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By DR KHAIRUN NIDZAM BIN RAMLI Solution for examples in notes Chapter !AM"
Example 3.1 A carrier signal with a peak voltage of 2.0 V is amplitude modulated with a 10 kH sine wave. !he modulation voltage has an effective value of "#0 mV. $ompute $ompute the following% a. !he !he per perce cent nt modul modulat atio ion& n& m '. !he instantaneous voltage of the positive and negative envelope when the 10 kH sine has completed () *s of its c+cle. c. ,llust ,llustrat ratee the resu result lting ing A- wavefo waveform rm Answer% a.
m
=
V m V c
× 100 =
"#0 × 10 2
−3
× 100 =
#3.03
2
'. V env V env
= V c + V m sin 2π f mt = V c 1 + m sin 2π f mt / 3 = 21 + 0.#303 sin 2π × 10 ×10 × () ×10−( // = ± 1.0V
c. A- wa waveform%
Example 3.2 A 00 carrier is modulated to a depth of "#. $alculate the total power in the modulated wave. Answer% P T
Example 3.3 An A- 'roadcast stations peak carrier voltage of 2 kV has 'een amplitude modulated to an index of "# with a 2 kH test tone. !he station 'roadcast fre4uenc+ is )10 kH. $ompute the following% a. !he lower and upper side'ands fre4uencies& f fre4uencies& f LSB and f and f USB USB '. !he peak modulation voltage& V m
1
c. !he peak lower and upper side'and voltages& V LSB and V USB d. !he maximum signal amplitude& V max Answer% a. f LSB = f c − f m = )10 − 2 = )0)kHz
= f c + f m = )10 + 2 = )12kHz
f USB
'. V m
= mV c = 0."#( 2 × 103 = 1.#kV
c. V LSB V USB
d. m
=
=
2 = V LSB
= V max
1.# ×10
3
mV c
=
2 0."#kV
= 0."#kV
− V c
V c
V max
= V c 1 + m/ = 2 ×103 (1 + 0."#) = 3.#kV
Example 3. A spectrum anal+er with an input impedance of #0 Ω is used to measure the power spectrum of an A- signal at the output of a preamplifier circuit. !he A- signal has 'een modulated with a sine wave. !he effective power P C is "# m& and each side'and& P USB and P LSB is 12# m. $ompute the following% !he total effective power& P T a. !he peak carrier voltage& V C '. !he modulation index& m& and the percentage of modulation index M c. !he modulation voltage V m d. !he lower and upper side'and voltages& V LSB and V USB e. 5ketch the waveform that +ou would see with an oscilloscope if it were placed f. in parallel with the spectrum anal+er Answer% a. P T = P C + P USB + P LSB = "# + 12# + 12# = 66#mW 2
'.
P C
=
V C 2 rms / R
V C peak / 2 = R
(V C peak / ) = 2 P C R 2
V C peak /
=
2 P C R
m2
c. P T = P C 1 + M
2
= m × 100 =
=
2 ( "# × 10
−3
)( #0) = ).(313V
66# − 1 = 0.)162 & m = 2 P T − 1 2 = "# P C 0.)162 ×100
=
)1.62
d. V m peak / = mV C peak / = 0 .)162).(313/ = ".0"0)V V m rms /
e. V LSB V USB
=
=
".0"0) 2
mV C 2
=
= V LSB =
= #V
mV C peak / 2 2 2.#V
=
0.)162( ).(313) 2 2
= 2.#V
2
V LSB 0 peak /
=
2 ( 2.# )
= 3.#3##V = V USB0 peak /
f.
