequency
 0 fc

18. Passband Signal with Different Carrier Frequencies

19. Pulse Width vs. Bandwidth
time
Perfect synchronization T
time
Offset causes severe ISI!

20. Raised Cosine Pulses
Minimum BW
frequency
time

21. Raised Cosine Pulses frequency time
Minimum BW
50% excess BW
100% excess BW
frequency
time
Excess bandwidth= (Total bandwidth – Nyquist bandwidth)/Nyquist bandwidth

22. Circle Vocoder
And for voice communications, this data rate depends on the data rate generated by the vocoder.

23. Blocks Transition
(Transitional Slide)

24. Blocks Stretched 8000 Bits Per Second
The vocoder selected by the Hughes engineers used the same technology that was selected for the North American digital cell phone standard IS54. That vocoder generates 8000 bits per second. But we actually need to transmit significantly more than this because...

25. Additional Bits Additional Bits Error Correction Control Information
... we need to insert additional bits to guard against errors and to provide overhead control information. The control information tells the transmitter and receiver operational information like which ground station to use, how much power to use, the identity of the person using the service, billing information, and so forth.
So, in reality, the system needs to support a data rate significantly larger than 8,000 bits per second – something in the range of approximately bits per second.
Channel
Ground Station
Power
Identity
Billing

26. 6000 Hz Available Bandwidth
In our case, the FCC has given us 6000 Hz. We can estimate the maximum number of symbols, which we can transmit every second, by applying one of the basic principles of digital communications, sometimes referred to as Nyquist's Theorem. Nyquist's Theorem says that, roughly speaking, we can count on being able to send about 4000 symbols per second over a 6000 Hz channel.
Nyquist’s Theorem: Can transmit 4000 symbols per second through a 6000 Hz channel

27. Colored Balls 2
5000 symbols per second would be quite difficult, and 4000 symbols per second is relatively easy.
Nyquist’s Theorem: Can transmit 4000 symbols per second through a 6000 Hz channel

28. 4000 bps < 8000 bps (Vocoder rate)
BPSK 4000 < 8000
BPSK: 1 Bit Per Symbol
4000 Bits Per Second (bps)
4000 bps < 8000 bps (Vocoder rate)
BPSK
So, let's assume that we are sending 4000 symbols per second. If we use Binary PSK, then we are transmitting 1 bit for each symbol, so the data rate in that case is 4000 bits per second. This is far below the 8000 bits per second generated by the vocoder, which means that we cannot use BPSK to transmit our voice signal.

29. QPSK 8000 BPS QPSK: 2 Bits Per Symbol
2 X 4000 = 8000 Bits Per Second (bps)
8000 bps = Vocoder rate
Need more bits for error correction and control!
QPSK
So now let's consider 4-PSK, or QPSK. In that case we are transmitting 2 bits per symbol, and with 4000 symbols per second, that gives a data rate of 2 times 4000, or 8000 bits per second. Recall that we need more than this, because on top of the vocoder rate of 8000 bits per second, we need to add additional bits to correct errors, and for control. This means that we cannot use QPSK either.

30. 8PSK 12000 BPS 8PSK: 3 Bits Per Symbol
3 X 4000 = 12,000 Bits Per Second (bps)
8000 bps bps
Vocoder rate + Error Correction and Control
8PSK
Moving then to 8-PSK, this gives us 3 bits per symbol, and with 4000 symbols per second, that gives a data rate of 3 times 4000, or 12,000 bits per second. This will give us the 8000 bits per second from the vocoder, plus 4000 bits per second for correcting and detecting errors, and for control. For this type of application, an additional 4000 bits per second beyond the vocoder rate gives us a reasonable margin for errors and control, so that we conclude that a voice service over this 6000 Hz channel can be supported with 8-PSK.

31. 16 phases: 4 Bits Per Symbol 4 x 4000 = 16,000 bps
16 PSK BPS
16 phases: 4 Bits Per Symbol
4 x 4000 = 16,000 bps
More than enough for vocoder rate + overhead
But couldn't we also transmit more than 8 phases, or equivalently, more than 3 bits for every symbol? For example, suppose that we transmit 4 bits per symbol. This would give us a bit rate of 4 times 4000, or 16,000 bits per second, which is more than we need to support a voice service with our 8000 bit per second vocoder.
We could do this, but it would make the system more complicated.

