1= ~ L(~3
HW
'vJ L It tJ
\.J /
-t-1-,
fc : : .
PL =-1ociB
FM = I' cf B
5 . 0 (rl-/-c
Ew : ?-.o {VIrh·
Cry. - :;.
Cr,. y
~
P,r
cl iJ ;
~ ::J o d 8...,
Q\1~
)_ d l3i
- :S. o J B
= ~cl5·
-~
Category
Transmitter
Transmitter Power
Transmit Antenna Gain
Transmitter Losses
~x
=20dBm+2dB-t dB
19 dBm
3
Path Loss
Diffraction Loss
Fading Margin
Shadowing Margin
Losses= PL + y
106 dB
Net Receiver Gain
RXGain
:::.
= 2.0dB
-85.0 dBm
5.0 dB
Minimum SNR
Max MDS
-
MDS
= kT,,B = P RX -
Given
Solve
SNR
s-.o d « :
kT,. 13
L> -r: :
I"'
/'A 'C..S
--
-:::.
kB
I~
JV F- =-
Net gain/loss provided by the receiver
PRX = Prx -Losses+ RXGain
Net Received Power
MD5 ::. - gs, o J &-.
Given
No cable, connector, etc. losses at RX
2.0 dB
0.0 dB
Receiver Antenna Gain
Receiver Losses
System
Performance
Given
No Diffraction
Mobility based fading. N/A in this example
Given
90 dB
N/A
N/A
16 dB
Net Propagation Losses
Receiver
Given
Given
Given
20 dBm
2.0 dBi
3.0 dB
Net Transmitted Power
Losses
Comment
Value
J 1.
J';J$k.. :::
:L'Io
/3,
L(
3
3J 3
•
\WoP$.
35500
G-r'rf" : +~
cr OtJ M H1:
f ( f<l- ;4-vr /-t ""..
T
-:=.
I~-... J,j 4
d
==- Sc b""1
'Prf ~ (oo .... Vv'
;'ADS=
F,·/l J :
b)
UP.in)
of anf-<11MS
:Jain)
!A.S;"j
~e..
PI)
bJ
C<tVJ
-tos-cJIS ...
~d f';, Cc. fc ... fc.fc /?._;< f'o-<.,
/,~ K {:,~., Lj -<-1-. / CL;....,.,f4rt- T~ lr« fC'-"<J',
Fr:,·s h e..
Cc.l {(, '" ,le
are ,,;f e-r.tf,'c,'f{J j
e:t-fec.-/,'~ ~·
1-1
ive11) ~ov-<'-"·<.r
( 7.S...,) l
( 6.zn ... P
PL
F ~""~ +t. e~1
==TOPS.
35500
f.; I{ /"
~ ~~-.
s- cl f!
f-t. e. L;,.., lc
c~,.., t:tffl'bX/_..,.-f.e
\!Ire.
.
- Lf 7r ( v
f3 "'Jj
t
i-
?
)
Category
Transmitter
Transmitter Power
Transmit Antenna Gain
Transmitter Losses
Net Transmitted Power
Channel
Losses
Path Loss
Diffraction Loss
Fading Margin
Shadowing Margin
Net Propagation Losses
Receiver
35500
Comment
20dBm
43 dB
0 dB
Given
Estimated based on dish size
63 dBm
85.5 dB
0
0
0
85.5 dB
43 dB
Net Received Power
20.5 dBm
Minimum SNR
Max MDS
~~ -
~0.5' ci gWJ
R.)( ::.
;;,l 0 c[ {5 (VI
Net TX Power
Free Space Path Loss
N/A
N/A
N/A
Losses
= L Losses
43 dB
0
Net Receiver Gain
System
Performance
-TOPS.
