Siti Mariah Ulfah 1147070071 Komunikasi Digital 1. a) What is the value value in deciels deciels of the the free!s"ace free!s"ace loss loss for a carrier carrier fre#uenc$ fre#uenc$ of 100 Mh% Mh% and a range of & miles' ) (he transmitter out"ut "oer is 10W. *ssume that oth the transmitting and receiving antennas are isotro"ic and that there are no other losses. +alculate the received "oer in d,W. c) -f in "art " art )/ the -2 is e#ual to 30 W/ calculate the received "oer in d,W/ d) -f the diameter of a dish antenna is douled/ calculate the antenna gain increase in d,. e) or the s$stem in "art a)/ a)/ hat must the diameter diameter of a dish antenna antenna e in order for the antenna gain to e 10d,' *ssume *ssume an antenna efficienc$ of 0.55 6*W*,* 8
a) λ
& x10 9
c =
f
=
=
10 9
&meters
; 3 miles x 1609 m/mile = 4827 meters
4π d
λ
)3
=
4π x 4937 &
)
=
4.0: x10 9
FSL = FSL(dB) = 10 log (409 x 108) = 861 dB P r
=
P t
3 ; ; 4π d d ) ; λ )
!)
"t(dB) = 10 log (10#/1#) "r (dB) = "t(dB) $ FSL(dB) = 10 d!# % 861 dB = $ 761 dB
&) "r (dB) = "t(dB) $ FSL(dB) = 13 dB# % 861 dB = $ 731 dB
d) ' =
4π Ae 3
λ
)
=
k d 3
"t(dB) = 10dB
ere d = a*te**a diameter i+ a*te**a diameter is do,!led' = .(4d2) = 4.d2 #i& mea*s te ' ill i*&reased ! a +a&tor o+ 4 or a* i*&reased o+ 6 dB
4π Ae
λ 3
)
=
4πη Ap
λ 3
)
=
π 3η d 3 ) λ 3
e) ' = 3
d
=
Gλ 3
π 3η )
)
=
&) 3 10)
π 3 0.55) = 168 m2
d = 407 m 2 ra*smitter as a* o,t,t o+ 2 # at a &arrier +re,e*& o+ 2 '5 ss,me tat te tra*smitti*g a*d re&eii*g a*te**as are ara!oli& dises ea& 3 +t i* diameter ss,me tat te e&ie*& o+ ea& a*te**a is 0 (a) eal,ate te gai* o+ ea& a*te**a (!) &al&,late te :" o+ te tra*smitted sig*al i* ,*its o+ dB# (&) i+ te re&eii*g a*te**a is lo&ated 2 miles +rom te tra*smitti*g a*te**a oer a +ree$sa&e at- *d te aaila!le sig*al oer o,t o+ te re&eii*g a*te**a i* ,*its o+ dB# <#B > a ' = 4?e = 4?@ A2 A2 = ?2 d 2 @ x ( 2x109) 2 eti.a d=3 ( 3x10 8) 2 d = 3+tx0-3048 m/+t = 0-9144m ' = 201-7 = 23 dB ! :" = "t't = 2 x 201-7 = 403-4 = 26 dB# 4 πd & Ls = ( )2 = ( λ
4 π x 25 x 1609 3 x 10 8 / 2 x 109
= 1-13 x 1013 = 130- dB
)2
"r = :" 'r = :" (dB#) C 'r(dB) % Ls Ls = 26 C 23 % 130- = $81- dB# &. from tale 5.1 e see that the "ro"osal from satellite television cor"oration called for direct roadcast satellite D,S) -2 of 57 d,W and donlink transmisiion fre#uenc$ of 13.5 <=%. *ssume that the onl$ loss is the donlink s"ace loss shon . su""ose that the donlink information consists of a digital signal ith a data rate of 5>107
its;s. *ssume that the
