S-72.3210 Channel modeling for radio communication systems
Additional demo problems
1
A1. In a fixed radio link design the
RF1
antenna height is selected to
d
st
h
h
guarantee that the 1 Fresnel
zone is free of obstacles during
standard propagation conditions
40 NU/km ) for the
( N h
situation depicted in Fig.1.
Req
a. Derive by geometrical considerReq
ations a symbolic expression for
the required antenna height h
above the reference ground.
Fig. 1
b. Calculate
the
approximate
numerical antenna height value
h on the frequency f = 3 GHz for the path lengths i) d = 50 km, ii) d =
100 km, and iii) d = 150 km. You may use the expression cos(2x) = 1
– 2sin2(x). The geometrical Earth radius is assumed to be 6370 km.
Link:geometry.dsf
SOLUTION
a) By considering a right-angled triangle we get
Req
h
Req
cos
h
RF1
d
2 Req
Req
d
4 Req
2 Req sin 2
d
4 Req
RF1
RF1
1 2sin 2
RF1
1 2sin 2
d2

8Req
Req
d2
8Req
Req
d
4 Req
1 2sin 2
d
4
d
4 Req
RF1
2
b) The wavelength at 3 GHz is
= 0.1 m
The equivalent Earth radius is
Ro
Req 
1 Ro
n
h
6370
8548 km
1 6370 0.000040
i) d = 50 km
d2
h
8Req
d
4
502
8 8548
0.05 50000
4
0.03656 km 25.00 m 61.56 m
ii) d = 100 km
d2
h
8Req
d
4
1002
8 8548
0.05 100000
4
0.14623 km 35.36 m 181.59 m
iii) d = 150 km
d2
h
8Req
d
4
1502
8 8548
0.05 150000
4
0.32902 km 43.30 m 372.32 m
A2. Determine from the geometry given in Fig. 1 an expression of the
minimum vertical thickness h of a super-refractive layer so that the
entire ray path is within the layer. Use suitable approximations and
calculate the layer thickness on a 40 km path for the refractivity
gradients i) –300 N-units/km, ii) –400 N-units/km, and iii) –500 Nunits/km. Calculate also the corresponding vertical changes of the
refractivity through the layer.
SOLUTION
Fig. 1
Req- h
Req
d/2Req
Req
h
h
d/2
h
Super_refr_layer.dsf
From the figure one can see that
Req
h
h
Req
h
h
Req
Req
h cos
h cos
d
2 Req
d
2 Req
Req
h 1 cos
d
2 Req
As the angle d 2 Req is small, the cosine-function can be replaced by the
two first terms of its Taylor-series, and as h  Req the expression can be
further simplified:
3
4
h  Req
h 1
1 d
1
2 2 Req
2
d2
d2
h

2
8Req
8R
Req
eq
The equivalent Earth radius is approximately depending on the vertical
gradient of the refractive index in the following way:
Req 
Ro
1 Ro
N
h
300
n
h
NU
km
Req
6370
1 0.0003 6370
402
h
0.0314 km,
8 6992
N
N
h
300 0.0314
h
N
h
400
h
N
h
402
8 4115
500
h
NU
km
NU
km
402
8 2915
Req
Req
9.42 N-units
6370
1 0.0004 6370
0.0486 km
N
N
4115 km
400 0.0486
6370
1 0.0005 6370
0.0686 km
6992 km
19.44 N-units
2915 km
500 0.0686
34.30 N-units
Note! If the 1st Fresnel zone should be inside the layer, the thickness would
be several tens of meters larger.
5
A3. How large is the diffraction loss (dB) caused by a knife-edge
obstacle, when it covers exactly the full diameter of the first Fresnel
zone
SOLUTION
When the 1st Fresnel zone is completely occupied by a knife-edge
obstacle the Fresnel diffraction parameter value is
h
As
2 l1 l2
l1l2
h
RF1
2
2
> –0.7 Eq. 2 – 56 can be used
Lknife edge
6.9 20log
6.9 20log
0.1
2
1
2
0.1
2 0.1
1 
2
0.1

 ????
