— Lecture Notes —

Wireless Communications I

- Part 1 -

The Physical Wireless Channel

A Course of the Undergraduate Programin Electrical and Computer Engineering

Jacobs University Bremen

Prof. Dr. Giuseppe Abreu

August 30, 2015

Lecture 1

Taking to the Air and Back to the Wire

Emission of Linear Antenna in Free Space

• A short linear antenna of length L excited with an alternate current I =I0e

j2πft, produces at a point at a distance r making an angle θ with theantenna, the following electro-magnetic waves:

E† =I0L cos θ

2πε0c

(1

r2+

c

j2πfr3

)ej2πf(t−r/c) (0.1)

E =I0L sin θ

4πε0c2

(j2πf

r+

c

r2+

c2

j2πfr3

)e−j2πf(t−r/c) (0.2)

H =I0L sin θ

4πε0c2

(j2πf

r+

c

r2

)ej2πf(t−r/c) (0.3)

where ε0 is the permittivity of the medium (vacuum/air).

In the near field things are “messy”!

• Far away (far-field), only E and H are significant, and simply to

E −−−−−→r1

jI0Lf sin θ

2rε0c2e−j2πf(t−r/c) (0.4)

H −−−−−→r1

jI0Lf sin θ

2rε0c2ej2πf(t−r/c) (0.5)

In the far field E and H are equivalent!

Response of Linear Antenna in Free Space

• Although it is clear from equations (0.4) and (0.5) that E and H aremathematically equivalent (in the far field), it is actually the magneticfield H that induces a current at the receiver’s antenna, specifically

Ir = −dH

dt=πI0Lf

2 sin θ

rε0c2exp

(j2πf(t− r/c)

)=α

rexp

(j2πf(t− r/c)

)(0.6)

where α = πI0Lf2 sin θ/ε0c

2.

Note: Many (or most!) text books are, strictly speaking, incorrect intalking about propagation over the electric field E. While not an issueon far field, in near field (e.g. body area networks) more care must betaken.

Large Scale Effects

Power Loss: Free space, stationary

Let us study some of the effects the physical environment impose over the ef-fectively received signal under stationary, free space conditions i.e.,

Er , <Ir =α

rcos (2πf(t− r/c)) (0.7)

• The amplitude of the signal reduces with 1/r:

• The power of the signal reduces with 1/r2:

E[E2r ] =

α2

2r2(0.8)

Free space is very similar to the wire channel!

E[I2] =E[V 2]

Ω2=

P

Ω2

Doppler Distortion: Free space, mobile

Consider a mobile receiver with velocity ~v. Let v denote the scalar speed ondirection of the line between transmitter and receiver. Then

r′ = r + vt (0.9)

Er =α

r + vtcos (2πf(1− v/c)t− 2πfr/c) (0.10)

• The power of the electric field stays the same:

E[E2r ] =

α2

E[(r + vt)2]E[cos2 (2πf(1− v/c)t− 2πfr/c)]

−−−→r1T∝1/f

α2

2r2(0.11)

2

• The frequency of the electric field varies with v:

f ′ = f · (1− v/c) (Doppler shift) (0.12)

Doppler shift causes distortion!

Shadowing: Free space, two paths, stationary

If two paths of lengths r1 and r2 exist between transmitter and receiver, then:

Er =α

r1cos (2πf(t− r1/c))−

α

r2cos (2πf(t− r2/c)) (0.13)

• A deterministic phase difference is observed:

φ1 = −2πfr1/c; φ2 = −2πfr2/c+ π (0.14)

∆φ = 2πf∆r

c+ π (0.15)

• The power of the electric field varies with the location:

E[E2r ] −−−→

r1T∝1/f

α2

r2− α2

r2E[cos (2πft) · cos (2πft+ ∆φ)] (0.16)

Shadowing is perceived as “blind spots.”The phenomenon is technically known as spatial selectivity!

Note: Simplistic definitions of delay spread and coherence band-width can be made at this point:

(delay spread) ∆τ ,∆r

c⇐⇒ Bc ,

1

∆τ(coherence bandwidth)

3

Figure 1: Interference of two sinusoidal waves of same frequency.

The reciprocal relationship between delay spread and coherencebandwidth is general. However, different proportionality constants

may be adopted (see e.g. [1, pp. 15]).

Path Loss Exponent: Ground plane reflection

Consider the special case where the second path is due to a ground planereflection, as depicted in figure . Then

hs

hs − hr

hs + hr

θθ

hr

r

r2

r1

Figure 2: Geometry of the ground plane two-path channel.

• The Delay spread varies with the inverse of r.

4

Homework 1: Show that

∆τ ,−−−−−−−→r(hs,hr)

2hshrrc

(0.17)

Hint: Use the asymptotic result√

1 + x −−−→x1

1 + x/2.

• The amplitude of the received signal decreases with 1/r2

Homework 2: Show that

Er =α

r1cos (2πf(t− r1/c))−

α

r2cos (2πf(t− r2/c))

−−−−−−−→rf/c

−4πfαhshrcr2

sin[2πf(t− r1/c)] (0.18)

Hint: Assume that r1 ≈ r2.

Homework 3: Show the same without assuming that r1 ≈ r2.

Hint: Use the asymptotic result (1 + x)−1 −−−→x1

1− x.

• The power of the received signal decreases with 1/r4

E[E2r ] −−−−−→

r1T∝1/f

β2

2r4(0.19)

where β =4πfαhshr

c.

Fresnel Diffraction: Knife-edge Obstacle

Consider the special case where the second path is due to diffraction at aknife-edge obstacle, as depicted in figure . Then

5

T

R

hs

hr

d1

d2

ψ1

ψ2

h h′

ψ

Figure 3: Geometry of knife-edge two-path channel.

