1 Small-scale Mobile radio propagation Small-scale Mobile radio propagation l Small scale propagation implies signal quality in a short distance or time range l In this small range, fading or rapid fluctuation of the signal amplitude is observed l One cause of fading is multipath or the process of signals reaching the receiver through different mechanisms such as LOS, reflection, diffration and scattering. l Small scale propagation implies signal quality in a short distance or time range l In this small range, fading or rapid fluctuation of the signal amplitude is observed l One cause of fading is multipath or the process of signals reaching the receiver through different mechanisms such as LOS, reflection, diffration and scattering.

2 Multipath effects l Rapid changes in signal amplitude over a small distance or time interval. l Rapid changes in signal phaseover a small distance or time interval. l Time dispersion (echoes) caused by multipath propagation delay. l Rapid changes in signal amplitude over a small distance or time interval. l Rapid changes in signal phaseover a small distance or time interval. l Time dispersion (echoes) caused by multipath propagation delay.

3 Causes of fading l In urban areas, fading occurs because the height of mobile is lesser than the height of surrounding structures, such as buildings and trees. l Existence of several propagation paths between transmitter and receiver. l In urban areas, fading occurs because the height of mobile is lesser than the height of surrounding structures, such as buildings and trees. l Existence of several propagation paths between transmitter and receiver.

4 Factors influencing small signal fading l Multipath propagations l Speed of mobile (Doppler shift) l Speed of surrounding objects l Bandwidths of signal and channel l Multipath propagations l Speed of mobile (Doppler shift) l Speed of surrounding objects l Bandwidths of signal and channel

5 Analysis of multipath channel ReceiverReceiver dd Spatial position TransmitterTransmitter

6 Convolution model for multipath signal Received signal: y(t) = A 0 x(t) + A 1 x(t -  1 ) + A 2 x(t -  2 ) +... Received signal: y(t) = A 0 x(t) + A 1 x(t -  1 ) + A 2 x(t -  2 ) +... T R, y(t) A 2 x(t-  2 ) A 1 x(t-  1 ), x(t), x(t) LOS LOS

7 System definition of multipath h(t) x(t)y(t)

8 Baseband signal definition j 2  f c t x(t) = Re { c(t) e } j 2  f c t x(t) = Re { c(t) e } l Transmitted signal c(t)- pulse

9 Received signal and system response j2  f c t y(t) = Re {r(t) e } j2  f c t j2  f c t h(t) = Re {h b (t) e } j2  f c t y(t) = Re {r(t) e } j2  f c t j2  f c t h(t) = Re {h b (t) e } l Received signal l Impulse response

10 Base band equivalent channel h b (t) c(t)r(t)

11 r(t) = c(t) * h b (t,  ) Modeling of the baseband multipath model h b (t,  ) t3t3t3t3 t3t3t3t3 t2t2t2t2 t2t2t2t2 t1t1t1t1 t1t1t1t1 t0t0t0t0 t0t0t0t0   N-2  N-1  o  1  2 o  1  2 o  1  2 o  1  2  o  1  2 o  1  2 o  1  2 o  1  2 Mathematicalmodel Mathematical model

12 Excess delay concept The delay axis ,  o <=  <=  n-1 is divided into equal time delay segments called excess delay bins. The delay axis ,  o <=  <=  n-1 is divided into equal time delay segments called excess delay bins.  0 = 0  1 =    1 =    2 = 2    2 = 2    N-1 = (N-1)    N-1 = (N-1)   The delay axis ,  o <=  <=  n-1 is divided into equal time delay segments called excess delay bins. The delay axis ,  o <=  <=  n-1 is divided into equal time delay segments called excess delay bins.  0 = 0  1 =    1 =    2 = 2    2 = 2    N-1 = (N-1)    N-1 = (N-1)  

13 Delay component design All multipath signals received within the bins are represented by a single resolvable multipath component having delay  i. All multipath signals received within the bins are represented by a single resolvable multipath component having delay  i. Design equation for bin width  Design equation for bin width  Bandwidth of signal = 2/  All multipath signals received within the bins are represented by a single resolvable multipath component having delay  i. All multipath signals received within the bins are represented by a single resolvable multipath component having delay  i. Design equation for bin width  Design equation for bin width  Bandwidth of signal = 2/ 