 1.3142
1.5414
6.9 20log 2.9656
7.37 dB
6
A4 Fig. 1 shows the geometry of a multiple screen diffraction path where
the screen height above the flat ground is hs and the distance between
the N screens is b. The transmitter antenna with the height htx above
ground is on the distance b/2 from the first screen and the receiver
antenna with the height hrx above ground is on the distance b/2 from
the Nth and last screen. Without loss of generalithy it is assumed that
the transmitter antenna is higher than the receiver antenna.
a) Derive using Deygot’s method the excess loss in symbolic form for
the given path.
b) Calculate a numerical value for the excess loss in dB relative to free
space loss when d = 2 km, b = 100 m, hs = 25 m, htx = 15 m, hrx = 1.5
m, and the frequency f = 2000 MHz.
screen 1
l1i
screen i
l2i
hi
htx
hs
hrx
b
d1i
screen N
Multi_screen_diffr_path.dsf
d2i
d
Fig. 1
SOLUTION
a) From the path geometry it can be seen that the path length along
ground is d = Nb.
In Deygot’s method one has first to calculate the Fresnel parameter ni
for each screen as it were a single obstacle.
i
hi
2
1
l1i
1
l2i
From Fig. 2 one can write the height of the screen above the
theoretical line-of-sight path to be
7
screen 1
screen i
l1i
l2i
hi
htx
x
screen N
hs
hrx
b
d1i
d2i
Multi_screen_diffr_path.dsf
d
Fig. 2
hi
hs
hrx
x
By considering similar triangles one gets
x
htx
d2i
d
hrx
From Fig. 1 it can be deduced that
d1i
i 0.5 b
i 0.5
d
N
d2i
d d1i
i 0.5
d
N
d
d
N
N i 0.5
The path lengths from the transmitter and receiver to Screen i are
obtained from the right-angled triangles in Fig. 3
l1i
d12i
l2i
d22i
hs
hs
htx
hrx
2
i 0.5
2
2 d
2
N2
N i 0.5
hs
2 d
htx
2
2
N2
hs
hrx
2
8
screen 1
l1i
screen i
l2i
hi
htx
x
screen N
hs
hrx
b
d1i
Multi_screen_diffr_path.dsf
d2i
d
Fig. 3
The Fresnel-parameters of the individual screens can now be written
as
hi
i
2
1
l1i
1
hs
l2i
2
hrx
htx
hrx
N i 0.5
N
1
i 0.5
2 d
1
2
hs
N2
htx
2
N i 0.5
2 d
2
hs
N2
hrx
2
From Figs. 1…3 it is clear that hi increases as the distance to the lower
antenna decreases. As the expression under the square root in the
Fresnel-paramter is a convex function of i with a minimum near to the
path midpoint it follows that the maximum Fresnel-parameter occurs
for that screen which is nearest to the lower antenna, in this case the
receiver antenna or
0.5
h
h
h
h
max
N
s
rx
tx
rx
N
2
1
N 0.5
2 d
1
2
N2
hs
htx
2
0.25
d2
N2
hs
hrx
2
9
According to Deygot’s method the excess heights for the remaining
individual diffraction paths from the higher antenna to the top of the
screen (nearest to the lower antenna) and the screen with the
maximum Fresnel-parameter should be determined,
screen 1
l1i
hi
htx
l2i
screen i
screen N
hs
hrx
b
d1i
Multi_screen_diffr_path.dsf
d2i
d
Fig. 4
Wtth the same reasoning as above the result is that the maximum
Fresnel-parameter occurs for the screen nearest to the higher antenna,
in this case Screen 1, Now
l11
b2
4
l21
N 1b
htx
htx
l21
d 0.5b
hs
htx
N 1
N 0.5
'
1
hs
htx
4N 2
hs
htx
2
N 1
d
N
hs
h1
max
hs
d2
2
hs
N 1
N 0.5
htx
N 1
d
N
d
d 0.5
N
2
hs
N 1
N
htx
1
1 0.5
N
1
d2
4N 2
hs
htx
2
1
N 1
d
N
10
screen 1
screen i
hi=0
htx
screen N
hs
hrx
b
d1i = l1i
Multi_screen_diffr_path.dsf
d2i
= l2i
d
Fig. 5
For the remaining subsets of screens in Deygot’s ,method the excess
height is zero as Fig. 5 shows. This also implies that all following
Fresnel-parameters take tha value zero. The total excess diffraction
loss is
Lmsd
L 1
L
L hs
htx
N
N
2 L 0
N 1
N 0.5
2
1
d2
4N 2
hs
htx
htx
hs
htx
2
1
N 1
d
N
N 0.5
N
hrx
L
2
1
N 0.5
N 2 L 0
