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Amplitude Modulation 1
1
01. Ans: (d)
Sol:
µ+=
21PP
2
CT
5.01400600
2
2
=−=µ
⇒
121
2
2
=µ⇒=µ
⇒
02. Ans: (b)
Sol:
µ+==
21PkW10P
2CT
For µ = 0.6, kW47.818.1
10PC ==
03. Ans: (c) Sol: Assuming single tone modulation, total
power
µ+=
21PP
2CT .
For µ= 0, PT = PC and for µ = 1, PT = 1.5PC. 04. Ans: (c) Sol: µ = 1 given
Pt = PC
µ+
21
2
Pt = 23 PC = 1.5PC
05. Ans (c) Sol: Ct II
1=
12 tt I
100115I =
2
μ1II2
Ct 2+=
2
μ1II2
tt 12+=
2
μ1100115 22
=−
µ = 0.803 06. Ans: (b)
Sol: (t)mK1
(t)mK)(ηEfficiency
22a
22a
+=
Where, (t)m2 → mean square value of m(t)
Ka = Amplitude sensitivity m(t) → square wave 2
m2 A(t)m =
100AK1
AKη% 2
m2a
2m
2a ×
+=
]AKμ[100μ1
μη% ma2
2
=×+
=
Maximum value of µ = 1
%5010011
1η% =×+
=
07. Refer Solution no: 4 08. Ans: (c)
Sol: s(t) =
ω−ω+ tsin
21tcos
211 21 cos tcω
∴ cos tcω +41 [ ]tcos.tcos2 1c ωω
−41 [ ]tsin.tcos2 2c ωω
Am
−Am
m(t)
LEVEL – 1 (Solutions)
ACE Engineering Publications (A unit of ACE Engg. Academy – Hyderabad, Vijayawada, Visakhapatnam, Tirupati, Delhi, Bhubaneswar, Bangalore, Pune &Chennai) (Copyrights Reserved)
2 Electronics & Communication Engg. ACE ∴ PSB = 1/8
PT = Pc + PSB = 21 +
81 = 5/8
∴Power efficiency =T
SB
PP
×100%= 20%
09. Ans: (a) Sol: Total modulation indices
22
21
2t µ+µ=µ = 0.32 + 0.42 = 0.25
∴ Total power Pt = PC
µ+
21
2t
= 10 × 103
+
225.01 = 11.25kW
10. Ans: (a) Sol: Efficiency (η)
= %11.111005.02
5.01002
2
2
2
=×+
=×µ+
µ
11. Ans: (d) Sol: Following frequency will be present in the
output 1000± 0.3 = 1000.3 kHz and 999.7 kHz and 1000±2 = 1002 kHz and 998 kHz
12. Ans: (d)
Sol: 2 fm = cf100
1
fc = 200 fm = 200 × 10 k = 2 MHz 13. Ans: (d) Sol: The given equation is compared with
[ ] tf2costf2cos1A cmC ππµ+
Where AC = 2 µ = 0.4 fm = 3 K fc = 1 M
56c 10
21010
21f ×=×=
= 5×105 fc = 500 K
2fm = 6000 fm = 3 K
Given cutoff frequency of ideal low pass filter i.e. fc = 8 K
So the spectral components at the output of the LPF are 3 KHz & 6 KHz.
14. Ans: (a)
Sol: 33.0510510
EEEE
minmax
minmax =+−
=+−
=µ
15. Ans: (c) Sol: Given equation is compared with
( )( )tff2cos2
Atf2cosA mc
CcC −π
µ+π
( )( )tff2cos2
Amc
c +πµ
+
fc = 2000 Hz, fc−fm = 1800 Hz fm = 200 Hz and Pc = 200 W and η = 0.2
2
2
2 µ+µ
=η
2
2
22.0
µ+µ
=
222.04.0 µ=µ+
0.4 = 0.8 µ2
707.02
1212 ==µ⇒=µ
16. Ans: (d) Sol: from the equation above
2002
AP
2c
C ==
400A2c = ⇒ Ac = 20 = M
and N = 2
AC µ
22
12M
N=
µ=
07.72
1022
2022
MN ====
fc = 8 K 0
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ACE Communications Postal Coaching Solutions. 3
17. Ans: (a)
Sol: given Pc = 1000R2
A2c =
10005002
A2c =
×⇒ Ac = 610
∴ Ac = 103 18. Ans: (c) Sol: given 6.0=µ
Amax = Ac (1+µ) = 103 (1+ 0.6) = 103 (1.6) = 1600V Amin = Ac (1−µ) = 103(1−0.6) = 400V 19. Sol: Given Amax = 110V and Amin = 90V
2
AAA minmax
C+
=
1002
2002
90110==
+=
minmax
minmax
AAAA
+−
=µ
1.020020
9011090110
==+−
=
The amplitude of each sideband is
No option is correct 20. Ans: (a) Sol: s(t) = 10 m(t) cos (2π x 106 t) m(t) = [1 + 0.5 cos (2π×103 t) + 0.4 cos
2π (10 × 103) t + 0.3 cos 2π(20×103) t]
It contains 7 spectral components Hence option 1 is not true 2. BW = 2×20×103
= 40×103 = 40 K Option 2 is true
3. ( ) ( ) ( )222t 3.04.05.0 ++=µ
707.05.0 == Option 3 is true
4. %202 2
t
2t =µ+
µ=η
Option 4 is not true 21. Ans: (d) Sol: ( ) ( )[ ] ( )t40costm20ts π+=
[ ]t4cos2t2cos424 π+π+=
π+π+= t4cos
121t2cos
61124
Here 121,
61
21 =µ=µ
∴ 22
21t µ+µ=µ
0347.0121
61 22
=
+
= 0.18 22. Ans: (d) Sol: The power of the AM signal is
µ+=
21PP
2t
Ct
( ) ( )
+=
218.01
224 22
= 292.66 ~ 293 W
2Ac
41.0100
4Ac ×
=µ =2.5V
fc−fm fc fc+fm
1220 1200 1180
fc fc+fm1 fc+fm2 fc+fm3 fc−fm3 fc−fm2 fc−fm1
4
1μcA
43μcA
2cA
4
1µcA
42µcA
43µcA 4
2µcA
1.