14 Final model for multipath response N-1 j  i N-1 j  i r(t) =  a i e c[t –  i ] i = 0  i = 0  l c(t) – Transmitted pulse l r(t) – Received pulse l N – Number of multipaths l a i – Amplitude of multipath i  i – Amplitude of multipath i  i – Amplitude of multipath i  i – Amplitude of multipath i  i – Amplitude of multipath i N-1 j  i N-1 j  i r(t) =  a i e c[t –  i ] i = 0  i = 0  l c(t) – Transmitted pulse l r(t) – Received pulse l N – Number of multipaths l a i – Amplitude of multipath i  i – Amplitude of multipath i  i – Amplitude of multipath i  i – Amplitude of multipath i  i – Amplitude of multipath i

15 Wideband multipath signals N-1 j  i Received signal r(t) =  a i e p[t–  i ] i = 0 i = 0 Instantaneous received power: N-1 N-1 |r(t)| 2 =  |a k | 2 |r(t)| 2 =  |a k | 2 k = 0 k = 0 =>Total received power = sum of the power of individual multipath components. N-1 j  i Received signal r(t) =  a i e p[t–  i ] i = 0 i = 0 Instantaneous received power: N-1 N-1 |r(t)| 2 =  |a k | 2 |r(t)| 2 =  |a k | 2 k = 0 k = 0 =>Total received power = sum of the power of individual multipath components.

16 Average wideband received power N-1 N-1 E a,  [P wB ] = E a,  [  |a i exp j  i | 2 ] i = 0 i = 0 N-1 _ N-1 _ =  a i 2 =  a i 2 i = 0 i = 0 E a,  =average power _ a i 2 = sample average signalusing multipath measurement equipment. N-1 N-1 E a,  [P wB ] = E a,  [  |a i exp j  i | 2 ] i = 0 i = 0 N-1 _ N-1 _ =  a i 2 =  a i 2 i = 0 i = 0 E a,  =average power _ a i 2 = sample average signalusing multipath measurement equipment.

17 Narrowband multipath signals N-1 j  i N-1 j  i Received signal: r(t) =  a i e p[t–  i ] i = 0 i = 0 Instantaneous received power: N-1 j  i N-1 j  i Received signal: r(t) =  a i e p[t–  i ] i = 0 i = 0 Instantaneous received power: N-1 j  i |r(t)| 2 = |  a i e | 2 i = 0 i = 0 N-1 j  i |r(t)| 2 = |  a i e | 2 i = 0 i = 0

18 Average narrowband received power N-1 j  i (t,  ) E a,  [P wB ]= E a,  [ |  a i e | 2 ] E a,  [P wB ]= E a,  [ |  a i e | 2 ] i = 0 i = 0 N-1 j  i (t,  ) E a,  [P wB ]= E a,  [ |  a i e | 2 ] E a,  [P wB ]= E a,  [ |  a i e | 2 ] i = 0 i = 0

19 ConclusionsConclusions l When the transmitted signal has  a wide bandwidth >> bandwidth of the channel multipath structure is completely resolved by the receiver at any time and the received power varies very little. l When the transmitted signal has a very narrow bandwidth (example the base band signal has a duration greater than the excess delay of the channel) then multipath is not resolved by the received signal and large signal fluctuations occur (fading). l When the transmitted signal has  a wide bandwidth >> bandwidth of the channel multipath structure is completely resolved by the receiver at any time and the received power varies very little. l When the transmitted signal has a very narrow bandwidth (example the base band signal has a duration greater than the excess delay of the channel) then multipath is not resolved by the received signal and large signal fluctuations occur (fading).

20 ExampleExample Assume a discrete channel impulse response is used to model urban radio channels with excess delays as large as 100  s and microcellular channels with excess delays not larger than 4  s. If the number of multipath bins is fixed at 64 find: (a)   (b)Maximum bandwidth, which the two models can accurately represent. Assume a discrete channel impulse response is used to model urban radio channels with excess delays as large as 100  s and microcellular channels with excess delays not larger than 4  s. If the number of multipath bins is fixed at 64 find: (a)   (b)Maximum bandwidth, which the two models can accurately represent.