= V C peak / + V LSB peak / + V USB peak / = ).(313 + 3.#3## + 3.#3## = 1#."023V V min peak / = V C peak / − V LSB peak / − V USB peak / = ).(313 − 3.#3## − 3.#3## = 1.#(03V
V max peak /
Example 3.# !he antenna current of an A- transmitter is ) A when onl+ the carrier is sent& 'ut it increases to ).63 A when the carrier is modulated '+ a single sine wave. 7ind the percentage of modulation and the antenna current when the percentage of modulation changes to 0.). Answer% m
=
m = I m
V m V C
=
I m R I C R
=
I m I C
).63 =
)
= 1.11(3 = 111.(3
I m I C
= mI C = 0.))/ = (. A
Example 3.( High89 tuned circuit is used to keep the : narrow to ensure that onl+ desired signal is passed. Assumed that 10 *H coil with resistance of 20 ; is connected in parallel with 101. p7 varia'le capacitor. a. !he circuit resonates at what fre4uenc+< '. hat is the inductive reactance< c. hat is the selectivit+ of the circuit< d. hat is the 'andwidth of the tuned circuit< e. 7ind the upper and lower cutoff fre4uencies<
Answer% a. f r =
1 2π LC
=
1 −(
2π 10 ×10
×101. ×10−12
= #MHz
3
'. X L c. Q =
= 2π f r L = 2π ( # ×10( (10 ×10−( = 31Ω X L R
d. BW =
=
f r Q
31 20
=
= 1#."
# ×10( 1#."
= 31)."kHz
e. =ne half on each side of center fre4uenc+ of # -H is 31).">2 ? 0.1#6 -H @pper& f 2 ? # 0.1#6 ? #.1#6 -H Bower& f 1 ? # 8 0.1#6 ? .)1 -H Example 3." Cetermine the image fre4uenc+ for a standard 'roadcast 'and receiver using ## kH ,7 and tuned to station at (20 kH. Answer% B= fre4uenc+ minus the desired stations fre4uenc+ of (20 kH should e4ua l the ,7. Hence& f B= D (20 kH ? ## kH f B= ? (20 kH ## kH f B= ? 10"# kH ow determine what other fre4uenc+& when mixed with 10"# kH& +ields a n output component at ## kH F D 10"# kH ? ## kH F ? 10"# kH ## kH F ? 1#30 kH !hus& 1#30 kH is the ima#e fre$uen%y . Example 3.) 7rom Example 2."& calculate the image reGection in deci'els& assuming that the input filter consists of one tuned circuit with a Q of 0. Answer% eGection ratio α
=
1 + Q 2 ρ 2
ρ
=
=
f imae f R!
−
f R! f imae
=
1#30 (20
−
(20 1#30
= 2.0(2#
1 + 0 2 ( 2.0(2# ) = )2.#0(1
20 log α = 20 log)2.#0(1/
2
= 3).33"B
Example 3.6 7or a 100 total availa'le power in the side'ands& compare the power in the side'ands when the modulation is standard A- with m ? 100& versus a 5$ design where 60 of the carrier power is suppressed. How man+ times greater is the side'and power in the suppressed carrier case< Answer% 7or m ? 1& P T ? P C 1 m>2/ ? P C 3>2/ P C ? 2>3/ P T ? 2>3/100/ ? ((." P SB ?100 D ((." ?33.3
C5:85$ P C ? ((." × 0.6 ? (0.0 reduced/ P SB $%W ? 33.3 (0.0 ? 63.3 !he power ratio ?
P SB P SB
$%W &L'
=
63.3 33.3
= 2.)
,n d: ? 10log2.)/ ? .# d: -eans in A-& information signal is transmitted onl+ 33.3 'ut in C5:85$ is 63.3.
Example 3.10 A #00 C5:85$ s+stem with 100 modulation suppresses #0 of the carrier and the suppressed carrier power goes to the side'ands. How much power is in the side'ands and how much is in the carrier< :+ how man+ d: has the side'and power increased< Answer% ith 100 modulation& final P C ? #00 and total P SB ? 2#0 ,f P C is diverted to 5:& ew P C ? #00 D 2#0 ? 2#0 ew P SB ? 2#0 2#0 ? #00 !he increase in power is #00>2#0 ? 2 ? 3 d:
