# Luigi Paura » 15.Statistical Channel Model

### Statistical Channel Model (1/2)

The range of αn(t) is generally large.

Since fc>>1 → Φn(t)=2πfcτ(t)mod 2 π $\in$ (0,2π) can be modeled as a uniform random variable,  namely, Φn(t) ≡ U(0,2π) for any fixed t.

Since z(t) is a sum of independent contributions, with roughly the same strength, it can be modeled as a complex Gaussian process thanks to the Central Limit Theorem.

### Statistical Channel Model (2/2)

|z(t)| can be modeled (for any fixed instant t) as a Rayleigh RV if the components.

zc(t) |z(t)| cos θ (t) and zs(t) |z(t)| sin θ (t) are zero mean, independent and with the same variance σ2

θ(t) ≡ U(0,2π)

If the mean of zc (t) and / or zs(t) are ≠ 0 → |z(t)| is a rice RV.

### Multipath time and coherence bandwidth (1/3)

h(t,τ) is a random process (τ is a parameter).

The statistical characterization in a wide sense is obtained by evaluating the mean and the autocorrelation of h(t, τ ). Alternatively, we can evaluate the mean and the correlations of the complex envelope: $\widetilde h (t, \tau) = \sum_n \alpha_n (t)e^{-j \psi_n (t)}\delta (\tau-\tau_n (t))$

### Multipath time and coherence bandwidth (2/3) $R_{\widetilde h} (\tau_1,\tau_2,\tau,\Delta t)~~~~~E[\widetilde h (\tau_2,t + \Delta t) \widetilde h^* (\tau_1,t)]=R_{\widetilde h}(\Delta t, \tau_1-\tau_2)\delta(\tau_1-\tau_2)=$ $=R_{\widetilde h} (\Delta t,\tau_1 - \tau_2)\delta (\tau_1-\tau_2)=R_{\widetilde h}(\Delta t,\tau) \delta (\tau)$

by assuming $\widetilde h (t,\tau)$ stationary and the scatterers τ1 and τ2 uncorrelated $R_{\widetilde h}(\tau, \Delta t)|_{\Delta t=0} = R_{\widetilde h} (\tau) =$ mean power associated with the scatters which introduces the delay τ.

### Multipath time and coherence bandwidth (3/3) $B_c~~~\frac 1 {T_M}~~~ \text{Coherence Bandwidth of the channel}$

If Bu<<Bc the channel is frequency non – selective namely, flat in frequency. $T_c= \frac 1 {B_d}$

if Ts<<Tc the channel is temporal non-selective namely, is flat in time.

### Flat-flat Channel

If both conditions are verified, then there isn’t Doppler (temporal variability) as well as Multipath (temporal dispersion): $h(t,\tau)=\sum_n \alpha_n (t)e^{-j\phi_n (t)}\delta(\tau-\tau_n(t))~~~=\alpha e^{-j\phi}\delta(\tau)$ $\widetilde c(t)\rightarrow \alpha e^{j\phi}\widetilde c (t)$

### Scattering Function $S_{\widetilde h} (\tau,\rho)~~~~~F_{\Delta t} \{R_{\widetilde h}(\tau, \Delta t)\}$

The scattering function describes the distribution of the spectral components associated with any path of delay Τ. $S_H(\Delta f, \rho)~~~~~ F_\tau \{S_{\widetilde h}(\tau,\rho)\}$

Doppler Power spectrum $S_H(\rho)~~~~~~S_H()\Delta f , \rho )|_{\Delta f = 0 }$

### Power Doppler Spettrum

Power Doppler spectrum $S_H(\rho)~~~~~~S_H()\Delta f , \rho )|_{\Delta f = 0 }$ $B_D~~~~~\underbrace {max}_{\rho} S_H(\rho)$

### Example (1/2)

Ionospheric Channel with two paths:

TM= 1 msec

Bu = 10 KHz

1/Bu =0.1 msec  1/Bu << TM → L =1msec/0.1 msec =10 delays

BDTM < 1 → underspread channel → TM<<Tc the phase estimation of the carrier is possible.

BDTM < 1 → overspread channel → TM>>Tc the phase estimation of the carrier is very difficult.

### Parameters of a fading channel

Ionospheric propagation with short waves:
TM =10-3-10-2 ; BD=10-1 ;SF=10-4-10-3

Mobile Telephony :
TM =10-5 ; BD=100 ;SF=10-3

### Diversity Techniques

How to counteract the fading effects ?

By introducing redundancy:

1. Channel Coding
2. Diversity namely utilizing multiple channels

Kinds of diversity:

1. Temporal Diversity
2. Polarization Diversity
3. Frequency Diversity
4. Spatial Diversity

### Exercise (1/11)

Determine for a channel with doppler spread BD =80 Hz the minimum temporal separation toensure  that the samples of the received signalare  roughly independent.

The coherence time is 1/BD =1/80 =12,5 msec

If the received signal is Gaussian , the samples separated by at least 12,5 msec are roughly independent.

### Exercise (2/11)

An HF channel with nominal bandwidth of 3200 Hz is utilized to transmitt digital information with bit rate of (1) 4800 bps or (2) 20bps.

Assuming that TM = 5msec, select a modulation technique for both bit rates and moreover state if an equalizer is necessary to counteract ISI.

Let us exclude in the presence of fading PAM or QAM the signalling since they carry information also on the amplitude.

If we consider MPSK one has: $R_b=\frac 1 {T_b}= \frac 1 {T/k}= W log M$ $logM\geq \frac{4800}{3200}\Rightarrow log M=2 \Rightarrow M=4$

### Exercise (3/11) $T=2T_b \Rightarrow \frac 1 T =\frac {4800}2 = 2400$ $W=\frac 1 T =2400 Hz \Rightarrow \text{bandwidth excess}=3200-2400=800HZ$

A transmission filter with spectral characteristics of the raised cosine can be utilized to obtain the desired shape. $\frac 1 {2400}=T=0.41 m~sec << 5 m~sec \Rightarrow \text{an equalizer is necessary}$

If the bit-rate is 20bps, by utilizing again a 4PSK 1/T =20/2 → T= 100 msec >> 5 msec → ISI is negligible

If we want to use all the available bandwidth we can resort to a Spread Spectrum technique.

### Exercise (4/11)

Let us consider an HF channel with nominal bandwidth 3200 Hz e TM =1ms. Determine the minimum number N of subcarriers operating to assure 4800bps and verify the condition: Tsc>>TM

Tsc = symbol duration for each subcarrier
Since TM = 1ms we choose Tsc =100·TM =100ms → Tsc>>TM namely: $B_{sc}=\frac 1 {T_{sc}}=10 Hz \Rightarrow B_{sc}<

If we utilize again a 4PSK for any subcarrier → Rb =2/100ms = =20bps →N = 4800/20 =240 carriers.

If we utilize 16PSK → 40bps for carrier →N = 120

### Exercise (5/11)

Determine the P(e) for a BPSK operating over a fading flat-flat channel Rayleigh distributed. We assume that the phase of the received signal is estimated and the optimum AWGN RX is utilized. $h(t,\tau)=\alpha \delta(t-\tau_0)$ $f(\alpha)=\left\{ \begin{array}{rl} \frac {\alpha}{\sigma^2}e^{-\alpha^2/2\sigma^2} \alpha \geq 0 \\ 0 \alpha <0 \end{array}\right.$ $r(t)=\aplha\sqrt{\frac{2E_b}{T_b}}cos(2\pi f_ct+m\pi + \phi)+n(t)$

### Exercise  (6/11)

The output of the demodulator is: $r=\alpha \sqrt{E_b}cos m \pi + n ~~~m=0,1$

For a fixed value of α: $P(e|\alpha)=Q\Biggl( \sqrt{\frac{2\alpha^2E_b}{N_0}}\Biggr)$ $P(e)=\int_0^\infty f(\alpha)Q\Biggr(\sqrt{\frac{2\alpha^2E_b}{N_0}}\Biggr) d\alpha$

### Exercise (10/11)

An antipodal signalling ± s(t) is utilized over a channel with only amplitude fading: r(t) ± αs(t)+ n(t) 0≤t≤T  with n(t) AWGN and

f(α)=0.1δ(α)+0.9δ(α-2)

Determine the P(e) for the demodulator which utilizes the matched filter to s(t).
P(e) → ? When E/N0 →∞
Assume that the signal is transmitted over two channels statistically indipendent with

f(αk)=0.1δ(αk)+0.9δ(αk-2) k=1,2

and noises are statistically independent.

