MITP 413: Wireless Technologies Week 4

1. MITP 413: Wireless TechnologiesWeek 4 Michael L. Honig Department of EECS Northwestern University January 2007

2. Channel Characterizations:Time vs. Frequency • Frequency-domain description • Time-domain description Multipath channel Amplitude attenuation, Delay (phase shift) input s(t) is a sinusoid “narrowband” signal r(t) s(t) Multipath channel time t time t multipath components input s(t) is an impulse (very short pulse) “wideband” signal (Note: an impulse has zero duration and infinite bandwidth!)

3. Two-Ray Impulse Response reflection (path 2) direct path (path 1) reflection is attenuated s(t) r(t)  time t time t = [(length of path 2) – (length of path 1)]/c

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

5. Bandwidth and Multipath Resolution reflection (path 2) direct path (path 1) multipath components are resolvable (delay spread) signal pulse signal pulse  T >  T <  T Wide bandwidth  high resolution Receiver can clearly distinguish two paths. Narrow bandwidth low resolution Receiver cannot distinguish the two paths.

6. Bandwidth and Multipath Resolution reflection (path 2) direct path (path 1) multipath components are resolvable signal pulse The receiver can easily distinguish the two paths provided that they are separated by much more than the pulse width T. Since the signal bandwidth B ≈ 1/T, this implies B >> 1/, or B >> Bc . .  Wide bandwidth  high resolution Receiver can clearly distinguish two paths.

7. Multipath Resolution and Diversity reflection (path 2) direct path (path 1) multipath components are resolvable Each path may undergo independent fading (i.e., due to Doppler). If one path is faded, the receiver may be able to detect the other path. In the frequency domain, this corresponds to independent fading in different coherence bands. signal pulse  Wide bandwidth  high resolution Receiver can clearly distinguish two paths.

8. Bandwidth and Geolocation reflection delay  = 2 x distance/c delay  s(t) s(t) r(t) r(t) time t Narrow bandwidth pulse time t High bandwidth pulse

9. Bandwidth and Geolocation reflection delay  = 2 x distance/c s(t) The resolution of the delay measurement is roughly the width of the pulse. Low bandwidth  wide pulse  low resolution High bandwidth  narrow pulse  high resolution r(t) time t Ex: If the delay measurement changes by 1 microsec, the distance error Is c x 10-6 = 300 meters!

10. Power-Delay Profile Received power vs. time in response to a transmitted short pulse. delay spread  For cellular systems (outdoors), the delay spread is typically a few microseconds.

11. Two-Ray Impulse Response reflection (path 2) direct path (path 1) reflection is attenuated s(t) r(t)  time t time t

12. Urban Multipath s(t) r(t) time t time t r(t) different location for receiver time t Spacing and attenuation of multipath components depend on location and environment.

13. Delay Spread and Intersymbol Interference s(t) r(t) Multipath channel time t time t Time between pulses is >> delay spread, therefore the received pulses do not interfere. r(t) s(t) Multipath channel time t Time between pulses is < delay spread, which causes intersymbol interference. The rate at which symbols can be transmitted without intersymbol interference is 1 / delay spread.

14. Coherence Bandwidth coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 The channel gain is approximately constant within a coherence bandwidth Bc. Frequencies f1 and f2 fade independently if | f1 – f2 | >> Bc. If the signal bandwidth < coherence bandwidth Bc, then the channel is called flat fading, and the transmitted signal is regarded as narrowband. If the signal bandwidth > Bc, then the channel is called frequency-selective and the signal is regarded as wideband.

15. Coherence Bandwidth and Diversity signal power (wideband) coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 Frequency-selective fading: different parts of the signal (in frequency) are affected differently by fading.

16. Coherence Bandwidth and Diversity signal power (wideband) coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 Frequency-selective fading: different parts of the signal (in frequency) are affected differently by fading. Wideband signals exploit frequency diversity. Spreading power across many coherence bands reduces the chances of severe fading. Wideband signals are distorted by the channel fading (distortion causes Intersymbol interference).

17. Narrowband Signal signal power (narrowband) coherence bandwidth Bc channel gain Frequencies far outside the coherence bandwidth are affected differently by multipath. frequency f1 f2 Flat fading: the narrowband signal fades uniformly, hence does not benefit from frequency diversity. For the cellular band, Bc is around 100 to 300 kHz. How does this compare with the bandwidth of cellular systems?

