- 1. Ch.2. Characterization of the Wireless Channel 2.1 Multipath Propagation Multipath propagation: A radio signal arrives at a receiver via two or more of many possible routes with the result that the arriving signals. although having a common time origin at the transmitter, arrive “out of step”. Channel characterization: • Time dispersion: different propagation paths have different propagation delays =⇒ received signal pulse is wider than the transmitted pulse, which introduces inter-symbol interference (ISI) to the received signal (mainly to wideband signals) Tx 0 t 00 0000 00 11 1111 11 000000000 000 111111111 111 line-of-sight (LOS) Rx t 0 Figure 1: An illustration of multipath propagation Many different propagation paths can have a common propagation delay as long as the paths have the same length. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 2. Wireless Communications and Networking Ch2 - Page 2 of 44 All the scatterers located on an ellipse with the transmitter and receiver as the foci cor- responds to a distinct propagation delay, and is therefore called a distinct scatterer. Copyright Prentice Hall Scatterer RxTx Figure 2: Ellipsoidal portrayal of scatterer location The receiver cannot distinguish the individual received signal component coming from path 1 from the component coming from path 2, etc. =⇒ The received signal component with propagation delay τ1 = d/c (where c is the velocity of light) is a result of the multiple propagation paths. • Fading: Consider transmitting a single-tone sinusoid. For simplicity, consider a two- path channel, where the delay of the line-of-sight (LOS) or direct path is assumed to be zero, and the delay of the non-line-of-sight (NLOS) or reﬂected path is τ. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 3. Wireless Communications and Networking Ch2 - Page 3 of 44 0 0 0 00 0 0 1 1 1 11 1 1 0 0 0 00 0 0 1 1 1 11 1 10101 0101 0 0 1 1 0 0 1 101 0 0 1 1 0 0 1 10101 0000000000000000000000 0000000000000000000000 1111111111111111111111 1111111111111111111111 Transmitter Receiver c Prentice Hall α1 cos(2πfct) α2 cos(2πfc(t − τ)) Figure 3: A channel with two propagation paths The received signal, in the absence of noise, can be represented as r(t) = α1 cos(2πfct) + α2 cos(2πfc(t − τ)), (1) where α1 and α2 are the amplitudes of the signal components from the two paths respec- tively. The received signal can also be represented as r(t) = α cos(2πfct + φ), (2) where α = α2 1 + α2 2 + 2α1α2 cos(2πfcτ) and φ = − tan−1 [ α2 sin(2πfcτ) α1 + α2 cos(2πfcτ) ] are the amplitude and phase of the received signal. Both α and φ are functions of α1, α2, and τ. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 4. Wireless Communications and Networking Ch2 - Page 4 of 44 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 2 2.5 3 3.5 4 fc τ α1 +α2 α1 −α2 © Prentice Hall Thereceivedsignalamplitudeα Figure 4: The amplitude ﬂuctuation of the two-path channel with α1 = 2 and α2 = 1 As the mobile user and/or scatterer moves, fcτ changes → the two received components add constructively (when fcτ = 0, 1, 2, . . .) or destructively (when fcτ = 0.5, 1.5, 2.5, . . .). =⇒ |r(t)| is a function of t, referred to as amplitude fading. At each deep fade, instantaneous SNR ↓ −→ BER ↑. In summary, multipath propagation in the wireless mobile environment results in a fading dispersive channel. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 5. Wireless Communications and Networking Ch2 - Page 5 of 44 2.2 The Linear Time-Variant Channel Model Consider a multipath propagation environment with N distinct scatterers, each charac- terized by amplitude ﬂuctuation αn(t) and propagation delay τn(t) , n = 1, 2, . . . , N. Consider a narrowband signal ˜x(t) transmitted over the wireless channel at a carrier frequency fc, such that ˜x(t) = ℜ{x(t)ej2πfct }, (3) where x(t) is the complex envelope of the signal. In the absence of background noise, the received signal at the channel output is ˜r(t) = ℜ{ N n=1 αn(t)x(t − τn(t))ej2πfc(t−τn(t)) } = ℜ{r(t)ej2πfct }, where r(t) is the complex envelope of the received signal and can be represented as r(t) = N n=1 αn(t)e−j2πfcτn(t) x(t − τn(t)). (4) As the mobile user moves, αn(t) and τn(t) are a function of t =⇒ the channel is linear time-variant. =⇒ The channel impulse response depends on the instant that the impulse is applied to the channel. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 6. Wireless Communications and Networking Ch2 - Page 6 of 44 Channel impulse response Linear time-invariant (LTI) channel: 0 0 0 0LTI channel c Prentice Hall δ(t) δ(t − t1) t1 r(t) = h(t) r(t) = h(t − t1) h(t) x(t) r(t) t t t t Figure 5: The linear time-invariant channel model We can use h(t) to describe the channel, where t is a variable describing propagation delay, assuming that the impulse is always applied to the channel at time zero. Linear time-variant (LTV) channel: 0 0 LTV channel 00 c Prentice Hall δ(t) δ(t − t1) t1 x(t) r(t) h(τ, t) r(t) = h1(t) r(t) = h2(t) = h1(t − t1) t t t t Figure 6: The linear time-variant channel model The channel impulse response is a function of two variables: one describing when the impulse is applied to the channel, the other describing the moment of observing the channel output or the associated propagation delay. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 7. Wireless Communications and Networking Ch2 - Page 7 of 44 Deﬁnition: The impulse response of an LTV channel, h(τ, t), is the channel output at t in response to an impulse applied to the channel at t − τ. The received signal is r(t) = ∞ −∞ h(τ, t)x(t − τ)dτ. (5) The channel impulse response for the channel with N distinct scatterers is then h(τ, t) = N n=1 αn(t)e−jθn(t) δ(τ − τn(t)), (6) where θn(t) = 2πfcτn(t) represents the carrier phase distortion introduced by the nth scatterer. Note: τn(t) changes by 1/fc −→ θn(t) changes by 2π =⇒ a small change in the propagation delay −→ a small change in αn(t) and τn(t), but a signiﬁcant change in θn(t) That is, the carrier phase distortion θn(t) is much more sensitive to user mobility than the amplitude ﬂuctuation αn(t). Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 8. Wireless Communications and Networking Ch2 - Page 8 of 44 Time-variant transfer function Linear time-invariant channel: H(f) = F[h(t)] (7) The channel output in the frequency domain is R(f) = H(f)X(f). (8) Linear time-variant channel: Deﬁnition The time-variant transfer function of an LTV channel is the Fourier transform of the impulse response, h(τ, t), with respect to the delay variable τ. H(f, t) = Fτ [h(τ, t)] = ∞ −∞ h(τ, t)e−j2πfτ dτ h(τ, t) = F−1 f [H(f, t)] = ∞ −∞ H(f, t)e+j2πfτ df where the time variable t can be viewed as a parameter. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 9. Wireless Communications and Networking Ch2 - Page 9 of 44 The received signal is r(t) = ∞ −∞ h(τ, t)x(t − τ)dτ = ∞ −∞ x(t − τ)[ ∞ −∞ H(f, t) exp(j2πfτ)df]dτ = ∞ −∞ H(f, t){ ∞ −∞ x(t − τ) exp[−j2πf(t − τ)]dτ} exp(j2πft)df = ∞ −∞ H(f, t)[ −∞ ∞ x(ξ) exp(−j2πfξ)(−dξ)] exp(j2πft)df = ∞ −∞ H(f, t)X(f) exp(j2πft)df △ = ∞ −∞ R(f, t) exp(j2πft)df where R(f, t) = H(f, t)X(f) and X(f) = F[x(t)]. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 10. Wireless Communications and Networking Ch2 - Page 10 of 44 Summary: c Prentice Hall x(t) r(t) R(f, t) = H(f, t)X(f)X(f) h(τ, t) Fτ −→ ←− F−1 f H(f, t) Figure 7: Frequency-time channel representation where r(t) = ∞ −∞ h(τ, t)x(t − τ)dτ R(f, t) = H(f, t)X(f) X(f) = ∞ −∞ x(t) exp(−j2πft)dt r(t) = ∞ −∞ R(f, t) exp(j2πft)df. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 11. Wireless Communications and Networking Ch2 - Page 11 of 44 Example 2.1: Consider an LTV channel with the impulse response given by h(τ, t) = 4 exp(−τ/T) cos(Ωt), τ ≥ 0, where T = 0.1 ms and Ω = 10π. a. Find the channel time-variant transfer function H(f, t). b. Given that the transmitted signal is x1(t) = 1, |t| ≤ T0 0, |t| > T0 , where T0 = 0.025 ms, ﬁnd the received signal in the absence of background noise. c. Repeat part (b) if the transmitted signal is x2(t) = x1(t − T1), where T1 = 0.05 ms. d. What do you observe from the results of parts (b) and (c)? Solution: a. The time-variant transfer function is H(f, t) = Fτ [h(τ, t)] = Fτ [4 exp(−τ/T) cos(Ωt)] = 4 cos(Ωt)Fτ [exp(−τ/T)] = 4T cos(Ωt) 1 + j2πfT . b. The received signal is calculated as follows: r1(t) = +∞ −∞ h(τ, t)x1(t − τ)dτ = ∞ 0 4 exp(−τ/T) cos(Ωt)x1(t − τ)dτ = 0, t ≤ −T0 4T cos(Ωt)[1 − exp(−t+T0 T )], −T0 < t < T0 4T cos(Ωt)[exp(−t−T0 T ) − exp(−t+T0 T )], t ≥ T0 Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 12. Wireless Communications and Networking Ch2 - Page 12 of 44 −0.1 −0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 −1.5 −1 −0.5 0 0.5 1 1.5 t (ms) x 1 (t)andr 1 (t)/T © Prentice Hall x 1 (t) r 1 (t) / T (a) 0 0.1 0.2 0.3 0.4 0.5 −1.5 −1 −0.5 0 0.5 1 1.5 t (ms) x 2 (t)andr 2 (t)/T © Prentice Hall x 2 (t) r 2 (t) / T (b) Figure 8: The transmitted signals xi(t) and normalized received signals ri(t)/T (i = 1 and 2) in Example 2.1 Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 13. Wireless Communications and Networking Ch2 - Page 13 of 44 The transmitted and received signals are plotted in Figure 8(a). c. Similar to part (b), the received signal is calculated as follows: r2(t) = +∞ −∞ h(τ, t)x2(t − τ)dτ = ∞ 0 4 exp(−τ/T) cos(Ωt)x1(t − T1 − τ)dτ = 0, t ≤ T1 − T0 4T cos(Ωt)[1 − exp(−t−T1+T0 T )], T1 − T0 < t < T1 + T0 4T cos(Ωt)[exp(−t−T1−T0 T ) − exp(−t−T1+T0 T )], t ≥ T1 + T0 The transmitted and received signals are plotted in Figure 8(b). d. From Figure 8, it is observed that: (1) the received signals have a larger pulse width than the corresponding transmitted signals because the channel is time dispersive; and (2) even though the transmitted signal x2(t) is x1(t) delayed by T1, the received signal r2(t) is not r1(t) delayed by T1 because the channel is time varying. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 14. Wireless Communications and Networking Ch2 - Page 14 of 44 2.3 Channel Correlation Functions Assumptions: (a) the channel impulse response h(τ, t) is a wide-sense stationary (WSS) process; (b) the channel impulse responses at τ1 and τ2, h(τ1, t) and h(τ2, t), are uncorrelated if τ1 = τ2 for any t. → wide-sense stationary uncorrelated scattering (WSSUS) channel Delay power spectral density (psd) The autocorrelation function of h(τ, t) is φh(τ1, τ2, ∆t) △ = 1 2 E[h∗ (τ1, t)h(τ2, t + ∆t)] = φh(τ1, ∆t)δ(τ1 − τ2) or equivalently, φh(τ, τ + ∆τ, ∆t) = φh(τ, ∆t)δ(∆τ), (9) where φh(τ, ∆t) = φh(τ, τ + ∆τ, ∆t)d∆τ. At ∆t = 0, we deﬁne φh(τ) △ = φh(τ, 0). (10) From Eqs. (9) – (10), we have φh(τ) = F∆τ [φh(τ, τ + ∆τ, ∆t)]|∆t=0 = F∆τ { 1 2 E[h∗ (τ, t)h(τ + ∆τ, t)]}. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 15. Wireless Communications and Networking Ch2 - Page 15 of 44 φh(τ) measures the average psd at the channel output as a function of the propagation delay, τ, and is therefore called the delay psd of the channel, also known as the multipath intensity proﬁle. The nominal width of the delay psd pulse is called the multipath delay spread, denoted by Tm. 0 c Prentice Hall φh(τ) Tm τ Figure 9: Delay power spectral density The nth moment of the delays ¯τn = τn φh(τ)dτ φh(τ)dτ . (11) The mean propagation delay, or ﬁrst moment, denoted by ¯τ, is ¯τ = τφh(τ)dτ φh(τ)dτ (12) and the rms (root-mean-square) delay spread, denoted by στ , is στ = (τ − ¯τ)2 φh(τ)dτ φh(τ)dτ 1/2 . (13) In calculating a value for the multipath delay spread, it is usually assumed that Tm ≈ στ . Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 16. Wireless Communications and Networking Ch2 - Page 16 of 44 Frequency and Time Correlation Functions The autocorrelation function of H(f, t)is φH(f1, f2, t, ∆t) △ = 1 2 E[H∗ (f1, t)H(f2, t + ∆t)] WSS =⇒ φH(f1, f2, ∆t) = 1 2 E[H∗ (f1, t)H(f2, t + ∆t)] US =⇒ φH(f1, f2, ∆t) = 1 2 E{[ h(τ1, t)e−j2πf1τ1 dτ1]∗ [ h(τ2, t + ∆t)e−j2πf2τ2 dτ2]} = 1 2 E[h∗ (τ1, t)h(τ2, t + ∆t)]e−j2π(f2τ2−f1τ1) dτ1dτ2 WSS = φh(τ1, τ2, ∆t)e−j2π(f2τ2−f1τ1) dτ1dτ2 US = φh(τ1, ∆t)δ(τ1 − τ2)e−j2π(f2τ2−f1τ1) dτ1dτ2 = φh(τ1, ∆t)e−j2π(f2−f1)τ1 dτ1 = φh(τ, ∆t)e−j2π(∆f)τ dτ (∆f = f2 − f1) △ = φH(∆f, ∆t) — time-frequency correlation function. Frequency correlation function: Let ∆t = 0, we have φH(∆f) △ = 1 2 E[H∗ (f, t)H(f + ∆f, t)] = φh(τ)e−j2π∆fτ dτ. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 17. Wireless Communications and Networking Ch2 - Page 17 of 44 Note: • φh(τ) and φH(∆f) is a pair of Fourier transform. • uncorrelated scattering =⇒WSS H(f, t) w.r.t. f • φH(∆f) characterizes the correlation of channel gains at f and f + ∆f for any t. 0 c Prentice Hall(∆f)c ∆f |φH(∆f)| Figure 10: Frequency-correlation function and channel coherence bandwidth φH(∆f) provides a measure of the “frequency coherence” of the channel. • coherence bandwidth: recall the channel with 2 propagation paths 0 00 0 00 0 1 11 1 11 1 0 00 0 00 0 1 11 1 11 10 0 1 1 0 0 1 1 0101 0 0 1 1 0 0 1 10101 00 0 11 1 00 0 11 1 0101 0000000000000000000000 0000000000000000000000 1111111111111111111111 1111111111111111111111 Transmitter Receiver c Prentice Hall α1 cos(2πfct) α2 cos(2πfc(t − τ)) Figure 11: A channel with two propagation paths The effect of the 2-path channel on the received signal depends on the signal fre- quency. With the same τ, two signals A cos(2πf1t) and A cos(2πf2t) will be affect differently. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 18. Wireless Communications and Networking Ch2 - Page 18 of 44 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 2 2.5 3 3.5 4 fc τ α1 +α2 α1 −α2 © Prentice Hall Thereceivedsignalamplitudeα Figure 12: The amplitude ﬂuctuation of the two-path channel with α1 = 2 and α2 = 1 =⇒ |f1 − f2| ↑ =⇒ |r1(t)| and |r2(t)| at any t will be uncorrelated. Deﬁnition: The maximum frequency difference for which the signals are still strongly correlated is called the coherence bandwidth of the channel, denoted by (∆f)c. =⇒ Two sinusoids with frequency separation larger than (∆f)c are affected differ- ently by the channel at any t. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 19. Wireless Communications and Networking Ch2 - Page 19 of 44 • Let Ws denote the bandwidth of the transmitted signal. If (∆f)c < Ws, the channel is said to exhibit frequency selective fading which introduces severe ISI to the received signal; If (∆f)c ≫ Ws, the channel is said to exhibit frequency nonselective fading or ﬂat fading which introduces negligible ISI. • φh(τ) ↔ φH(∆f) =⇒ (∆f)c ≈ 1/Tm. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 20. Wireless Communications and Networking Ch2 - Page 20 of 44 Time correlation function φH(∆t): Letting ∆f = 0 in the time-frequency correlation function φH(∆f, ∆t), we have φH(∆t) △ = φH(0, ∆t) = 1 2 E[H∗ (f, t)H(f, t + ∆t)]. (14) • φH(∆t) characterizes, on average, how fast the channel transfer function changes with time at each frequency. • The nominal width of φH(∆t), (∆t)c, is called the coherence time of the fading channel. • If the channel coherence time is much larger than the symbol interval of the trans- mitted signal, the channel exhibits slow fading. • φH(∆t) is independent of f due to the US assumption =⇒ US in the time domain is equivalent to WSS in the frequency domain. 0 c Prentice Hall (∆t)c ∆t |φH(∆t)| Figure 13: Time-correlation function and channel coherence time Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 21. Wireless Communications and Networking Ch2 - Page 21 of 44 Doppler Power Spectral Density Doppler shifts: An LTV channel introduces Doppler frequency shifts: Given the transmitted signal fre- quency fc, the received signal frequency is fc+ν(t), where ν(t) is the Doppler frequency shift and is given by ν(t) = V fc c cos θ(t). 