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4 Electronics & Communication Engg. ACE 23. Ans: (c) Sol: Given equation is compared with
[ ]tf2costf2costf2cos1A3m32m21m1c πµ+πµ+πµ+
so AC = 10, µ1 = 0.5 , µ2 = 0.8, µ3= 0.9 fm1 = 1K, fm2 = 10K, fm3 = 20K and given fc = 1M
The band width of the signal = 2(maximum frequency) = 2(20K) = 40 kHz 24. The power taken by the component 990
kHz is (a) 8 (b) 10 (c) 20 (d) 30 24. Ans: (a) Sol:
8.0
K990K10)(
K1000→−
Power taken by the component
990 kHz = ( ) 88
8.0108
A 2222
2C =
×=
µ
25. Ans: (a)
Sol: 502
10020
2A
P22
cC ==
1==
23
22
21t µ+µ+µ=µ
222t 9.08.05.0 ++=µ
3.1=
µ+=
21PP
2t
Ct
( )
+=
23.1150
2
= 92.25
PSB = Pt−Pc = 42.25 PUSB = PLSB = 21.125
The ratio 2289.025.92125.21
PP
t
USB ==
26. Ans: (a)
Sol:
µ+=
21PP
2t
Ct
( )
+=
23.1150
2
= 92.25 27. Ans: (c)
Sol: 23
22
21t µ+µ+µ=µ
222t 9.08.05.0 ++=µ
3.1=
459.02 2
t
2t =µ+
µ=η
28. Ans: (c) Sol: S(t) = AC[1 + Ka m(t)] cos ωct
Ka m(t) > 1 Envelope is |AC(1 + Ka m(t))|
29. Ans: (b) Sol: The Time constant RC > Tc (= 1/fc) and RC < Tm (= 1/fm).
Since fc = 1MHz, RC > 1 µsec, and fc = 2 KHz, RC < 500 µsec
30. Ans: (a) Sol: V(t) = Ac.cosωct + 2 cosωmt . cosωct.
Comparing this with the AM−DSB−SC signal A cosωct + m(t).cosωct, it implies that m(t) = 2cosωmt ⇒ Em = 2
To implement Envelope detection, Ac ≥ Em
∴ (Ac)min = 2 31. Ans: (c) Sol: x(t) = 5(1 + 2 cos 200πt) cos(2000πt) compare with standard single tone AM x(t) = AC(1 + µ cos 2πfmt) cos 2πfct µ = 2 For µ > 1 only synchronous demodulator
is used to demodulate the signal. 32. Ans: (a)
Sol: To avoid diagonal clipping RC < W1
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(Copyrights Reserved)
ACE Communications Postal Coaching Solutions. 5
33. Ans: (c) Sol: τ = 5 µsec
f = 5
1010511 6
6 =×
=τ −
55
10251010
×=×
=
= 200k = 0.2M 34. Ans: (b) Sol: An Envelope detector is a diode half wave
rectifier followed by an RC-low pass detector. The output of the detector represents the envelope of the incoming high frequency signal. It is used for FSK signals.
Envelope detector also called as non synchronous detector
35. Ans: (d) 36. Ans: (c) Sol: Given Pc = 400 W and R = 50 Ω µ = 0.8 The power developed across the load is
µ+=
21PP
2
Ct
( )
+=
28.01400
2
[ ] W52832.1400 ==
01. Ans: (a) Sol: µ =1, at 100% modulation in spectrum
analyzer the amplitude of upper and lower frequencies are each equal to one half the
amplitude of carrier
2AC (or) 6 dB
power 6 dB means (3dB + 3dB) =
21.
21 =
21 (half of carrier amplitude)
02. Ans (b) Sol: S(t) = 10cos2π106t + 8cos2π5×103tcos2π106t
S(t) = 0.8 ×10cos2π106t + 0.5×8cos2π5000tcos2π106t
= 8(1+ t50002cos84
π )cos2π106t
µ = 84 =
21 = 0.5
S(t) = 10cos2π106t+0.5×8cos2π5000tcos2π106t
= 8(1+84 cos2π5000t)cos2π106t
µ = 84 =
21 = 0.5
03. Sol: fC = 1 MHz = 1000 kHz The given m(t) is symmetrical square
wave period T = 100 µ/sec
fm = 0T
1 =10 kHz
LEVEL – 2 (Solutions)
Tuned ckt
Tuned ckt Carrier message
= 1 MHz + 5 kHz Gain = 0.5
Gain = 0.8
Carrier = 1 MHz
µ = ?
