Module 1

— Lecture Notes —
Wireless Communications I

- Part 1 The Physical Wireless Channel

A Course of the Undergraduate Program
in 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 =
I0 ej2πf t , produces at a point at a distance r making an angle θ with the
antenna, the following electro-magnetic waves:


c
I0 L cos θ 1
†
+
ej2πf (t−r/c)
(0.1)
E =
2πε0 c
r2
j2πf r3


c
c2
I0 L sin θ j2πf
e−j2πf (t−r/c)
(0.2)
+
+
E=
4πε0 c2
r
r2
j2πf r3


I0 L sin θ j2πf
c
H=
+ 2 ej2πf (t−r/c)
(0.3)
4πε0 c2
r
r
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

H −−−−−→
r1

jI0 Lf sin θ −j2πf (t−r/c)
e
2rε0 c2

(0.4)

jI0 Lf sin θ j2πf (t−r/c)
e
2rε0 c2

(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 are
mathematically equivalent (in the far field), it is actually the magnetic
field H that induces a current at the receiver’s antenna, specifically
Ir = −



dH
πI0 Lf 2 sin θ
=
exp j2πf (t − r/c)
2
dt
rε0 c


α
= exp j2πf (t − r/c)
r

where α = πI0 Lf 2 sin θ/ε0 c2 .

(0.6)

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

Large Scale Effects
Power Loss: Free space, stationary
Let us study some of the effects the physical environment impose over the effectively received signal under stationary, free space conditions i.e.,
α
Er , < {Ir } = cos (2πf (t − r/c))
(0.7)
r
• The amplitude of the signal reduces with 1/r:
• The power of the signal reduces with 1/r2 :
E[Er2 ] =

α2
2r2

(0.8)

Free space is very similar to the wire channel!
E[I 2 ] =

P
E[V 2 ]
= 2
Ω2
Ω

Doppler Distortion: Free space, mobile
Consider a mobile receiver with velocity ~v . Let v denote the scalar speed on
direction of the line between transmitter and receiver. Then
r0 = r + vt
Er =

α
cos (2πf (1 − v/c)t − 2πf r/c)
r + vt

(0.9)
(0.10)

• The power of the electric field stays the same:
E[Er2 ] =

α2
E[cos2 (2πf (1 − v/c)t − 2πf r/c)]
E[(r + vt)2 ]
α2
−−
−
→
r1
2r2
T ∝1/f

2

(0.11)

• The frequency of the electric field varies with v:
f 0 = 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
α
cos (2πf (t − r1 /c)) −
cos (2πf (t − r2 /c))
r1
r2

(0.13)

• A deterministic phase difference is observed:
φ1 = −2πf r1 /c;

φ2 = −2πf r2 /c + π

∆φ = 2πf

∆r
+π
c

(0.14)
(0.15)

• The power of the electric field varies with the location:
E[Er2 ] −−
−→
r1
T ∝1/f

α2
α2
−
E[cos (2πf t) · cos (2πf t + ∆φ)]
r2
r2

(0.16)

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

Note: Simplistic definitions of delay spread and coherence bandwidth can be made at this point:
(delay spread)

∆τ ,

∆r
1
⇐⇒ Bc ,
c
∆τ

3

(coherence bandwidth)

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

The reciprocal relationship between delay spread and coherence
bandwidth 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 plane
reflection, as depicted in figure . Then
r1
hs − hr
hs

r2

hs + hr

hr
θ

θ
r

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 )

Hint: Use the asymptotic result

√

2hs hr
rc

(0.17)

1 + x −−−→ 1 + x/2.
x1

• The amplitude of the received signal decreases with 1/r 2
Homework 2: Show that
Er =

α
α
cos (2πf (t − r1 /c)) −
cos (2πf (t − r2 /c))
r1
r2
4πf αhs hr
−−−−−−−→ −
sin[2πf (t − r1 /c)]
cr2
rf /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 −−−→ 1 − x.
x1

• The power of the received signal decreases with 1/r 4
E[Er2 ] −−−
−−→
r1
T ∝1/f

where β =

β2
2r4

(0.19)

4πf αhs hr
.
c

Fresnel Diffraction: Knife-edge Obstacle
Consider the special case where the second path is due to diffraction at a
knife-edge obstacle, as depicted in figure . Then

5

ψ
ψ1

h0

h

T
d1

ψ2
d2
R

hs
hr

Figure 3: Geometry of knife-edge two-path channel.
• A deterministic phase difference is observed:
π
∆φ = ν 2
2
with
d1 + d2
ψ=h
d d
r 1 2
2d1 d2
ν=ψ
λ(d1 + d2 )

(0.20)