32. MOTION BLUR
As explained in the last video, the problem with increasing the number of bits per symbol is that we have to increase the number of phases or symbols to transmit, and these become harder to distinguish at the receiver. Namely, 4 bits per symbol means that we have to choose from among 16 possible phases, and to avoid confusing these symbols at the receiver, we need to transmit with more power.

33. 90˚
QPSK w/Bit Labels 0˚
180˚
270˚
Because we can choose from one of four phases, each phase can be used to represent two bits, instead of one as before. This is shown in the figure by labeling each phase with two bits. Namely, zero phase corresponds to transmitting 00, shown in green, 90 degrees corresponds to transmitting 11, shown in purple, 180 degrees corresponds to 10, shown in orange, and 270 degrees corresponds to 01, shown in red.
(Bit labels fade-in one at a time)

34. 90˚
8PSK
45˚
135˚ 0˚
180˚
225˚
270˚
315˚
Notice that the angle of each flag position again corresponds directly to the starting phase of the radio wave.
(Each pair of colored dots connected by a line wipe down successively)

35. QPSK Signal Constellation
amplitude = 1 x 1 x x x

36. “Rotated” QPSK Signal Constellation

37. “Rotated” QPSK Signal Constellation
amplitude = 1 x x 1 x x

38. In-Phase/Quadrature Componentsx
(a,b)  a sin 2fct + b cos 2fct 1 x x
b is the “in-phase” signal component
a is the “quadrature” signal component x

39. In-Phase/Quadrature Componentsx x 1 x x

40. Example Constellations
QPSK
BPSK x x x x
16-QAM
quadrature
8-PSK x x x x
For the 16-QAM signal constellation, what signal does a particular point represent? x x x x x x x x
in-phase x x x x

41. Example Constellations
quadrature
QPSK
BPSK
in-phase x x x x x x
16-QAM
quadrature
8-PSK x x x x
For the 16-QAM signal constellation, what signal does a particular point represent? x x x x x x x x x x x x x x x
in-phase x x x x x

42. Quadrature Modulation
in-phase signal
even bits
Baseband
Signal X
Split:
Even/Odd
source
bits
transmitted (RF)
signal +
Baseband
Signal X
odd bits
quadrature signal

43. Modulation for Fading Channels
Problems:
1. Amplitude variations (shadowing, distance, multipath)
2. Phase variations
3. Frequency variations (Doppler)
Solution to 1:
Avoid amplitude modulation
Power control

44. Modulation for Fading Channels
Problems:
1. Amplitude variations (shadowing, distance, multipath)
2. Phase variations
3. Frequency variations (Doppler)
Solution to 1:
Avoid amplitude modulation
Power control
Solution to 2 & 3:
Avoid phase modulation (use FSK)
“Noncoherent” demodulation: does not use phase reference
Differential coding/decoding
“Coherent” demodulation: Estimate phase shifts caused by channel.
Increase data rate/Doppler shift ratio

45. Binary Frequency-Shift Keying (FSK)
Bits:

46. Minimum Shift Keying (MSK)
Bits:
Frequencies differ by ½ cycle
Used in GSM

47. Binary Differential Modulation
(i+1)st bit = 0:
0o phase shift
waveform for ith symbol
(i+1)st bit = 1:
180o phase shift

48. Binary Differential Modulation
(i+1)st bit = 0:
0o phase shift
waveform for ith symbol
(i+1)st bit = 1:
180o phase shift
Drawback: a detection error for the ith bit propagates to the (i+1)st bit.

49. Example: DQPSK x x x x x x x x constellation for ith symbol
bits: 00 x 10 x x 01 x x x 11 x x
constellation for ith symbol
constellation for (i+1)st symbol
Used in IS-136

50. Coherent Phase Modulation
Receiver estimates phase offset
More complicated than noncoherent (e.g., differential) modulation.
Receiver requires a pilot signal.
Transmit known symbols, measure phase.
Pilot symbols are overhead (not information bits).