0.5
fC−fm fC fc+fm
0.5
0.8
2AC =4
4ACµ =0.5
100µsec
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6 Electronics & Communication Engg. ACE
This frequencies 980k, 1020k are not present because the symmetrical square wave it consists of half wave symmetrices only odd harmonics are present, even harmonics are dismissed
04. Sol: m(t) = sinC(200t)sinc2(1000t) = sinc(200t)sinc(1000t)sinc(1000t)
BW = 2 × 1100 BW = 2200 Hz
05. Ans: (a) Sol: P(t) = u(t) −u(t−1) ⇒
g(t) = P(t) *P(t) =
=
X(t) = 100(P(t) +0.5g(t))cosωct = 100(1+0.5t)cosωct = Ac(1+Kam(t))cosωct Ka = 0.5, m(t) = t µ = ka[m(t)]max µ = 0.5 ×1 = 0.5
06. Ans (c) Sol: m(t) = 2cos(2πfmt) C(t) = Accos(2πfct)
AM signal without over modulation? a) X(t) = ACm(t)cos2πfct (not AM
signal) b) X(t) AC[1+ m(t)]cos2πfct (µ = 1 × 2 µ = 2 Over modulation)
c) X(t) = Ac[1 + 41 m(t)]cos2πfct, ka =
41 ,
µ = 42 = 0.5
d) Not an AM signal 07. Ans: (c) Sol: m(t) = −0.2 + 0.6sinω1t, ka = 1, Ac = 100 S(t) = Ac[1−0.2 + 0.6sinω1t]cosωct = 100[0.8 + 0.6sinω1t]cosωct Vmax = Ac[1 + µ] = 100[0.8 + 0.6] = 140 Vmin = Ac[1−µ] = 100[0.8 −0.6] = 20 = 20V to 140 V 08. Ans: (a) Sol: m(t) = 2cos2πf1t +cos2πf2t C(t) = Accos2πfct
S(t) = [Ac + m(t)]cos2πfct
S(t) = Ac[1 + CA
1 m(t)]cos2πfct
Ka = cA
1
Am1 = 2, Am2 = 1
µ1 = KaAm1 = CA
2 , µ2 = KaAm2 = CA
1
µ = 22
21 µ+µ ⇒ 0.5 = 2
c2c A
1A4
+
⇒ AC = 20
09. Ans: (d) Sol: Amax = 10V Amin = 5V µ = 0.1 V
fc−3fm =970K
fc−2fm =980K
fc−fm =990K
fc =1000K
fc+fm 1010K
fc+2fm =1020K
fc+3fm =1030K
500 −500 0 500 0 −500 100 0 −100
−1100 0 1100 =
P(t) 1
0 1
1
0 1
1
0 1
0 1 2
g(t)
*
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ACE Communications Postal Coaching Solutions. 7
µ = minmax
minmax
AAAA
+−
= 31 = 0.33
AC =2
AA minmax + = 2
510 + = 7.5 V
Amplitude deviation ACµ = 7.5×31
= 2.5 V
µ2 = 0.1 ⇒ Ac2µ2 = 2.5 Ac2 = 25 V Which must be added to attain = 17.5 Ans = 17.5
10. Ans: (a) Sol: The given signal can be AM-DSB-SC or
AM – DSB – Full carrier. If it is an AM – DSB – SC, demodulation requires a local signal of frequency 400 KHz. If it is AM – DSB – FC, It can be demodulated by using envelope detector.
11. Ans: (c)
12. Ans (d) Sol: m(t) = (Ac + Am cosωmt)cosωct.
= Ac(1 + AcAm cosωmt)cosωct.
Given Ac = 2Am
= Ac(1 +21 cosωmt)cosωct.