(0.21)
(0.22)

where ν, ψ are the Fresnel coefficient and Fresnel diffraction angle,
respectively.
• The amplitude of the received signal varies with ν:
Er −−−−−−−→ F (ν) cos(2πf t − φ)

(0.23)

h(d1 ,d2 )

where F (ν) is the Fresnel diffraction loss


Z
1+j ∞
−jπx2
F (ν) =
exp
dx
2
2
ν

6

(0.24)

Homework 4: Show that for ν > 1, the Fresnel loss |F (ν)| is tightly
approximated by
|F (ν)| ≈ √

1
2πν

(0.25)

Hint: Start by showing that






Z ∞
−jπx2
1
1
exp
dx =
− C(ν) − j
− S(ν)
2
2
2
ν
where

Z
C(ν) =

ν

cos x2 dx;

Z
S(ν) =

0

(0.26)

ν

sin x2 dx

(0.27)

0

Fresnel Diffraction Loss

1.2
1.1
1
0.9
0.8
Loss: |F (ν)|

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
−5

−4

−3

−2

−1

0
1
Coefficient: ν

2

3

Figure 4: Behavior of Fresnel Diffraction Loss/Gain.

7

4

5

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 transmitter
and receiver, and between the specular image of the transmitter and the receiver,
respectively. Then
r10 = r1 + v1 t;
Er =

r20 = r2 + v2 t

(0.28)

α
cos (2πf (1 − v1 /c)t − 2πf r1 /c)
r1 + v1 t
α
−
cos (2πf (1 − v2 /c)t − 2πf r2 /c)
r2 + v2 t

(0.29)

• The power of the signal varies rapidly with time:
E[Er2 ] −r−−
→
1
i
T ∝1/f

α2 α2
− E[cos (2πf t(1 − v1 /c))·cos (2πf t(1 − v2 /c) + ∆φ)] (0.30)
r2 r2

Note: Simplistic definitions of Doppler spread and coherence time can be
made 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
X


Er =
βi (t, f ) cos 2πf (t − τi (t))
i

(
= <

)
X
i

βi (t, f ) exp j2πf (t − τi (t))




For example, for the two-path model:
α
α
β1 =
; β2 =
r1 + v1 t
r2 + v2 t
φ1
v2 t + r2
φ2
v1 t + r1
τ1 =
−
; τ2 =
−
c
2πf
c
2πf
φ1 = 0; φ2 = π

9

(0.31)

Model 2: Ray tracing in narrowband complex domain
Consider further the following simplification and generalization to the complex
domain
X


Er =
βi (t) exp j2πf (t − τi (t))
(0.32)
i

whereby
• We neglect the antenna’s frequency selectivity
βi (t, f ) −→ βi (t)

(0.33)

• We neglect the bandwidth
F[cos 2πf t] = 21 [δ(2πf ) + δ(−2πf )]
jF[sin 2πf t] = 21 [δ(2πf ) − δ(−2πf )]
F[exp(j2πf t)] = 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 in
terms of Fourier transform is
Z∞
h(f, t) =

h(τ, t) exp(−j2πf τ )dτ

(0.34)

−∞
F −1

h(f, t) ←−−−→ h(τ, t)
F

(0.35)

Attention to the short-hand notation: F{h(t)} = h(f )
• The response of the channel to an arbitrary input:
F −1

x(t) ←−−−→ x(f )
F

10

(0.36)

is given by
Z∞
x(f )h(f, t) exp(j2πf t)df

Er (t) =

(0.37)

−∞

Z∞
Er (t) =



Z∞

x(f ) 
−∞


h(τ, t) exp(−j2πf τ )dτ  exp(j2πf t)df

−∞

Z∞
Er (t) =



Z∞

h(τ, t) 
−∞

−∞
Z∞

Er (t) =
−∞


x(f ) exp(j2πf (t − τ ))df  dτ
h(τ, t)x(t − τ )dτ

(0.38)

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

• For the ray tracing case:
h(f, t) ,

X
i



βi (t) exp − j2πf τi (t)

h(τ, t) −−−−−→

X

F

i



βi (t)δ τ − τi (t)

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


Er (t) =
βi (t)x τ − τi (t)

(0.39)

(0.40)

(0.41)

i

Z

Where did the frequency dependence go?!
The impulse response model arises as a consequence of ignoring
frequency dependence in each path response!

Homework 5: Explain why the absence of frequency dependence on
the 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 the
former is indeed the response to the input x(t) = exp(j2πf t)!