51. Probability of Error

52. Probability of Error with Fading

53. Orthogonal Frequency Division Multiplexing (OFDM)
Modulate
Carrier f1
substream 1
Split into
M substreams
Modulate
Carrier f2
substream 2
source
bits
substream M
OFDM
Signal +
Modulate
Carrier fM

54. OFDM Spectrum … … f1 f2 f3 f4 f5 f6  0
Total available bandwidth
Data spectrum for
a single carrier
Power … … f1 f2 f3 f4 f5 f6
frequency
 0
subchannels
M “subcarriers, or subchannels, or tones”
“Orthogonal” subcarriers  no cross-channel interference.

55. OFDM Example: 802.11a 20 MHz bandwidth, M=64 (48 for data payload)
Subchannel bandwidth = 20 MHz / 64 = kHz
Symbol rate / subchannel = 250 kilosymbols/sec
Total symbol rate = 64 x 250 x 103 = 16 Msymbols/sec
Bit rate?
16 QAM/subchannel  4 bits/symbol x 250 x 103 = 1 Mbps/subchannel, or 64 Mbps total
64 QAM/subchannel  6 bits/symbol x 250 x 10^3 = 1.5 Mbps/subchannel, or 96 Mbps total
Includes overhead (synchronization, error correction, control) Actual data rate: 36 / 54 Mbps

56. Why OFDM?
Single-carrier transmission also possible: 250 x 10^3 symbols/sec in kHz means 16 Msymbols/sec would be transmitted in 20 MHz.

57. Why OFDM? Exploits frequency diversity channel gain
Flat fading on each subchannel simplifies receiver (no multipath/ISI)
Frequency agility: can avoid “bad” parts of channel (requires feedback).
Drawback: high peak-to-average power.
subcarrier bandwidth < coherence bandwidth Bc
signal power
(wideband)
channel
gain
Frequencies far outside the coherence
bandwidth are affected differently by multipath. f1
frequency f2
bits are coded across subcarriers
Single-carrier transmission also possible: 250 x 10^3 symbols/sec in kHz means 16 Msymbols/sec would be transmitted in 20 MHz.

58. The Multiple Access Problem
How can multiple mobiles access (communicate with)
the same base station?
Frequency-Division (AMPS)
Time-Division (IS-136, GSM)
Code-Division (IS-95, 3G) Direct Sequence/Frequency-Hopped
Random Access (Wireless Data)

59. Duplexing (Two-way calls)
Frequency-Division Duplex (FDD)
Channel 1
Channel 2
Time-Division Duplex (TDD)
Time slot (frame) 1
Time slot (frame) 2

60. Combinations FDMA/FDD (AMPS) TDMA/FDD (GSM) TDMA/TDD (IS-136 or USDC)
CDMA/FDD (IS-95, CDMA2000)
CDMA/TDD (3G/UMTS)
Frequency-Hopped CDMA/TDD (Bluetooth)

61. Cellular Spectrum (50 MHz)
uplink
824
825
835
845
846.5
849 A* A B A* B*
869
870
880
890
891.5
894
downlink
AMPS (1G): 30 kHz Channels
416 FDD Channels:
395 FDD voice channels
21 FDD control channels

62. Properties of FDMA

63. Properties of FDMA Can be analog or digital (AMPS is analog).
Narrowband: channel contained within coherence bandwidth – undergoes flat fading.
Low capacity
Best for circuit-switched (dedicated) connections.
Requires guard channels for adjacent channel interference.

64. Time Division Multiple Access
Channel f1
N time slots
H: Frame Header
Frame
. . . H 1 2 N H
Channel f2
Time slots
. . . H 1 2 N H
Channel fK
. . .
. . . H 1 2 N H

65. Time Division Multiple Access
Frame
. . .
N time slots
H: Frame Header H 1 2 N H
Time Slot
Preamble
and synch
Data to or from user K
+ control information
Guard
time

66. TDMA/Time-Division DuplexH 1 2 3 4 5 6 H { {
Uplink time slots
Downlink time slots

67. Properties of TDMA

68. Properties of TDMA Data transmission occurs in bursts.
Must ensure small delays for speech.
High peak to average power on reverse link.
Can measure signal strength in idle time slots (e.g., for handoff).
Can assign multiple time slots for higher data rates.
Significant overhead/complexity for synchronization.
Guard times needed between time slots for delay spread.
May require an equalizer to mitigate intersymbol interference.