18. channel gain coherence bandwidth Bc frequency Coherence Bandwidth and Delay Spread delay spread  channel gain delay spread  coherence bandwidth Bc frequency Coherence bandwidth is inversely proportional to delay spread: Bc≈ 1/.

19. Fading Experienced by Wireless Systems Standard Flat/Freq.-Sel. Fast/Slow AMPS Flat Fast IS-136 Flat Fast GSM F-S Slow IS-95 (CDMA) F-S Fast 3G F-S Slow to Fast (depends on rate) 802.11 F-S Slow Bluetooth F-S Slow

20. Propagation and Handoff Received Signal Strength (RSS) from right BST from left BST unacceptable (call is dropped) time

21. Propagation and Handoff Received Signal Strength (RSS) from right BST with handoff handoff threshold from left BST unacceptable (call is dropped) time

22. Propagation and Handoff Received Signal Strength (RSS) from right BST with handoff handoff threshold from left BST RSS margin unacceptable (call is dropped) time time needed for handoff

23. Propagation and Handoff Received Signal Strength (RSS) from right BST handoff threshold from left BST RSS margin unacceptable (call is dropped) time time needed for handoff

24. Handoff Threshold • Handoff threshold too high  too many handoffs (ping pong) • Handoff threshold too low  dropped calls are likely • Threshold should depend on slope on vehicle speed (Doppler). Received Signal Strength (RSS) from right BST handoff threshold from left BST RSS margin unacceptable (call is dropped) time time needed for handoff

25. Handoff Measurements (3G) • Mobile maintains a list of neighbor cells to monitor. • Mobile periodically measures signal strength from BST pilot signals. • Mobile sends measurements to network to request handoff. • Handoff decision is made by network. • Depends on available resources (e.g., channels/time slots/codes). Handoffs take priority over new requests (why?). • Hysteresis needed to avoid handoffs due to rapid variations in signal strength.

26. Handoff Decision • Depends on RSS, time to execute handoff, hysteresis, and dwell (duration of RSS) • Proprietary methods • Handoff may also be initiated for balancing traffic. • 1G (AMPS): Network Controlled Handoff (NCHO) • Handoff is based on measurements at BS, supervised by MSC. • 2G, GPRS: Mobile Assisted Handoff (MAHO) • Handoff relies on measurements at mobile • Enables faster handoff • Mobile data, WLANs (802.11): Mobile Controlled Handoff (MCHO) • Handoff controlled by mobile

27. Soft Handoff (CDMA) ”Make before break” DURING AFTER BEFORE MSC MSC MSC BSC BSC BSC BSC BSC BSC Hard Handoff (TDMA) MSC MSC MSC BSC BSC BSC BSC BSC BSC

28. SINR Measurements: 1xEV-DO drive test plots

29. 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)!

30. Why Digital Communications? • Can combine with sophisticated signal processing (voice compression) and error protection. • Greater immunity to noise/channel impairments. • Can multiplex different traffic (voice, data, video). • Security through digital encryption. • Flexible design possible (software radio). 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)! VLSI + special purpose digital signal processing  digital is more cost-effective than analog!

31. Binary Frequency-Shift Keying (FSK) Bits: 10110

32. Quadrature Phase Shift Keying (QPSK) Bits: 0001 10 11

33. Binary Phase Shift Keying (BPSK) Bits: 101 10 Baseband signal

34. Amplitude Shift Keying (4-Level ASK) Bits: 0001 10 11 Baseband signal symbol duration

35. Baseband  RF Conversion Passband (RF) signal Baseband signal sin 2fct time X T fc is the carrier frequency Power Power signal bandwidth is roughly 1/T frequency frequency fc 0  0

36. Why Modulate?

37. Why Modulate? • The baseband spectrum is centered around f=0. Without modulation all signals would occupy low frequencies and interfere with each other. • It is difficult to build effective antennas at low frequencies since the dimension should be on the order of a wavelength. • Low frequencies propagate further, causing more interference.

38. Selection Criteria How do we decide on which modulation technique to use?

39. 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.

40. Example: Binary vs. 4-Level ASK 3A A A -A -A -3A 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?

41. Noisy Baseband Signals 3A A A -A -A -3A 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?

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

43. How to Increase Bandwidth Efficiency?

44. 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.

45. 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.)

46. 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

47. 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.

48. Shannon Capacity 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

49. 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.

50. 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?