Model 4: Baseband Impulse Response
Define the baseband time-variant response function in frequency


h(f + f c, t) if f + fc > 0
hb (f, t) =
0
if f + fc 6 0

(0.42)

F −1

hb (f, t) ←−−−→ hb (τ, t)

(0.43)

F

• The response of the channel to an arbitrary input is:
Z∞
Er,b (t) =
x(f )hb (f, t) exp(j2πf t)df

(0.44)

−∞

Z∞
Er,b (t) =



Z∞

x(f ) 
−∞

hb (τ, t) exp(−j2π(f + fc )τ )dτ  exp(j2πf t)df

−∞

Z∞
Er,b (t) =





Z∞

h(τ, t) 
−∞

−∞


x(f ) exp(j2πf (t − τ ))df  exp(−j2πfc τ )dτ

Z∞
Z∞
Er,b (t) = h(τ, t)x(t − τ ) exp(−j2πfc τ )dτ = hb (τ, t)x(t − τ )dτ
−∞

(0.45)

−∞

Note: Recall the Fourier transform frequency shift property
x(t) exp(−j2πfc t) ←−−−→ x(f + fc )
F

• For the ray tracing case:
X


hb (f, t) ,
βi (t) exp − j2π(f + fc )τi (t)

(0.46)

i

hb (τ, t) −−−−−→
F

Er,b (t) =

X
i

X
i



βi (t)δ τ − τi (t) exp(−j2πfc τi (t))



βi (t)x τ − τi (t) exp(−j2πfc τi (t))

12

(0.47)
(0.48)

• Recall that the propagation delay is


vi t
ri
φi
fD
τi (t) =
+
−
= τi0 + i t
c
c
2πf
fc

(0.49)

where fDi (fc , vi ) , fc vi /c is the Doppler shift.
• For the ray tracing case:
X


Er,b (t) =
βi (t) x τ − τi (t) exp(−j2π(fc τi0 + fDi t))

(0.50a)

i

=

X

=

X

i



βi (t) exp(−j2πfc τi0 ) x τ −τi (t) exp(−j2πfDi t) (0.50b)


βi0 (t, τi0 ) x τ − τi (t) exp(−j2πfDi t)

(0.50c)

where βi0 (t, τi0 ) , βi (t) exp(−j2πfc τi0 ) with βi (t) ,

α
such that
ri + vi t

i

|βi0 (t)| = |βi (t)|

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

Note: Of course different frequencies have different absorption
factors, but this is not a small scale effect!
• Impulse response, i.e., response to input x(t) = exp(j2πf t)
X


Er,b (t) =
βi0 (t, τi0 ) δ τ − τi (t) exp(−j2πfDi (fc , vi ) t)

(0.51)

i

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 carrier
frequency 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 previous
lecture, consider the more abstract, if simpler, baseband multi path channel
model below
X
hb (t) =
βi exp(−jφi (t))
(0.52)
i

where the dependence of βi with time ignored, while all the other physical parameters of interest such as Doppler shifts, propagation delay etc. are embedded
in the time-varying phase φi (t)
φi (t) , 2πfc τi0 + 2πfDi t

(0.53)

Autocorrelation Function under Multipath Narrowband Model
Let us furthermore split hb (t) into I and Q components
P
hI (t) = βi cos(φi (t))
i
P
hQ (t) = − βi sin(φi (t))

(0.54)
(0.55)

i

• The autocorrelation function of the in-phase component is given by
X  
AI (τ ) , E [hI (t)hI (t + τ )] =
E βi2 E [cos(φi (t)) cos(φi (t + τ ))]
i

(0.56)
which can be simplified to
AI (τ ) =

1 X  2
E βi E [cos(2πfDi τ )]
2 i

(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 βi2 = 2P/N

(0.58)

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

fc vi
fc v
≈
cos θi
c
c
14

(0.59)

• The angles-of-arrival are uniformly distributed:
θi ≈

2πi
= i∆θ
N

(0.60)

where ∆θ , 2π/N .
• Substituting the approximations above we obtain



P X
2πfc vτi0
AI (τ ) =
E cos
cos i∆θ
N i
c


P X
2πfc vτi0
=
cos
cos i∆θ ∆θ
2π i
c

(0.61)

• Taking the number of terms to infinity (N → ∞)
P
AI (τ ) =
2π

Z2π


cos


2πfc vτ
cos θ dθ = P J0 (2πfD τ )
c

(0.62)

0

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), and
the integration performed in equation (0.62), fD represents the aggregate contribution 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 Doppler
Spread.
• The Coherence Time of a (dense) multi path channel is defined as the
time it takes for the channel to become uncorrelated, i.e.
Tc = min{τ }|A(τ ) = 0

Homework 7: Prove that Tc =

3
.
8fD
r

Hint: Use the approximation Jη (x) ≈

15



2
cos x − π2 η − π4 .
πx

(0.63)

Bessel Function of the First Kind

1

0.8

J0 (2πfD τ )

0.6

0.4

0.2

0

−0.2

−0.4
0

0.5

1

1.5
fD τ

2

2.5

3

Figure 5: Normalized autocorrelation function of uniformly dense wireless channel (Clarke’s) model.
• The Fourier Transform of the Autocorrelation Function of a channel is referred to as the Power Spectral Density. For the narrowband
dense multi path model above (Clarke’s):

1
 P p
for |f | 6 fD
2πfD 1 − (f /fD )2
SI (f ) , F{AI (τ )} =
(0.64)

0
otherwise

Notice how as fD increases, the Power Spectral Density SI (f ) tends
to 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 a
rectangular function at the origin. What does that mean?
sinc(x) =

sin(πτ )
πτ

−−−−−→ rect(f ) = u (f + fD ) − u (f − fD )
r1

16

Power Spectral Density

3.5

3

fD = 2

2.5

fD = 4

S(f )

2

1.5
fD = 8

1

0.5

0
−9

−7

−5

−3

−1

1

3

5

7

9

f

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
X


hb (τ, t) =
βi (t)δ τ − τi (t) exp(−j2πfc τi (t))
(0.65)
i

I
β1

i1
q1
τ1
βN

τN −1

Q

τN
Delay

Figure 7: Illustration of a baseband impulse response.

• The function P(τ ) below is known as the Power Delay Profile of the
channel, at the location z.
P(τ ) , γEt|z [|hb (τ, t)|2 ]
where γ is normalizing constant.

18

(0.66)

P
β1

βN
τ1

τN

Delay

Figure 8: Illustration of the Power Delay Profile function.
Note: Recognize that the Power Delay Profile is actually a large scale
characteristic of the channel!
• Given a certain PDP, its corresponding average and rms delays can also
be defined as
Z ∞
τ P(τ )dτ
µτ = Z0 ∞
(0.67)
P(τ )dτ
0

Z

στ = 


0

∞
2

 21

(τ − µτ ) P(τ )dτ 

Z ∞

P(τ )dτ

(0.68)

0

• A widely used PDP model (supported by empirical and analytical evidence) is the exponential profile
P(τ ) =

1
exp(−τ /γ).
γ

(0.69)

Homework 8: Prove that the mean and rms delays of an exponential
PDP 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 in
equation (0.69), the number of paths N that arrive in a time T is a
Poisson random variable with intensity 1/γ. Or, mathematically
P {no. paths in T = N } =

e−T /γ
N!

 N
T
.
γ

nP
N

o
τi < T .

(0.70)

What is the average number of paths?
Hint 1: Study P

nP
N +1
i

τi > T

o

and P

i

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

T

τ1

τ2

τn
Tm =

τn+1

Pm

i=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
X
i=1

hi δ τ − τi




(0.71)

where we have neglected the temporal dependence of τi with time, and included
the time-varying phase rotations caused by Doppler shifts into the complex
gains 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 ]

20

(0.72)

• If the channel gains hi are independent, the associated covariance matrix
is
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 can
also be assumed to be zero mean.
Next, consider the frequency domain response function of the channel, given
by the Fourier transform of hb (τ ), i.e.
Z

∞

hb (τ ) exp(−j2πf τ )dτ =

hb (f ) ,
−∞

N
X

hi exp(−j2πf τi )

(0.74)

i=1

• Consider a pair of sinusoidal signals with frequencies f1 and f2 , and define
the associated vectors
wk , [exp(j2πfk τ1 ), · · · , exp(j2πfk τN )]

(0.75)

Notice the positive arguments!

• The responses to each of these sinusoidal signals in the frequency domain can be written in matrix form as
yk = wk† · h

(0.76)

• Let us define also the quantities
y = [y1 , y2 ]
rk = |yk |

(Envelope Response)
∆f = f1 − f2

(0.77)
(0.78)
(0.79)

• The cross-correlation coefficient between the envelope responses in
the frequency domain is given by.
C(r1 , r2 ) =

E[r1 r2 ] − E[r1 ] E[r2 ]
E[r12 ] − E[r1 ]2

21

(0.80)

• It can be shown [5] that

√

πP
2
E[r12 ] = P

E[r1 ] =

(0.81)
(0.82)
2 2

E[r1 r2 ] = 2 F1 − 12 , − 12 ; 1; λ2


πP (1 − λ )
4

(0.83)

P
where P ,
Pi , 2 F1 (a, b; c; z) is the Gaussian hypergeometric function and λ2 is the Fourier transform of the sample correlation function
of the Power Delay Profile, specifically
2

PN PN

λ =

n

m

Pn Pm exp(j2π∆f (τn − τm ))
P2

(0.84)

• From equation (0.83) one obtains [5]
C(r1 r2 ) =

2 F1



− 12 , − 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-correlation
Coefficient is the same as the Fourier transform of the Power Delay
Profile.
Homework 10: Show from equation (0.85) that C(r1 r2 ) ≈ λ2 .
Hint 1: What is the maximum value of λ2 ?.