69. Second Generation (2G) Cellular: TDMA Standards
GSM
IS-136
Global System for Mobile Communications
Originated in Europe
Incompatible with 1G systems
More than an air-interface standard: specifies wireline interfaces/functions
North Americal Digital Cellular (NADC)
Fits into existing AMPS standard
Air-interface only
Another standard, IS-41, specifies networking functions
TDMA/FDMA, FDD
Dynamic frequency assignment
50 MHz allocated ( MHz)
200 kHz channels
kbps
TDMA/FDMA, TDD
Fixed frequency assignment
50 MHz allocated ( MHz)
30 kHz channels
48.6 kbps

70. North Americal Digital Cellular (IS-136)
One frame = 1944 bits (972 symbols)
= 40 ms; 25 frames/sec
Slot 1
Slot 2
Slot 3
Slot 4
Slot 5
Slot 6
30 kHz channels
TDD: 2 time slots per user
3 full-duplex traffic channels per 30 kHz channel
Data rate 48.6 kbps
324 bits per msec time slot
1 slot
ramp-up bits 6 6 16 28
122 12 12
122 G R
data
sync
data
SACCH
CDVCC
data
Reverse-Link
guard bits
SACCH: Slow Associated Control Channel
Monitors signal strength, handoff requests
CDVCC: Coded Digital Verification Color Code
Verifies proper connection 28 12
130 12
130 12
sync
SACCH
data
CDVCC
data
reserved
Forward-Link

71. ... … GSM Frame Structure Frame Tn: nth TCH frame
bits ms
Frame
TS0
TS1
TS2
TS3
TS4
TS5
TS6
TS7
Tn: nth TCH frame
S: Slow Associated Control Channel frame
I: Idle frame
4.615 ms T0 T1 T2
...
T10
T11
T12 S
T13
T14
T15 …
T22
T23
T24
I/S
120 ms
Speech Multiframe = 26 TDMA frames
200 kHz FDD channels divided into 8 time slots per frame
Total number of available channels = (25 MHz – 2 X Guard Band)/200 kHz
100 kHz guard bands  124 channels
Total number of traffic channels = 8 X 124 = 992
Channel data rate = kbps
Without overhead, data rate/user = 24.7 kbps

72. Hyperframe = 2048 superframes
GSM Time Slots
Hyperframe = 2048 superframes
lasts ~3 hrs 28 min 54 sec
6.12 s
Superframe
120 ms
Multiframe
4.615 ms
“Normal Burst” Traffic Channel (TCH)
148 bits/time slot
114 coded information bits
Frame efficiency %
(total bits – overhead bits)/(total bits)
Frame 1 2 3 4 5 6 7 ms 3 57 1 26 1 57 3
8.25
Time
slot
Tail bit
Tail bit
Coded Data
Stealing flag
Midamble
Stealing flag
Coded Data
Guard period

73. TDMA Capacity IS-136 GSM AMPS: 395 channels / N=7  56 users/cell
Total bandwidth = 12.5 MHz, 395 channels (30 kHz)
Number of channels/cell = 395/N ~ 56 with N=7
Number of time slots/channel = 6, 2 time slots/user
Number of users/cell = 56 x 3 = 168
N=4  capacity = (7/4) x 168 ~ 294 users/cell (S/I may not be adequate)
GSM
Total bandwidth = 12.5 MHz, 200 kHz channels  62 channels
With cell cluster size N=3 (typical), capacity is (62/3) x 8 ~ 165 users/cell
AMPS: 395 channels / N=7  56 users/cell

74. The Multiple Access Problem
How can multiple mobiles access (communicate with)
the same base station?
Use different frequencies (FDMA)
Use different time slots (TDMA)
Use different pulse shapes (CDMA)

75. Two-User Example User 1: s1(t) time s2(t) User 2: time received signal
bits: 1 1 1 1
User 1:
s1(t)
T/2 T
time T 2T 3T 4T 5T -1 1 1 1
s2(t)
User 2: T T 2T 3T 4T 5T
T/2
time 2
received signal
r(t)= s1(t)+s2(t) T 2T 3T 4T 5T
How to recover each users’ bits? -2

76. Correlation Given two sequences: a1, a2, a3, …, aN b1, b2, b3, …, bN
The correlation between the sequences is defined as:
(a1 x b1) + (a2 x b2) + (a3 x b3) + … + (aN x bN)
Examples:
correlated with = 5
correlated with = 1
correlated with = = 14
If the correlation between two sequences is zero, they are said
to be orthogonal.

