MODULATION
ul
1.1
1.4
Phas Phas
Modu Modula lati tion on
1. arri arri in ccor ccorda danc nc with with th nsta nstant nt neou neou va ue of he nt llig llig nc sign signal al de
AMPL AMPLIT ITUD UD
ODUL ODULAT ATIO ION' N' (AM)
sma
:t
instantaneousamplirude
o,f theintell theintellig~nc ig~nc~. ~.
fo eAM'
ntell ntellig igen ence ce Signa Signa
'e~
E,
AM Signa Signa
inst instan anta tane neou ou valu valu peak peak valu valu angu angula la velo veloci city ty phas phas
angl angl
e, Carri Carrier er Signa Signa
arri arri
sign signal al
are: ud qu
na
MODULATION,~ "-"
,'-_
ud freq freque uenc nc
du modul modulat atio io
1. arri arri in ccor ccorda danc nc with with th nsta nstant nt neou neou va ue of he nt llig llig nc sign signal al de
AMPL AMPLIT ITUD UD
ODUL ODULAT ATIO ION' N' (AM)
sma
:t
instantaneousamplirude
o,f theintell theintellig~nc ig~nc~. ~.
fo eAM'
ntell ntellig igen ence ce Signa Signa
'e~
E,
AM Signa Signa
inst instan anta tane neou ou valu valu peak peak valu valu angu angula la velo veloci city ty phas phas
angl angl
e, Carri Carrier er Signa Signa
arri arri
sign signal al
are: ud qu
na
MODULATION,~ "-"
,'-_
ud freq freque uenc nc
du modul modulat atio io
Resu Result ltin in AM sign signal al
usin usin
th trig trigon onom omet et ic
iden identi ti
and
where AM ampl amplit itud ud eAM
also also th
ampl amplit itud ud
will be
MODU MODULA LATI TION ON INDE INDE ampl amplit itud ud
eAM
(E
eAM
EesinO)et+ EesinO)et+Emsi EmsinO)et nO)etsinO) sinO)mt mt
is vari varied ed
inCO inCO t) inO) inO)et et
modulat
where
EXAMPLE:
(E~E
Boar Problem, November 2002 What V~
," ,a
.6 an
min
0. 9, espe ti el
SOLUTION: max
'+
max
Vmin
2.6-0.29
max
2.6+0.29 0.7993
where modulation max
min
inde signal
PERCENTAGE OF MODULATION
where
EXAMPLE:
AM SPECTRUM
signal. mEe
Ee
mEe
d)
e)
fm
fm
r,
frequenc
domain
SOLUTION:
fe +f
a)
Ole= 21tf Ole =46.Sxl0 rad/s
=~
where full
ar ie AM signal m=
r,
46.Sxl0
fe
740070SH
21t
b) Olm= 21tfin Olm=44xl0 rad/s
ation
e) time domain fm
44x10
30
=-2-n-
t,
25
'.
c)
,VJ,
m=-
m=-
("fA
25-
Ee
,Y
-30
25 0.
d) carrie
.... ,v
frequenc 2.5V
signal
lowe sideband
domain 25V
39.37kHz 740.07kH
2.5V
740.77kH
fLSB=fe-fm 7400705Hz-7003H
uppe
EXAMPLE:
(EeE
Board Problem, November 1999)
de
ne
sideband
fUSB fc +f 40 705H
00 Hz
nt
uenc
of he
modula
on
SOLUTION:
b)
25k
25k
+--,
"~
m=O.5 %M=mxl00% EXAMPLE: pk pk
%M=50%
Calculate
modulation. SOLUTION:
m=l
a)
m=
max max
%M=mxl00%
rnin
%M=100%
EXAMPLE:
What
(E
m=0.25
V r n a x andVmin
of2.
an
.29, es ective y?
=m l00% %M=25%
modulat
SOLUTION:
SIMULTANEOUS TONE MODULATION Vr
ax
Vrnin
2.6-0.29
tones.
