Mobile Communications

Transcription

1 Mobile Communications Wen-Shen Wuen Trans. Wireless Technology Laboratory National Chiao Tung University WS Wuen Mobile Communications 1 Outline Outline 1 Small-Scale Multipath Propagation 2 Impulse Response Model for a Multipath Channel 3 Small-Scale Multipath Measurements 4 Parameters of Mobile Multipath Channels 5 Types of Small-Scale Fading 6 Rayleigh and Ricean Distributions 7 Statistical Models for Multipath Fading Channels WS Wuen Mobile Communications 2

2 Introduction Small-Scale Multipath Propagation Small-Scale Fading rapid fluctuations of the amplitudes, phases or multipath delays of a radio signal over a short period of time or travel distance from interference between two or more versions of TX signal which arrive at the RX at slightly different times. Small-Scale Fading Effects Rapid changes in signal strength over a small travel distance or time interval Random frequency modulation due to varying Doppler shifts on different multipath signals Time dispersion (echoes) caused by multipath propagation delays WS Wuen Mobile Communications 4 Small-Scale Multipath Propagation Factors Influencing Small-Scale Fading Multipath propagation Constantly changing environment (reflecting and scattering objects) Random phase and amplitude of the different multipath components signal fading and/or distortion Lengthens the time required for the baseband portion of the signal to reach the RX intersymbol interference Speed of mobile terminals Relative motion between BS and MS different Doppler shifts random frequency modulation WS Wuen Mobile Communications 5

3 Small-Scale Multipath Propagation Factors Influencing Small-Scale Fading, cont d Speed of surrounding objects Moving objects in the radio channel time varying Doppler shifts on multipath components Coherence time defines the staticness of the channel and is directly impacted by the Doppler shift. Transmission bandwidth of the signal Signal bandwidth > channel bandwidth distortion in RX signal Coherence bandwidth: a measure of the maximum frequency difference for which signals are still strongly correlated in amplitude. WS Wuen Mobile Communications 6 Doppler Shift Small-Scale Multipath Propagation Phase change due to difference in path lengths φ = 2π l λ = 2πv t λ cos θ (1) Change in frequency, Doppler shift, f d f d = 1 2π φ t = v cosθ (2) λ Mobile is moving toward (away from) the transmitter, Doppler shift is positive (negative), the apparent received frequency is increased (decreased) WS Wuen Mobile Communications 7

4 Small-Scale Multipath Propagation Example 1 consider a TX which radiates a sinusoidal carrier frequency of 1850 MHz. Fro a vehicle moving 60 mph, compute the received carrier frequency if the mobile is moving (a) directly toward the TX (b) directly away from the TX and (c) in a direction which is perpendicular to the direction of arrival of the TX signal. Solution: f c = 1850 MHz, λ = c/f c = /( ) = m speed v = 60 mph = m/s (a) toward TX: f = f c + f d = /0.162 = MHz (b) away from TX: f = f c f d = /0.162 = MHz (c) perpendicular to the direction of arrival of the TX signal: θ = 90 cosθ = 0 no Doppler shift. The received signal frequency is the same as the TX frequency of 1850 MHz. WS Wuen Mobile Communications 8 Impulse Response Model for a Multipath Channel Review: Linear Time-Invariant (LTI) System A System S maps an input signal x(t) in to an output signal y(t): y(t) = S{x(t)} (3) A System S is called linear, if: { } S a n x n (t) = a n S{x n (t)} = n n n a n y n (t) (4) A System S is called time-invariant, if: S{x(t t 0 )} = y(t t 0 ) (5) WS Wuen Mobile Communications 10

5 Impulse Response Model for a Multipath Channel Impulse Response of a LTI System Input x(t) LTI System h(t) Output y(t) Frequency domain: where y(t) = h(τ)x(t τ)dτ = x(t) h(t) (6) Y (f ) = H(f )X(f ) (7) H(f ) = X(f ) = Y (f ) = h(τ)e j2πf τ dτ (8) x(τ)e j2πf τ dτ (9) y(τ)e j2πf τ dτ (10) WS Wuen Mobile Communications 11 Impulse Response Model for a Multipath Channel Fourier Transform of Bandpass Signals Transmitted s(t) Re [s ] l (t)e j2πf ct Channel h(t) Received r(t) Re [r ] l (t)e j2πf ct Fourier transform of transmitted signal s(t): S(f ) = s(t)e j2πft dt = Re [ s l (t)e j2πf ct ] e j2πft dt = 1 ( ) s l (t)e j2πfct + s l 2 (t)e j2πf ct e j2πft dt = 1 ( Sl (f f c ) + S l 2 ( f f c) ) (11) Fourier transform of received r(t): R(f ) = 1 2 ( Rl (f f c ) + R l ( f f c) ) WS Wuen Mobile Communications 12