77. Chip Sequence chips User 2:
User 2’s chip sequence (1, -1) is called a signature. T
T/2
time
chip duration Tc
symbol duration T=2Tc
What is user 1’s signature?
bits: 1 1 1
s2(t) T 2T 3T 4T 5T
Transmitted chips: 1 -1 -1 1 1 -1 1 -1 -1 1

78. Chip Sequence chips User 1: time chip duration Tc
symbol duration T=2Tc
What is user 1’s signature?
bits: 1 1 1 1
s_1(t) T 2T 3T 4T 5T -1
Transmitted chips: 1 1 1 1 -1 -1 1 1 -1 -1

79. Chip Sequence chips User 1: User 1’s signature is 1, 1. time
chip duration Tc
symbol duration T=2Tc
What is user 1’s signature?
bits: 1 1 1 1
s1(t) T 2T 3T 4T 5T -1
Transmitted chips: 1 1 1 1 -1 -1 1 1 -1 -1

80. Two-User Example s1(t) s2(t) r(t)= s1(t)+s2(t) 1 1 1 1 T 2T 3T 4T 5T1 1
s1(t) T 2T 3T 4T 5T -1
Transmitted chips: 1 1 1 1 -1 -1 1 1 -1 -1 1 1 1
s2(t) T 2T 3T 4T 5T
Transmitted chips: 1 -1 -1 1 1 -1 1 -1 -1 1 2
r(t)= s1(t)+s2(t) T 2T 3T 4T 5T -2
Received chips: 2 2 -2 2 -2

81. Correlator Receiver Bit Decision < 0  0 > 0  1 r(t) Sample
Chips
Correlate with
desired user’s signature
estimated
bits

82. Why Does This Work? A1 s1 Correlate with User 2’s signature amplitude
amplitude
signature (1,1)
A2 s2
Correlate with
User 1’s signature
The user signatures are orthogonal.
Furthermore:
A1 s1
Correlate with
User 1’s signature
2A1

83. Correlator, or “Matched Filter” Receiver
Bit Decision
< 0  0
> 0  1
A1 s1+A2 s2
Correlate with
User 1’s signature
User 1’s bits
Bit Decision
< 0  0
> 0  1
Correlate with
User 2’s signature
User 2’s bits
The correlator is “matched” to user 1’s signature s1, and rejects s2
(and vice versa).

84. Observations Users transmit simultaneously (not TDMA).
Users overlap in frequency (not FDMA).
Spectrum:
User 1
signal bandwidth is roughly 1/T
frequency
Why is user 2’s spectrum “flatter” than user 1’s spectrum?
Spectrum:
User 2
signal bandwidth is roughly 1/Tc = 2/T
frequency
Bandwidth expansion (factor of 2)  “spread spectrum” signaling.

85. Users and Bandwidth Expansion
To guarantee orthogonal signatures (no interference), the length
of the signatures must be ≥ the number of users.
Example (4 users):
signature:
signature:
User 2:
3T/4
3T/4
User 1:
T/4
T/2 T
time
T/4
T/2 T
time
signature:
Verify that the correlation between pairs of signatures is zero.
signature:
User 3:
3T/4
User 4:
3T/4
T/4
T/2 T
time
T/4
T/2 T
time
The chip rate is 4 times the symbol rate, hence the bandwidth
expansion is a factor of 4.

