1. MITP 413: Wireless Technologies Week 4
Michael L. Honig
Department of ECE
Northwestern University
April 2005

2. Why Digital Communications?
1G (analog)  2G (digital)  3G (digital)
Digitized voice requires about 64 kbps, therefore the required bandwidth
is >> the bandwidth of the voice signal (3—4 kHz)!

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

4. Quadrature Phase Shift Keying (QPSK)
Bits:

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

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

7. 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

8. Why Modulate?

9. Selection Criteria
How do we decide on which modulation technique to use?
Which of these criteria are especially important for wireless channels (as opposed to wired channels)?

10. Example: Binary vs. 4-Level ASK
Rate = 1/T symbols/sec
Bandwidth is roughly 1/T Hz
Bandwidth efficiency = 1 bps/Hz
Rate = 2/T symbols/sec
Bandwidth is roughly 1/T
Bandwidth efficiency = 2 bps/Hz
What about power efficiency?

11. Noisy Baseband Signals
Rate = 1/T symbols/sec
Bandwidth is roughly 1/T Hz
Bandwidth efficiency = 1 bps/Hz
Power =A2 (amplitude squared).
Rate = 2/T symbols/sec
Bandwidth is roughly 1/T
Bandwidth efficiency = 2 bps/Hz
Power = (A2 + 9A2)/2 = 5A2
What about probability of error vs transmitted power?

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

13. How to Increase Bandwidth Efficiency?

14. The Fundamental Question
Given: B Hz of bandwidth
S Watts of transmitted signal power
N Watts per Hz of background noise
(or interference) power
What is the maximum achievable data rate?
(Note: depends on Pe.)

15. Claude Shannon (1916-2001) Father of “Information Theory”
His 1948 paper “A Mathematical Theory of
Communications” laid the foundations for modern communications and networking.
“Shannon was the person who saw that the binary digit was the fundamental element in all of communication. That was really his discovery, and from it the whole communications revolution has sprung.”
-- R. Gallager (MIT)
Other contributions and interests: digital circuits, genetics, cryptography, investing, chess-playing computer, roulette prediction, maze-solving, unicycle designs, juggling

16. Shannon’s Channel Coding Theorem (1948)
noise
Information
Source
Encoder
Channel
Decoder
bits
input
x(t)
output
y(t)
Estimated
bits
Information rate: R bits/second
Channel capacity: C bits/second
R < C  There exists an encoder/decoder combination that
achieves arbitrarily low error probability.
R > C  The error probability cannot be made small.

17. 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

18. Observations “There exists” does not address complexity issues.
As the rate approaches Shannon capacity, to achieve small error rates, the transmitter and (especially) the receiver are required to do more and more computations.
The theorem does not say anything about delay.
To achieve Shannon capacity the length of the transmitted code words must tend to infinity!
The previous formula does not apply with fading, multipath, frequency-selective attenuation.
It has taken communications engineers more than 50 years to find practical coding and decoding techniques, which can achieve information rates close to the Shannon capacity.

19. Example: GSM Bandwidth = 200 kHz, S/I = 9 dB = 7.943
C = 200,000 x log(8.943) ≈ 632 kbps
This is what would be achievable in the absence of fading, multipath, etc. Currently, the rate is about 270 kbps.
Is the Shannon formula relevant?

21. Data Rates for Deep Space Applications
Mariner: 1969 (Mars)
Pioneer 10/11: 1972/3 (Jupiter/Saturn fly-by)
Voyager: (Jupiter and Saturn)
Planetary Standard: 1980’s (military satellite)
BVD: “Big Viterbi Decoder”
Galileo: 1992 (Jupiter) (uses BVD)
Turbo Code: 1993

22. TYPES ALL PSK
(PSK Type Summary Slide)

23. 1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second
BPSK
Bitream 0
1000 Bits
Per Second
1000 Symbols
(Phases) Per Second
Now in an actual radio system, the modulator may transmit thousands of symbols each second.
1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second

24. 1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second
BPSK
Bitream 1
1000 Bits
Per Second
1000 Symbols
(Phases) Per Second
For example, if the modulator transmits 1000 symbols, or phases, each second, then with Binary PSK, since each symbol represents 1 bit, the data rate is 1000 bits per second.
1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second

25. 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second
QPSK
Bitream 00
2000 Bits
Per Second
1000 Symbols
(Phases) Per Second
But if we use 4-PSK, or QPSK, then each symbol represents 2 bits, and the data rate is 2000 bits per second.
1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

26. 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second
QPSK
Bitream 10
2000 Bits
Per Second
1000 Symbols
(Phases) Per Second
(3-bit bitstream animation continues at random)
1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

27. 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second
QPSK
Bitream 01
2000 Bits
Per Second
1000 Symbols
(Phases) Per Second
(2-bit bitstream animation continues at random)
1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

28. 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second
QPSK
Bitream 11
2000 Bits
Per Second
1000 Symbols
(Phases) Per Second
(2-bit bitstream animation continues at random)
1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

29. 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second
8PSK
Bitstream 000
3000 Bits
Per Second
1000 Symbols
(Phases) Per Second
And if we use 8-PSK, which conveys 3 bits for each symbol, then the data rate would be 3000 bits per second.
1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

30. 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second
8PSK
Bitstream 001
3000 Bits
Per Second
1000 Symbols
(Phases) Per Second
(3-bit bitstream animation continues at random)
1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

31. 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second
8PSK
Bitstream 101
3000 Bits
Per Second
1000 Symbols
(Phases) Per Second
(3-bit bitstream animation continues at random)
1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

32. 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second
8PSK
Bitstream 110
3000 Bits
Per Second
1000 Symbols
(Phases) Per Second
(3-bit bitstream animation continues at random)
1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

33. 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second
8PSK
Bitstream 111
3000 Bits
Per Second
1000 Symbols
(Phases) Per Second
(3-bit bitstream animation conlcudes)
1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

34. 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.

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

36. 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”.

37. Pulse Width vs. Bandwidth
Power
signal pulse
bandwidth B = 1/T
Narrowband
frequency
time 2T
signal pulse
Power
bandwidth B = 1/T
Wideband
time
frequency 2T

38. Shifted Nyquist Pulses
Bits:

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

40. 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
 0 fc
frequency

41. Passband Signal with Different Carrier Frequencies

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

43. Raised Cosine Pulses
Minimum BW
frequency
time

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

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

46. Blocks Transition
(Transitional Slide)

47. 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...

48. 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

49. 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

50. 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

51. 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.

52. 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.

53. 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.

54. 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.

55. 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.

56. 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)

57. 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)

58. QPSK Signal Constellation
amplitude = 1 x 1 x x
What happens if we rotate the points? x

59. 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

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

61. 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

62. 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

63. 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

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

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

66. Binary Differential Modulation
(i+1)st bit = 0:
0o phase shift
What is the drawback of differential modulation?
waveform for ith symbol
(i+1)st bit = 1:
180o phase shift

67. 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

68. 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).

69. Probability of Error

70. Probability of Error with Fading

71. 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

72. 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.

73. 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

74. 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.