The demodulator employ two matched filters and summ the two outputs to obtain the decision variable. Evaluate the P(e)

### Exercise (11/11) $P_e(a)=Q\Biggl(\sqrt{\frac{2\alpha^2\epsilon}{N_0}}\Biggr)$ $P(e)=\frac{0.1}2 + 0. 9 Q\Biggl(\sqrt{\frac{8E_b}{N_0}}\Biggr)$ $\frac E {N_0}\rightarrow \infty \Rightarrow P(e)\rightarrow \frac{0.1}2=0.05$ $r=\alpha_1 E+\alpha_2 E+n_1+n_2=(\alpha_1+\alpha_2)E+n$

with n a Gaussian RV with variance N0E $P(e|\alpha_1,\alpha_2)=Q\Biggl(\sqrt{\frac{(\alpha_1+\alpha_2)^2E}{N_0}}\Biggr)$

### Exercise $P(e)=\sum_{\alpha_1,\alpha_2}p(\alpha_1,\alpha_2)=Q\Biggl(\sqrt{\frac{(\alpha_1+\alpha_2)^2E}{N_0}}\Biggr)$ $=\frac{0.01}2 + 0.09 Q\Biggl(\sqrt{\frac{4E_b}{N_0}}\Biggr)+0.81Q\Biggl(\sqrt{\frac{16E_b}{N_0}}\Biggr)$ $\frac E{N_0}\rightarrow \infty P(e)\rightarrow \frac{0.01}2=0,05 \Rightarrow P(e)~~\text{is 10 times maller}$

### Exercise

A binary comunication system transmits the same information over two channels utilizing an antipodal signalling. The received signals are: $r_1=\pm \sqrt{E_b}+n_1$ $r_2=\pm \sqrt{E_b}+n_2$

With n1 and n2 statistically independent, zero mean and variances σ12 e σ22

The detector performs the decision based on: r r1+kr2

Determine the value of k that minimizes the P(e)

### Exercise

r is Gaussian RV with $E(r)=E(r_1)+E(r_2)=(1+k)\sqrt{\epsilon_b}=m_b$

If it has been transmitted $+\sqrt{E_b}$ $\sigma_r^2=\sigma_1^2+k^2\sigma_2^2$ $f(r)=\frac 1 {\sqrt{2\pi}\sigma_r}e^{-\frac{(r-m_r^2)}{2\sigma_r^2}}$ $P(e=P(e|+\sqrt{E_b})=\int_{-\infty}^0 f(r)dr$ $=\frac 1 {\sqrt{2\pi}}\int_{-\infty}^{0} e^{-\frac{x^2}2}dx$ $=Q\Biggr(\sqrt\frac{m_r^2}{\sigma_r^2}\Biggr)$

where: $\frac{m^2_r}{\sigma_r^2}=\frac{(1+k)^\epsilon_b}{\sigma_1^2+k^2\sigma_2^2}$

### Exercise $k_{ott}:\frac d {dr}\Biggr(\frac {m_r^2}{\sigma_r^2}\Biggl)=0 ~~~\Rightarrow ~~~~k_{ott}=\frac {\sigma_1^2}{\sigma_2^2}$ $k_{ott}>1 ~\text{se}~ \sigma_1^2>\sigma_2^2 \Rightarrow r_2~~\text{is weight more than} ~ r_1$ $k_{ott}<1 ~\text{se}~ \sigma_1^2<\sigma_2^2 \Rightarrow r_1~~\text{is weight more than} ~ r_2$ $k_{ott}=1 ~\text{se}~ \sigma_1^2=\sigma_2^2 \Rightarrow r_2~~\text{is weight as} ~ r_1$ $\text{If}~ \sigma_2^2=3\sigma_1^2 \Rightarrow k=1/3 \Rightarrow ~~~~~\frac{m_r^2}{\sigma_1^2}=\frac {(1+\frac 1 3 )^2\epsilon_b}{\sigma_1^2+\frac 1 9 (3\sigma_1^2)}=\frac 4 3 (\frac{\epsilon_b}{\sigma_1^2})$ $\rext{If}~k=1$ $\frac{m_r^2}{\sigma_r^1}=\frac{4\epsilon}{\sigma_1^2+3\sigma_1^2}=\frac{\epsilon_b}{\sigma_1^2}~~\Rightarrow ~~ 10 log \frac 4 3 =1.25~~ dB$

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