Hint 2: Use the approximation 2 F1 − 21 , − 12 ; 1; λ2 ≈ (4/π − 1)λ2 + 1,
for 0 6 λ2 6 1.
The Envelope Spectral Cross-correlation Coefficient is the
Fourier Transform of the Power Delay Profile!

22

Gaussian Hypergeometric Function

1.3

1.25

2 F1 (−1/2, −1/2; 1; λ

2

)

1.2

1.15

1.1

1.05

1
0

0.1

0.2

0.3

0.4

0.5
λ2

0.6

0.7

0.8

0.9

1



Figure 10: Gaussian Hypergeometric Function 2 F1 − 12 , − 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 over
which the envelope spectral cross-correlation coefficient is greater than a given
threshold η.




f = f C(r1 , r2 ) = η
∗

(0.86)

• From the above, we may replace the envelope spectral cross-correlation coefficient by the Fourier Transform of the PDP. Consider the Exponential
PDP of equation (0.69)
P(f ) =

|F {exp(−τ /στ )} |
1
=p
.
στ
1 + (2πf στ )2

(0.87)

• Assuming that 2πf στ  1, we have
f∗ =

23

1
.
2πηστ

(0.88)

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

Statistical Fading Channel Models
Let us now pursue a large scale statistical perspective of the small scale
variations of amplitude (envelope) and phase observed in the wireless channel.
Again, start form the baseband multipath channel model below
X
hb (t) =
βi exp(−jφi (t))
(0.89)
i

Then, consider the extension of the model above to the following
P
z(t) = αi cos(φi (t)) + jβi sin(φi (t))

(0.90)

i

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

(0.91)

where in (0.90) we accommodate for the possibility (so far ignored) that the
path 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 number
of sinusoidal functions with random phases and amplitudes amount to
the 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 often
referred to as the Random Phase Problem.

The Central Limit Theorem
Let {x1 , · · · , xN } be independent, random variates with arbitrary distribu2
tions. Define µX , EN [xi ] and σX
, EN [x2i ] − E2N [xi ], where EN denotes
expectation over the set of N samples. Then, the quantity
N

ZN =

X Yi
N X N − N µX
√
√
=
σX N
N
i=1

(0.92)

converges to the standard Normal distribution.
Proof: Notice that Yi implicitly defined above has a zero-mean and unit variance. By force of Taylor’s theorem,
ϕY (t) = 1 −

t2
+ o(t2 ),
2
24

t→0

(0.93)

where ϕY (t) denotes the characteristic function of the distribution of Y and
o(t2 ) denotes a term that decreases faster then t2 as t → 0.
Since the characteristic function of a sum is the product of characteristic functions, and due to a simple change of variables, it follows that
n 


 2 n
2
t2
t
t
= 1−
→ e−t /2 , n → ∞.
(0.94)
ϕY √
+o
2n
n
n
2

It is then identified that the e−t /2 is the characteristic function of the standard Normal distribution, which concludes the proof.


If paths are independent, in large quantities, and statistically
equivalent (that is, no dominant), then by force of the
Central Limit Theorem, equation (0.91) describes
a complex Gaussian process.

Side Note: Maximum Entropy Distributions
Let us momentarily depart from the discussion about the Random Phase Problem and highlight the important property of entropy maximality that a few
important distributions (two of which already appeared so far) exhibit.
In what follows we shall make use of the following optimization technique known
as Lagrange Multiplier method.

Lagrange Multiplier Method
Consider the following constrained optimization problem
minimize/maximize f (x)
subject to g(x) = c
Define the function



Λ(x, λ) , f (x) + λ · g(x) − c

(0.95)
(0.96)

(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

Z

p(x) dx = 1.

(0.99)

a

Due to the non-negativity of p(x) and the normalization constraint given by
equation (0.99), p(x) is a probability distribution1 .
Next, consider the entropy of p(x)
H(p) , −

Z
a

b

p(x) ln p(x) dx = −

Z

b

p ln p dx.