2 d
1
2
N2
hs
htx
2
0.25
d2
N2
hs
hrx
2
11
where
L
6.9 20log
0.1
2
1
0.1
12
b)
'
1
hs
N 1
N 0.5
htx
2
1
d2
hs
4N 2
20 1
20 0.5
25 15
2
0.15
htx
19
19.5
9.744
N
hs
2
0.15
2
0.15
hrx
20002
2
25 15
25 1.5
2
19
2000
20
1
1
50.990 1900
9.744
0.26851 5.049
0.5
N
hrx
2 d
1
20 1
2000
20
1
1
N 0.5
2
0.15
1
2500 100
htx
2
1
4 202
10
1
N 1
d
N
1
2
hs
N2
htx
2
0.25
d2
N2
hs
hrx
2
0.5
20
15 1.5
1
2 2000
20 0.5
2
20
1
2
25 15
2
0.25
20002
202
25 1.5
2
13
23.5 13.5
0.5
20
2
0.15
1
1
19.52 1002 102
0.25 1002 23.52
23.1625
2
0.15
1
3802500 100
1
23.1625
2
1
1
0.15 1950.026 55.247
2500 552.25
23.1625
0.24818 11.539
The excess loss due to multiple screen diffraction is
Lmsd
L 1
L 20
6.9 20log
6.9 20log
18 L 0
5.049 0.1
2
1 5.049 0.1
11.539 0.1
18 6.9 20log
0 0.1
2
2
1 11.539 0.1
1 0 0.1
6.9 20log 9.998
6.9 20log 22.922
18 6.9 20log 0.9050
26.90 34.10 18 6.03 169.54 dB
14
A5 This task deals with the calculation of excess loss due to diffraction
over a knife edge obstacle without and with a reflected ray at 2 GHz.
The paths geometries are shown in Fig. 2.
a) Determine the excess diffraction loss due to a single knife edge obstacle at the distance d = 1000 m from the transmitter. The distance of
the receiver from the obstacle is b = 2 m. The transmitter antenna and
obstacle height above the flat ground is 20 m and 1.5 m respectively.
b) In this case there is a reflecting wall at the distance w = 20 m behind
the obstacle. How large is the excess loss of the diffracted and
reflected ray at the receiver if the reflection loss is 3 dB?
c) Between which values will the excess loss vary of the sum pf the
directly diffracted ray and the reflected ray depending on the phase
difference between the two rays?
Notice that the approximate heights of the obstacle for the two rays
above corresponding lines of sight are given in Fig, 2.
l1
l2
h h
hr
ht = h
Fig. 2a
d
l2
h
h
d
Fig. 2b
d
b
b2
l2
ht
ht
hr
hr
2
b
l1
ht = h
d
h
b
hr
w
diff_reflx_model.dsf
h
l2
d
w
w 2w b
w2
w
2w b
d
ht
hr
2
ht
hr
A6. The
delay power spectrum of a radio channel
Po
P( )
exp
u( ) , where u( ) is the unit step function.
is
15
a) Determine the delay spread and coherence bandwidth of the channel
as these are defined as the interval over which respective absolutevalued spectrum falls 20 dB under its maximum value.
F exp
u( )
1 j2 f
b) If the coherence bandwidth is defined as 1/delay spread in task a, how
many dB has the absolute value of the frequency correlation function
fallen below its maximum value at this frequency difference?
SOLUTION
a)
P( ) Po
exp
Pmax
Po
ln 100
F
Po
exp
P Bcoh
Pmax
Bcoh
0.01
exp
100
4.605
u( )
Po
j2 f
1
Po
1
9999
2
2 Bcoh
2
Po
Po
P( f )
1
0.01
1
2 Bcoh
15.91
b)
P Bcoh
Pmax
Po
1
1 2
4.605
0.591
2.28 dB
2
1
Po
1
2
4.605
2
2 f
2
2
10000
16
A7 What is the maximum bandwidth of an ideal bandpass signal that it
could be considered to be a narrow-band signal in the equalizer test
channel model in GSM. This model has 6 taps of equal average
strength at 0, 3.2, 6.4, 9.6, 12.8, and
s. The required signal to
error signal ratio is 20 dB.
SOLUTION
2
rms
1 02 1 3.22 1 6.42 1 9.62 1 12.82 1 162
1 1 1 1 1 1
1
0 10.24 40.96 92.16 163.84 256 93.87 s 2
6
1
3
2
2
4 2 Brms
rms
3
B
2
rms
2
4 2 B 2 rms
3
MHz
2 10
93.87
2.845 kHz
17
A8. Calculate the average path loss on a 2 km path on 2.1 GHz with
worst street orientation using the COST237 Walfisch Ikegami model
with the following parameters: average roof height 20 m, base station
antenna height 35 m, street width 20 m, average building spacing
100m, and mobile station antenna height 1.5m.
