1) The document discusses propagation models for wireless communication systems, including free space propagation and the Fresnel zone model. 2) Free space propagation assumes a clear line-of-sight path between the transmitter and receiver. Received power is calculated using the Friis transmission equation and depends on transmitted power, antenna gains, wavelength, and distance between antennas. 3) The Fresnel zone model describes diffraction of radio waves around obstructions. It represents successive regions between transmitter and receiver where the secondary path length increases in increments of half the wavelength. Blocking of the first Fresnel zone can significantly increase diffraction loss.

1. Sistem Transmisi Komunikasi
Fresnell Zone
Pertemuan 6

2. REFLEKSI (pemantulan)

3. REFRAKSI (pembiasan)

4. DIFRAKSI (pembelokan)

5. SCATERING (penyebaran)

6. Free space loss
Attenuation of signal due to distance alone.
Lbf = 20 log (4πd / λ) dB
or in practical units
Lbf = 32.5 + 20 log(f) + 20 log (d) dB
where f is in MHz and d is in distance
For most practical situations, free space loss is the minimum loss
 worst case interference.
Applicable to interference from aircraft, satellites.
Apparent “line-of-sight” paths not necessarily free space loss only!
ITU-R Recommendation P.525

8. Classical 2-ray ground bounce model

9. 9
4.2 Free Space propagation Model
• Used to predict received signal strength
when the transmitter and receiver have a
clear, unobstructed line-of- sight path
between them.
• For examples: Satellite comm. systems and
microwave line-of-sight radio links.

10. 10
• The power received by a receiver antenna at a
distance d is given by the Friis free space equation:
(4.1)
where Pt: transmitted power
Pr: received power
Gt, Gr: antenna gain
L: the system loss factor not related to propagation. (miscellaneous
loss, and L 1)
: wavelength in metersλ
≥

11. 11
• The gain of an antenna
(4.2)
where Ae: the effective aperture related to the physical size of
antenna.
• The wavelength is related to the carrier frequency
(4.3)
where f: the carrier frequency in Hertz
: the carrier frequency in radians per second.
c: the speed of light in meters/sec
cω

12. 12
• Path Loss
– Represents signal attenuation in dB between the effective
transmitter power and the receiver power.
– For free space
which is valid only in the far-field of transmitting antenna region.
That is, the far-field distance
and df must satisfy
df>>D and df>>
where D is the largest physical linear dimension of antenna.
λ
2
2D
d f =

13. Fresnel Zone

14. Daerah Fresnel Pertama
 Daerah Fresnel pertama merupakan hal yang patut diperhatikan
dalam perencanaan lintasan gelombang radio line of sight.
 Daerah ini sebisa mungkin harus bebas dari halangan pandangan
(free of sight obstruction), karena bila tidak, akan menambah
redaman lintasan.
 Gambar menunjukkan 2 (dua) bekas lintasan propagasi
gelombang radio dari pemancar (T x) ke penerima (Rx), yaitu
berkas lintasan langsung (direct ray) dan berkas lintasan pantulan
(reflected ray), yang mempunyai radius F1 dari garis lintasan
langsung.

15.  Jika berkas lintasan pantulan mempunyai panjang setengah
kali lebih panjang dari berkas lintasan langsung, dan dianggap
bumi merupakan pemantul ya ng sempurna (koefisien pantul =
-1, artinya gelombang datang dan gelombang pantul berbeda
fasa 180 derajat), maka pada saat tiba di penerima akan
mempunyai fasa yang sama dengan gelombang langsung.
Akibatnya akan terjadi intensitas kedua gelombang pada saa t
mencapai antena penerima akan saling menguatkan.
 Berdasarkan Gambar dan keterangan di atas, F1 disebut
sebagai radius daerah Fresnel pertama , yang dirumuskan
dengan:
 dimana : F1 = radius daerah Fresnel pertama (m)
 f = frekuensi kerja (GHz)
 d1 = jarak antara Tx dengan halangan (km)
 d2 = jarak antara Rx dengan halangan (km)
 d = d1+ d 2 = jarak antara Tx dan Rx (km)

16. • Untuk daerah Fresnel pertama di tengah lintasan d = d1+ d2, dan d1 = d2
=1/2 d, sehingga:

17. • Sedangkan untuk radius daerah Fresnel kedua , daerah Fresnel ketiga, dan
seterusnya seperti diilustrasikan pada Gambar 6-11, dinyatakan dengan
rumusan berikut:
• n = 1,2,3, … . Atau secara singkat dinyatakan:
• dimana F1 = radius daerah Fresnel pertama (m)