0 0 0 00 1 1 1 110 0 1 1 00 00 11 11 0 0 1 1 00110011 00 00 11 11 00 00 11 11 0011001100110101 010 0 0 00 1 1 1 11 c Prentice Hall θ(t) VMS BS x Figure 14: The Doppler effect Consider the wireless channel with N distinct scatterers where the propagation delay can be approximated by its mean value ¯τ. h(τ, t) ≈ N n=1 αn(t) exp[−j2πfcτn(t)]δ(t − ¯τ) △ = Z(t)δ(τ − ¯τ), (15) r(t) = +∞ −∞ h(τ, t)x(t − τ)dτ = +∞ −∞ [Z(t)δ(τ − ¯τ)]x(t − τ)dτ = Z(t)x(t − ¯τ). In the frequency domain, the received signal is R(f) = F[r(t)] = F[Z(t)x(t − ¯τ)] = F[Z(t)] ⋆ F[x(t − ¯τ)] = F[Z(t)] ⋆ [X(f)e−j2πf ¯τ ], Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 22. Wireless Communications and Networking Ch2 - Page 22 of 44 =⇒ The channel indeed broadens the transmitted signal spectrum by introducing new frequency components, a phenomenon referred to as frequency dispersion. Doppler-spread function H(f, ν): H(f, ν) is the channel gain associated with Doppler shift ν to the input signal component at frequency f. R(f) = +∞ −∞ X(f − ν)H(f − ν, ν)dν. (16) Relation between H(f, t) and H(f, ν): H(f, ν) = Ft[H(f, t)] = +∞ −∞ H(f, t)e−j2πvt dt H(f, t) = F−1 ν [H(f, ν)] = +∞ −∞ H(f, ν)e+j2πνt dν =⇒ being time-variant in the time domain can be equivalently described by having Doppler shifts in the frequency domain. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 23. Wireless Communications and Networking Ch2 - Page 23 of 44 Autocorrelation function of H(f, ν): ΦH(f1, f2, ν1, ν2) △ = 1 2 E[H∗ (f1, ν1)H(f2, ν2) = 1 2 E[H∗ (f1, t1)H(f2, t2)]ej2πν1t1 e−j2πν2t2 dt1dt2 WSSUS = φH(∆f, ∆t)e−j2π[ν2(t1+∆t)−ν1t1] d∆tdt1 (where ∆f = f2 − f1 and ∆t = t2 − t1) = φH(∆f, ∆t)e−j2πν2∆t d∆t e−j2π(ν2−ν1)t1 dt1 = ΦH(∆f, ν2)δ(ν2 − ν1) where ΦH(∆f, ν) = φH(∆f, ∆t)e−j2πν∆t d∆t is the Fourier transform of φH(∆f, ∆t) w.r.t. ∆t. At ∆f = 0, ΦH(ν) △ = ΦH(0, ν) = ∞ −∞ φH(∆t)e−j2πν∆t d∆t. • ΦH(ν) is the Fourier transform of the channel correlation function φH(∆t). =⇒ ΦH(ν) is psd as a function of the Doppler shift ν. =⇒ ΦH(ν) is called Doppler power spectral density function. 0 c Prentice Hall Bd ν ΦH(ν) Figure 15: Doppler power spectral density and Doppler spread Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 24. Wireless Communications and Networking Ch2 - Page 24 of 44 • The nominal width of the Doppler psd, Bd, is called the Doppler spread. • Since φH(∆t) ↔ ΦH(ν), (∆t)c ≈ 1 Bd . • The mean Doppler shift is ¯ν = νΦH(ν)dν ΦH(ν)dν and the rms Doppler spread is σν = [ (ν − ¯ν)2 ΦH(ν)dν ΦH(ν)dν ]1/2 . As an approximation, it is usually assumed that Bd ≈ σν. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 25. Wireless Communications and Networking Ch2 - Page 25 of 44 Example 2.3 Delay psd and frequency-correlation function Consider a WSSUS channel whose time-variant impulse response is given by h(τ, t) = exp(−τ/T)n(τ) cos(Ωt + Θ), τ ≥ 0, where T and Ω are constants, Θ is a random variable uniformly distributed in [−π, +π], and n(τ) is a random process independent of Θ, with E[n(τ)] = µn and E[n(τ1)n(τ2)] = δ(τ1 − τ2). a. Calculate the delay psd and the multipath delay spread. b. Calculate the frequency correlation function and the channel coherence bandwidth. c. Determine whether the channel exhibits frequency-selective fading for GSM sys- tems with T = 0.1 ms. Solution: a. From Eq. (11), we have φh(τ) = F∆τ { 1 2 E[h∗ (τ, t)h(τ + ∆τ, t]} = F∆τ { 1 2 E[n(τ)n(τ + ∆τ)]E[e−(2τ+∆τ)/T cos2 (Ωt + Θ)]}, τ ≥ 0 = F∆τ { 1 4 e−(2τ+∆τ)/T δ(∆τ)E[1 + cos(2Ωt + 2Θ)]}, τ ≥ 0 = 1 4 e−2τ/T , τ ≥ 0, where E[cos(2Ωt + 2Θ)] = +π −π cos(2Ωt + 2θ) 1 2π dθ = 0. For τ < 0, φh(τ) = 0. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 26. Wireless Communications and Networking Ch2 - Page 26 of 44 The mean propagation delay is ¯τ = ∞ 0 τφh(τ)dτ ∞ 0 φh(τ)dτ = ∞ 0 τ 1 4e−2τ/T dτ ∞ 0 1 4e−2τ/T dτ = T 2 and the multipath delay spread is Tm ≈ στ = [ (τ − ¯τ)2 φh(τ)dτ φh(τ)dτ ]1/2 = [ τ2 φh(τ)dτ φh(τ)dτ − ¯τ2 ]1/2 = T 2 . b. The frequency correlation function is φH(∆f) = F[φh(τ)] = ∞ 0 1 4 e−2τ/T e−j2π(∆f)τ dτ = T 8 + j8πT(∆f) . The coherence bandwidth is (∆f)c ≈ 1/Tm = 2 T . c. With T = 0.1 ms, we have (∆f)c = 20 kHz. The GSM channels have a bandwidth of 200 kHz. Since (∆f)c ≪ 200 kHz, the channel fading is frequency selective. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 27. Wireless Communications and Networking Ch2 - Page 27 of 44 Example 2.4 Doppler psd For the channel speciﬁed in Example 2.3 with Ω = 10π, ﬁnd a. the Doppler psd, b. the mean Doppler shift and the rms Doppler spread, c. the channel coherence time, and d. whether the channel exhibits slow fading for GSM systems. Solution: a. The Doppler psd can be calculated by taking the Fourier transform of the time cor- relation function φH(∆t). In this way, we need to calculate the correlation function φh(τ, ∆t) ﬁrst. For the WSS channel, we have for τ ≥ 0 φh(τ, ∆t) = F∆τ { 1 2 E[h∗ (τ, t)h(τ + ∆τ, t + ∆t)]} = F∆τ { 1 2 E[e−τ/T n(τ) cos(Ωt + Θ) · e−(τ+∆τ)/T n(τ + ∆τ) cos(Ωt + Ω∆t + Θ)]} = F∆τ { 1 4 e−(2τ+∆τ)/T E[n(τ)n(τ + ∆τ)]E[cos(Ω∆t) + cos(2Ωt + Ω∆t + 2Θ)]} = F∆τ { 1 4 e−(2τ+∆τ)/T δ(∆τ) cos(Ω∆t)} = 1 4 e−2τ/T cos(Ω∆t), τ ≥ 0. The time correlation function is then φH(∆t) = φH(∆f, ∆t)|∆f=0 = +∞ −∞ φh(τ, ∆t)dτ = 1 4 cos(Ω∆t) +∞ 0 e−2τ/T dτ = T 8 cos(Ω∆t). Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 28. Wireless Communications and Networking Ch2 - Page 28 of 44 The Doppler psd is ΦH(ν) = F[φH(∆t)] = F[ T 8 cos(Ω∆t)] = T 16 [δ(2πν − Ω) + δ(2πν + Ω)]. That is, the channel introduces two Doppler shifts, ±Ω/2π = ±5 Hz, with equal psd. b. The mean Doppler shift is zero as the two Doppler shifts are negative of each other and have the same psd. The rms Doppler spread is σν = { +∞ −∞ ν2 · T 16[δ(2πν − Ω) + δ(2πν + Ω)]dν +∞ −∞ T 16[δ(2πν − Ω) + δ(2πν + Ω)]dν }1/2 = { T 16[( Ω 2π )2 + (− Ω 2π )2 ] T 16[1 + 1] }1/2 = Ω 2π , which is 5 Hz. c. The coherence time is (∆t)c ≈ 1 σν = 0.2 s. d. In GSM systems, the data rate Rs = 270.833 kbps, which corresponds to a symbol interval Ts = 1 Rs ≈ 3.7 × 10−6 s. Since Ts ≪ (∆t)c, the channel exhibits slow fading. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 29. Wireless Communications and Networking Ch2 - Page 29 of 44 2.4 Large-Scale Path Loss and Shadowing Consider a ﬂat fading channel with the channel impulse response h(τ, t) ≈ h(¯τ, t) △ = g(t)δ(τ − ¯τ). The received signal is r(t) = +∞ −∞ h(τ, t)x(t − τ)dτ ≈ g(t)x(t − ¯τ). Flat fading channel c Prentice Hall x(t) r(t) = g(t)x(t − τ) h(τ, t) = g(t)δ(τ − τ) Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 30. Wireless Communications and Networking Ch2 - Page 30 of 44 50 100 150 200 250 300 350 400 450 500 −120 −115 −110 −105 −100 −95 −90 −85 −80 −75 −70 Amplitudefading|g(t)|(dB) time (ms) © Prentice Hall local mean (a) Overall amplitude fading |g(t)| (dB) 50 100 150 200 250 300 350 400 450 500 −35 −30 −25 −20 −15 −10 −5 0 5 10 15 Short−termamplitudefading|Z(t)|(dB) time (ms) © Prentice Hall local mean (b) Short-term amplitude fading |Z(t)| (dB) Figure 16: Representation of long-term and short-term fading components Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 31. Wireless Communications and Networking Ch2 - Page 31 of 44 Free Space Propagation Model When the distance between the transmitting antenna and receiving antenna is much larger than the wavelength of the transmitted wave and the largest physical linear dimension of the antennas, the power Pr at the output of the receiving antenna is given by Pr = PtGtGr( λ 4πd )2 , where Pt = total power radiated by an isotropic source, Gt = transmitting antenna gain, Gr = receiving antenna gain, d = distance between transmitting and receiving antennas, λ = wavelength of the carrier signal = c/fc, c = 3 × 108 m/s (velocity of light), fc = carrier frequency, and PtGt △ = effective isotropically radiated power (EIRP). The term (4πd/λ)2 is known as the free-space path loss denoted by Lp(d), which is Lp(d) = EIRP × Receiving antenna gain Received power = −10 log10[ λ2 (4πd)2 ] (dB) = −20 log10( c/fc 4πd ) (dB). In other words, the path loss is Lp(d) = 20 log10 fc + 20 log10 d − 147.56 (dB). Note that the free-space path loss increases by 6 dB for every doubling of the distance and also for every doubling of the radio frequency. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 32. Wireless Communications and Networking Ch2 - Page 32 of 44 Log-Distance Path Loss with Shadowing Let ¯Lp(d) denote the log-distance path loss. Then, ¯Lp(d) ∝ ( d d0 )κ , d ≥ d0 or equivalently, ¯Lp(d) = ¯Lp(d0) + 10κ log10( d d0 ) dB, d ≥ d0 • d0: 1 km for macrocells, 100 m for outdoor microcells, and 1 m for indoor picocells Table 1: Path loss exponents for different environments Environment Path Loss Exponent, κ free space 2 urban cellular radio 2.7 to 3.5 shadowed urban cellular radio 3 to 5 in building with line of sight 1.6 to 1.8 obstructed in building 4 to 6 Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 33. Wireless Communications and Networking Ch2 - Page 33 of 44 • Shadowing: As the mobile moves in uneven terrain, it often travels into a propagation shadow behind a building or a hill or other obstacle much larger than the wavelength of the transmitted signal, and the associated received signal level is attenuated signiﬁcantly. • A log-normal distribution is a popular model for characterizing the shadowing process. Let ǫ(dB) be a zero-mean Gaussian distributed random variable (in dB) with standard de- viation σǫ (in dB). The pdf of ǫ(dB) is given by fǫ(dB)(x) = 1 √ 2πσǫ exp(− x2 2σ2 ǫ ). Let Lp(d) denote the overall long-term fading (in dB). Then, Lp(d) = ¯Lp(d) + ǫ(dB) = ¯Lp(d0) + 10κ log10( d d0 ) + ǫ(dB) (dB). • ǫ(dB) follows the Gaussian (normal) distribution =⇒ ǫ in linear scale is said to follow a log-normal distribution with pdf given by fǫ(y) = 20/ ln 10 √ 2πyσǫ exp[− (20 log10 y)2 2σ2 ǫ ]. • σǫ: 8 dB for an outdoor cellular system and 5 dB for an indoor environment. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 34. Wireless Communications and Networking Ch2 - Page 34 of 44 2.5 Small-Scale Multipath Fading Consider a ﬂat fading channel with N distinct scatterers: r(t) = N n=1 αn(t)e−j2πfcτn(t) x(t − τn(t)) ≈ [ N n=1 αn(t)e−j2πfcτn(t) ]x(t − ¯τ). The complex gain of the channel is Z(t) = N n=1 αn(t)e−j2πfcτn(t) = Zc(t) − jZs(t) where Zc(t) = N n=1 αn(t) cos θn(t) Zs(t) = N n=1 αn(t) sin θn(t) and θn(t) = 2πfcτn(t). Also, Z(t) = α(t) exp[jθ(t)] where α(t) = Z2 c (t) + Z2 s (t), θ(t) = tan−1 [Zs(t)/Zc(t)]. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 35. Wireless Communications and Networking Ch2 - Page 35 of 44 c Prentice Hall Transmitter Receiver LOS path α0(t)e−jθ0(t) NLOS path Figure 17: NLOS versus LOS scattering Rayleigh fading (NLOS propagation) E[Zc(t)] = E[Zs(t)] = 0. (17) Assume that, at any time t, for n = 1, 2, . . . , N, a. the values of θn(t) are statistically independent, each being uniformly distributed over [0, 2π]; b. the values of αn(t) are identically distributed random variables, independent of each other and of the θn(t)’s. =⇒ According to the central limit theorem, Zc(t) and Zs(t) are approximately Gaussian random variables at any time t if N is sufﬁciently large. Zc and Zs are independent Gaussian random variables with zero mean and equal variance σ2 z = 1 2 N n=1 E[α2 n]. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 36. Wireless Communications and Networking Ch2 - Page 36 of 44 =⇒ fZcZs (x, y) = 1 2πσ2 z exp[− x2 + y2 2σ2 z ], − ∞ < x < ∞, − ∞ < y < ∞. =⇒ The amplitude fading, α, follows a Rayleigh distribution with parameter σ2 z, fα(x) = x σ2 z exp(− x2 2σ2 z ), x ≥ 0 0, x < 0 , (18) with E[α] = σz π/2 and E(α2 ) = 2σ2 z; The phase distortion follows the uniform distribution over [0, 2π], fθ(x) = 1 2π , 0 ≤ x ≤ 2π 0, elsewhere ; The amplitude fading α and the phase distortion θ are independent. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 37. Wireless Communications and Networking Ch2 - Page 37 of 44 Rician Fading (LOS propagation) Z(t) = Zc(t) − jZs(t) + Γ(t), where Γ(t) = α0(t)e−jθ0(t) is the deterministic LOS component. E[Z(t)] = Γ(t) = 0. The distribution of the envelope at any time t is given by the Rayleigh distribution mod- iﬁed by a. a factor containing a non-centrality parameter, and b. a zero-order modiﬁed Bessel function of the ﬁrst kind. The resultant pdf for the amplitude fading at any t, α, is known as the Rician distribution, given by fα(x) = x σ2 z exp(− x2 2σ2 z ) Rayleigh · exp{− α2 0 2σ2 z } · I0( α0x σ2 z ) modiﬁer = x σ2 z exp(− x2 + α2 0 2σ2 z )I0( α0x σ2 z ), x ≥ 0, where α0 is α0(t) at any t, α2 0 is the power of the LOS component and is the non-centrality parameter, I0(·) is the zero-order modiﬁed Bessel function of the ﬁrst kind and is given by I0(x) = 1 2π 2π 0 exp(x cos θ)dθ. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 38. Wireless Communications and Networking Ch2 - Page 38 of 44 • The K factor: K ∆ = Power of the LOS component Total power of all other scatterers = α2 0 2σ2 z . K → 0 =⇒ the Rician distribution → the Rayleigh distribution; K → ∞ =⇒ the Rician fading channel → an AWGN channel. 0 1 2 3 4 5 6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 x fα (x) © Prentice Hall K=−∞ dB K=−5 dB K=0 dB K=5 dB Figure 18: Rayleigh and Rician fading distributions with σz = 1 Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 39. Wireless Communications and Networking Ch2 - Page 39 of 44 Second-Order Statistics - LCR and AFD Level Crossing Rate (LCR): The crossing rate at level R of a ﬂat fading channel is the expected number of times that the channel amplitude fading level, α(t), crosses the speciﬁed level R, with a positive slope, divided by the observation time interval. α Copyright Prentice Hall t T t t t t2 3 51 4 0 (t) t R Figure 19: Level crossing rate and average duration of fades NR = E[upward crossing rate at level R]. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 40. Wireless Communications and Networking Ch2 - Page 40 of 44 Let ˙α = dα(t)/dt denote the amplitude fading rate and let fα ˙α(x, y) denote the joint pdf of the amplitude fading α(t) and its derivative ˙α(t) at any time t. Then, NR = ∞ 0 yfα ˙α(x, y)|x=Rdy. For the Rayleigh fading environment, fα ˙α(x, y) = x 2πσ2 ˙ασ2 z exp[− 1 2 ( x2 σ2 z + y2 σ2 ˙α )], x ≥ 0, − ∞ < y < ∞, where σ2 ˙α = 1 2 (2πνm)2 σ2 z and νm is the maximum Doppler shift. The LCR is NR = ∞ 0 y · R 2πσ2 ˙ασ2 z exp[− 1 2 ( R2 σ2 z + y2 σ2 ˙α )]dy = √ 2πνm( R √ 2σz ) exp(− R2 2σ2 z ). Letting ρ = R √ 2σz we have NR = √ 2πνmρ exp(−ρ2 ). Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 41. Wireless Communications and Networking Ch2 - Page 41 of 44 −10 −8 −6 −4 −2 0 2 4 6 8 10 10 −4 10 −3 10 −2 10 −1 10 0 ρ (dB) Normalizedlevelcrossingrate,ρexp(−ρ2 ) © Prentice Hall Figure 20: The normalized level crossing rate of the ﬂat Rayleigh fading channel Copyright Prentice Hall α t 0 b c (t) Figure 21: An example of amplitude fading level versus time Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 42. Wireless Communications and Networking Ch2 - Page 42 of 44 Average Fade Duration (AFD) The average fade duration at level R is the average period of time for which the channel amplitude fading level is below the speciﬁed threshold R during each fade period. Let χR denote the AFD. It is a statistic closely related to the LCR. Mathematically, the AFD can be represented as χR = E[the period that the amplitude fading level stays below the threshold R in each upward crossing]. =⇒ NR · χR = lim T→∞ MT T · MT i=1 ti MT = lim T→∞ MT i=1 ti T = P(α ≤ R). For the Rayleigh fading environment, the cdf of α is P(α ≤ x) = x 0 fα(y)dy = 1 − exp(− x2 2σ2 z ). The corresponding AFD is χR = P(A ≤ R) NR = 1 − exp(−R2 /2σ2 z) √ 2πνm(R/ √ 2σz) exp(−R2/2σ2 z) = exp(ρ2 ) − 1 √ 2πνmρ . Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 43. Wireless Communications and Networking Ch2 - Page 43 of 44 −10 −8 −6 −4 −2 0 2 4 6 8 10 10 −1 10 0 10 1 10 2 10 3 10 4 ρ (dB) Normalizedaveragefadeduration,[exp(ρ2 )−1]/ρ © Prentice Hall Figure 22: The normalized average fade duration of the ﬂat Rayleigh fading channel Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra
- 44. Wireless Communications and Networking Ch2 - Page 44 of 44 Example 2.8 The LCR NR and AFD χR Consider a mobile cellular system in which the carrier frequency is fc = 900 MHz and the mobile travels at a speed of 24 km/h. Calculate the AFD and LCR at the normalized level ρ = 0.294. Solution: At fc = 900 MHz, the wavelength is λ = c fc = 3×108 900×106 = 1 3 m. The velocity of the mobile is V = 24 km/h = 6.67 m/s. The maximum Doppler frequency is νm = V/λ = 6.67 1/3 = 20 Hz. The average duration of fades below the normalized level ρ = 0.294 is χR = eρ2 − 1 √ 2πνmρ = e(0.294)2 − 1 √ 2π × 20 × 0.294 = 0.0061 s. The level crossing rate at ρ = 0.294 is NR = √ 2πνmρe−ρ2 = √ 2π × 20 × 0.294e−(0.294)2 = 16 upcrossings/second. Provide by Mr. Achyuta Nanda Mishra Provided by Mr. Achyuta Nanda Mishra