6 Impulse Response Model for a Multipath Channel Representation of Linear Bandpass System, h(t) Linear bandpass channel, h(t) H(f ) = H l (f f c ) + H l ( f f c) { H(f ) f > 0 H l (f f c ) = 0 f < 0 { H 0 f > 0 l ( f f c) = H ( f ) f < 0 h(t) = h l (t)e j2πfct + h l (t)e j2πf ct (12) [ ] = 2Re h l (t)e j2πf ct (13) WS Wuen Mobile Communications 13 Impulse Response Model for a Multipath Channel Response of a Bandpass Channel to a Bandpass Signal r(t) = s(τ)h(t τ)dτ = Re [r ] l (t)e j2πf ct R(f ) = S(f )H(f ) = 1 ( Sl (f f c ) + S l 2 ( f f c) )( H l (f f c ) + H l ( f f c) ) R(f ) = 1 ( Sl (f f c )H l (f f c ) + S l 2 c)h l ( f f c) ) (14) = 1 ( Rl (f f c ) + R l 2 c) ) (15) where r l (t) = R l (f ) = S l (f )H l (f ) s l (τ)h l (t τ)dτ WS Wuen Mobile Communications 14

7 Impulse Response Model for a Multipath Channel Linear Time-Variant Mobile Radio Channel The received signal r(d,t) at position d r(d,t) = s(t) h(d,t) = For causal system h(d,t) = 0 for t < 0 thus r(d,t) = t s(τ)h(d,t tτ)dτ s(τ)h(d,t τ)dτ Receiver moves at a constant velocity v and the position of the mobile is d = vt t r(vt,t) = s(τ)h(vt,t τ)dτ t r(t) = s(τ)h(vt,t τ)dτ = s(t) h(vt,t) = s(t) h(d,t) WS Wuen Mobile Communications 15 Impulse Response Model for a Multipath Channel Channel Impulse Response Model Transmitted s(t) Bandpass Channel h(t, τ) Received r(t) [ s(t) = Re h(t, τ) = 2Re r(t) = Re ] s l (t)e j2πf ct (16) [ ] h l (t,τ)e j2πf ct (17) [ r l (t)e j2πf ct ] (18) s l (t) Lowpass Eqivalent Channel h l (t,τ) r l (t) r l (t) = s l (t) h l (t,τ) h(t, τ) is the channel impulse response which completely characterizes the channel and is a function of t and τ: t represents the time variations due to motion. τ represents the the channel multipath delay for a fixed value of t. WS Wuen Mobile Communications 16

8 Impulse Response Model for a Multipath Channel Discrete-Time Impulse Response Model Excess delay bins Time delay resolution: τ = τ i+1 τ i, τ 0 = 0, τ 1 = τ, τ i = i τ for i = 0,1,...,N Useful frequency span of model: 2/ τ Maximum excess delay: N τ WS Wuen Mobile Communications 17 Impulse Response Model for a Multipath Channel Baseband Impulse Response of a Multipath Channel N 1 h l (t,τ) = a i (t,τ)e j (2πf c τ i (t)+φ i (t,τ)) δ(t τi (t)) i=0 a i (t,τ): real amplitude of i-th multipath component at time t. τ i (t): excess delays of i-th multipath component at time t. θ i (t,τ) 2πf c τ i (t) + φ i (t,τ): phase shift due to free-space propagation of the i-th multipath component plus additional phase shifts encountered in the channel. Time invariant or at least wide sense stationary over a small-scale time or distance interval N 1 h l (τ) = a i e jθ i δ(τ τ i ) i=0 only valid for the time delay resolution of the channel impulse response model accurately and uniquely resolves every multipath component over the local area. WS Wuen Mobile Communications 18