86. Correlator Receiver (4 users)
Bit Decision
< 0  0
> 0  1
A1 s1+A2 s2+A3 s3 + A4 s4
Correlate with
User 1’s signature
User 1’s bits
Bit Decision
< 0  0
> 0  1
Correlate with
User 2’s signature
User 2’s bits
Bit Decision
< 0  0
> 0  1
Correlate with
User 3’s signature
User 3’s bits
Bit Decision
< 0  0
> 0  1
Correlate with
User 4’s signature
User 4’s bits

87. Processing Gain
The PG is essentially the bandwidth expansion factor, and is often given by (1/Tc)/(1/T) = T/Tc (chips per symbol), which is the length of the signature.
The signature (sequence of chip values) is also called a spreading code. The signature may be randomly generated, in which case it is called a pseudo-noise (PN) sequence.
“Direct-Sequence” CDMA refers to a spread spectrum signaling scheme in which the signal is spread by transmitting a sequence of chips at a rate faster than the symbol rate.
What is the processing gain for the preceding examples?

88. DS-CDMA Transmitter X time Source bits chips Spreader (generate chips)
Modulator
(e.g., QPSK)
RF signal
Ex: 100 source bits, processing gain = 10  1000 chips
Nyquist chip shape
sin 2fct X
Baseband signal
Passband (RF) signal Tc
time

89. Orthogonality and Asynchronous Users1 1 1 1
s1(t) T 2T 3T 4T 5T -1 1 1 1
s2(t) T 2T 3T 4T 5T
time
Asynchronous users can
start transmissions at different times.
Chips are misaligned
 signatures are no longer orthogonal!

90. Orthogonality and Asynchronous Users1 1 1 1
s1(t) T 2T 3T 4T 5T -1 1 1 1
s2(t) T 2T 3T 4T 5T
time
Asynchronous users can
start transmissions at different times.
Chips are misaligned
 signatures are no longer orthogonal!
Orthogonality among users requires:
Synchronous transmissions
No multipath

91. Correlator, or “Matched Filter” Receiver
delay
2A1 +
multiple acess interference (MAI)
From user 2
A1 s1(t) +A2 s2(t-)
Bit Decision
< 0  0
> 0  1
Correlate with
User 1’s signature
2A1 +
multiple acess interference (MAI)
From user 1
Bit Decision
< 0  0
> 0  1
Correlate with
User 2’s signature
The multiple access interference adds to the background noise and can cause
errors. For this reason, CDMA is said to be interference-limited.
Because CDMA users are typically asynchronous, and because of multipath,
it is difficult to maintain orthogonal signatures at the receiver. Consequently,
in practice, the signatures at the transmitter are randomly generated.

92. Spreading Bandwidth and MAI
Example: PG=4
s1:
s2:
Correlation = -2
Energy in s1 (or s2) is 12 + (-1)2 + (-1) = 4
Normalized correlation = correlation/energy = -2/4 = -1/2
Example: PG=10

93. Spreading Bandwidth and MAI
Example: PG=4
s1:
s2:
Correlation = -2
Energy in s1 (or s2) is 12 + (-1)2 + (-1) = 4
Normalized correlation = correlation/energy = -2/4 = -1/2
Example: PG=10
Conclusion: On average, the correlation between signatures
decreases as the signature length (PG) increases.

94. Closed-Loop Power Control
Solves near-far problem: crucial part of CDMA cellular systems (IS-95, 3G).
Minimizes battery drain.
Complicated (increases cost)
Requires overhead: control bits in feedback channel to tell transmitter to lower/raise power.
Cannot compensate for fast fading.

95. 2G CDMA: IS-95 or cdmaOne Introduced by Qualcomm (San Diego)
Direct-Sequence Spread Spectrum signaling
FDD
Wideband channels (1.25 MHz)
Tight, closed-loop power control
Sophisticated error control coding
Multipath combining to exploit path diversity
Noncoherent detection
Soft handoff
High capacity
Air-interface only: uses IS-41

96. CDMA vs. TDMA (early 1990s) TDMA CDMA Proven technology
Large investment in research, development
Earlier military applications
Near-far problem
Enticing (exaggerated?) performance claims

97. TDMA vs. CDMA: Performance Critera
Capacity:
Users per Hz per km2
Channel conditions
System assumptions
Perfect power control?
Modulation and coding?
Complexity
Power control (CDMA)
Synchronization (TDMA)
Equalization
Frequency assignment
Flexibility
Integrated services
(voice/data)
Multimedia
Variable rate/QoS

