1. MITP 413: Wireless Technologies Week 5
Michael L. Honig
Department of ECE
Northwestern University
April 2004

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

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

4. Quadrature Phase Shift Keying (QPSK)
Bits:

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

6. Probability of Error
4-ASK
BPSK
7 dB

7. 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

8. 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.

9. 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. 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

12. TYPES ALL PSK
(PSK Type Summary Slide)

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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.

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

26. Minimum Bandwidth (Nyquist) Pulse Shape

27. Shifted Nyquist Pulses
Bits:

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

29. Passband Signal with Different Carrier Frequencies

30. Raised Cosine Pulses
Minimum BW
frequency
time

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

32. Blocks Transition
(Transitional Slide)

33. 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...

34. 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

35. 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

36. 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

37. 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.

38. 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.

39. 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.

40. 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.

41. 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.

42. 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)

43. 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)

44. 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

45. 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

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

47. Probability of Error

48. Probability of Error with Fading

49. 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