21 SolutionSolution For urban radio channel Maximum excess delay of channel  N = N   = 100  s. N = 64  =  N /N = 100  s /64 =  s Maximum bandwidth represented accurately by model = 2/  = 1.28 MHz = 2/  = 1.28 MHz For urban radio channel Maximum excess delay of channel  N = N   = 100  s. N = 64  =  N /N = 100  s /64 =  s Maximum bandwidth represented accurately by model = 2/  = 1.28 MHz = 2/  = 1.28 MHz

22 For microcellular channel Maximum excess delay of channel  N = N   = 4  s. N = 64   =  N /N = 4  s /64 = 62.5 ns Maximum bandwidth represented accurately by model = 2/   = 32 MHz Maximum excess delay of channel  N = N   = 4  s. N = 64   =  N /N = 4  s /64 = 62.5 ns Maximum bandwidth represented accurately by model = 2/   = 32 MHz

23 ExampleExample Assume a mobile traveling at a velocity of 10m/s receives two multipath components at a carrier frequency of 1000 MHz. The first component is assumed to arrive at  = 0 with an initial phase of 0  and a power of –70dBm. The first component is assumed to arrive at  = 0 with an initial phase of 0  and a power of –70dBm. The second component is 3dB weaker than the first one and arrives at  = 1  s, also with the initial phase of 0 . The second component is 3dB weaker than the first one and arrives at  = 1  s, also with the initial phase of 0 . Assume a mobile traveling at a velocity of 10m/s receives two multipath components at a carrier frequency of 1000 MHz. The first component is assumed to arrive at  = 0 with an initial phase of 0  and a power of –70dBm. The first component is assumed to arrive at  = 0 with an initial phase of 0  and a power of –70dBm. The second component is 3dB weaker than the first one and arrives at  = 1  s, also with the initial phase of 0 . The second component is 3dB weaker than the first one and arrives at  = 1  s, also with the initial phase of 0 .

24... Example If the mobile moves directly in the direction of arrival of the first component and directly away from the direction of arrival of the second component, compute the following: (a) The narrow band and wide band received power over the interval 0-0.5s (b) The average narrow band received power. If the mobile moves directly in the direction of arrival of the first component and directly away from the direction of arrival of the second component, compute the following: (a) The narrow band and wide band received power over the interval 0-0.5s (b) The average narrow band received power.

25 Solution (a) Narrow band instantaneous power N-1 j  i (t,  ) N-1 j  i (t,  ) |r(t)| 2 = |  a i e | 2 i = 0 i = 0 Now –70dBm => 100 pw so a 1 = √ 100 pw and –73dBm => 50 pw so a 2 = √ 50 pw  i = 2  d/ = 2  vt/  i = 2  d/ = 2  vt/ = (3*10 8 )/(100*10 6 ) = 0.3 m = (3*10 8 )/(100*10 6 ) = 0.3 m  1 = 2  *10*t/0.3 = t rad. N-1 j  i (t,  ) N-1 j  i (t,  ) |r(t)| 2 = |  a i e | 2 i = 0 i = 0 Now –70dBm => 100 pw so a 1 = √ 100 pw and –73dBm => 50 pw so a 2 = √ 50 pw  i = 2  d/ = 2  vt/  i = 2  d/ = 2  vt/ = (3*10 8 )/(100*10 6 ) = 0.3 m = (3*10 8 )/(100*10 6 ) = 0.3 m  1 = 2  *10*t/0.3 = t rad.

26  2 = -  1 = t rad. t = 0 |r(t)| 2  = | √100  + √50 | 2 = 291pw t = 0.1 |r(t)| 2 = |√100 e j209.4 x √50 e -j209.4 x 0.1 | 2 = 78.2pw = 78.2pw t = 0.2 |r(t)| 2 = |√100 e j209.4 x √50 e -j209.4 x 0.2 | 2 = 81.5pw = 81.5pw  2 = -  1 = t rad. t = 0 |r(t)| 2  = | √100  + √50 | 2 = 291pw t = 0.1 |r(t)| 2 = |√100 e j209.4 x √50 e -j209.4 x 0.1 | 2 = 78.2pw = 78.2pw t = 0.2 |r(t)| 2 = |√100 e j209.4 x √50 e -j209.4 x 0.2 | 2 = 81.5pw = 81.5pw