=
+=
4μ
2AP
2μ1
2AP
22c
s
22c
T
1689
161
811
4
21
2
2
×=+
=+
=µ
µ
s
TPP
13. Ans (b)
Output of filter = Accos(2πfct + 900) = A sin (2π106t)
Envelope detector detects only DC component (A)
PT = 18 Ps
A m(t)=0
AC(1+µ)=AC+Acµ ⇒10V = 7.5 + 2.5
AC(1-µ)=AC - ACµ 5V = 7.5 −2.5
1
999K 1000k 1001K
900
ED
ACE Engineering Publications (A unit of ACE Engg. Academy – Hyderabad, Vijayawada, Visakhapatnam, Tirupati, Delhi, Bhubaneswar, Bangalore, Pune &Chennai) (Copyrights Reserved)
01. Ans: (b) Sol: sinc2 (1000t) cos 2π fct 2× 1000 = 2000 = 2kHz 02. Ans: (a)
Sol: 2fm = cf100
2
2 × 10k = cf100
2
⇒ fc = 1MHz 03. Ans: (a) Sol:
2×10k = 20k 04. Ans: (b) Sol: Band width = 2fm (max frequency)
= 2(2ω 1) = 4ω 1
4
]21[14
]AA[A 222m
21m
2c +
=+
= Power
W25.145==
05. Ans: (c) Sol: XAM(t)=10[1+0.5sin2πfmt]cos (2πfct)
Average side band power = 2
P 2Cµ
PC = 2)10( 2
= 50 W
∴ PSB = W25.62
)5.0(50 2
=×
06. Ans: (c)
Sol: % Power saving = 100powertotalsavedPower
×
= 100
21P
P2
C
C ×
µ+
=
28.01
12
+× 100
= 75.76% 07. Ans: (c) Sol: The required frequency components are fc
(= 1 MHz) ± fm V0 = a0 [Ac
1 cos 2πf 1c t + m(t)]
+a1 [A1c cos 2π f 1
c t + m(t)]3
= a0 [Ac 1 cos 2πf 1
c t + m(t)]
+a1 [ (Ac1)3cos32π
fc 1t + m3(t) + 3Ac
1m2(t).cos 2π fc1t
+ 3(Ac1.cos2π fc t )2.m(t)]
The AM – DSB – SC signal lies in a1.3m(t) . ( Ac
1. cos2 π fc t)2 = 3a1(Ac
1)2.m(t) [1 + cos 2π (2f 1c )t ]
The side band frequencies are 2f 1c ± fm,
which can be filtered by a BPF. 2 fc
1 = 1 MHz, fc1 = 0.5 MHz
08. Ans: (b) Sol: e(t) = 50[1+0.89cos 5000t +0.3sin 9000t]
cos (6 ×106)t ωc = 6 × 106 ; ω1 = 5000; ω2 = 9000
The corresponding sidebands are 6 × 106 ± 5000 = 6.005 × 106 and 5.995 ×
106 6 × 106 ± 3000 = 6.003 × 106 and 5.997 × 106
5 10
Double Side Band Modulation 1
2 LEVEL – 1 (Solutions)
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ACE Communication Postal Coaching Solutions 9
System
system
o/p m(t) k
+ +
+
+ ∑
∑
∑
V2
V1
bV22
aV12
A cos ωC
1
2
09. Ans: (a) Sol: A Balanced modulator generates a DSB-
SC signal, by multiplying the baseband signal and the carrier signal.
10. Ans (b) Sol: νo = a ν i + b ν i
3
s(t) = m(t) + c(t) v0 = a [m(t) + c(t)] + b [ m(t) + c(t)]3 v0 = a[m(t) + c(t)] + b[m3(t) + c3(t)+
3m(t) c2(t) + 3m2(t)c(t)]
π+
π=2
tcf4cos1cAtmf2cosm3Ab
t)cf2(2costmf2cos2
mAcAbtmf2cos
2
mAcAb ππ+π=
fc| = 2fc = 4 MHz
11. Ans (b) Sol: kHz15f
1m =
kHz10f2m =
kHz20f3m =
USB frequencies = f’c + fm
4010 kHz, 4015 kHz, 4020 kHz 01. Ans: (d) Sol: Given M(f) = 1 f1≤ f ≤f2 f1 = 1 kHz
f2 = 10 kHz
= 0 else where
y(t) = m(t) cos2πfct
y(f) = M(f)
−++
2)ff(S)ff(S cc
2
)ff(M)ff(M cc −++=
f + fc = 1010kHz / f = 10kHz
=1001 kHz /f = 1 kHz
fc-f = 990kHz / f = 10 kHz
= 999kHz / f = 1kHz
∴Range of frequencies for which y(t) has non zero spectral components is 990 kHz – 999 kHz and 1001 kHz to 1010 kHz
02. Ans (c) Sol: V1 = k [m(t) + c(t)] V2 = [m(t) – c(t)] V0 = aV1
2-b V22
= ak2[m(t)+c(t)]2-b[m(t)-c(t)]2 = ak2 [m2(t)+c2(t)+2m(t)c(t)]
−b[m2(t)+c2(t)-2m(t)c(t)] = [ak2-b]m2(t)+[ak2 −b]c2(t)
+2[ak2+b][m(t)c(t)]
on verification if K = ab
S(t) = 4bm(t)c(t)→DSBSC Signal 03. Ans (a) Sol: Given A = 10 m(t) = cos1000πt b =1 B.W=? and power = ? s(t) = 4b.A cos2πfct. cos2π (500)t = 40.cos2πfct. cos2π (500)t B.W = 2 fm = 2 (500) = 1 kHz
f1 f2
1
LEVEL – 2 (Solutions)
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10 Electronics and Communication Engg. ACE
Power = W4004
116004AA 2
m2
C =×
=
04. Ans: (c) Sol: Carrier = cos2π (100 × 106)t
Modulating signal = cos(2π × 106)t o/p of Balanced modulator = 0.5[cos 2π (101 × 106)t + cos 2π(99×106)t]
The o/p of HPF is 0.5 cos 2π(101 × 106)t
o/p of the adder is
= 0.5 cos 2π (101×106) t + sin 2π (100×106)t = 0.5 cos 2π[(100+1)106t]+ sin 2π(100×106)t = 0.5[cos 2π (100 ×106)t. cos 2π (106)t − sin 2π(100 × 106)t. sin 2π (106)t] + sin 2π(100 ×106)t]
= 0.5 cos 2π (100 ×106)t. cos 2π (106)t + sin 2π(100×106)t [1−0.5 sin2π (106)t]
Let 0.5 cos 2π (106)t = r(t) cos θ(t) 1−0.5 sin 2π (106)t = r(t).sin θ(t)
The envelope is r(t) = [ 0.25 cos2 2π (106)t
+ 1− 0.5 sin 2π (106)t2]1/2 = [1.25 − sin 2π(106)t]1/2
= [45 − sin 2π (106)t]1/2
05. Ans: (b)
Sol: O/p of 1st balanced modulator is
o/p of HPF is
The o/p of 2nd balanced modulator is consisting of the following +ve frequencies.