• A deterministic phase difference is observed:

∆φ =π

2ν2 (0.20)

with

ψ = hd1 + d2

d1d2(0.21)

ν = ψ

√2d1d2

λ(d1 + d2)(0.22)

where ν, ψ are the Fresnel coefficient and Fresnel diffraction angle,respectively.

• The amplitude of the received signal varies with ν:

Er −−−−−−−→h(d1,d2)

F (ν) cos(2πft− φ) (0.23)

where F (ν) is the Fresnel diffraction loss

F (ν) =1 + j

2

∫ ∞ν

exp

(−jπx2

2

)dx (0.24)

6

Homework 4: Show that for ν > 1, the Fresnel loss |F (ν)| is tightlyapproximated by

|F (ν)| ≈ 1√2πν

(0.25)

Hint: Start by showing that∫ ∞ν

exp

(−jπx2

2

)dx =

(1

2− C(ν)

)− j

(1

2− S(ν)

)(0.26)

where

C(ν) =

∫ ν

0

cosx2dx; S(ν) =

∫ ν

0

sinx2dx (0.27)

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2Fresnel Diffraction Loss

Loss:|F

(ν)|

Coefficient: ν

Figure 4: Behavior of Fresnel Diffraction Loss/Gain.

7

Mandatory Reading- Goldsmith [2]: Chapter 2

Advised Reading- Tse&Viswanath [1]: Chapter 2 (Section 2.1)

Recommended- MIT Course: Principles of Digital Communications [3],- Lecture 21 by Prof. Gallager (MIT)

8

Lecture 2

Small Scale Effects

Two-path Fading: Free space, two paths, mobile

Let v1 and v2 denote scalar velocities projected on the lines between transmitterand receiver, and between the specular image of the transmitter and the receiver,respectively. Then

r′1 = r1 + v1t; r′2 = r2 + v2t (0.28)

Er =α

r1 + v1tcos (2πf(1− v1/c)t− 2πfr1/c)

− α

r2 + v2tcos (2πf(1− v2/c)t− 2πfr2/c) (0.29)

• The power of the signal varies rapidly with time:

E[E2r ] −−−→

ri1

T∝1/f

α2

r2−α

2

r2E[cos (2πft(1− v1/c))·cos (2πft(1− v2/c) + ∆φ)] (0.30)

Note: Simplistic definitions of Doppler spread and coherence time can bemade at this point:

(Doppler spread) ∆f , f · (v1 − v2)/c⇐⇒ Tc ,1

∆f(coherence time)

Time selectivity (fading)!

Model 1: Ray tracing

Consider the folowing generalization of the two-path model

Er =∑i

βi(t, f) cos(2πf(t− τi(t))

)= <

∑i

βi(t, f) exp(j2πf(t− τi(t))

)(0.31)

For example, for the two-path model:

β1 =α

r1 + v1t; β2 =

α

r2 + v2t

τ1 =v1t+ r1

c− φ1

2πf; τ2 =

v2t+ r2

c− φ2

2πf

φ1 = 0; φ2 = π

9

Model 2: Ray tracing in narrowband complex domain

Consider further the following simplification and generalization to the complexdomain

Er =∑i

βi(t) exp(j2πf(t− τi(t))

)(0.32)

whereby

• We neglect the antenna’s frequency selectivity

βi(t, f) −→ βi(t) (0.33)

• We neglect the bandwidth

F [cos 2πft] = 12 [δ(2πf) + δ(−2πf)]

jF [sin 2πft] = 12 [δ(2πf)− δ(−2πf)]

F [exp(j2πft)] = δ(2πf)

Attention to the short-hand notation: δ(2πf) is a spike at 2πf

Model 3: Channel impulse response

Define the time-variant channel response function in frequency, which interms of Fourier transform is

h(f, t) =

∞∫−∞

h(τ, t) exp(−j2πfτ)dτ (0.34)

h(f, t)F−1

←−−−→F

h(τ, t) (0.35)

Attention to the short-hand notation: Fh(t) = h(f)

• The response of the channel to an arbitrary input:

x(t)F−1

←−−−→F

x(f) (0.36)

10

is given by

Er(t) =

∞∫−∞

x(f)h(f, t) exp(j2πft)df (0.37)

Er(t) =

∞∫−∞

x(f)

∞∫−∞

h(τ, t) exp(−j2πfτ)dτ

exp(j2πft)df

Er(t) =

∞∫−∞

h(τ, t)

∞∫−∞

x(f) exp(j2πf(t− τ))df

dτ

Er(t) =

∞∫−∞

h(τ, t)x(t− τ)dτ (0.38)

A signal in the wireless channel undergoes a time-varyingconvolution with the channel’s response function!

• For the ray tracing case:

h(f, t) ,∑i

βi(t) exp(− j2πfτi(t)

)(0.39)

h(τ, t) −−−−−→F

∑i

βi(t)δ(τ − τi(t)

)(0.40)

such that the response to an input x(t) is

Er(t) =∑i

βi(t)x(τ − τi(t)

)(0.41)

Where did the frequency dependence go?!

∫

The impulse response model arises as a consequence of ignoringfrequency dependence in each path response!

Homework 5: Explain why the absence of frequency dependence onthe ray tracing response to a narrowband signal as shown in (0.41)is intuitively correct.

Hint: Study equation (0.37) qualitatively.

11

Note: Compare equation (0.41) with equation (0.32), and see that theformer is indeed the response to the input x(t) = exp(j2πft)!

Model 4: Baseband Impulse Response

Define the baseband time-variant response function in frequency

hb(f, t) =

h(f + fc, t) if f + fc > 00 if f + fc 6 0

(0.42)

hb(f, t)F−1

←−−−→F

hb(τ, t) (0.43)

• The response of the channel to an arbitrary input is:

Er,b(t) =

∞∫−∞

x(f)hb(f, t) exp(j2πft)df (0.44)

Er,b(t) =

∞∫−∞

x(f)

∞∫−∞

hb(τ, t) exp(−j2π(f + fc)τ)dτ

exp(j2πft)df

Er,b(t) =

∞∫−∞

h(τ, t)

∞∫−∞

x(f) exp(j2πf(t− τ))df

exp(−j2πfcτ)dτ

Er,b(t) =

∞∫−∞

h(τ, t)x(t− τ) exp(−j2πfcτ)dτ =

∞∫−∞

hb(τ, t)x(t− τ)dτ (0.45)

Note: Recall the Fourier transform frequency shift property

x(t) exp(−j2πfct)←−−−→F

x(f + fc)

• For the ray tracing case:

hb(f, t) ,∑i

βi(t) exp(− j2π(f + fc)τi(t)

)(0.46)

hb(τ, t) −−−−−→F

∑i

βi(t)δ(τ − τi(t)

)exp(−j2πfcτi(t)) (0.47)

Er,b(t) =∑i

βi(t)x(τ − τi(t)

)exp(−j2πfcτi(t)) (0.48)

12

• Recall that the propagation delay is

τi(t) =vit

c+

(ric− φi

2πf

)= τ ′i +

fDifc

t (0.49)

where fDi(fc, vi) , fcvi/c is the Doppler shift.

• For the ray tracing case:

Er,b(t) =∑i

βi(t)x(τ − τi(t)

)exp(−j2π(fcτ

′i + fDit)) (0.50a)

=∑i

βi(t) exp(−j2πfcτ ′i)x(τ−τi(t)

)exp(−j2πfDit) (0.50b)

=∑i

β′i(t, τ′i)x(τ − τi(t)

)exp(−j2πfDit) (0.50c)

where β′i(t, τ′i) , βi(t) exp(−j2πfcτ ′i) with βi(t) ,

α

ri + vitsuch that

|β′i(t)| = |βi(t)|

Frequency carrier affects the Doppler shift, butdoes not affect the path coefficient!

Note: Of course different frequencies have different absorptionfactors, but this is not a small scale effect!

• Impulse response, i.e., response to input x(t) = exp(j2πft)

Er,b(t) =∑i

β′i(t, τ′i) δ(τ − τi(t)

)exp(−j2πfDi(fc, vi) t) (0.51)

Doppler shift causes violent and random phase rotations!

Specifically, exp(−j2πfDi(fc, vi) t) goes from 1 to 0 every t =1

4|fDi |

Doppler shift is random, since it is affected by the carrierfrequency and the relative velocity between the

transmitter/receiver pair!

Multiple paths with mismatched Doppler shifts cause fading!

13

Statistical Characteristics of Fading Channels

Let us now pursue a small scale statistical perspective of the wireless channel(valid for short periods of time). Inspired in the models discussed in the previouslecture, consider the more abstract, if simpler, baseband multi path channelmodel below

hb(t) =∑i

βi exp(−jφi(t)) (0.52)

where the dependence of βi with time ignored, while all the other physical pa-rameters of interest such as Doppler shifts, propagation delay etc. are embeddedin the time-varying phase φi(t)

φi(t) , 2πfcτ′i + 2πfDit (0.53)

Autocorrelation Function under Multipath Narrowband Model

Let us furthermore split hb(t) into I and Q components

hI(t) =∑i

βi cos(φi(t)) (0.54)

hQ(t) = −∑i

βi sin(φi(t)) (0.55)

• The autocorrelation function of the in-phase component is given by

AI(τ) , E [hI(t)hI(t+ τ)] =∑i

E[β2i

]E [cos(φi(t)) cos(φi(t+ τ))]

(0.56)which can be simplified to

AI(τ) =1

2

∑i

E[β2i

]E [cos(2πfDiτ)] (0.57)

Homework 6: Prove equation (0.57).

Hint: Substitute equation (0.53) into (0.56) and simplify.

Autocorrelation Function under Clarke’s Multipath Model

• The average powers of all paths are the same:

E[β2i

]= 2P/N (0.58)

• The Doppler shift of each path is proportional to its angle-of-arrival:

fDi =fcvic≈ fcv

ccos θi (0.59)

14

• The angles-of-arrival are uniformly distributed:

θi ≈2πi

N= i∆θ (0.60)

where ∆θ , 2π/N .

• Substituting the approximations above we obtain

AI(τ) =P

N

∑i

E[cos

(2πfcvτ

′i

ccos i∆θ

)]=

P

2π

∑i

cos

(2πfcvτ

′i

ccos i∆θ

)∆θ (0.61)

• Taking the number of terms to infinity (N →∞)

AI(τ) =P

2π

2π∫0

cos

(2πfcvτ

ccos θ

)dθ = PJ0(2πfDτ) (0.62)

where we finally dropped the “ i ” and “ ’ ” from the notation for simplicity,and since those no longer have meaning in the continuous model above.

Note: It is trivial to show that AQ(τ) = AI(τ).

Coherence Time and Doppler Spread

Let us start by emphasizing that the expectation taken in equation (0.61), andthe integration performed in equation (0.62), fD represents the aggregate con-tribution of all phase mismatches caused by the Doppler shifts of all paths.In other words, fD is now to be interpreted (and referred to) as a DopplerSpread.

• The Coherence Time of a (dense) multi path channel is defined as thetime it takes for the channel to become uncorrelated, i.e.

Tc = minτ|A(τ) = 0 (0.63)

Homework 7: Prove that Tc =3

8fD.