27 t = 0.3 |r(t)| 2 = 291pw t = 0.4 |r(t)| 2 = 78.2pw t = 0.5 |r(t)| 2 = 81.5pw t = 0.3 |r(t)| 2 = 291pw t = 0.4 |r(t)| 2 = 78.2pw t = 0.5 |r(t)| 2 = 81.5pw

28 Wideband instantaneous power N-1 N-1 |r(t)| 2 =  |a k | 2 = = 150 pW k = 0 k = 0 N-1 N-1 |r(t)| 2 =  |a k | 2 = = 150 pW k = 0 k = 0

29 (b) Average narrow band received power E a,  [P CW ] = [2(291) + 2(78.2) +2(81.5)] /6 = pw = pw l The average narrow band power and wideband power are almost the same over 0.5s. l While the narrow band signal fades over the observation interval, the wideband signal remains constant. E a,  [P CW ] = [2(291) + 2(78.2) +2(81.5)] /6 = pw = pw l The average narrow band power and wideband power are almost the same over 0.5s. l While the narrow band signal fades over the observation interval, the wideband signal remains constant.

30 Small-scale multipath measurements l Direct Pulse Measurements l Spread Spectrum Sliding Correlator Measurement l Swept Frequency Measurement l Direct Pulse Measurements l Spread Spectrum Sliding Correlator Measurement l Swept Frequency Measurement

31 Types of Small Scale Fading Multipath time delay Doppler Spread Flat fading Frequency Selective Fading Frequency Selective Fading FastFadingFastFading Slow fading

32 Mechanisms that cause fading l 2 main propagation mechanisms: l Multipath time delay spread l Doppler spread l These two mechanisms are independent of each other. l 2 main propagation mechanisms: l Multipath time delay spread l Doppler spread l These two mechanisms are independent of each other.

33 Multipath terms associated with fading T s = Symbol period or reciprocal bandwidth B s = Bandwidth of transmitted signal B c = Coherence bandwidth of channel B c = 1/(50   )  where   is rms delay spread T s = Symbol period or reciprocal bandwidth B s = Bandwidth of transmitted signal B c = Coherence bandwidth of channel B c = 1/(50   )  where   is rms delay spread

34 Calculation of Delay Spread __ _ __ _   2 =  2 - (  ) 2 Where:_  = (  a k 2   ) / (  a k 2 ) __  2 = (  a k 2   2 ) / (  a k 2 ) __ _ __ _   2 =  2 - (  ) 2 Where:_  = (  a k 2   ) / (  a k 2 ) __  2 = (  a k 2   2 ) / (  a k 2 )

35 Fading effects due to Doppler spread f c = frequency incident signal Received signal spectrum = f c +/- f d f d = Doppler shift f c = frequency incident signal Received signal spectrum = f c +/- f d f d = Doppler shift fcfcfcfc V

36 Doppler spread and coherence time Doppler frequency shift : f d = (v / ) cos , Wavelength = c / f c Maximum Frequency deviation = f m = v / Maximum Frequency deviation = f m = v / Doppler Spread B D = f m Coherence time = T c = / f m Doppler frequency shift : f d = (v / ) cos , Wavelength = c / f c Maximum Frequency deviation = f m = v / Maximum Frequency deviation = f m = v / Doppler Spread B D = f m Coherence time = T c = / f m

37 Mathematical estimation of fading Flat fading l Mobile channel has constant gain and linear phase response. l Spectral characteristics of the transmitted signal are maintained at receiver l Condition: B s << B c => T s >>   => T s >>   Flat fading l Mobile channel has constant gain and linear phase response. l Spectral characteristics of the transmitted signal are maintained at receiver l Condition: B s << B c => T s >>   => T s >>  

38 Frequency selective fading l Mobile channel has a constant gain and linear phase response over a finite bandwidth l Condition: B s > B c => T s T s <   Frequency selective fading l Mobile channel has a constant gain and linear phase response over a finite bandwidth l Condition: B s > B c => T s T s <  

39 Flat fading or frequency selective fading? Common rule of thumb l If Ts ≥ 10   => Flat fading l If Ts Frequency selective fading Common rule of thumb l If Ts ≥ 10   => Flat fading l If Ts Frequency selective fading