9 Two-Path Model Impulse Response Model for a Multipath Channel Example 2 Consider two-path model with distance d 1 and d 2, therefore the delay are τ 1 = d 1 /c and τ 2 = d 2 /c and assume τ 1 and τ 2 does not change with time. h(τ) = a 1 δ(τ τ 1 ) + a 2 δ(τ τ 2 ), where a i = a i e jφ i. H(f ) = h(τ)e j2πf τ dτ = a 1 e j2πf τ 1 + a 2 e j2πf τ 2 H(f ) = a a a 1 a 2 cos ( 2πf τ φ ), where τ = τ 2 τ 1 and φ = φ 2 φ 1 and f north = 1/ τ WS Wuen Mobile Communications 19 Impulse Response Model for a Multipath Channel Power Delay Profile Probing a channel with a pulse p(t) to approximate δ(t): p(t) δ(t) For small-scale channel modeling, the power delay profile of the channel is the spatial average of h l (t;τ) 2 over a local area. If p(t) has a time duration much smaller than the impulse response of the multipath channel, the received power delay profile in a local area can be approximate by P(τ) k h l (t;τ) 2 WS Wuen Mobile Communications 20

10 Impulse Response Model for a Multipath Channel Wideband Probing of Channel A pulsed, transmitted RF signal: s(t) = Re { p(t)e j2πf ct } The pulse p(t) = τmax T for 0 t T bb bb is a repetitive baseband pulse train with very narrow pulse width T bb and repetition period T rep τ max. Lowpass channel output r l (t) N 1 r l (t) = a i e jθ N 1 i p(t τ i ) = i=0 i=0 a i e jθ i [ τmax Π t T ] bb T bb 2 τ i The received power r l (t) 2 at time t 0 and τ j τ i > T bb r l (t 0 ) 2 = = = 1 τ max 1 τ max 1 τ max τmax r(t) r (t)dt 0 { } τmax N 1 N 1 Re a j (t 0 )a i (t 0 )p(t τ j )p(t τ i )e j(θ j θ i ) 0 j=0 i=0 ( ) N 1 a 2 k (t 0)p 2 (t τ k ) dt (19) k=0 WS Wuen Mobile Communications 21 Impulse Response Model for a Multipath Channel Wideband Probing of Channel, cont d r l (t 0 ) 2 = = 1 τ max N 1 k=0 N 1 k=0 τmax ( [ a 2 k (t τmax 0) Π t T ]) 2 bb 0 T bb 2 τ k dt a 2 k (t 0) (20) The total received power is simply related to the sum of the powers in the individual multipath components and is scaled by the ratio of the pulse s width and amplitude, and the maximum observed excess delay of the channel. WS Wuen Mobile Communications 22

11 Impulse Response Model for a Multipath Channel Wideband Probing of Channel, cont d Assume received power from multipath components is a random process, average small-scale received power: [ ] N 1 E a,θ [P WB ] = E a,θ a i e jθ i 2 i=0 N 1 a 2 i i=0 Summary: If a transmitted signal is able to resolve the multipath components, then the average small-scale received power is simply the sum of the average powers received in each multipath component. WS Wuen Mobile Communications 23 Impulse Response Model for a Multipath Channel Narrowband Probing of Channel A CW transmitted RF signal: s(t) = Re { c(t)e j2πf ct }, c(t) = 1 Received baseband envelope: N 1 r l (t) = a i e jθ i(t,τ) i=0 Instantaneous power: r l (t) 2 N 1 = a i e jθ i i=0 2 Average received power over a local area: [ N 1 E a,θ [P CW ] = E a,θ i=0 2 ] a i e jθ N 1 i i=0 a 2 N 1 i + 2 i=0 i,j i N r ij cos(θ i θ j ) where r ij = E a [a i a j ] is the path amplitude correlation coefficient WS Wuen Mobile Communications 24