re#uired ;0 is 10 d,/ the s$stem tem"erature at $our home receiver is ?00K/ and that $our roofto" dish has an efficienc$ of 0.55. hat is the minimum dish diameter that $ou can use in order to close the link ' Do $ou think the neighors ill o@ect '
<#B > 0
M =
EIRPG /T
( ) 0
E R K Ls L0 ; D=1 N 0
( ) 0
'r = E
E 0 R . K . L s . L0 .T s / :" N 0
'r(dB) = 10C76-99$228-6C206-1C27-78$7 =3-27 dB = 336-12 2
¿
ʎ Gr
4 πb
=
( 3 x 10
8
9
)
/ 12,5 x 10 2❑ x 3365
4 πx 0,55
= 0-28 m2; diameter = 0-98 meter 4 a* amlier as a* i*,t a*d o,t,t resista*&e o+ 0om- a 60$dB gai*a*d a a*didt o+ 10.G5 #e* a 0om resistor at 290 is &o**e&ted to te i*,t- te o,t,t rms *oise oltage is 100mi&roH Ietermi*e te eJe&tie *oise temerat,re o+ te amlier <#B > "o,t = Ho2 = . (oo $ o: ) #'
: = 10$8 = 1-38x10 $23(290C o:) 104 x 106 o: C290 = 2x10$20 = 119 $23 1-38 x 10 * amlier as a *oise g,re o+ 4 dB- a Ba*dit o+ 00 .G5- a*d a* i*,t resista*&e o+ 0 K &al&,late te i*,t sig*al oltage *eeded to ield a* o,t,t S: = 1 e* te amlier is &o**e&ted to a sig*al so,r&e o+ 0 K at 290
i = .# = 138x10$23 x 290 x x 10 = 2 x10$# F = 4dB = 21 = (S/);/(S/) Let (S/)o = 1 ; (S/) = 21 Si = 21 i = 02 x 10 $1 # 2
Xm − 15 =5.02 x 10 W R
Mm2 =21 x 10 $3 Holt2 Mm = 0 m# 6 &o*sider a &omm,*i&atio* sstem it te +olloi*g se&i&atio*s> tra*smissio* +re,e*&=3'G5- mod,latio* +ormat is B"S- !it$error ro!a!ilit=10$3- data rate=100 !its/s- li*. margi* =3dB- :"=100re&eier a*te**a gai*=10dB-dista*&e !etee* tra*smitter a*d re&eier=40000.m ass,me tat te li*e loss !etee* te re&eii*g a*te**a a*d te re&eier is *egligi!le (a) &al&,late te maxim,m ermissi!le *oise oer se&tral de*sit i* atts/ert5 re+ere*&ed to te re&eier i*,t (!)at te maxim,m ermissi!le eJe&tie *oise temerat,re i* eli* +or te re&eier i+ te a*te**a temerat,re is 290.N (&) at is te maxim,m ermissi!le *oise g,re i* dB +or te re&eierN <#B >
a "B = 10$3 = O (
√ 2 E ! ) o
√ 2 E
Iari ta!el B1 -
¿
! ) = 309
o ! % 4-77 = 6-8dB o 4 πd Ls = ( )2 = ( 4? x 109 λ
) 2
( 3 x 108/ 3 x 10 9) o = .o = :" 'r = D(!/o)regd :Ls .o(dB#/Gg) = 20C10 % ( 3C68C20C194) .o = $ 193-8 dB#/Gg = 4-17x 10$20 #att/Gg ! os = o/: = 4-17x 10$20 = 3022 1-38x10$23 o: = os % o = 3022$290 =2732 & o: = (F$1)290 F = o: C1 = 2732 C 1 = 10-42 = 10-2 dB 290 290 7 re&eier reamlier as a *oise g,re o+ 13dB - a gai* o+ 60 dB- a*d !a*dit o+ 2 DG5 e *te**a temerat,re is 490 a*d te i*,t sig*al oer is 10$12 # a) Fi*d te eJe&tie temerat,re - i* .eli* - o+ te reamlier !) *d te sstem temerat,re i* .eli* &) *d te o,t,t S: de&i!els <#B > 0
a
T R=¿ (F$1)290= (20$1)290= 10
!
T s =T A + T R= 490 C10 = 6000
0
0
0
❑ 6 23 6 & NewT = ': T s W =10 x 1,38 x 10 x 6000 x 2 x 10
7
10
= 1-66x
watt 6
10
So,t = ' si* = =
−6
10
watt 10
(S/)o,t =
−12
x 10
−6 −7
1,66 x 10
= 6-02= 7-8 dB
8 ss,me tat a re&eier as te +olloi*g arameters> gai* = 0 dB- *oise g,re =10 dB- !a*didt = 00 DG5- i*,t sig*al oer = 0 x 10 $12 # So,r&e temerat,re- P = 10- li*e loss =0dB Qo, are as.ed to i*sert a reamlier !etee* te a*te**a a*d te re&eier e reamler is to ae a gai* o+ 20dB a*d a !a*didt o+ 00 DG5 Fi*d te reamliRier *oise g,re tat o,ld !e re,ired to roider a 10$dB imroeme*t i* oerall sstem S: <#B >
P = 10 Si* = 0 x 10
$12
So,t = 0 x 10$12 x 10 = m#
#
#GT ":D" > P : = (F$1)290 = 2610 P: = P C P: = 10 C 2610 = 2620 o*t = '/: PS # = 10 x 128x10 181 m# (S/)o,t = m# / 181m# = 276 :e,ired imreme*t is 10dB- ;read (S/)o,t = 276 #G ":D">
P = 10
$23
x 2620 x 00 x 10 6
So,t = 0 x 10 $12 x 102 x 10 = 00m# 500 mW
(S/)o,t = 276 =
Nout
500 mW
o,t =
= 1812m#
27.6
{
permiseable oise power out
o,t = ' $ . PS# = 1812 m# 18.12 x 10
PS =
2
5
10 x 10 x 1.38 x 10
T ⁰ R 2
PUD" = P:1 C
G1
−6
−23
x 500 x 10
6
= 2622
= 226
2610
P:1 = 226 $
100
= 226
F1 = 178 = 2 dB 9 *d te maxim,m alloa!le eJe&tie sstem temerat,re P S re,ired to V,st &lose a arti&,lar li*. arameters are as+ollos> tra*smissio* +re,e*& = 12'G5- :"= 10dB#- re&eier a*te**a gai* = 0 dBmod,latio* te is *o*&oere*tl dete&ted BFS- oter losser = 0 dB- a*d te dista*&e !etee* tra*smitter a*d re&eier = 100 .m <#B > 1
"B =
2
−1
e
2
Ep No
= 10$
E 6 $ No = $2l* (2x10 ) = 2164
= 134 dB
( )
5 2
( ) 4 πd
Ls =
!
4 π x 10
2
=
8
3 x 10
9
12 x 10
= 14 dB
E b
PS (dB) = :" C 'r % (D C
N o C : C . C Ls C Lo)
= 10 C 0 % (0 C 134 C 40 % 2286 C14) = 312 dB = 1318 10 Uo*sider a :e&eier made , o+ te +olloi*g tree stages; te i*,t stage is a re amlier it a gai* o+ 20dB a*d a *oise g,re o+ 6 dB e Se&o*d stage is a 3$dB loss *etor. e o,t,t stage is a* amlier it a gai* o+ 60 dB a*d a *oise g,re o+ 16 dB (a) Fi*d te &omosite *oise g,re +or te re&eier (!):eeat art (a) it te reamlier remoed <#B > 1 = 20 dB F1 = 6dB
3dB Loss
3 = 60 dB F3 = 16 dB
" 2−1
a ) F = F1 C
G1
" 3−1
C
2 −1
F=4C
100
3.98 −1
C
= 479 = 68 dB
3.98 −1
!) F = 2 C
1/ 2
796 = 19 dB
G 1 G2
100 x 1 / 2