40 Fast fading channel l The channel impulse response changes rapidly within the symbol duration. l This causes frequency dispersion due to Doppler spreading, which leads to signal distortion. l Condition: T s > T c T s > T c B s < B D l The channel impulse response changes rapidly within the symbol duration. l This causes frequency dispersion due to Doppler spreading, which leads to signal distortion. l Condition: T s > T c T s > T c B s < B D

41 Slow fading channel l The channel impulse response changes at a much slower rate than the transmitted signal l Velocity of mobile (or velocity of objects in channel) l Condition: T s << T c B s >> B D l The channel impulse response changes at a much slower rate than the transmitted signal l Velocity of mobile (or velocity of objects in channel) l Condition: T s << T c B s >> B D

42 Rayleigh and Ricean distributions l In mobile radio channels, the Rayleigh distribution is commonly used to describe the statistical time varying nature of the received fading signal l When there is a dominant (non-fading) signal component present such as LOS propagation path, the small scale fading envelope distribution is Ricean l In mobile radio channels, the Rayleigh distribution is commonly used to describe the statistical time varying nature of the received fading signal l When there is a dominant (non-fading) signal component present such as LOS propagation path, the small scale fading envelope distribution is Ricean

43 Statistical Models l Probability density function: p(r) = (r/  2 ) e –(r 2 + A 2 ) I o (Ar/  2 ), for r  0 (2  2 ) (2  2 ) p(r)= 0 for r < 0 I o  Modified Bessel function r  Received fading signal voltage A  LOS amplitude (A=0 for Rician)  2  Variance of fading signal l Probability density function: p(r) = (r/  2 ) e –(r 2 + A 2 ) I o (Ar/  2 ), for r  0 (2  2 ) (2  2 ) p(r)= 0 for r < 0 I o  Modified Bessel function r  Received fading signal voltage A  LOS amplitude (A=0 for Rician)  2  Variance of fading signal

44 Level crossing and fading statistics l Level crossing rate (LCR) is the rate at which the normalized Rayleigh fading envelope crosses a specified level in a positive going direction. LCR = N R =  (2  ) f m  e -  2 f m = Maximum Doppler frequency  = R/R ms = specified level R, normalized to the rms value of fading signal l Level crossing rate (LCR) is the rate at which the normalized Rayleigh fading envelope crosses a specified level in a positive going direction. LCR = N R =  (2  ) f m  e -  2 f m = Maximum Doppler frequency  = R/R ms = specified level R, normalized to the rms value of fading signal

45 Average fade duration l Average period of time for which the received signal is below a specified level R. __ __   = e  2 – 1 ________ ________  f m  2   f m  2  __ __   = e  2 – 1 ________ ________  f m  2   f m  2 

46 ExampleExample (a) For a Rayleigh fading signal, compute the positive going level crossing rate for  = 1, when the maximum Doppler frequency (f m ) is 20 Hz. (b) What is the maximum velocity of the mobile for this Doppler frequency if the carrier is 900 MHz?

47 Solution  = 1 f m = 20 Hz The number of zero level crossings is: N R =  2  (20) e -1 N R =  2  (20) e -1 = Crossings/Sec = Crossings/Sec Maximum velocity of mobile = f d = 20 (3 X 10 8 )/(900X10 6 ) = 6.66 m/s

48 ExampleExample Find the average fade duration for threshold level  = 0.01,  = 0.1 and  = 1, when the Doppler frequency is 20 Hz.

49 SolutionSolution   = e  2 – 1  f m  2   f m  2   s  s ms   = e  2 – 1  f m  2   f m  2   s  s ms

50 Statistical methods for fading channels l Clark’s model for Flat Fading l Two-Ray Rayleigh Fading Model l Saleh and Valenzuela Indoor statistical Model l SIRCIM (Simulation of Indoor Radio Channels Impulse Response Models) l SMRCIM (Simulation of Mobile Radio Channel Impulse-Response Models) l Clark’s model for Flat Fading l Two-Ray Rayleigh Fading Model l Saleh and Valenzuela Indoor statistical Model l SIRCIM (Simulation of Indoor Radio Channels Impulse Response Models) l SMRCIM (Simulation of Mobile Radio Channel Impulse-Response Models)