Hint: Use the approximation Jη(x) ≈√

2

πxcos(x− π

2 η − π4

).

15

0 0.5 1 1.5 2 2.5 3−0.4

−0.2

0

0.2

0.4

0.6

0.8

1Bessel Function of the First Kind

J0(2πf D

τ)

fDτ

Figure 5: Normalized autocorrelation function of uniformly dense wireless chan-nel (Clarke’s) model.

• The Fourier Transform of the Autocorrelation Function of a chan-nel is referred to as the Power Spectral Density. For the narrowbanddense multi path model above (Clarke’s):

SI(f) , FAI(τ) =

P

2πfD

1√1− (f/fD)2

for |f | 6 fD

0 otherwise(0.64)

Notice how as fD increases, the Power Spectral Density SI(f) tendsto a pair of spikes at −fD and fD, implying that the

Autocorrelation Function tends to a sinusoidal waveform.What does that mean?

Conversely, as fD → 0, the Power Spectral Density SI(f) tends to arectangular function at the origin. What does that mean?

sinc(x) = sin(πτ)πτ −−−−−→

r1rect(f) = u (f + fD)− u (f − fD)

16

−9 −7 −5 −3 −1 1 3 5 7 90

0.5

1

1.5

2

2.5

3

3.5Power Spectral Density

f

S(f)

fD = 2

fD = 4

fD = 8

Figure 6: Power Spectral Density of Narrowband Dense Multipath Channel.

Question: What is the integral of SI(f) over its entire domain,and what does it represent?

Mandatory Reading- Goldsmith [2]: Chapter 3

Additional Reading- Rappaport [4]: Chapter 5- Tse&Viswanath [1]: Chapter 2

Recommendend- Principles of Digital Communications [3],- Lecture 21 by Prof. Gallager (MIT)- At: http://www.youtube.com/watch?v=2DbwtCePzWg

17

Lecture 3

Statistical Characteristics of Fading Channels (Cont.)

Let us return to the baseband channel’s time-domain response function

hb(τ, t) =∑i

βi(t)δ(τ − τi(t)

)exp(−j2πfcτi(t)) (0.65)

Q

Delay

I

q1

i1

τ1

β1

τN−1τN

βN

Figure 7: Illustration of a baseband impulse response.

• The function P(τ) below is known as the Power Delay Profile of thechannel, at the location z.

P(τ) , γEt|z[|hb(τ, t)|2] (0.66)

where γ is normalizing constant.

18

P

Delayτ1

β1

τN

βN

Figure 8: Illustration of the Power Delay Profile function.

Note: Recognize that the Power Delay Profile is actually a large scalecharacteristic of the channel!

• Given a certain PDP, its corresponding average and rms delays can alsobe defined as

µτ =

∫ ∞0

τP(τ)dτ∫ ∞0

P(τ)dτ

(0.67)

στ =

∫ ∞

0

(τ − µτ )2P(τ)dτ∫ ∞0

P(τ)dτ

12

(0.68)

• A widely used PDP model (supported by empirical and analytical evi-dence) is the exponential profile

P(τ) =1

γexp(−τ/γ). (0.69)

Homework 8: Prove that the mean and rms delays of an exponentialPDP shown in equation (0.69) as are both equal, i.e., µτ = στ = γ.

19

Homework 9: Prove that under an exponential PDP as shown inequation (0.69), the number of paths N that arrive in a time T is aPoisson random variable with intensity 1/γ. Or, mathematically

Pno. paths in T = N =e−T/γ

N !

(T

γ

)N. (0.70)

What is the average number of paths?

Hint 1: Study P∑N+1

i τi > T

and P∑N

i τi < T

.

Hint 2: Recall that the sum of exponential variates follow theErlang distribution which yield a Poisson.

τ1 τ2

T

τn τn+1

Tm =∑mi=1 τi

Figure 9: Illustration of exponential arrivals.

Envelope Spectral Crosscorr. and Power Delay Profile

Let us now consider the following simplification of the channel response function

hb(τ) =N∑i=1

hiδ(τ − τi

)(0.71)

where we have neglected the temporal dependence of τi with time, and includedthe time-varying phase rotations caused by Doppler shifts into the complexgains hi.

Recall that N is typically also a random variable!

• Let the Power Delay Profile of the channel be given by

P , [|h1|2, · · · , |hN |2] = [P1, · · · , PN ] (0.72)

20

• If the channel gains hi are independent, the associated covariance matrixis

Rh , E[hh†] = diag[P1, · · · , PN ] (0.73)

where † denotes transpose conjugate, and h , [h1, · · · , hN ].

• Due to the fast rotation of the phases of hi, these random variates canalso be assumed to be zero mean.

Next, consider the frequency domain response function of the channel, givenby the Fourier transform of hb(τ), i.e.

hb(f) ,∫ ∞−∞

hb(τ) exp(−j2πfτ)dτ =

N∑i=1

hi exp(−j2πfτi) (0.74)

• Consider a pair of sinusoidal signals with frequencies f1 and f2, and definethe associated vectors

wk , [exp(j2πfkτ1), · · · , exp(j2πfkτN )] (0.75)

Notice the positive arguments!

• The responses to each of these sinusoidal signals in the frequency do-main can be written in matrix form as

yk = w†k · h (0.76)

• Let us define also the quantities

y = [y1, y2] (0.77)

rk = |yk| (Envelope Response) (0.78)

∆f = f1 − f2 (0.79)

• The cross-correlation coefficient between the envelope responses inthe frequency domain is given by.