(0.100)

a

So, we are interested in the problem:
maximize −
Z

Z

b

p ln p dx

(0.101)

a
b

subject to

p dx = 1

(0.102)

a

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

Λ(p) = −

b

p ln p dx + λ
a

a

p dx − 1

(0.103)

In this case the gradient reduces to a simple derivative on p, namely
∂Λ(p)
= − ln p(x) − 1 + λ
∂p(x) dx

(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 λ
that satisfied the normalization constraint yields
λ = 1 − ln(b − a)

(0.106)

1 Rigorously speaking these are only two of the three Kolmogorov axioms required of probability distributions. The third (countable additivity) is implied in the (assumed) integrability
of 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] without
prior 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 and
variance?
In other words, what is the solution of the problem:
maximize

−

Z∞
p(x) ln p(x) dx

(0.108)

−∞
Z∞

subject to
−∞
Z∞

and
−∞
Z∞

and
−∞

p(x) dx = 1

(0.109)

x · p(x) dx = µ

(0.110)

(x − µ)2 p(x) dx = σ 2

(0.111)

Before we continue, notice that
Z
Z
p(x) ln p(x) dx =
p(x − µ) ln p(x − µ) dx
R

(0.112)

R

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 that
the maximization problem reduces to In other words, what is the solution of the

27

problem:
maximize

−

Z∞
p(x) ln p(x) dx

(0.113)

−∞

Z∞
subject to
−∞
Z∞

and

p(x) dx = 1

(0.114)

x2 p(x) dx = σ 2

(0.115)

−∞

The Lagrange function, taking into account this constraint is then
 ∞

 ∞

Z∞
Z
Z
Λ(p) , − p(x) ln p(x) dx+λ1 p(x) dx − 1 −λ2 x2 p(x) dx − σ 2
−∞

−∞

−∞

(0.116)
And the derivative on p is
∂Λp
= − ln p(x) − 1 + λ1 − λ2 x2 = 0,
∂p(x) dx
such that

2

2

p(x) = e(λ1 −1)−λ2 x = e(λ1 −1) · e−λ2 x .

(0.117)

(0.118)

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

(λ1 −1)

Z∞

−∞
Z∞

e(λ1 −1)

−∞

e

−λ2 x2

r
dx =

2

x2 e−λ2 x dx =

π
· e(λ1 −1) = 1
λ2
r

π
· e(λ1 −1)
4λ32











1


 λ2 =
2σ 2
=⇒


 e(λ1 −1) = √ 1




= σ2 
2πσ


(0.119)

Substituting equation (0.119) into (0.118) we have
p(x) = √

x2
1
· e− 2σ2 .
2πσ

(0.120)

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

28

Example 3: Entropy Maximality of Exp. Distribution
Homework 11: Prove that the maximum entropy distribution in
R+ (non-negative Real support) with a known mean is the Exponential Distribution.
Hint: Use the Lagrance Multiplier method.

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

Back to the Random Phase Problem...
Let us return to the Random Phase Problem
X
z(t) =
αi cos(φi (t)) + jβi sin(φi (t)) = x(t) + jy(t).
i

With the conviction that x(t) and y(t) can be assumed to be Gaussian random functions, let us attempt to derive corresponding Fading distributions.
In so-doing, pay close attention to the impact of different choices that µ
and 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 σ 2
x ∼ p(x) =

2

−x2
1
√ e 2σ2
σ 2π

y ∼ p(y) =

−y
1
√ e 2σ2
σ 2π

(0.121)

• Due to independence, the joint distribution of x and y is
2

p(x, y) =

−(x +y
1
2σ 2
e
σ 2 2π

2

)

(0.122)

• Going from density to probability
P {x, y} = p(x, y)dxdy

29

(0.123)

• Going from cartesian to polar coordinates
p
r = x2 + y 2
θ = atan xy
dxdy ←→ rdr dθ
P {r, θ} =

(0.125)

−r 2
r
2σ 2 dr dθ
e
2πσ 2

p(r, θ) =

(0.124)

(0.126)

−r 2
r
e 2σ2
2
2πσ

(0.127)

• Marginalizing
Z2π
p(r) =

−r 2
r
r −r22
2σ 2 dθ =
e
e 2σ , r > 0
2πσ 2
σ2

(Rayleigh PDF) (0.128)

0

Z∞
p(θ) =

−r 2
1
r
2σ 2 dr =
e
, 0 6 θ 6 2π
2
2πσ
2π

(0.129)

0

Notice that the envelope and phase distributions turn up to be
independent since p(r, θ) = p(r) · p(θ).
Rayleigh channels arise when no line-of-sight between transmitter
and receiver exists, and waves are scattered equally in I and Q.

30

Rayleigh Fading Distribution

0.7

0.6
σ=1

0.5

p(r; σ)

0.4

σ=2

0.3

σ=3

0.2

0.1

0
0

1

2

3

4

5
r

6

7

8

9

10

Figure 11: Examples of Rayleigh PDFs.