98. Service Providers and Technologies (2005)
Verizon
Cellular & PCS
(850 & 1900 MHz)
1X CDMA2000;
rolling out EV-DO
8-128 Kbps
up to 2.5 Mbps
Cingular1
Cellular
U.S. TDMA &
GSM/GPRS3
8 Kbps
8-85 Kbps
Sprint
PCS
(1900 MHz)
T-Mobile
GSM/GPRS
NexTel2
Public service band (800 MHz)
iDEN (TDMA) &
WiDEN4
25-64 kbps
near 100 kpbs
U. S. Cellular 1X
CDMA 2000
1Merged with AT&T.
2Currently merging with Sprint.
3Plans to roll out UMTS in N. America.
4Wideband version of iDEN.

99. 3G Air Interfaces cdma2000 Wideband (W)-CDMA
Also referred to as “multicarrier” CDMA
1X Radio Transmission Technology (RTT): 1.25 MHz bandwidth (1 carrier)
Supports 307 kbps instantaneous data rate in packet mode
Expected throughput up to 144 kbps
3X RTT: 3.75 MHz bandwidth (3 carriers)
Data rates can exceed 2 Mbps
1xEV (Evolutionary): High Data Rate standard introduced by Qualcomm
1xEV-DO: data only, 1xEV-DV: data and voice
Radio channels assigned to single users (not CDMA!)
2.4 Mbps possible, expected throughputs are a few hundred kbps
1xEV-DV has twice as many voice channels as IS-95B
Also referred to as Universal Mobile Telecommunications System (UMTS)
European proposal to ITU (1998)
Backwards compatibility with 2G GSM and IS-136 air interfaces
Network and frame structure of GSM
``Always on’’ packet-based data service
Supports packet data rates up to 2 Mbps
Requires minimum 5 MHz bandwidth, FDD, coherent demodulation
6 times spectral efficiency of GSM

100. CDMA Capacity Performance depends on
Let S= Transmitted power (per user), R= information rate (bits/sec),
W= Bandwidth, K= Number of users
Eb= S/R
N0= (Number of interferers x S)/W = ((K-1) x S)/W
Therefore Eb/N0 = (W/R)/(K-1) = (Processing Gain)/(K-1)
For a target Eb/N0, the number of users that can be supported
is K = (Processing Gain)/(Eb/N0) + 1

101. Interference and CDMA Capacity
If interference is reduced by a factor 1/g, then the
target Eb/N0 can be reduced by 1/g:
If W/R is large, then reducing interference by 1/g
(approximately) increases the capacity by a factor of g.

102. Properties of CDMA High capacity with power control.
Power control needed to solve near-far problem.
Robust with respect to interference.
Benefits from voice inactivity and sectorization.
No loss in trunking efficiency.
Wideband: benefits from frequency/path diversity.
No frequency assignments (eases RF planning).
Soft capacity: performance degrades gradually as more users are added.
Asynchronous
Soft handoff

103. Frequency Hopping Properties: Applications
Exploits frequency diversity (can hop in/out of fades)
Can avoid narrowband interference (hop around)
No near-far problem (Can operate without power control)
Low Probability of Detect/Intercept
Spread spectrum technique – can overlay
Cost of frequency synthesizer increases with hop rate
Must use error correction to compensate for erasures due to fading and collisions.
Applications
Military (army)
Part of original standard
Enhancement to GSM
Bluetooth

104. Bluetooth: A Global Specification for Wireless Connectivity
Wireless Personal Area Network (WPAN).
Provides wireless voice and data over short-range radio links via low-cost, low-power radios (“wireless” cable).
Initiated by a consortium of companies (IBM, Ericsson, Nokia, Intel)
Standards are being developed (IEEE ).

105. Bluetooth Specifications
Allows small portable devices to communicate together in an ad-hoc “piconet” (up to eight connected devices).
Frequency-hopped spread-spectrum in the 2.4 GHz UNII band.
Packet switching with 1600 hops/s over 1 MHz channels.
Range set at 10m.
Gross data rate of 1 Mbps (TDD), with second generation plans for 2 Mbps.
64 kbps voice channels
Maximum asymmetric data transfer rate of 721 kbps in either direction, or kbps symmetric link
Interferes with
Competes with ?