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JU_SmallScaleFading(1)

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JU_SmallScaleFading(1)

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  1. EN253 205 (2/2562) Mobile Communication Characteristic of Radio Propagation Small Scale Fading Asst. Prof. Nararat Ruangchaijatupon Electrical Engineering Khon Kaen University Office: EN04325A, Email: nararat@kku.ac.th Small-Scale Fading • Or only “Fading” • Fluctuation of amplitude, phase, multipath delay over a short period of time • Caused by interference between 2 or more versions of transmitted signal arrived at slightly different time – Movement of mobile terminal or surround objects – Speed/Direction of movement respect to BS 2

  2. Effects of Multipath 1. Rapid changes in signal strength over a small travel distance or time interval 2. Random frequency modulation due to Doppler shifts on different multipaths 3. Time dispersion (echoes) caused by multipath propagation delays • Variation depends on signal intensity, relative propagation time, bandwidth of the transmitted signal 3 Factors on Small-Scale Fading • Multipath propagation – Cause fading, distortion, ISI • Speed of mobile – Speed/direction  Doppler shift • Speed of surrounding objects – If speed of surrounding objects is higher than speed of mobile terminal  Doppler shift • Transmission bandwidth of signal 4

  3. Doppler Shift • The shift in received signal frequency due to motion is called the Doppler shift Source: T. Rappaport, Wireless Communications: Principles and Practice 5 Doppler Shift (cont.) Difference in path length from S to mobile terminal  cos d l      v t cos Phase change in received signal due to difference path lengths 2 v l      v t  2     cos • a mobile moving at constant v (X to Y) during time t • S = signal source (very far away) Change in frequency  Doppler Shift 1 2    v       f cos d t 6

  4. Doppler Shift - Example Consider a transmitter which radiates a sinusoidal carrier frequency of 1850 MHz. For a vehicle moving 60 mph, compute the received carrier frequency if the mobile is moving (a) directly toward the transmitter, (b) directly away from the transmitter, and (c) in a direction which is perpendicular to the direction of the transmitted signal. Solution Given: Carrier frequency fc= 1850 MHz  8 c 3 10  Therefore, wavelength m     . 0 162 6 cf 1850 10 Vehicle speed v = 60 mph = 26.82 m/s 7 Doppler Shift – Example (cont.) (a) The vehicle is moving directly toward the transmitter. The Doppler shift in this case is positive. 26.82 0.162       6 f f f MHz 1850 10 1850.00016 c d (b) The vehicle is moving directly away from the transmitter. The Doppler shift in this case is negative. 26.82 1850 10 0.162       6 f f f MHz 1849.999834 c d (c) The vehicle is moving perpendicular to the angle of arrival of the transmitted signal. In this case, and there is no Doppler shift. The received signal frequency is the same as the transmitted frequency of 1850 Mhz.      90 ,cos 0, 8

  5. Parameters of Mobile Multipath Channels • Power delay profile • Time dispersion parameters – Mean excess delay – rms delay spread – Maximum excess delay or excess delay spread – Coherence bandwidth • Doppler spread – Coherence time 9 Power Delay Profile Plot of relative power as a function of excess delay with respect to a fixed time delay reference From 900 MHz cellular system in San Francisco 10 Source: T. Rappaport, Wireless Communications: Principles and Practice

  6. Power Delay Profile (cont.) Inside a grocery store at 4 GHz 11 Source: T. Rappaport, Wireless Communications: Principles and Practice Power Delay Profile (cont.) • Used to derive channel multipath parameters • Represented as plots of relative received power ( ) as a function of excess delay ( ) with respect to a fixed time delay reference 2 k a  Source: T. Rappaport, Wireless Communications: Principles and Practice 12

  7. Mean Excess Delay • Mean excess delay is the first moment of the power delay profile   k a     P   k    2 a P k k k k   k  k  k 2 k • These delays are measured relative to the first detectable signal arriving at the receiver at 0  0 13 RMS Delay Spread RMS delay spread is the square root of the second central moment of the power delay profile       2  2 where     P   k  k    2 2 2 a P k k k k    2  k  k 2 a k k 14