SOLUTION
Lo
32.5 20log d km
32.5 20log 2
Lrts
20log f MHz
20log 2100
16.9 10log w m
6.02 66.44 104.96 dB
10log f MHz
20log hroof m
16.9 10log 20
32.5
hrx m
Lori
10log 2100
20log 20 1.5
4
16.9 13.01 33.22 25.34 4 32.65 dB
Lmsd
Lbsh
Lbsh
ka
kd log d km
18log 1 htx m
hroof m ,
18log 1 35 20
ka
54 dB, htx
kd
18 dB, htx
k f log f MHz
htx
9log b m
hroof
21.67 dB
hroof
hroof
2100
1
3.11 dB, small cities
925
2100
4 1.5
1
2.09 dB, large cities
925
Lbsh ka kd log d km k f log f MHz 9 log b m
4 0.7
kf
Lmsd
21.67 54 18log 2
21.67 54 5.42
L
Lo
Lrts
Lmsd
3.11
log 2100
2.09
10.33
6.94
104.96 32.65
18
9 log 100
9.42 dB
12.81 dB
9.42
12.81
147.0 dB
150.4 dB
18
A9. Coverage planning in a cellular network shows that 50% of the
locations have coverage at a 5 km distance from the base station. The
path loss exponent is 3.5. How large is the cell radius if the coverage
requirement at the cell border is i) 90%, ii) 95%, and iii) 99%? The
standard deviation of the shadow facing is 8 dB.
SOLUTION
The additional loss from shadow fading is given by
x2
2
exp
P L
Li
L
2
L
L
dx Q
L
L
L
L INVQ P L
Li
L
L INVQ 1 Pcoverage
Pcoverage
0.9
L 8 INVQ 0.1
8 1.282 10.26 dB
Pcoverage
0.95
L 8 INVQ 0.05
8 1.645 13.16 dB
Pcoverage
0.99
L 8 INVQ 0.01
8 2.326 18.61 dB
When the other link arameters stay constant the increased path loss
can only be compensated by a shorter cell range. With a single slope
path loss model this implies that
L1
Lo
10n lg d1 do Lo
L 10n lg d 2 do
10n lg d1 do
L 10n lg d1 do
L 10n lg d1 do
d2
lg d 2 do
10n lg d1 d 2
d110 0.1 L n
5 10 0.110.26 3.5
Pcoverage
0.9
L 10.26 dB d2
Pcoverage
0.95
L 13.16 dB d2
Pcoverage
0.99
L 18.61 dB
d2
5 10 0.113.16 3.5
2.546 km
2.104 km
5 10 0.118.61 3.5 1470 km
19
A10.
A good DVB-T receiver needs a field strength of 20 dB V/m
to deliver a satisfying picture. The transmitter network is planned for
80 km range with 90 % coverage with the receiver antenna height
corresponding to average terrain cover height. How large transmit
power (EIRP) is needed to fulfill the requirement at 600 MHz
according to the ITU-R Recommendation P1546? The effective
transmitter antenna height is 300 m..
SOLUTION
From the actual ITU-R Recommendation P1546 graph the field
strength level at 80 km distance is 30 dB V/m with a transmitter
power of 1 kW EIRP and 50 % location probability. This is 10 dB
higher than the required level.
The shadow fading margin (SFM) for 90 % location probability is
SFM
Eq
E50
INVQ
q
100
L f
q is the probability that the field strength level exceeds a given value,
in this case 100 – 90 = 10 %
The standard deviation of shadow fading for signals having a
bandwidth larger than 1 MHz is according the Rec. P1546 5.5 dB.
Thus
SFM
INVQ
10
100
5.5 1.282 5.5 7.05 dB
The needed transmitter EIRP level is
Ptx 10lg(1 kW)
Erequired
E1 kW
30 dBW 10 7.05 27.05 dBW
SFM
0.507 kW EIRP
20
A11.
Which receiver antenna height will give a field strength equal
to the free space value on an over-land 30 km path at 600 MHz
according to the ITU-R Rec. P1546. The transmitter antenna height is
300 m and the path is in rural environment.
SOLUTION
Received field strength at 600 MHz over land paths as function
of distance for different transmitter antenna heights exceeded
50% of time. Receiver antenna height 10 m (equal to representative height of ground cover)
100
free space propagation
77.6 80
Field strength in dB V/m for 1 kW e.r.p.
60.1 60
htx = 10 m
20 m
37.5 m
75 m
40
150 m
300 m
20
600 m
1200 m
0
20
40
60
80
1
ITU_prop_models.dsf
10 distance/km 100
1000
21
From the graph can be seen that the free space field is 77.8
dB V/m, and the field with 10 m receiver antenna height is 60.1
dB V/m.