18. 18
Excess Path Length = difference between direct path & diffracted path
 = d – (d1+d2)
3.7.1 Fresnel Zone Geometry
• consider a transmitter-receiver pair in free space
• let obstruction of effective height h & width   protrude  to page
- distance from transmitter = d1
- distance from receiver = d2
- LOS distance between transmitter & receiver = d = d1+d2
Knife Edge Diffraction Geometry for ht = hr
h
TX RX
hr
ht
d2d1
hobs
d
d =  d1+  d2, where , 22
idh + di =
2
1
2
dh + = 2
2
2
dh ++ – (d1+d2)

19. 19
Phase Difference between two paths given as
3.54





 +
21
21
2
2 dd
ddh 
Assume h << d1 , h << d2 and h >>  then by substitution and Taylor
Series Approximation
Knife Edge Diffraction Geometry ht > hr

d2
d1


h
TX
RX
hr
ht
hobs
h’





 +
=
∆
21
21
2
2
22
dd
ddh
λ
π
λ
π
 = 3.55
=
( )





 +
21
212 2
2 dd
dd
h
λ
π

20. 20
(0.4 rad ≈ 23o
)
x = 0.4 rad  tan(x) = 0.423
tan(x)
x
when tan x  x   =  + 
 





 +
=+
21
21
21 dd
dd
h
d
h
d
h
Equivalent Knife Edge Diffraction Geometry with hr subtracted from all other heights


d2d1
TX
RX
ht-hr

hobs-hr
180-
tan  =  
1d
h
tan  =  
2d
h

21. 21
Eqn 3.55 for  is often normalized using the dimensionless Fresnel-Kirchoff diffraction
parameter, v
)(
2)(2
21
21
21
21
dd
dd
dd
dd
h
+
=
+
λ
α
λ
v = (3.56)
when  is in units of radians  is given as
 = 2
2
v
π (3.57)
from equations 3.54-3.57   , the phase difference, between LOS & diffracted path is
function of
• obstruction’s height & position
• transmitters & receivers height & position
simplify geometry by reducing all heights to minimum height

22. 22
(1) Fresnel Zones
• used to describe diffraction loss as a function of path difference, 
around an obstruction
• represents successive regions between transmitter and receiver
• nth
region = region where path length of secondary waves is n/2
greater than total LOS path length
• regions form a series of ellipsoids with foci at Tx & Rx
λ/2 + d
1.5λ + d
d
λ + d
at 1 GHz λ = 0.3m

23. 23
destructive interference
•  = /2
•d = /2 + d1+d2
For 1st
Fresnel Zone, at a distance d1 from Tx & d2 from Rx
• diffracted wave will have a path length of d
d1 d2
d
Tx Rx
constructive interference:
• d =  + d1+d2
•  = 
For 2nd
Fresnel Zone

24. 24
Fresnel Zones
• slice the ellipsoids with a transparent plane between transmitter &
receiver – obtain series of concentric circles
• circles represent loci of 2nd
ry wavelets that propagate to receiver
such that total path length increases by /2 for each successive circle
• effectively produces alternatively constructive & destructive
interference to received signal
T
R
O
d1
d2
h
Q
• If an obstruction were present, it could block some of the Fresnel
zones

25. 25
Assuming, d1 & d2 >> rn  radius of nth
Fresnel Zone can be given in terms of n, d1,d2, 
21
21
dd
ddn
+
λ
rn = (3.58)
Excess Total Path Length,  for each ray passing through nth
circle
2
3/23
/21
 =n/2n
Tx
Rx

26. 26
Fresnel zones: ellipsoids with foci at transmit & receive antenna
• if obstruction does not block the volume contained within 1st
Fresnel
zone  then diffraction loss is minimal
• rule of thumb for LOS uwave:
if 55% of 1st
Fresnel zone is clear  further Fresnel zone clearing
does not significantly alter diffraction loss
d2d1
 and v are positive, thus h is positive
TX RX
h
 excess path length
/2

3/2
)(
2)(2
21
21
21
21
dd
dd
dd
dd
h
+
=
+
λ
α
λ
v =e.g.