• Discrete-time Rayleigh variates can be generated via the Box-M¨
uller
Transform
√
x = − ln u1 sin(2πu2 )
(0.130a)
√
(0.130b)
y = − ln u2 cos(2πu1 )
where ui are i.i.d. standard uniform random variates.
• Continuous-time Rayleigh variates can be generated via the so-called Jakes
Method
#
"N
√ X
1
jβn
R(t, k) = 2 2
(0.131)
e
cos (2πfn t + θn,k ) + √ cos 2πfD t
2
n=1
where fD is the Doppler spread and the remaining parameters are
πn
N +1
βn + 2π(k−1)
N +1

βn =
θn,k =

There are many improved variations of the Jakes Model!

31

(0.132)
(0.133)

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 the
same variance σ 2 , but unequal means
2

x ∼ p(x) =

2

−(x−µ)
1
√ e 2σ2
σ 2π

y ∼ p(y) =

−y
1
√ e 2σ2
σ 2π

(0.134)

• Due to independence, the joint distribution and joint probability of
x and y are
2

p(x, y) =

2

2

−(x +y ) −(µ −2xµ)
1
2σ 2
e 2σ2 e
2
σ 2π

P {x, y} = p(x, y)dxdy (0.135)

• Going from cartesian to polar coordinates
2

P {r, θ} =

2

) 1 rµ cos θ
r −(r +µ
e 2σ2
e σ2 dr dθ
2
σ
2π
2

p(r, θ) =

2

) 1 rµ cos θ
r −(r +µ
e 2σ2
e σ2
2
σ
2π

32

(0.136)

(0.137)

• Marginalizing
Z2π
p(r) =

2

2

r −(r +µ ) 1
p(r, θ)dθ = 2 e 2σ2
σ
2π

0

Z2π
e

rµ cos θ
σ2
,

r>0

0
2

=

2

 rµ 
)
r −(r +µ
2σ 2
e
I
0
σ2
σ2

(Rice PDF)

(0.138)

K cos(θ−φ))),

(0.139)

Z∞
p(θ) =

p(r, θ)dr,

0 6 θ 6 2π

0

=

√
2
1+2 πK·cos(θ−φ)·e4K cos (θ−φ)
4K
2πe

√

· (1+erf(2

where K , µ2 /2σ 2 (i.e., ratio between the power of direct component and the power of indirect component)
is referred
to as the Rice


5π 7π
Factor, and the the phase shift φ = π4 , 3π
4 , 4 , 4 .
Notice that if we define Ω = µ2 + 2σ 2 (total channel power), the
following variation of the Rice PDF can be written
!
r
2
K(1 + K)
2(1 + K)r −K− (1+K)r
Ω
p(r) =
e
I0 2r
(0.140)
Ω
Ω

• Rice variates can be generated similarly to Rayleigh simply by adding
the constant µ.
Rice fading occurs when a dominant LOS path
exists amidst the scattered paths.

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

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

33

Rice Fading Distribution

0.7

0.6
µ = 0; σ = 1

0.5

p(r; µ, σ)

µ = 1; σ = 1

0.4

0.3
µ = 1; σ = 2
µ = 2; σ = 2

0.2

0.1

0
0

1

2

3

4

5
r

6

7

8

9

10

Figure 13: Examples of Rice PDFs.

Hoyt Solution of the Random Phase Problem
• Assume that x and y are independent zero-mean Gaussian variates
with the distinct variances

2

x ∼ p(x; b, σ ) = p

1
π(1 + b)σ 2
1

2

x2
− (1+b)σ 2

(0.141a)

y2
− (1−b)σ 2

(0.141b)

e

y ∼ p(y; b, σ ) = p
e
π(1 − b)σ 2
where 0 6 b 6 1.
• The variances of the I and Q components are
σ2
(1 + b)
2
σ2
(1 − b)
σy2 =
2

σx2 =

such that σ 2 = σx2 + σy2 .

34

(0.142)
(0.143)

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


r2
2r
br2
− σ 2 (1−b2 )
√
p(r; b, σ) =
e
I0
σ 2 (1 − b2 )
σ 2 1 − b2
√
1 − b2
p(θ; b) =
2π(1 − b cos(2θ))

(Hoyt PDF)
(0.144)
(0.145)

Homework 12: Prove equations (0.144) and (0.145).
Hint: Start by showing that the joint PDF is
p(r, θ) =

1−b cos(2θ) 2
r
−
r
√
e σ2 (1−b2 )
2
2
σ π 1−b

• Alternatively, Hoyt PDFs can be expressed as


2 2 2
) r
(1 + q 2 )r − (1+q
(1 − q 4 )r2
2 σ2
4q
p(r; q, σ) =
e
I0
qσ 2
4q 2 σ 2
2 2
q σ
p(θ; q) =
2
2π(1 + q )(sin2 θ + q 2 cos2 θ)

(0.146)
(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 relationship
between b and q?.