C(r1, r2) =E[r1r2]− E[r1]E[r2]

E[r21]− E[r1]2

(0.80)

21

• It can be shown [5] that

E[r1] =

√πP

2(0.81)

E[r21] = P (0.82)

E[r1r2] = 2F1

(− 1

2 ,− 12 ; 1;λ2

) πP (1− λ2)2

4(0.83)

where P ,∑Pi, 2F1(a, b; c; z) is the Gaussian hypergeometric func-

tion and λ2 is the Fourier transform of the sample correlation functionof the Power Delay Profile, specifically

λ2 =

∑Nn

∑Nm PnPm exp(j2π∆f(τn − τm))

P 2(0.84)

• From equation (0.83) one obtains [5]

C(r1r2) =2F1

(− 1

2 ,− 12 ; 1;λ2

)− 1

4/π − 1(0.85)

Note: The result holds for arbitrary Power Delay Profiles!

• Finally, let us show that the Envelope Spectral Cross-correlationCoefficient is the same as the Fourier transform of the Power DelayProfile.

Homework 10: Show from equation (0.85) that C(r1r2) ≈ λ2.

Hint 1: What is the maximum value of λ2?.

Hint 2: Use the approximation 2F1

(− 1

2 ,− 12 ; 1;λ2

)≈ (4/π − 1)λ2 + 1,

for 0 6 λ2 6 1.

The Envelope Spectral Cross-correlation Coefficient is theFourier Transform of the Power Delay Profile!

22

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11

1.05

1.1

1.15

1.2

1.25

1.3

Gaussian Hypergeometric Function

2F1(−

1/2,−1/2;1;λ2)

λ2

Figure 10: Gaussian Hypergeometric Function 2F1

(− 1

2 ,− 12 ; 1;λ2

)and a linear

approximation (4/π − 1)λ2 + 1.

Coherence Bandwidth and RMS Delay

The coherence bandwidth of the channel is defined as the bandwidth overwhich the envelope spectral cross-correlation coefficient is greater than a giventhreshold η.

f∗ =

f

∣∣∣∣ C(r1, r2) = η

(0.86)

• From the above, we may replace the envelope spectral cross-correlation co-efficient by the Fourier Transform of the PDP. Consider the ExponentialPDP of equation (0.69)

P(f) =|F exp(−τ/στ ) |

στ=

1√1 + (2πfστ )2

. (0.87)

• Assuming that 2πfστ 1, we have

f∗ =1

2πηστ. (0.88)

23

The coherence frequency is proportional to theinverse of the rms delay!

Statistical Fading Channel Models

Let us now pursue a large scale statistical perspective of the small scalevariations of amplitude (envelope) and phase observed in the wireless channel.Again, start form the baseband multipath channel model below

hb(t) =∑i

βi exp(−jφi(t)) (0.89)

Then, consider the extension of the model above to the following

z(t) =∑i

αi cos(φi(t)) + jβi sin(φi(t)) (0.90)

z(t) = x(t) + jy(t) (0.91)

where in (0.90) we accommodate for the possibility (so far ignored) that thepath gains may be distinct on the different complex dimensions (polarization),while in (0.91) we assume that the contributions of the sum of a large numberof sinusoidal functions with random phases and amplitudes amount tothe independent random processes x(t) and y(t).

The problem of determining the probability density functions(PDF’s) of the amplitude r , |z| and phase θ , ∠z of random

processes with the structure shown above is oftenreferred to as the Random Phase Problem.

The Central Limit Theorem

Let x1, · · · , xN be independent, random variates with arbitrary distribu-tions. Define µX , EN [xi] and σ2

X , EN [x2i ] − E2

N [xi], where EN denotesexpectation over the set of N samples. Then, the quantity

ZN =NXN −NµX

σX√N

=

N∑i=1

Yi√N

(0.92)

converges to the standard Normal distribution.

Proof: Notice that Yi implicitly defined above has a zero-mean and unit vari-ance. By force of Taylor’s theorem,

ϕY (t) = 1− t2

2+ o(t2), t→ 0 (0.93)

24

where ϕY (t) denotes the characteristic function of the distribution of Y ando(t2) denotes a term that decreases faster then t2 as t→ 0.

Since the characteristic function of a sum is the product of charac-teristic functions, and due to a simple change of variables, it follows that[

ϕY

(t√n

)]n=

[1− t2

2n+ o

(t2

n

)]n→ e−t

2/2, n→∞. (0.94)

It is then identified that the e−t2/2 is the characteristic function of the stan-

dard Normal distribution, which concludes the proof.

If paths are independent, in large quantities, and statisticallyequivalent (that is, no dominant), then by force of the

Central Limit Theorem, equation (0.91) describesa complex Gaussian process.

Side Note: Maximum Entropy Distributions

Let us momentarily depart from the discussion about the Random Phase Prob-lem and highlight the important property of entropy maximality that a fewimportant distributions (two of which already appeared so far) exhibit.In what follows we shall make use of the following optimization technique knownas Lagrange Multiplier method.

Lagrange Multiplier Method

Consider the following constrained optimization problem

minimize/maximize f(x) (0.95)

subject to g(x) = c (0.96)

Define the functionΛ(x, λ) , f(x) + λ ·

(g(x)− c

)(0.97)

Then, the solution of the problem is amongst the solutions of

∇Λ(x, λ) = 0 (0.98)

where ∇ denotes gradient.

The solutions of equation (0.98) are in general inflection points.One must therefore verify/choose an admissible solution.

25

Example 1: Entropy Maximality of Uniform Distribution

Let p(x) be a non-negative function in [a, b] such that∫ b

a

p(x) dx = 1. (0.99)

Due to the non-negativity of p(x) and the normalization constraint given byequation (0.99), p(x) is a probability distribution1.