27. 27
h = 0   and v =0
TX RX
d2
d1
d2d1
 and v are negative h is negative
h
TX RX

)(
2)(2
21
21
21
21
dd
dd
dd
dd
h
+
=
+
λ
α
λ
v =

28. 28
3.7.2 Knife Edge Diffraction Model
Diffraction Losses
• estimating attenuation caused by diffraction over obstacles is
essential for predicting field strength in a given service area
• generally not possible to estimate losses precisely
• theoretical approximations typically corrected with empirical
measurements
Computing Diffraction Losses
• for simple terrain  expressions have been derived
• for complex terrain  computing diffraction losses is complex

29. 29
Knife-edge Model - simplest model that provides insight into order of magnitude for
diffraction loss
• useful for shadowing caused by 1 object  treat object as a knife edge
• diffraction losses estimated using classical Fresnel solution for field
behind a knife edge
Knife Edge Diffraction Geometry, R located in shadowed region
Huygens 2nd
dry
source
d2
d1
T
R
h’

30. 30
Gd(dB) = Diffraction Gain due to knife edge presence relative to E0
• Gd(dB) = 20 log|F(v)| (3.60)
Gd(dB)
-3 -2 -1 0 1 2 3 4 5
Graphical Evaluation
5
0
-5
-10
-15
-20
-25
-30 v

31. 31
Table for Gd(dB)
[0,1]20 log(0.5 e- 0.95v
)
[-1,0]20 log(0.5-0.62v)
> 2.420 log(0.225/v)
[1, 2.4]20 log(0.4-(0.1184-(0.38-0.1v)2
)1/2
)
 -10
vGd(dB)

32. 32
e.g. Let:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m
2. diffraction loss
• from graph is Gd(dB)  -22dB
• from table Gd(dB)  20 log (0.225/2.74) = - 21.7dB
)10(333.0
)2000(2
25
)(2
6
21
21
=
+
dd
dd
h
λv = = 2.74
1. Fresnel Diffraction Parameter
3. path length difference between LOS & diffracted rays
m
dd
ddh
625.0
10
2000
2
25
2 6
2
21
21
2
=





=




 +
 
4. Fresnel zone at tip of obstruction (h=25)
• solve for n such that  = n/2
• n = 2· 0.625/0.333 = 3.75
• tip of the obstruction completely blocks 1st
3 Fresnel zones
Compute Diffraction Loss at h = 25m

33. 33
e.g. Let:  = 0.333 (fc = 900MHz), d1 = 1km, d2 = 1km, h = 25m
2. diffraction loss from graph is Gd(dB)  1dB
)10(333.0
)2000(2
25
)(2
6
21
21
−=
+
dd
dd
h
λ
v = = -2.74
1. Fresnel Diffraction Parameter
3. path length difference between LOS & diffracted rays
m
dd
ddh
625.0
10
2000
2
25
2 6
2
21
21
2
=




−
=




 +
 
4. Fresnel zone at tip of the obstruction (h = -25)
• solve for n such that  = n/2
• n = 2· 0.625/0.333 = 3.75
• tip of the obstruction completely blocks 1st
3 Fresnel zones
• diffraction losses are negligible since obstruction is below LOS path
Compute Diffraction Loss at h = -25m

34. 34
f = 900MHz   = 0.333m
 = tan-1
(75-25/10000) = 0.287o
 = tan-1
(75/2000) = 2.15o
 = +  = 2.43o
= 0.0424 radians
find diffraction loss
)(
2
21
21
dd
dd
+λ
αv =
from graph, Gd(dB) = -25.5 dB
24.4
)12000(333.0
)2000)(10000(2
0424.0 ==
find h if Gd(dB) = 6dB
• for Gd(dB) = 6dB  v ≈ 0
• then  = 0 and  = - 
• and h/2000 = 25/12000  h = 4.16m

2km10km

T
R25m

75m
2km10km
100m
T
25m
50m
R
=0
2km10km

T
R25m
 h

35. 35
3.7.3 Multiple Knife Edge Diffraction
• with more than one obstruction  compute total diffraction loss
(1) replace multiple obstacles with one equivalent obstacle
• use single knife edge model
• oversimplifies problem
• often produces overly optimistic estimates of received signal
strength
(2) wave theory solution for field behind 2 knife edges in series
• Extensions beyond 2 knife edges becomes formidable
• Several models simplify and estimate losses from multiple obstacles

36. Diffraction (knife-edge) geometry

37. 37
Fresnel diffraction geometry
Illustration of Fresnel zones for different knife-edge diffraction scenarios.
Fresnel diffraction or near-field
diffraction occurs when a wave passes
through an aperture and diffracts in
the near field, causing any diffraction
pattern observed to differ in size and
shape, depending on the distance
between the aperture and the
projection

38. 38
Multiple knife-edge diffraction