2.6+0.29 0.7993
m3
m2
rt2
EXAMPLE:
BW=2f BW=2(Sk) BW
10kHz
where BW
tota bandwidt
fmN
maximu
intelligence frequenc
mod~lat
__
._
where
EXAMPLE:
m,
th to al index
ef ec iv
mo ulatio
.5 kHz, with modulation
m,
ef ective modulation index.
m2 mN
SOLUTION:
~ r - (0-.25-)2-+~(-.5-)2
EXAMPLE:
=0.56
simultaneously?
EXAMPLE:
1996) Three
rd
(EeE
SOLUTION:
he AM spectrum will be espectively, si ultaneou ly mzE
m3
mjE
Ec
odulat
450-
mjE
1-2 SOLUTION:
cBW
2fmmax
BW
2(5k)
fc -l
fc
+l
fc +3
r, +Sk
;lOCH
--- --
=1Jm
+m +m3
100V
200V
modulation
modulatroIi<
m2
=0.44
300V 450V
ffi3
also
=0.67
~(0.2if
(0.44
(o.67f
=0.83
where P,
%M=m xl00% %M
Pc
(0.83)(100
tota tr mitted ow mo late arri modulation inde
AM POWER CALCULATIONS
Pc PUSB
PLSB
_m
LSB-~ fLSB
fc
fuss
Ee
er
EXAMPLE:
November
carrier? SOLUTION:
where
E,
carrie amplitud load resistance
PUSB
PI =P{1 P=~ 1+-
LS
Pc
~2
12
66.67W
EXAMPLE:
EXAMPLE: SOLUTION:
powe
contains th inte ligence?
SOLUTION:
PI
I
1451.2
=P{1+
1997) An
~(o.lf
SOk
7 " "
(O.2f
(oAf
=0.68
(1+ (O.~Sf)
b) Pc
36.73kW SB,
=P
~t
Then, SB
(0.68f
SB,
=P --
SB
(c.ss)'
SB
36.73k~-
SB
13.3kW
11S.6
c)
=P{l+
~2
EXAMPLE:
=50{1+ unmodulate
arrier power, Pc
SOOW
modulated simultaneously by ou tones, rn mz
0.2, m,
0. an
0.1,
(0.~8Y)
61S.6W
O.S, determine
EXAMPLE:
(ECE Boar Problem Aprill~98) Ifa SOLUTION:
a)
'k is un odul ted, determin powe
he
it is
odulat
th tota radiated at'30%
niciCiula
SOLUTION:
%Pincrease 50
=p(+m2) 8k(1
CURR
(O.~f)
=8.36kW
where
EXAMPLE:
100% modulation
in
mean
correspo ding
SOLUTION:
EXAMPLE:
P=P(I+ =P +f p.
otal transmit ed curren nm du ated ar ie cu re modulation inde
I,
SOLUTION:
-P
14
ll
1+-z
100%
modul
Vt=VcH·· :l
(0.75)2 %M=mxl00% 56.6VRM %M=111.34%
VACUUM-TUBE AM MODULATOR AM VOLTAG
EQUATION
hi late lt transmissions.
ca
ilit
fo
is where
v,
tota tr smitte lt late carrie volt modulation inde
PEP
ak
P, m=l
EXAMPLE:
50
MS
75%
lo
po
ig -p
er
100% modulation.
Pc =1 c)
SOLUTION:
.) 2.66P
39.9k
=15kW b)
,c
12.Sk
dc max max
(audio
6.2SkW
Ptransmitted audio)
P m a x(audio
Ptransmitted audio)
6.2Sk(0.8)
PtransmitteAaudio)
kW
PERCENTAGE POWE
SAVING
1]
EXAMPLE:
Pc 0.25)2 an
100o/~ 1+ (0.~5)
SOLUTION:
100% Tota Powe c+
SB
100%
MODULATION
lf3E SE
er Sing ideban Full Ca rier Independen Sideband
R3E C3F
Sing ideban Suppre se Ca rier Sing ideban Reduce Carrie Vestigia Sideband
EXAMPLE:
Pc +Pe~ S =
FORM OF AMPLITUD
100% +1 a)
AM. A3E
%PS=83.3% c)
d)
H3E DSB
mo~ula
SOLUTION:
a)
A3E: p = p -
500(0.8f =160W
INGL SI EB ND (S B) TRANSMITTE RATING P=P~
(0.8f =80W where
H3E:
PEP pk
PI=P +P
EXAMPLE:
ation
ea ea envelope voltag antenn resistance
SOLUTION:
rrlo signal pk
PEP=
72 110
PEP= Intell ge ce
si al em
1.3 FR QU NCY MODULATION
l1erator, eP
Intelligence Signal
FMSignal
J.7Msignal
e,
Carrier Signal
moaMa_····
Em sinwmt
FM signal in frequenc vs. time
fe+Op fe-o r,
where
eFM (o
""--/------.-->
(Om
an
ar
an ular velocity of he intell genc modulation
inde
E,
MODULATION INDE
Mathematically,
where
where eFM
fm
th intelligence signal intelligence frequency
also Note: intelligence amplitude.
Mathematically,
ation
modul
EXAMPLE:
where
frequenc
deviati6rt(Hz
deviatio sensitivit '(Hz/V peak intelligence 'signahimplitud
;"
(V)
EXAMPLE:
eFM
50 in(9
10
c)
Find th
od lati
10sin5000t)V.
in
SOLUTION:
modulator with
a) (OC
21tfc
modulating signal of 5cos(21t(2500)t) (Oc
21t
SOLUTION:
o=kE
95xl0
(10~Z
)?V),
21t fc
15119720Hz
50kHz
fm
(Om
21t
50k 2500
5000 21t
=20
fm
ati6n
796Hz
moduia
c)
FMSPECTRUM
d)
(lOX796) 7960Hz
EXAMPLE:
eFM
EJrO(mf)sinroct
Jt(mfXsin(ro s 1 SOLUTION:
mf=-
fm 50kHz
Jz(
si (r
J3(m
si ro mf
+romft-sin(ro ~(i)
si
-romft] ro r; 2ro
ft _:sin(ro si
Nromft±sin(r
ft] -,-NromftD
Note:
10kHz only th se si an it amplitudes ar transmitted.
coefficients si ific
modul
bu
where =(fc
BW fm
signal intelligence frequenc
where
(highest fm
te
th mo lati in intelligence frequenc
where frequenc
deviatio
where approximat ba:ndwidth frequenc deviatio fm
intelligence frequenc
modula
EXAMPLE:
this system usin th modu atin
SOLUTION:
voltag
imul aneo sl
BW BW=2Nf
SOLUTION:
a) m[=-
fm 45k 0=(40XSOO)
m[=3
20kHz
b) '= 1.3V
'=800Hz from th
relationship
Ou:.Em o=kE
ation
modula
ed ce
15 -6 (Em .E
fm =1.5kHz
{,\Z.6 !2)
c S ' = (ZOk
c S ' = 10kHz
mf'=~
(j'
where
,10k mf 800
DR max
fm(';'ax)
deviatio
rati
maximu
allowabl
deviatio
maximu
intelligence frequenc
EXAMPLE:
EXAMPLE;
(BCE
2001) What
(BCE
2002) FM igna if th ma im
fr
nc
iati
ll wa le intelligen
eviati is 12 kH an fr en is kHz?
SOLUTION:
SOLUTION:
Usin
15 kH kHz?
Carson's Equation fm)
ation
DR
max
th
imum
where
12k
DR=-
%M actual frequenc
actual
DR=3
maximu
OaX
EXAMPLE:
TV transmission employ
deviatio
allowabl
deviatio
EXAMPLE:
percentage modulation 15 kHz.
SOLUTION:
SOLUTION:
%M max
DR
actual
100%
°max
fm(max)
25k
DR=-
15k DR =1.67
PERCENTAGE MODULATION
Note:
omax
fo
carrie
ta ar
swing.
CS=20
60kHz
modula
ro ca ti
Consid
75k
Phas
odulator
em
%M=80%
epM
Intelligence Signal
PMSignai
EXAMPLE:
(ECE oa Pr bl m, November March 1996) fo FM broadcasting
00 an Carrie
Signal
Carriersignal
SOLUTION:
%M
} ac tu a
100%
(}max
Oactual
(0.8)(75k
0actual
60kH
CS
1.4
2(60kHz)
PHAS
"-r=sc;m
Intelligence signal
CS=28
MODULATION
=EmrunID
-a,
ation modula
where
PM signal
deviatio sensitivit (rad/V peak intelligence amplitud (V)
EXAMPLE:
2rad/V Mathematically,
twic th
amplitude?
SOLUTION:
8=kE where
epM
8=e~d)clOV)
E, ro
carrie angula velocity
ro
intelligence signal intelligence signal angula velocity
Em'=2Em Em'= 2(10 =20V
modula
8=40rad
25k 250
mfJr
r:
mf ~100 EXAMPLE:
is
am
and
m=100 PM. hi give th equation
SOLUTION:
Ecsin(roct+msinromt) el oc roc
2n(100M) 628318531rad
eFM
6sin(2n(100M}
100sin2n(250~)
epM
6sin(2n(100M}+
100sin2n(250})
Note: ec intelligence frequency.
2n(250}
el intelligence frequenc
will be mf=-
'·
6f
fni'~6(250Hz) fm
'= 1500Hz
modula
mp'oc will
mp'=mp=100 will be
intelligence. eFM :;:6sin(21t(100M~
16.67sin21t(1500~)
'=21tf will be
'= 21t(250
ep
will be mf'=r
since mf
0'
is constant
8'
6sin(21t(100M~
10 si 21t(1500 ~)
EXAMPLE:
deviation.
6(rm)
SOLUTION:
CS=28 8= t O O k
mf '= .!.(100)
is constant
ation
modula
6.oc
6=kE an
k=-=-
6'
SOLUTION:
eq CI
3xfc
fel
3(7.5M)
fCI
22.5MHz
3x 3(5k)
8'= 10kHz
15kHz
EXAMPLE:
ca et et eq
a)
modulatio
of'an
ar
nc
qu
fC2 ct
4{22.5M}
multiplier
=90MHz
02 =4xo
inus idal
igna by om
60kHz signal-to-nois re-e
amount 'equal Assu in is fC3
th
PRE-EMPHASIS
4{15k} 02
ultiplie
fC2
hasi
ratio. Circ it
th oscillat frequency. re uency)
fo'
A u d io ' o it t
fC3 =90M+6M
'(pre,-e~phasized) R2
fC3=96MHz 60kHz Note:
deviation.
FREQUENCY MULTIPLICATIONVS. HETERODYNING ultiplyi
ha in heterodyning
ation
th sinusoidal
Jl\9dula
FrequencyRespons
ofPre-emphasis ,:
/',.v
,!
RC
75fls "L
21tRC
2122Hz
Freque'llcy Response' of De-emphasis /',.v
DE-EMPHASIS
thehigh audio
of th pre-emphasis
PRACTICE PROBLEMs: 1. pre-empha-siz-ed--'VV'v~~---'I'~,-.,-:
,M ig al ha
Carn r, fr
-:"A,udi ou
~nc¥ '1 Th carrie pfak
Audio alculate he
ulatio
in
(A wers
eAM=1Osin(21t(1530k)t)+, 1.25cos(21t(1529k)t):'" 1.25cos(21t(1531k)t),O.25)
modtila
/V
0..7416) frequenc
inusoida
gn 'is
that ca
kHz)
4r
W)
.H .k
,15
.( 6.
.M
An
frequencies w i l l 2.5 ~.9955 MHz)
th
ation
carrie is sinusoidally modulated,
modulat