Thus, the spectral peaks occur at 2 kHz and 24 kHz
06. Ans (d) Sol: Given SSB AM is used, LSB is transmitted )10f(f cLO +=
t]ff[2cos2AA
T/)t(S mcmc
X −π=
t)10cf(2cost)mfcf(2cos2
mAcAXR/)t(S +π−π=
]t)f10cos(t)f10f2[cos(4AA
mmcmc ++−+⇒
i.e. from 310 Hz to 1010 Hz 07. Ans: (d) Sol: The given Circuit. is a ring modulator,
which is also called double balanced modulator. The corresponding output will contain no base band and carrier frequencies.
Given fc = 1 MHz, fm = 400 Hz The Output contains fc+fm = 1000.4 KHz fc − fm = 999.6 KHz.
-13 -11 -10 -9 -7 7 9 11 13 10 f(kHz)
-13 -11 -10 10 11 13 f(kHz)
0 2 3 23 24 26 f(kHz) ACE Engineering Publications (A unit of ACE Engg. Academy – Hyderabad, Vijayawada, Visakhapatnam, Tirupati, Delhi, Bhubaneswar, Bangalore, Pune &Chennai)
(Copyrights Reserved)
SSB and VSB 1
3
01. Ans: (c) 02. Ans: (c) 03. Ans: (b) Sol: Balanced modulator is used to generate
AM-DSB-SC. Remaining all the methods are used for generating AM-SSB-SC.
04. Ans: (c)
Sol: In FDM, AM-SSB-SC is used, since it occupies less Band width.
05. Ans: (a) Sol: Ac = 10
( )tff2cos2AA
mcmc +π
52
110,102
210=
×=
×
06. Ans: (b)
Sol: tf2sin2
)t(mAtf2cos2
)t(mAccc
c π±π
∧
20A102
Ac
c =⇒=
After passing through envelope detector, the output is
1004
4004
)t(mA2c ==
07. Ans: (a)
Sol: 2AA
,2AA 2mc1mc
2
2100,2
1100 ××
− → LSB. fc − fm1, fc − fm2
08. Ans: (c) Sol: An AM-SSB Signal occupies a
Bandwidth of 5KHz Thus the total bandwidth is 12×5 = 60 KHz. In addition between two adjacent multiplexed signals, there is guard band of 1KHz. i.e. total
BW = 60+11 = 71 KHz. 09. Ans: (d) Sol: for VSB fmax < BW<2fmax 3k < BW < 2 × 3k 10. Ans: (b) 01. Ans (c) Sol: Given
KHz02.100f,KHz100f
,Hz400f,Hz200f,Hz100f
0L
321
cc
mmm
==
===
]t)3mfcfcos(t)2mfcfcos(
t)1mfcf[cos(2
mAcAxT/)t(S
+++
++=
tf2cosA]T/)t(S[R/)t(S0Lccxx π=
)]20fcos()fffcos(
)20fcos()fffcos(
)20fcos()fff[cos(4AA
33Lo
22Lo
11Lo
mmcc
mmcc
mmccm
2c
−+++
+−+++
+−+++⇒
Detector output frequencies: 80Hz, 180Hz, 380Hz 02. Ans (d) Sol: Given 11 voice signals B.W. of each signals = 3kHz Guard Band Width = 1kHz Lowest fc = 300kHz Highest fc =
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LEVEL – 2 (Solutions)
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12 Electronics and Communication Engg. ACE
)kHz1(10)kHz3(11kHz300fflostH mc ++=+⇒
= 343kHz
kHz340
kHz3kHz343fHc
=
−=
03. Ans: (a) Sol: USB = Upper side band fc = 60kHz fc to fc + fm Lowest USB = 60K + 300Hz (ω = 300Hz) = 60.3 kHz Highest USB = 60K+3000Hz = 63 kHz (ω = 3000Hz) ∴The range of upper side band is
60.3 to 63 kHz 04. Ans: (d) Sol: Output of synchronous detector with
phase shift is φcos)t(m2
A 2c
When φ = 0 → Original signal retrieved φ = 90 → quadrature Null effect 0 < φ < 90 → Output cannot be detected
properly 05. Ans: (b) Sol: Given, Amplitude (sidebands)
41
= Carrier amplitude
41
2A
4A
4A ccc ×=
µ+
µ
2
AA2 c
c =µ
41
=µ⇒
06. Ans (c)
Sol: % power saved = ]
21[
2A
]4
1[2
A
22c
22c
µ+
µ+
21
41
2
2
µ+
µ+
=
= 0.9848 = 98.48 % ≈ 99% 07. Ans: (b) Sol: Given m(t) = cos 2πfmt fc = 200kHz m(t) is modulated by passing through
DSB and SSB modulators.
?AA
s
d =⇒
(DSB)avg power = (SSB)avg power if it is a DSBSC signal then,
8
A4
A 22s
22d µ
=µ
21
AA
2s
2d =
707.0AA
s
d =
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01. Ans: (b) Sol: θi (t) = 10πt+2πt2
[ ] t410)t(dtd
i π+π=θ
fi = ( ))t(dtd
21
iθπ=
ππ+π
2t410
at t = 2.5 sec
fi = ( ) 102
5.2410=
ππ+π
02. Ans: (c) Sol: Instantaneous Angle is ψi(t) = cos(2×108πt+75 sin2×103πt)
Instantaneous freq. dt
td ii
)(ψω =
= 2×108+75(2×103π) cos(2×103)πt
∴Peak freq. deviation δω
=π
π××2
10275 3
⇒ δf = 75 kHz
03. Ans: (c)
Sol: fi = ( ))t(dtd
21
iθπ
= [ ]3t200t200dtd
21
π+ππ
×π+π
π= 2
1
t23200200
21
= [ ] Hz25050021
=ππ
04. Ans: (a) Sol: Am
| = 2Am
∆f = kf Am ∆f | = 2 [ kf Am ] = 2 [∆f]
05. Ans: (d) Sol: Given fm
| = 2fm
∆f = kf Am ∆f is independent of fm ∴∆f remains same 06. Ans: (d) 07. Sol: 2 (∆f + fm) = 2(50 × 103 + 500) = 101 kHz 08. Ans:(b) Sol:
)tωcos(2
0.1A)tωcos(ω2
0.1AtcosA)t(V mcmccAM −ω+++ω=
2
A0.1tcosA(t)v cFM +ω=
( ) ( )tωωcos2
A0.1.tωωcos mcmc −−+
=+ )t(v)t(v FMAM 2 A cos ωct
+ 0.1 A cos ( )tmc ω+ω
∴The resulting signal is SSB with carrier
NBFM is similar to AM, except the phase of the lower side band 2
09. Ans: (a) Sol: VFM(t) = Ac cos[ωc t + kf ∫m(t).dt]
∫ )t(m .dt = ∫ ω dt.tcosE mm = m
mm tsin.Eω
ω
∴ VFM(t) = AC. cos
ω
ωω tsin Ek+t m
m
mf c
VFM(t) = AC. cos[ωct + mf.sinωmt] 10. Ans: (a)
Sol: 4
π210π
21010fAKβ 4
3
m
mf =××
==
Frequency Modulation 1
4 LEVEL – 1 (Solutions)
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14 Electronics and Communication Engg. ACE 11. Ans: (c) Sol: S(t) = Ac cos [2π fct + 2π kf ∫ dt)t(m ]
and β = m
mf
fAk
0.4 = 5.0A5A.4
mm =⇒
cossin =∫
= 0.5 cos10πt
12. Ans: (a)
Sol: A squaring circuit acts as a frequency doubler and doubles the frequency deviation.
∴ ∆f2 = 2. ∆f1 = 180 kHz. B.W = 2(∆f2 + fm) = 2(180 + 5) = 370 kHz
13. Ans: (b)
Sol: LSB signal is of frequency (1000−10) KHz = 990 KHz Considering this as Base band signal frequency, the B.W of the o/p of NBFM is 1.98×106 Hz
≅ 2 MHz
14. Ans: (a)
Sol: v(t) = 10 cos[2π ×105 t +5 sin (2π ×1500t)
+ 7.5 sin (2π ×1000t)]
Instantaneous Angle ψi(t) = 2π×105 t + 5 sin (2π×1500t)
+7.5sin (2π×1000t)
Instantaneous frequency ωi = ψdtd
i(t)
ωi = 2π×105 + 5 [2π×1500 cos (2π×1500t)]
+ 7.5[2π×1000 cos (2π×1000t)]
∴∆ω = 2π(7500) + 2π (5000)
∴∆f = 12500 Hz.
fm = 1000 (Maximum of the two)
∴Modulation Index = 5.12f
f
m
=∆
15. Ans: (c) Sol: x(t) = cosωct + 0.5 cosωct.sinωct
Let 1 = r(t).cosθ(t) 0.5cos ωt = r(t).sinθ(t) ∴x(t) = r(t). cos [ωct − θ(t)]
= 2)tcos5.0(1 ω+
× cos [ωct− tan-1 (0.5 cos ωt)] Hence x(t) is both AM and FM 16. Ans: (d) 17. Ans: (c) 18. Ans: (b) 19. Ans: (d) 20. Ans: (c) Sol: From the given FM signal, modulation
index is 15. As per Carson’s rule BW = 2(mf + 1)fm = 2 × 16 × 100 = 3.2 KHz. 21. Ans: (c) Sol: Ac [1+ µ cos 2 π fmt] cos 2π fct a) fm = 104 ⇒ BW = 2 ×104 = 20 k b) fm = 2×104⇒ BW = 2 ×2×104 = 40k c) fm = 103 ⇒ BW = 2×103 = 2k d) 2fm = 10,000 fm = 5,000 ⇒ BW = 2fm = 10k 22. Ans: (c)
Sol: Power = 2
A2c
a) 2
102
= =2
100 50
b) 2
32
= 29 = 4.5
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ACE Communication Postal Coaching Solutions 15
c) 5.24249
272
==
d) 22
22
=
a > c > b > d 50 > 24.5 > 4.5 > 2
23. Ans: (d) 24. Sol: P. 10 cos [2πfct + 5 sin 2π×103t] B.W = 2 (β+1) fm = 2 (5+1)×103 Hz
B.W = 12 kHz Q. 10 cos 2πfct-7sin2πfct sin4π×103t
→ Narrow Band FM Ac β = 7 ⇒ β = 0.7 < 1 B.W = 2fm = 2 (2kHz) = 4kHz R. [1+0.6sin2π104t] cos2πfct
→ AM signal B.W = 2fm = 2 (104Hz) = 20kHz S. (cos4π×103t+cos 8π×103t)cos2πfctB.W
= 2 (fmHighest) = 2 × 4 kHz = 8 kHz None of the option is correct 25. Ans: (c)
Sol: β = β=φ∆∆ ,f
f
m
= 8
8 = 500
f∆
∆f = 4000 26. Ans: (b) Sol: The spectrum of AM signal / FM signal
with modulation index < 0.5 consists of the carrier component and the first order side band pair.
27. Ans: (c) 28. Ans: (c) Sol: Given S(t) = 100 cos[2π 107t + 4sin 2000πt] R = 50Ω Am = 1V, fm 1 kHz ∆f = β × fm = 4 kHz 29. Ans: (a)
Sol: WR
AP c
total 100502100100
2
2
=××
==
30. Ans: (b) Sol: % of total power at 10 kHz? Jo(4) = -0.4
)(JR2
AP 2
o
2c
c β=⇒
)(JPP 2
oT
c β=
= (-0.4)2
16.0PP
T
c =
%16PP
T
c =
Pc = 16% (PT) 31. Ans: (c) 32. Ans: (c) Sol: Given
fc = 5KHz. and m(t) = 100 cos 2000πt. 2fm = 2000 ⇒ fm = 1k fc fc − fm fc + fm fc − 2fm = 3k fc + 2fm = 7k 33. Ans: (b) Sol: c(t) = 10 cos ωct m(t) = cos 20 πt kf = 50Hz/V ∆f = kf Am
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16 Electronics and Communication Engg. ACE = 50 × 1 = 50
β = 51050
fm
f ==∆
∴The harmonics that contains 99% of the total power is = 2(β+1) +1
= 2(5+1) +1 = 2(5+1) +1 = 13 harmonics None of the option is correct 34. Ans: (b) Sol: c(t) = 10 cos (2π×10×103)t m(t) = cos (20πt) kf = 40 πHz/v Am = 1
∆ω = 2πkf Am = 40π ∆f = 20 Highest frequency fmax = fc+∆f = 10×103+20 = 10.02 kHz 35. Ans: (b) Sol: Given m(t) = ),t10x12x2cos(.2 3π fc = 100×106Hz , Ac = 10V 2πkf = 2π(12)×103Hz/V kf = 12×103Hz/V ∆f = kf Am = 24×103Hz
210121024
ff
3
3
m
=××
=∆
=β
Wide Band FM signal is represented as
S(f) = Ac t)nff(2cos)(J mcn
n +πβ∑∞
−∞=
Amplitude and lowest frequency of the resulting FM signal is 10J3(2) ( from the given options)
36. Ans: (d) Sol: A Frequency Tripler increases the
modulation index by 3 37. Ans: (a)
Sol: In an FM signal, adjacent spectral components will get separated by fm = 5 kHz Since BW = 2(∆f + fm) = 1MHz
=1000 × 103 ∆f + fm = 500 kHz, ∆f = 495 kHz The nth order non-linearity makes the carrier frequency and frequency deviation increased by n-fold, with the base-band signal frequency (fm) left unchanged since n = 3, ∴ (∆f)New = 1485 kHz & (fc)New = 300 MHz New BW = 2(1485 + 5) ×103
= 2.98 MHz = 3 MHz 38. Ans: (d) 39. Ans: (a) Sol: FM Generation: 1) Direct method 2) Indirect method 1. Direct method: In reactance modulator, varactor diode
and FET can be used as a voltage variable capacitor.