35

Hoyt Fading Distribution

0.7
b = 0.8; σ = 0.5

0.6

0.5

p(r; b, σ)

b = 0.6; σ = 1

0.4
b = 0.2; σ = 2

0.3

0.2

0.1

0
0

0.5

1

1.5

2

2.5

3
r

3.5

4

4.5

5

5.5

6

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 the
Half-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


2
m m 2m−1
p(r; m, σ 2 ) = Γ(m)
r
exp(− σm2 r2 )
σ2
p(θ; m) =

Γ(m)
m−1
2m Γ(m/2)2 | sin(2θ)|

(Nakagami PDF)

(Nakagami Phase PDF)

• Discrete-time Nakagami-m variates can be generated with
r
Ω
√
√
z=
· (ux γx + juy γy ),
2m

(0.148)

where ux and uy are independent and uniformly distributed random numbers 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 fading
figure m can be constructed via


v
v
s
ubmc
udme
u
u
X
X
Ω


t
x2i (t) + juy (t|T )t
yi2 (t) ,
z(t|T ) =
ux (t|T )
dme + bmc
i=1
i=1
(0.150)
where d·e and b·c denote the ceil and floor operators, and the processes
x2i (t) and yi2 (t) are as prescribed in the Jakes method.
• For arbitrary real m, the best approximate method so far is [6]
(
zL with probability ρ
z=
(0.151)
zU with probability 1 − ρ
where mL , b2mc/2, mU , mL + 0.5 and
zL ∼ pz (z; mL , σ)

(0.152)

zU ∼ pz (z; mU , σ)

(0.153)

ρ=2

mL
(mU − m)
m

38

(0.154)

Nakagami-m Fading Distribution

1.5
m = 0.2; σ = 0.5

1.25
m = 1.5; σ = 0.5

p(r; m, σ)

1

m = 0.5; σ = 1

0.75
m = 1; σ = 2

0.5

0.25

0
0

0.5

1

1.5

2
r

2.5

3

3.5

4

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

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

1.5

1

Imag(z)

0.5

0

−0.5

−1

−1.5

−2
−2

−1.5

−1

−0.5

0
Real(z)

0.5

1

1.5

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 the
Gaussian distribution for circular variates
- Closed-form
40

pT (x; α, ξ) ,

1
· eα cos(x−ξ) , x ∈ S , [−π, π]
2πI0 (α)

- Jacobi-Anger infinite series representation
pT (x; α, ξ) =

∞
X
Ik (α)
1
+2
· cos(k(x − ξ))
2π
I (α)
k=1 0

• Tikhonov variates can be generated efficiently as described in [7].
The Tikhonov distribution is also known as
von Mises distribution.
It is associated with the phase error at the output of
a Phase Locked Loop (PLL) with input power
proportional to the parameter alpha.

Tikhonov Phase Distribution

1
0.9

α = 5; ξ = −π/4

0.8
0.7

α = 3; ξ = 0

p(θ; α, ξ)

0.6
0.5
0.4
α = 1; ξ = π/2

0.3
0.2
0.1
0
−1

−0.8

−0.6

−0.4

−0.2

0
θ

0.2

0.4

0.6

Figure 20: Examples of Tikhonov PDFs.

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0.8

1

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 the
coherence 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 and
random mixture of Nakagami-m variates,” IEEE Transactions on
Communications, vol. 58, no. 9, pp. 2561 - 2575, Sep. 2010.
- Article: G. T. F. de Abreu, “On the generation of Tikhonov
variates,” 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.
2005.

Cambridge University Press,

[3] Gallager, http://www.youtube.com/watch?v=2DbwtCePzWg.
[4] T. S. Rappaport, Wireless Communications:
Prentice-Hall, 2002.

Principles and Practice.

[5] Q. T. Zhang and S. H. Song, “Exact expression for the coherence bandwidth
of 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 of
Nakagami-m variates,” IEEE Trans. Commun., vol. 58, no. 9, pp. 2561 –
2575, Sep. 2010.
[7] ——, “On the generation of Tikhonov variates,” IEEE Trans. Communications, vol. 56, no. 7, pp. 1157–1168, Jul. 2008.
[8] J. G. Proakis, Digital Communications, Fourth Edition.
Mc-Graw-Hill, 2000.

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