Receiver Antenna Gain
Receiver Losses
Value
N/A
- 105 dBm
Net gain/loss provided by the receiver
PRx
= PTx -Losses+ RXGain
PRX
= MDS + SNR
N/A
Given
EE433
Given:
Homework
Fall2013
The Navy's SPN-43 Air Traffic Control Radar operates at 3.5 GHz with a transmit power of 1.0
MW and a Transmitter Antenna Gain of 32 dB. Bandwidth is 1.6 MHz, TX Height is 16 meters,
RX Height is 1.5 meters. If you have a receiver with a noise figure of 3.0 dB, antenna gain of 2
dB, and a 10 dB SNR is required to received the signal, generate a link budget to calculate the
following:
Find: Develop a link budget to calculate the following:
(a)
(b)
(c)
(d)
(e)
Maximum distance for free-space propagation.
Maximum distance for two-ray propagation.
Maximum distance for Log-Distance propagation with parameters n = 3.8 and a = 15 dB.
Maximum distance for the Hata Model, using the standard Hata parameters.
If you were the Radar Operating Officer, and the Radar was located at coordinates of (3 8.144818, 76.434097), which model would you choose and why?
MDS=kTnB
Tn =290(n/-1)=290(1.995-1)=288.6K
23
MDS = kTnB = (1.38 x 10- J I K)(288.6K)(1.6MHz)
MDS
= -112 .0dBm
EE433
Homework
Category
Transmitter
Channel
Losses
Value
Transmitter Power
90 dBm
Transmit Antenna Gain
32 dB
Transmitter Losses
0 dB
Net Transmitted Power
Fall2013
Comment
1 MW Radar TX Power
122 dBm
Path Loss
Calculated
Diffraction Loss
N/A
Fading Margin
N/ A: No Mobility in this scenario.
Shadowing Margin
N/A
Net Propagation Losses
226 dB
Maximum based on Net RX Power
Receiver Antenna Gain
2.0 dB
Given
Receiver Losses
0 dB
Receiver
Net Receiver Gain
Net Received Power
System
Performance
2.0 dB
-102 dBm
Minimum SNR
10 dB
Given
Max MDS
-112.0 dBm
Based on NF and 2.0 MHz BW
EE433
Homework
Fa112013
Free Space Path Loss
?I(d)=226 dB
(0.0857m )
_
10
2
d
226 8
= 1, 360,000 km
10
Two Ray Path Loss
PI (d)= 226dB
PL( d)= 40 log 10 (d)+ 20 log 10 ( h,) + 20log 10 ( hr)
-20 1og"1(h, )-201ogw(h, )+PL( d)
d max
= 10
40
-20 logw(h1 )-201ogw(h, )+PL(d)
d max = 10
40
-20logw(l6m )-20log 1n(1.5m )+226dB
40
= 10
= 91.2 km
Average Path Loss (Log-Distance Model)
PL(d0 =1m)=-10log 10
PL(d)-PL(dn)
d max = 10
IOn
_.1, 2
((4trdt J
?
=-10log 10
226dB-43.3dB
= 10
IO(J.s)
= 64.2km
(( 0.0857 m ) J=43.3dB
(4tr(1m))
2
2
EE433
Fall 2013
Homework
Average Path Loss (Hata Model)
PL( d)= 69.55- 26.16log 10 ( / ) -13.82log 10 ( h1 ) -a( h,) +
( 44.9- 6.55logl0 ( h,) )logiO (d)- 2 ( logiO (
{s)
r-
a( h,) = (l.llog 10 ( / ) - 0.7) h,- (1.56log 10 ( / ) - 0.8)
PL(d)= 69.55- 26.161og 10 ( 3500) -13.821og 10 (16)- a (1.5) +
( 44.9- 6.551og 10 (1.5)) log 10 (d)- 2 ( log 10 ( 3 ;~o)
f -5.4
a ( hr) = (1.11og 10 (3500)- 0.7) (1.5)- (1.561og 10 (3500)- 0.8) = 0.07
PL( d)= -39.80 + 43.751og 10 (d) -14.19
+43.75log 10 (d)= +54.0 + PL (d)
lo
g,o
dmax
(d)=
54.0 + 226
43.75
= 2,510,000 km
5.4