Thus a receiver antenna height giving 77.8 – 60.1 = 17.7 dB higher
field strength is the largest appropriate value as that gives the free
space field which cannot be exceeded.
The increase of field strength vs. receiver antenna height is given by
E
hrx
3.2 6.2log( f ) log hrx hcm
hcm10
E
3.2 6.2log( f )
17.7
10 10 20.42
73.55 m
10 10
17.7
3.2 6.2log(600)
22
Received field strength at 600 MHz over land paths as function
of distance for different transmitter antenna heights exceeded
50% of time. Receiver antenna height 10 m (equal to representative height of ground cover)
100
free space propagation
Field strength in dB V/m for 1 kW e.r.p.
80
60
htx = 10 m
20 m
37.5 m
75 m
40
150 m
300 m
20
600 m
1200 m
0
20
40
60
80
1
ITU_prop_models.dsf
10 distance/km 100
1000
23
A12 In long-haul fixed radio tropospheric multipath propagation is the
main cause for signal fading and enhancement. The ITU-R Recommendation P530 gives expressions for the fading and enhancement
distributions.
The radio link parameters are the following: f = 7.5 GHz, d = 45 km,
dN1 = –350 NU/km, and hL htx hrx 80 m
a) Which fade depth is exceeded 0.005% of time?
b) Which enhancement is exceeded 0.005% of time?
SOLUTION
a) The percentage of time a certain fade depth A is exceeded is given by
the expression whem the link path topography is unknown
pw
Kd 3.0 1
1.2
p
100.033 f 0.001hL A /10
where K 10 6.2 0.0029dN1
Insertion of given values gives
K 10
6.2 0.0029
350
10 5.185
pw 10 5.185 45.03.0 1 0
1.2
100.033 7.5 0.001 80 A /10
10 5.185 104.9596 100.2475 0.08 10 A /10
10 0.0579 10 A /10
Now the fade depth can be solved
NB. The above expressions give the fade probability value, not the
probability percentage.
pw 10 0.0579 A /10
A 10
10
0.0579 A /10 lg pw
0.0579 lg pw
10 0.0579 lg 0.00005
0.0579
4.3001
10
0.0579 4.3001
42.43 dB
As A > 25 dB the probability expression and thus the result is valid.
24
b) The enhancement probability is given by the expression
pew
P
L
E
0.01 10
1.7 0.2 A0.01 E / 3.5
First the value of A0.01 must be obtained (0.01 means pw = 0.01 %),
A 10
10
0.0579 lg pw
0.0579
4
10
10
0.0579 lg 0.0001
0.0579 4
40.06 dB
Now the actual enhancement value can be solved
pew
0.01 10
10
2
6.312 E / 3.5
0.688 E
lg pew
3.5
0.688 3.5 lg pew
0.688 3.5 lg 0.00005
10
E
1.7 0.2 40.06 E / 3.5
0.688 E / 3.5
14.37 dB
As E > 10 dB the enhancement probability expression and thus the
result are valid,
25
A13.The typical urban channel model for GSM1800 contains six flat
Rayleigh-fading taps having a Doppler-spectrum according to Clarke’s
model. A case with a mobile station moving 125 km/h is performed.
a) How many times in a second does a tap signal pass below –10 dB and
pass above +10 dB relative to the r.m.s. value?
b) How long is the average duration of a signal fade below –10 dB and
the average duration of a signal enhancement above +10 dB relative to
the r.m.s. value?
SOLUTION
a) The level crossing rate for a Rayleigh-fading narrow-band signal in a
channel with Clarke’s Doppler-shift model is
NA
2
2 f D max exp
The maximum Doppler-shift is
f D max
v
f
c
125 3.6
8
1.8 109
208.33 Hz
3 10
The absolute value of the faded signal level is
2
10 10 10
0.1
The absolute value of the enhanced signal level is
2
1010 10 10
Then the level crossing rates upwards or downwards through the –10
dB level is
N A 10 dB
2
208.33
0.1 exp
0.1
149.42 s 1
Then the level crossing rates upwards or downwards through the +10
dB level is
N A 10 dB
2
208.33
10 exp
10
0.07497 s 1 4.498 min 1
26
b) The average duration of a fade deeper than a given value is
2
exp
TA
f D max
1
2
exp 0.1
208.33
0.1
1
2
6.369 10 4 s 636.9 s
The duration of an enhancement is defined as
TA
P a A
NA
exp
f D max
2
2 exp
1
2
f D max
Insertion of the actual values gives
TA
1
208.33 10
2
1.9150 10 4 s 191.5 s
2