12 Impulse Response Model for a Multipath Channel Narrowband Probing of Channel, cont d When cos(θ i θ j ) = 0 and/or r ij = 0 the average power for a CW signal is equivalent to the average received power for a wideband signal in a small-scale region. multipath phases are identically and independently distributed (i.i.d. uniform) over [0,2π] or the path amplitude r i,j are uncorrelated. The received local ensemble average power of wideband and narrowband signals are equivalent. WS Wuen Mobile Communications 25 Impulse Response Model for a Multipath Channel Example 3 Assume a discrete channel impulse response is used to model the following channels and the number of multipath bins is fixed at 64, find (a) τ (b) the maximum RF bandwidth the models can accurately represent. Solution: Channel types urban 100 µs microcell 4 µs indoor 500 ns Max. excess delays τ max = N τ,bw = 2/ τ (a) τ max = 100µs, N = 64, τ = µs, BW = 2/ τ = 1.28 MHz (b) τ max = 4µs, N = 64, τ = 62.5ns, BW = 2/ τ = 32 MHz (c) τ max = 500ns, N = 64, τ = ns, BW = 2/ τ = 256 MHz WS Wuen Mobile Communications 26

13 Impulse Response Model for a Multipath Channel Example 4 A mobile of traveling velocity of 10 m/s receives two multipath components at a carrier frequency of 1 GHz. The 1 st component is arrived at τ = 0 with an initial phase of 0 and a power of 70 dbm and the 2 nd which is 3 db weaker arrives at τ = 1µs also with initial phase of 0. If the mobile moves directly toward the direction of arrival of the 1 st component and directly away from that of the 2 nd component, compute the narrowband instantaneous power at time intervals of 0.1s from 0s to 0.5s. Solution: λ = c/f = /( )0.3m, and 70dBm=100 pw The narrowband instantaneous power at t = 0s: r l (t = 0) 2 = N 1 i=0 a ie jθ i(t,τ) 2 = 100pW e j0 + 50pW e j0 2 = 291pW at t = 0.1s: θ 1 = 2πd/λ = 2πvt/λ = 2π /0.3 = 20.94rad = 2.094rad = 120 θ 2 = 120, r l (t = 0.1) 2 = 100pW e j pW e j120 2 = 79.3pW at t = 0.2s: θ 1 = 240,θ 2 = 240, r l (t = 0.2) 2 = 100pW e j pW e j240 2 = 79.3pW Similarly, at t = 0.3s, r l (t = 0.3) 2 = 291pW, t = 0.4s, r l (t = 0.4) 2 = 79.3pW, t = 0.5s, r l (t = 0.5) 2 = 79.3pW. WS Wuen Mobile Communications 27 Impulse Response Model for a Multipath Channel Example 5 Continue from previous example: Compute the average narrowband and wideband received powers over the interval, assuming the amplitudes of the two multipath components do not fade over the local area. Solution: The average narrowband received power is = 149pW 6 The wideband received power using E a,θ [P WB ] is [ N 1 E a,θ [P WB ] = E a,θ N 1 ] a i e jθ i 2 a i 2 i=0 i=0 = 100pW + 50pW = 150pW WS Wuen Mobile Communications 28

14 Small-Scale Multipath Measurements Measurement of Small-Scale Multipath Channel Channel sounding: Measurement of the properties (impulse response) of wireless channels Direct RF pulse system Minimum resolvable delay between multipath component = pulse width Subject to interference and noise due to wide passband filter required for multipath time resolution Phases of the individual multipath components are not received due to the use of an envelope detector. WS Wuen Mobile Communications 30 Small-Scale Multipath Measurements Spread Spectrum Sliding Correlator Channel Sounding Spread spectrum: a carrier signal is spread over a large bandwidth by mixing it with a binary pseudo-noise (PN) sequence having a chip duration T c and a chip rate R c = 1/T c Hz. Power spectrum envelope of the TX spread spectrum signal: ( ) sinπ(f fc )T 2 c S(f ) = = sinc 2 (π(f f c )T c ) π(f f c )T c RF bandwidth (null-to-null): BW = 2R c Sliding correlator: TX chip clock is run at a slightly faster rate thatn the RX chip clock Processing gain PG = 2R c R bb = 2T bb T c = SNR out SNR in where T bb = 1/R bb is the period of the baseband information. WS Wuen Mobile Communications 31

15 Small-Scale Multipath Measurements Spread Spectrum Sliding Correlator Channel Sounding, cont d Time resolution ( τ): τ = 2T c = 2/R c typically the rms pulse width of a chip is smaller than the absolute width of the triangular correlation pulse, and the resolution is on the order of T c = 1/R c. The time between maximal correlation T T = T c γl = γl R where c γ = α α β is the slide factor and l = 2n 1 is the PN sequence length able to reject passband noise and improving the coverage range for a given transmitter power WS Wuen Mobile Communications 32 Small-Scale Multipath Measurements Frequency Domain Channel Sounding WS Wuen Mobile Communications 33