Next, consider the entropy of p(x)

H(p) , −∫ b

a

p(x) ln p(x) dx = −∫ b

a

p ln p dx. (0.100)

So, we are interested in the problem:

maximize −∫ b

a

p ln p dx (0.101)

subject to

∫ b

a

p dx = 1 (0.102)

Applying the Lagrange Multiplier method, the Lagrangean (on p) is

Λ(p) = −∫ b

a

p ln p dx+ λ

(∫ b

a

p dx− 1

)(0.103)

In this case the gradient reduces to a simple derivative on p, namely

∂Λ(p)

∂p(x) dx= − ln p(x)− 1 + λ (0.104)

from which it follows (solving the root) that

p(x) = eλ−1. (0.105)

Notice from equation (0.105) that p(x) does not depend on x!

Finally, substituting equation (0.105) into (0.99) so as to find the value of λthat satisfied the normalization constraint yields

λ = 1− ln(b− a) (0.106)

1Rigorously speaking these are only two of the three Kolmogorov axioms required of proba-bility distributions. The third (countable additivity) is implied in the (assumed) integrabilityof p(x).

26

Thus, finally we obtain

p(x) =1

b− a, a 6 x 6 b (Uniform Distribution) (0.107)

The maximum entropy distribution in [a, b] withoutprior information is the Uniform Distribution!

Example 2: Entropy Maximality in R with Known µ and σ

Now, what is the maximally entropic distribution in R with known mean andvariance?In other words, what is the solution of the problem:

maximize −∞∫−∞

p(x) ln p(x) dx (0.108)

subject to

∞∫−∞

p(x) dx = 1 (0.109)

and

∞∫−∞

x · p(x) dx = µ (0.110)

and

∞∫−∞

(x− µ)2p(x) dx = σ2 (0.111)

Before we continue, notice that∫Rp(x) ln p(x) dx =

∫Rp(x− µ) ln p(x− µ) dx (0.112)

which can be proved by simple substitution of variables.

The mean constraint is irrelevant!

In other words, we may consider, w.l.g., the spacial case where µ = 0, such thatthe maximization problem reduces to In other words, what is the solution of the

27

problem:

maximize −∞∫−∞

p(x) ln p(x) dx (0.113)

subject to

∞∫−∞

p(x) dx = 1 (0.114)

and

∞∫−∞

x2p(x) dx = σ2 (0.115)

The Lagrange function, taking into account this constraint is then

Λ(p) , −∞∫−∞

p(x) ln p(x) dx+λ1

∞∫−∞

p(x) dx− 1

−λ2

∞∫−∞

x2 p(x) dx− σ2

(0.116)

And the derivative on p is

∂Λp

∂p(x) dx= − ln p(x)− 1 + λ1 − λ2x

2 = 0, (0.117)

such thatp(x) = e(λ1−1)−λ2x

2

= e(λ1−1) · e−λ2x2

. (0.118)

Substituting equation (0.118) into the normalization and variance constraints,respectively, yields

e(λ1−1)

∞∫−∞

e−λ2x2

dx =

√π

λ2· e(λ1−1) = 1

e(λ1−1)

∞∫−∞

x2e−λ2x2

dx =

√π

4λ32

· e(λ1−1) = σ2

=⇒

λ2 =

1

2σ2

e(λ1−1) =1√2πσ

(0.119)Substituting equation (0.119) into (0.118) we have

p(x) =1√2πσ

· e−x2

2σ2 . (0.120)

The maximum entropy distribution in R with prior knowledge ofvariance is the Gaussian Distribution!

28

Example 3: Entropy Maximality of Exp. Distribution

Homework 11: Prove that the maximum entropy distribution inR+ (non-negative Real support) with a known mean is the Exponen-tial Distribution.

Hint: Use the Lagrance Multiplier method.

Recall from Lecture 4 that the Exponential model is commonly usedto describe the Power Delay Profile of wireless channels.

Back to the Random Phase Problem...

Let us return to the Random Phase Problem

z(t) =∑i

αi cos(φi(t)) + jβi sin(φi(t)) = x(t) + jy(t).

With the conviction that x(t) and y(t) can be assumed to be Gaussian ran-dom functions, let us attempt to derive corresponding Fading distributions.

In so-doing, pay close attention to the impact of different choices that µand σ exercise on the resulting fading models!

Rayleigh Solution of the Random Phase Problem

• Assume that x and y are independent and identically distributed(i.i.d) Gaussian variates with zero mean and variance σ2

x ∼ p(x) =1

σ√

2πe−x2

2σ2 y ∼ p(y) =1

σ√

2πe−y22σ2 (0.121)

• Due to independence, the joint distribution of x and y is

p(x, y) =1

σ22πe−(x2+y2)

2σ2 (0.122)

• Going from density to probability

Px, y = p(x, y)dxdy (0.123)

29

• Going from cartesian to polar coordinates

r =√x2 + y2 θ = atan yx (0.124)

dxdy ←→ rdr dθ (0.125)

Pr, θ =r

2πσ2e−r22σ2 dr dθ (0.126)

p(r, θ) =r

2πσ2e−r22σ2 (0.127)

• Marginalizing

p(r) =

2π∫0

r

2πσ2e−r22σ2 dθ =

r

σ2e−r22σ2 , r > 0 (Rayleigh PDF) (0.128)

p(θ) =

∞∫0

r

2πσ2e−r22σ2 dr =

1

2π, 0 6 θ 6 2π (0.129)

Notice that the envelope and phase distributions turn up to beindependent since p(r, θ) = p(r) · p(θ).

Rayleigh channels arise when no line-of-sight between transmitterand receiver exists, and waves are scattered equally in I and Q.

30

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Rayleigh Fading Distribution

r

p(r;σ)

σ = 1

σ = 2

σ = 3

Figure 11: Examples of Rayleigh PDFs.

• Discrete-time Rayleigh variates can be generated via the Box-MullerTransform

x =√− lnu1 sin(2πu2) (0.130a)

y =√− lnu2 cos(2πu1) (0.130b)

where ui are i.i.d. standard uniform random variates.

• Continuous-time Rayleigh variates can be generated via the so-called JakesMethod

R(t, k) = 2√

2

[N∑n=1

ejβn cos (2πfnt+ θn,k) +1√2

cos 2πfDt

](0.131)

where fD is the Doppler spread and the remaining parameters are

βn = πnN+1 (0.132)

θn,k = βn + 2π(k−1)N+1 (0.133)

There are many improved variations of the Jakes Model!

31

Figure 12: Illustration of probability regions of Rayleigh fading.

Rice Solution of the Random Phase Problem

• Assume that x and y are independent Gaussian variates with thesame variance σ2, but unequal means

x ∼ p(x) =1

σ√

2πe−(x−µ)2

2σ2 y ∼ p(y) =1

σ√

2πe−y22σ2 (0.134)

• Due to independence, the joint distribution and joint probability ofx and y are

p(x, y) =1

σ22πe−(x2+y2)

2σ2 e−(µ2−2xµ)

2σ2 Px, y = p(x, y)dxdy (0.135)

• Going from cartesian to polar coordinates

Pr, θ =r

σ2e−(r2+µ2)

2σ21

2πerµ cos θσ2 dr dθ (0.136)

p(r, θ) =r

σ2e−(r2+µ2)

2σ21

2πerµ cos θσ2 (0.137)

32

• Marginalizing

p(r) =

2π∫0

p(r, θ)dθ =r

σ2e−(r2+µ2)

2σ21

2π

2π∫0

erµ cos θσ2 , r > 0

=r

σ2e−(r2+µ2)

2σ2 I0

(rµσ2

)(Rice PDF) (0.138)

p(θ) =

∞∫0

p(r, θ)dr, 0 6 θ 6 2π

= 1+2√πK·cos(θ−φ)·e4K cos2(θ−φ)

2πe4K· (1+erf(2

√K cos(θ−φ))), (0.139)

where K , µ2/2σ2 (i.e., ratio between the power of direct compo-nent and the power of indirect component) is referred to as the RiceFactor, and the the phase shift φ =

π4 ,

3π4 ,

5π4 ,

7π4

.

Notice that if we define Ω = µ2 + 2σ2 (total channel power), thefollowing variation of the Rice PDF can be written

p(r) =2(1 +K)r

Ωe−K−

(1+K)r2

Ω I0

(2r

√K(1 +K)

Ω

)(0.140)

• Rice variates can be generated similarly to Rayleigh simply by addingthe constant µ.

Rice fading occurs when a dominant LOS pathexists amidst the scattered paths.

The Rice PDF reduces to the Rayleigh PDF is µ = 0.

Notice that equation (0.139) implies that Rice fadingis quadrant-stable for large µ.

33

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Rice Fading Distribution

r

p(r;µ,σ

)µ = 0; σ = 1

µ = 1; σ = 1

µ = 1; σ = 2

µ = 2; σ = 2

Figure 13: Examples of Rice PDFs.

Hoyt Solution of the Random Phase Problem

• Assume that x and y are independent zero-mean Gaussian variateswith the distinct variances

x ∼ p(x; b, σ2) =1√

π(1 + b)σ2e− x2

(1+b)σ2 (0.141a)

y ∼ p(y; b, σ2) =1√

π(1− b)σ2e− y2

(1−b)σ2 (0.141b)

where 0 6 b 6 1.

• The variances of the I and Q components are

σ2x =

σ2

2(1 + b) (0.142)

σ2y =

σ2

2(1− b) (0.143)

such that σ2 = σ2x + σ2

y.

34

• Proceeding as before, the PDFs of the envelope and phase are

p(r; b, σ) =2r

σ2√

1− b2e− r2

σ2(1−b2) I0

(br2

σ2(1− b2)

)(Hoyt PDF)

(0.144)

p(θ; b) =

√1− b2

2π(1− b cos(2θ))(0.145)

Homework 12: Prove equations (0.144) and (0.145).

Hint: Start by showing that the joint PDF is

p(r, θ) =r

σ2π√

1− b2e−

1−b cos(2θ)σ2(1−b2) r2

• Alternatively, Hoyt PDFs can be expressed as

p(r; q, σ) =(1 + q2)r

qσ2e−

(1+q2)2r2

4q2σ2 I0

((1− q4)r2

4q2σ2

)(0.146)

p(θ; q) =q2σ2

2π(1 + q2)(sin2 θ + q2 cos2 θ)(0.147)

where −1 6 q 6 1.

Homework 13: Prove equation (0.146).

Hint: Compare equations (0.144) and (0.146). What is the relationshipbetween b and q?.

35

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Hoyt Fading Distribution

r

p(r;b,σ)

b = 0.8; σ = 0.5

b = 0.6; σ = 1

b = 0.2; σ = 2

Figure 14: Examples of Hoyt PDFs.

Figure 15: Examples of the phase distributions of Rician and Hoyt fading.

36

(a) Hoyt (b = 0.5) (b) Hoyt (b = −0.75)

Figure 16: Probability regions of signals under Hoyt fading.

Hoyt channels are found in satellite communications.

Notice the pola rization effect of Hoyt channels.

The Hoyt PDF also reduces to theHalf-normal PDF when b = 0.

Nakagami Solution of the Random Phase Problem

• No assumption over the statistics of x and y!

• Derivation is very hard but yields

p(r;m,σ2) = 2Γ(m)

(mσ2

)mr2m−1 exp(−m

σ2 r2) (Nakagami PDF)

p(θ;m) = Γ(m)2mΓ(m/2)2 | sin(2θ)|m−1 (Nakagami Phase PDF)

• Discrete-time Nakagami-m variates can be generated with

z =

√Ω

2m· (ux√γx + juy

√γy), (0.148)

where ux and uy are independent and uniformly distributed random num-bers belonging to the set −1, 1, while γx and γy are i.i.d. Gamma

37

variates with shape parameter m/2 and scale parameter 2, i.e.,

γi ∼ Γ(m/2, 2) (0.149)

The parameter m is known as the fading figure

• Piecewise continuous Nakagami-m processes with half-integer fadingfigure m can be constructed via

z(t|T ) =

√Ω

dme+ bmc

ux(t|T )

√√√√dme∑i=1

x2i (t) + juy(t|T )

√√√√bmc∑i=1

y2i (t)

,

(0.150)where d·e and b·c denote the ceil and floor operators, and the processesx2i (t) and y2

i (t) are as prescribed in the Jakes method.

• For arbitrary real m, the best approximate method so far is [6]

z =

zL with probability ρ

zU with probability 1− ρ(0.151)

where mL , b2mc/2, mU , mL + 0.5 and

zL ∼ pz(z;mL, σ) (0.152)

zU ∼ pz(z;mU, σ) (0.153)

ρ = 2mL

m(mU −m) (0.154)

38

0 0.5 1 1.5 2 2.5 3 3.5 40

0.25

0.5

0.75

1

1.25

1.5Nakagami-m Fading Distribution

r

p(r;m,σ

)

m = 0.2; σ = 0.5

m = 0.5; σ = 1

m = 1; σ = 2

m = 1.5; σ = 0.5

Figure 17: Examples of the Nakagami-m PDFs.

(a) Nakagami (m = 1.25) (b) Nakagami (m = 2.5)

Figure 18: Probability regions of signals under Nakagami-m fading.

39

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Imag

(z)

Real(z)

Trajectory Plots of Piecewise Continuous Complex−ValuedNakagami− m Channels (4 Snapshots; m = 2)

Figure 19: Snapshots of the piecewise continuous Nakagami-m processes.

Nakagami-m channels reduce to Rayleigh for m = 1,approximate Hoyt for 0.5 6 m < 1, and Rice for m > 1.

Due to this flexibility and also to the simple expression,the Nakagami-m model is the most popular after Rayleigh.

Circular Central-Limit Theorem: Tikhonov PDF

Circular variates are variates with 2π-periodic density

r ∼ p(mod(r, 2π)) (0.155)

• The Tikhonov or von Mises distribution is the equivalent of theGaussian distribution for circular variates

- Closed-form

40

pT (x;α, ξ) ,1

2πI0(α)· eα cos(x−ξ), x ∈ S , [−π, π]

- Jacobi-Anger infinite series representation

pT (x;α, ξ) =1

2π+ 2

∞∑k=1

Ik(α)

I0(α)· cos(k(x− ξ))

• Tikhonov variates can be generated efficiently as described in [7].

The Tikhonov distribution is also known asvon Mises distribution.

It is associated with the phase error at the output ofa Phase Locked Loop (PLL) with input power

proportional to the parameter alpha.

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Tikhonov Phase Distribution

θ

p(θ;α,ξ)

α = 5; ξ = −π/4

α = 3; ξ = 0

α = 1; ξ = π/2

Figure 20: Examples of Tikhonov PDFs.

41

Mandatory Reading- Goldsmith [2]: Chapter 3 (Sections 3.2 and 3.3)

Additional Reading- Rappaport [4]: Chapter 5 (Sections 5.4 to end)- Tse&Viswanath [1]: Chapter 2 (Section 2.3 to end)- Proakis [8]: Chapter 2 (Section 2.1.4)

Recommendend- Website: Lagrange Multiplier Method

. http://en.wikipedia.org/wiki/Lagrange_multiplier

- Article: Q. T. Zhang and S. H. Song, “Exact expression for thecoherence bandwidth of Rayleigh fading channels,” IEEE Trans.Commun., vol. 55, no. 7, pp. 1296 – 1299, 2007.

- Article: G. T. F. de Abreu, “On the moment-determinance andrandom mixture of Nakagami-m variates,” IEEE Transactions onCommunications, vol. 58, no. 9, pp. 2561 - 2575, Sep. 2010.

- Article: G. T. F. de Abreu, “On the generation of Tikhonovvariates,” IEEE Transactions on Communications, vol. 56, no. 7, pp.1157 - 1168, Jul. 2008.

42

References

[1] D. Tse and P. Viswanath, Fundamentals of Wireless Communications.Cambridge University Press, 2008.

[2] A. Goldsmith, Wireless Communications. Cambridge University Press,2005.

[3] Gallager, http://www.youtube.com/watch?v=2DbwtCePzWg.

[4] T. S. Rappaport, Wireless Communications: Principles and Practice.Prentice-Hall, 2002.

[5] Q. T. Zhang and S. H. Song, “Exact expression for the coherence bandwidthof rayleigh fading channels,” IEEE Trans. Commun., vol. 55, no. 7, pp. 1296– 1299, 2007.

[6] G. T. F. de Abreu, “On the moment-determinance and random mixture ofNakagami-m variates,” IEEE Trans. Commun., vol. 58, no. 9, pp. 2561 –2575, Sep. 2010.

[7] ——, “On the generation of Tikhonov variates,” IEEE Trans. Communica-tions, vol. 56, no. 7, pp. 1157–1168, Jul. 2008.

[8] J. G. Proakis, Digital Communications, Fourth Edition. New York, NY:Mc-Graw-Hill, 2000.

43