  8. Typical Measured Values of RMS Delay Spread 15 RMS Delay Spread - Example Solution 0 d B 0 d B P  ( )  (1)(0) (1)(1) 1 1  2 (1)(0) 1 1    0.5 (0.5)  1 2      s 0.5  2 (1)(1) 1 2      2 2 s 0.5 1 s  0 2 (a) Compute the RMS delay spread (b) If BPSK modulation is used, what is the maximum bit rate without the need of an equalizer?            2 2 0.5 s 0.25   0.1 s T  1 . 0  s 5 . 0     5 TS s 1 . 0 1 T   0.2 10   6 R sps ksps 200 s s  R kbps 200 b 16

  9. Maximum Excess Delay • Time delay during which multipath energy falls to X dB below the maximum Depends on Noise Threshold Example of an indoor power delay profile 17 Source: T. Rappaport, Wireless Communications: Principles and Practice Coherence Bandwidth (1) • A statistical measure of the range of frequencies over which the channel can be considered “flat”, i.e., a channel which passes all spectral components with approximately equal gain and linear phase • Derive from rms delay spread • 2 sinusoids with frequency separation greater than Bcare affected differently by the channel 18

  10. Coherence Bandwidth (2) • If transmission bandwidth of signal is greater than coherence bandwidth of the channel, the signal will be distorted • If less than coherence BW, the amplitude will change rapidly, but will not be distorted in time and fading can be ignored • Coherence BW is a measurement of the maximum frequency difference for which signals are still strongly correlated in amplitude 19 Coherence Bandwidth (3) • If the frequency correlation function is above 0.9  1 c B  50  • If the frequency correlation function is above 0.5 c B 1   5  20

  11. Coherence Bandwidth - Example Calculate the mean excess delay, rms delay spread, and the maximum excess delay (10 dB). Estimate the 50% coherence bandwidth of the channel. Would this channel be suitable for AMPS or GSM service without the use of an equalizer? 21 Coherence Bandwidth – Example (cont.) From the figure, the maximum excess delay is 5 us The mean excess delay (1)(5) (0.1)(1) [0.01 0.1 0.1 1]  2 2 2 (1)(5) (0.1)(1) 1.21     (0.1)(2)  (0.1)(2) (0.01)(0)     s 4.38  2    (0.01)(0)     2 s 21.07  21.07 (4.38)    2 The rms delay spread The coherence bandwidth (50%) 1.37 s  1  1    c B kHz 146  s 5 5(1.37 )  Bc> 30 kHz, AMPS will work without an equalizer but GSM (requires 200 kHz) will need an equalizer 22

  12. Doppler Spread • Doppler spread (BD) is the measure of spectral broadening caused by the time rate of change of the mobile radio channel • The range of frequencies which the received Doppler spectrum is non-zero • Doppler spectrum: to • If the transmitted signal bandwidth is much greater than BD, the effects of Doppler spread are negligible at the receiver  f f  f f c d c d 23 Coherence Time • Coherence time (Tc) is the time domain dual of Doppler spread • A statistical measure of the time duration over which the channel impulse response is invariant, quantifies the similarity of the channel response at different time • Used to characterize the time varying nature of the frequency dispersiveness  1 f T c m 24

  13. Coherence Time (cont.) • If the time correlation function of the channel is above 0.5, then T 16 9  . 0 179   c f f m m mf  v  mf • • In modern digital communication, a popular rule is that 9 16  is the maximum Doppler shift, 0.423 f   T c 2 f m m • Two signals arriving at a time separation greater than Tc(symbol rate less than Tc) are affected differently by the channel 25 Coherence Time - Example • Determine the proper spatial sampling interval required to make small-scale propagation measurements which assume that consecutive samples are highly correlated in time. How many samples will be required over 10 m travel distance if fc= 1900 MHz and v = 50 m/s. How long would it take to make these measurements, assuming they could be made in real time from a moving vehicle? What is the Doppler spread for the channel? Solution Time between samples is Tc/2 9 9 9 16 16 16 16 3.14 50 1900 10 m c f v vf     9 3 10    8 c       T s 565 c    6 26

  14. Coherence Time – Example (cont.)  Taking time samples at less than half Tc, at 282.5 s, corresponds to a spatial sampling interval of 50 565 2 2   vT s      c x m cm 0.014125 1.41 Therefore, the number of samples required over a 10 m travel distance is 10 10 0.014125 x     N samples 708 x m m s 10 50 The time taken to make this measurement equals to s 0.2   6 vf c 50 1900 10 3 10      The Doppler spread is B f Hz c 316.66 D m 8 27 Question & Discussion Assignment

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