16 Parameters of Mobile Multipath Channels Power Delay Profile Presented as a function of excess delay with respect to a fixed time delay reference. Found by averaging instantaneous power delay profile measurements over a local area. Often choose to sample at spatial distance of λ/4 Outdoor (450 MHz to 6 GHz): receiver movement < 6 m Indoor (450 MHz to 6 GHz): receiver movement < 2 m WS Wuen Mobile Communications 35 Parameters of Mobile Multipath Channels Parameters of Mobile Multipath Channels mean excess delay τ: the first moment of the power delay profile. k a 2 k τ = τ k k P(τ k )τ k k a 2 = k k P(τ k ) rms delay spread σ τ : the square root of the second central moment of the power delay profile. where τ 2 = σ τ = k a 2 k τ2 k k a 2 k τ 2 τ 2 = k P(τ k )τ 2 k k P(τ k ) maximum excess delay (X db): the time delay during which multipath energy falls to X db below the maximum, τ X τ 0. excess delay spread (X db), τ X WS Wuen Mobile Communications 36

17 Parameters of Mobile Multipath Channels Example of an Indoor Power Delay Profile WS Wuen Mobile Communications 37 Parameters of Mobile Multipath Channels Multipath Channel Parameters Multipath channel parameters can be determined from power delay profile. τ,τ 2 and σ τ depend on the choice of noise threshold These delays are measured relativel to the first detactable signal arriving at τ 0 = 0. Delays do not rely on the absolute power level of P(τ), but only the relative amplitudes of the multipath components within P(τ). Analogous to the delay spread parameters in the time domain, coherence bandwidth is used to characterize the channel in frequency domain. WS Wuen Mobile Communications 38

18 Parameters of Mobile Multipath Channels Typical Measured Values of RMS Delay Spread WS Wuen Mobile Communications 39 Parameters of Mobile Multipath Channels Example 6 (a) Compute the RMS delay spread for the following power delay profile. (b) If BPSK modulation is used, what is the maximum bit rate that can be sent through the channel without needing an equalizer? P(τ) 0 db 0 db 0 1 µs τ Solution: (a) τ = (1)(0)+(1)(1) 1+1 = 1 2 = 0.5µs. τ 2 = (1)(0)2 +(1)(1) == 1 2 = 0.5µs2. σ τ = τ 2 τ 2 = 0.5 (0.5) 2 = 0.25 = 0.5µs. (b) σ τ T s 0.1 T s σ τ 0.1 = = 5µs R s = 1 T s = sps = 200ksps R b = 200kbps WS Wuen Mobile Communications 40

19 Parameters of Mobile Multipath Channels Coherence Bandwidth Time domain: delay spread Frequency domain: coherence bandwidth The RMS delay spread and coherence bandwidth are inversely proportional to each other. σ τ 1 B c Coherence bandwidth B c is 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). B c is the range of frequencies over which two frequency components have a strong potential for amplitude correlation. The bandwidth over which the frequency correlation is above 0.9: B c 1 50σ τ The bandwidth over which the frequency correlation is above 0.5: B c 1 5σ τ WS Wuen Mobile Communications 41 Parameters of Mobile Multipath Channels Example 7 Calculate the mean excess delay, RMS delay spread, and the maximum excess delay (10 db) for the multipath profile below. Estimate the 50% coherence bandwidth of the channel. Would the channel be suitable for AMPS or GSM service without the use of an equalizer? P(τ) [db] τ [µs] Solution τ 10dB = 5µs, τ = (1)(5)+(0.1)(1)+(0.1)(2)+(0.01)(0) = 4.38µs τ 2 = (1)(5)2 +(0.1)(1) 2 +(0.1)(2) 2 +(0.01)(0) = 21.07µs 2 σ τ = (4.38) 2 = 1.37µs, B c 5σ 1 1 = τ 5(1.37µs) = 146 khz B c > 30 khz AMPS will work without an equalizer; B c < 200 khz GSM will not work without an equalizer. WS Wuen Mobile Communications 42

20 Parameters of Mobile Multipath Channels Doppler Spread and Coherence Time Delay spread and coherence bandwidth describe time dispersive nature of channel in a local area. Doppler spread and coherence time describe the time varying nature (due to relative motion between MS and BS or movements of objects in the channel) of the channel in a small-scale region. Doppler Spread B D is a measure of the spectral broadening caused by the time rate of change of the mobile radio channel. B D is the range of frequencies over which the received Doppler spectrum is essentially non-zero. Slow fading channel: If the baseband signal bandwidth B bb > B D, the effect of Doppler spread are negligible at the receiver. WS Wuen Mobile Communications 43 Parameters of Mobile Multipath Channels Doppler Spread and Coherence Time, cont d Coherence Time T c is the time domain dual of Doppler spread, characterizing the time varying nature of the frequency dispersiveness of the channel in the time domain. T c is a statistical measure of the time duration over which the channel impulse response is essentially invariant and quantifies the similarity of the channel response at different times. T c is the time duration over which two received signals have a strong potential for amplitude correlation. The Doppler spread and coherence time are inversely proportional to one another: T c 1 f m or T c 1 f m WS Wuen Mobile Communications 44

21 Parameters of Mobile Multipath Channels Doppler Spread and Coherence Time, cont d the time over which the time correlation function is above 0.5 T c 9 16πf m geometric mean of the two definition: T c = 9 16πf 2 m = f m where f m is the maximum Doppler shift (f m = v/λ) If the symbol rate of transmission is greater than the 1/T c, the channel will not cause distortion due to motion. WS Wuen Mobile Communications 45 Parameters of Mobile Multipath Channels Example 8 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 f c = 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 B D for the channel? Solution: Let T sample T c 2 and use the smallest value of T c for conservative design. T c 9 16πf m = 9λ 16πv = 9c 16πvf c = = 565µs 2 = µs Spatial sampling interval: x = v T c 2 = 1.41cm Number of samples required over 10 m traveling distance: N x = x 10 = 708, 10m Time required to make these measurements: 50m/s = 0.2s Doppler spread: B D = f m = vf c c = = Hz WS Wuen Mobile Communications 46

22 Types of Small-Scale Fading Types of Small-Scale Fading Multipath delay spread leads to time dispersion and frequency selective fading Doppler spread leads to frequency dispersion and time selective fading Small-scale fading (based on multipath time delay spread) Flat fading BW of signal < BW of channel Delay spread < Symbol period Frequency selective fading BW of siganl > BW of channel Delay spread > Symbol period Small-scale fading (based on Doppler spread) Fast fading High Doppler spread Coherence time < Symbol period Channel variations faster than baseband signal variations Slow fading Low Doppler spread Coherence time > Symbol period Channel variations slower than baseband signal variations WS Wuen Mobile Communications 48 Flat Fading Types of Small-Scale Fading Flat fading if B s B c and T s σ τ. Also known as amplitude varying channels and narrowband channels The most common amplitude distribution is the Rayleigh distribution. WS Wuen Mobile Communications 49

23 Types of Small-Scale Fading Frequency Selective Fading Frequency selective fading if B s > B c and T s < σ τ. Received signal includes multiple versions of the transmitted waveform which are faded and delayed in time introducing intersymbol interference (ISI) Frequency selective: channel gain is different for different frequency components. Also known as wideband channels If T s 10σ τ, a channel is flat fading; and T s < 10σ τ for frequency selective. WS Wuen Mobile Communications 50 Types of Small-Scale Fading Fast Fading and Slow Fading Due to Doppler Spread Depending on how rapidly the transmitted baseband signal changes as compared to the rate of change of the channel: fast/ slow fading Fast Fading Channel impulse response changes repidly within the symbol duration. T c < T s and B s < B D frequency dispersion (time selective fading) due to Doppler spreading Slow Fading Channel impulse response changes at a rate much slower than the transmitted baseband signal. T s T c and B s B D Channel may be assumed to be static over one or several reciprocal bandwidth intervals WS Wuen Mobile Communications 51

24 Types of Small-Scale Fading Type of Fading Channels WS Wuen Mobile Communications 52 Rayleigh and Ricean Distributions Rayleigh Fading Distribution To describe the statistical time varying nature of the received envelope of a flat fading signal, or the envelope of an individual multipath components Envelope of the sum of two quadrature Gaussian noise signal obeys a Rayleigh distribution: r = I 2 + Q 2 where I,Q are zero-mean Gaussian random variable has the pdf: p(x) = 1 2πσ exp ( x2 2σ 2 Rayleigh distribution has a probability density function (pdf) given by { ( ) r exp r2 (0 r ) p(r) = σ 2 2σ 2 0 (r < 0) σ: the rms value of the received voltage signal before envelope detection σ 2 : the time average power of the received signal before envelope detection. ) WS Wuen Mobile Communications 54

25 Rayleigh and Ricean Distributions Rayleigh Fading Distribution, cont d Probability that the envelope of the received signal does not exceed a specified value R is given by the cumulative distribution function (CDF) Mean of r: r P(R) = Pr(r R) = r = E[r] = 0 R 0 ) p(r)dr = 1 exp ( R2 2σ 2 π rp(r)dr = σ 2 = σ Variance of r: σ 2 r = E[r2 ] E 2 [r] (ac power in the signal envelope) σ 2 r = Median value of r: r median 0 rmedian 0 r 2 p(r)dr σ2 π ( 2 = σ2 2 π ) = σ 2 2 p(r)dr = 0.5 r median = σ 2ln2 = 1.177σ WS Wuen Mobile Communications 55 Rayleigh and Ricean Distributions Rayleigh Fading Distribution, cont d Rayleigh distribution is widely used in wireless communications due to An excellent approximation in a large number of practical scenarios A worst case scenario: there is no dominant signal component and thus there is a large number of fading dips. Depends only on a single parameter, the mean received power Mathematical convenience: computation of error probabilities can often be done in a closed form. WS Wuen Mobile Communications 56

26 Rayleigh and Ricean Distributions Rayleigh Fading Distribution, cont d WS Wuen Mobile Communications 57 Rayleigh and Ricean Distributions Ricean Fading Distribution If there is a dominant stationary (nonfading) signal component present, the small-scale fading envelope distribution is Ricean. Ricean distribution: { ( ) ( ) r exp r2 +A 2 I ra p(r) = σ 2 2σ 2 0 (r 0,A 0) σ 2 0 (r < 0) where A is the peak amplitude of the dominant signal and I 0 ( ) is the modified Bessel function of the first kind, zero order. Ricean factor K = A 2 /(2σ 2 ): the ratio of the power in the dominant (LOS) signal to the variance of the multipath. K [db] = 10log A2 2σ 2 WS Wuen Mobile Communications 58

27 Rayleigh and Ricean Distributions Ricean Fading Distribution, cont d A 0,K 0 ( db), the Ricean distribution degenerates to a Rayleigh distribution. K 1, the Ricean pdf is approximately Gaussian about the mean value A. WS Wuen Mobile Communications 59 Statistical Models for Multipath Fading Channels Clarke s Model for Flat Fading Doppler shift: f n = v λ cosα n E and H field components: N E z = E 0 C n cos ( ) 2πf c t + θ n H x = E 0 η H y = E 0 η n=1 N C n sinα n cos ( ) 2πf c t + θ n n=1 N C n cosα n cos ( ) 2πf c t + θ n n=1 where θ n = 2πf n t + φ n Normalize E and H fields such that N Cn 2 = 1 n=1 WS Wuen Mobile Communications 61

28 Statistical Models for Multipath Fading Channels Clarke s Model for Flat Fading Assume f n f c and E z,h x,h y can be approximated as Gaussian random variable if N is sufficiently large. Express the E-field in in-phase and quadrature form: E z (t) = I(t)cos ( 2πf c t ) Q(t)sin ( 2πf c t ) where I(t) = E 0 Nn=1 C n cos ( 2πf n t + φ n ) and Q(t) = E 0 Nn=1 C n sin ( 2πf n t + φ n ) I(t) and Q(t) are uncorrelated zero-mean Gaussian random variables with equal variance: I 2 = Q 2 = E z 2 = E 2 0 /2 The envelope of the received E-field, E z (t) is E z (t) = I 2 (t) + Q 2 (t) = r(t) The random received signal envelope r has a Rayleigh distribution: ) p(r) = ( r2 σ 2 exp r2 2σ 2 (r 0) where σ 2 = E 2 0 /2 WS Wuen Mobile Communications 62 Doppler Spectra Statistical Models for Multipath Fading Channels Total received power P r = 2π 0 AG(α)p(α)dα Instantaneous frequency of the received signal component arriving at an angle α f (α) = f = v λ cosα + f c = f m cosα + f c The received power spectrum S(α) = A[p(α)G(α) + p( α)g( α)] and S(f ) = 0 for f f c > f m df = dα sinα f m ( ) ( f α = cos 1 fc f fc,sinα = 1 f m f m A[p(α)G(α) + p( α)g( α)] S(f ) = ( ) f fc 2 f m 1 f m ) 2 WS Wuen Mobile Communications 63

29 Statistical Models for Multipath Fading Channels Doppler Spectra, cont d For a vertical λ/4 antenna (G(α) = 1.5) and a uniform distribution p(α) = 1/2π: 1.5 S Ez (f ) = ( ) f fc 2 πf m 1 f m Classical Doppler spectrum (Jakes spectrum) Describes frequency dispersion A measure for the temporal variability of the channel. WS Wuen Mobile Communications 64 Statistical Models for Multipath Fading Channels Simulating Doppler Fading WS Wuen Mobile Communications 65

30 Statistical Models for Multipath Fading Channels Simulating Doppler Fading, cont d WS Wuen Mobile Communications 66 Statistical Models for Multipath Fading Channels Simulating Multipath with Doppler-Induced Rayleigh Fading WS Wuen Mobile Communications 67

31 Statistical Models for Multipath Fading Channels Simulating Two-Ray Multipath Impulse response h b (t) = α 1 exp(jφ 1 )δ(t) + α 2 exp(jφ 2 )δ(t τ) Two-ray Rayleigh fading model WS Wuen Mobile Communications 68 Statistical Models for Multipath Fading Channels Level Crossing and Fading Statistics Level crossing rate (LCR): the expected rate at which the Rayleigh fading envelope (normalized to the local rms signal level) crosses a specified level in a positive-going direction. The average number of level crossing per second N R at a specified R: N R = 0 ṙp(r,ṙ)dṙ = 2πf m ρe ρ2 where ṙ = dr(t) dt, p(r,ṙ) is the joint density function of r and ṙ at r = R. f m is the maximum Doppler frequency and ρ = R/R rms WS Wuen Mobile Communications 69

32 Statistical Models for Multipath Fading Channels Level Crossing and Fading Statistics, cont d Average fading duration τ: the average period of time for which the received signal is below a specified level R. τ = 1 Pr[r R] = 1 1 N R N R T where τ i is the duration of the fade and T is the observation interval of the fading signal Pr[r R] for Rayleigh distribution is Pr[r R] = R 0 i τ i p(r)dr = 1 exp ( ρ 2) τ = eρ2 1 ρf m 2π WS Wuen Mobile Communications 70 Statistical Models for Multipath Fading Channels Example 9 For a Rayleigh fading signal, compute the occurrence rate of fading dips for ρ = 1, when the maximum Doppler frequency is 20 Hz. What is the maximum velocity of the mobile for the Doppler frequency if the carrier frequency is 900 MHz? Solution: N R = 2π(20)(1)e ( 1) = dips/sec f m = v/λ v = f m /λ = 20/(1/3) = 6.66m/s = 24km/hr Example 10 Find the average fade duration for threshold levels ρ = 0.01,ρ = 0.1, and ρ = 1 when the Doppler frequency is 200 Hz. Solution: e For ρ = 0.01: τ = (0.01)(200) = 2π 19.9µs For ρ = 0.1: τ = For ρ = 1: τ = e (0.1)(200) 2π = 200µs e12 1 (1)(200) 2π = 3.43ms WS Wuen Mobile Communications 71

33 Statistical Models for Multipath Fading Channels Example 11 Find the average fading duration for a threshold level of ρ = when the Doppler frequency is 20 Hz. For a binary digital modulation with bit duration of 50 bps, is the Rayleigh fading slow or fast? What is the average number of bit errors per second for the given data rate? Assume a bit error occurs whenever any portion of a bit encounters a fad for which ρ < 0.1. Solution: τ = e (0.707)(20) 2π = 18.3ms 50bps T b = 1/50 = 20ms > τ fast Rayleigh fading τ ρ=0.1 = 2ms only one bit on average will be lost during a fade N ρ=0.1 = 2π(20)(0.1)e (0.1)2 = 4.96 dips/sec BER = 5/50 = 0.1 WS Wuen Mobile Communications 72