01. Ans: (a) Sol: s(f) = 10 cos (20πt+πt2)
)t(dtd
21fi π
=
]t220[21
π+ππ
=
fi(t) = 10+t
1)t(fdtd
i =
VCO FM m(t)
LEVEL – 2 (Solutions)
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ACE Communication Postal Coaching Solutions 17 02. Ans: (d)
Sol: S(t) = Ac ∑ β∞
−∞−nn )(J cos(2πfc +βsint)
∆f = 3(2fm) = 12kHz
β = mff∆ = 6
∴S(t) = ∑∞
−∞=
+n
cn tfJ )sin2cos()6(.5 βπ
fc = 1000kHz, fm = 2 kHz = cos(2π(1008 ×103)t = cos(2π(1000 +4×2)×103t] i.e n = 4 The required coefficient is 5.J4(6) 03. Ans: (c) Sol: 2πfm = 4π 103 ⇒ fm = 2k J0 (β) = 0 at β = 2.4
β = m
mf
fAk
2.4 = k2
2k f ×
kf = 2.4 KHz /v at β = 5.5
mf
2k4.25.5 ×= ⇒ 872.72
04. Ans (c) Sol: β = 6
J0(6) = 0.1506 ; J3(6) = 0.1148 J1(6) = 0.2767 ; J4(6) = 0.3576 J2(6) = 0.2429 ;
?P
P
T
cf mf4=
±
PT =
R2A 2
c
β+β+β+β+β=± )(2
4J)(23J)(2
2J)(21J)(
2
2oJ
R
2cA
mc f4fP
β+β+β+β=
±)(J)(J)(J)(
2J
RA
P 24
22
21
2o
2c
f4fmc
5759.02
12879.0
PP
T
f mf4c ==±
= 57.6 % 05. Ans: (c) Sol: m(t) = 10cos20πt fm = 10 Hz inserting correct signal and frequency
β = 10
105×=
m
mf
fAk
= 5
From fc to fc + 4sin pass through ideal
BPF Powers in these frequency components
)(22J
R2
2CA
2)(21J
R2
2CA
2)(20J
R2
2CA
P β+β+β=
( )β+β+ 24J
R1
2CA
223J
R2
2CA
2
++
+−+−=
22
2222C
)391.0(2)365.0(2)049.0(2)328.0(2)178.0(
R2A
= 41.17 watts 06. Ans: (d)
Sol: Pt = R2
A2C (R = 1Ω)
= 2
100 = 50 W
2)(JA 3C β−
2)(JA 2C β−
2)(JA 1C β−
2)(JA 0C β
2)(JA 1C β
2)(JA 2C β
2)(JA 3C β
fC-3fm fC-2fm fC-fm fC fC+fm fC+2fm fC+3fm
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18 Electronics and Communication Engg. ACE
% Power = powertotal
componentsinPower× 100
= 10050
17.41×
= 82.35% 07. Ans: (b) Sol: x(t) = Ac cos ωct − Am cos(ωc – ωm)t +
Am cos(ωc + ωm)t is an NBFM signal, Under Tone Modulation. It should be demodulated by a discriminator
08. Ans (d)
09. Ans (a) Sol: 6410200f 3
e|
i ××= = 128×105
65
c| 109.1010128f ×−×=
= 19×105
MHz
fpc
2.91
481019 5
=
××=
∆fi = 25 ∆fi = 25×64 ∆fN = 25×64×48
= 76.8kHz
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01. Ans: (d) Sol: [ ]t150cos40t15030sint102π2(t)ψ 6
i ++×=
∴ ∆φ = 2π(30 sin 150t + 40 cos 150 t) Let 30 = r sin α ; 40 = r cos α ∴ ∆ φ = 2 π r cos (150 t -α),
where r = 50)40()30( 22 =+
∴ ∆ φ = 100 π cos (150 t - α) ∴ Maximum phase deviation = 100 π
∴ωL = dt
(t)Ψd i
[ ]150tsin(150)(40)150tos(30)(150)c1022 6 −+×= π
( )( )[ ]t150sin(150)(40)t150cos15030π2ω −=∆∴ = 3000 π [ ]tt 150sin4150cos3 −
Let 3 = r. cos α & 4 = r. sin α ∴ ∆ω = 3000 π . r . cos (150 t + α)
where r = 22 )4()3( + = 5
∴ ∆ ω = 15000 π . cos (150 t + α) Maximum frequency deviation,
∆ω = 15000 π
∆f =π2ω∆ = 7.5 kHz
02. Ans: (d)
Sol: Under Tone Modulation, frequency deviation in PM is ∆ω = Kp Em ωm
where phase deviation = Kp Em Since, phase deviation remains unchanged, ∆ω α ωm
New deviation ∆ω2 = 2. ∆ω1 ∆f2 = 2. ∆f1 = 20 kHz B.W = 2(∆f2 + fm2) = 2(20 + 2) = 44 kHz
03. Ans: (c) Sol: A PM signal is given as
( ))t(mktcosA pc +ω .The instantaneous angle )t(mkt)t( pci ×+ω=ψ . Instantaneous frequency
dt
)t(dmkdt
)t(dpc
ii ⋅+ω=
ψ=ω
∴Frequency deviation
( )tmdtdk f∝δω
If dt
)t(md is large, m(t) is said to be of
larger frequency. Hence, mωδω ∝
Consider m(t) = Em⋅Sin ωmt.
tcosE)t(mdtd
mmm ω⋅ω=∴
mmmE ω∝δω⇒ω=δω⇒ 04. Ans: (a) Sol: ∆ φ = kp Am
kp = 22/1
1=
05. Ans: (c) Sol: given fc = 1MHz fmax = fc + kf Am kp = 2π kf
kf = π
π=
π 22k p
= 21
=
×+ 56 10
2110
( )56 105.010 ×+=
( )46 10510 ×+=
= ( ) 33 105010 + = (103 + 50) k
Phase Modulation 1
5 LEVEL – 1 (Solutions)
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20 Electronics and Communication Engg. ACE = 1050 kHz. fmin = fc −kf Am
=
×− 56 10
2110
( )56 105.010 ×−=
( )46 10510 ×−=
= ( ) 33 105010 − = (103−50) k = 950 kHz 01. Ans: (d)
Sol: mff∆
=β
∆φ = mff∆
∆f = ∆φ fm = kp Am fm 02. Ans: (d) Sol: θi (t) = 2πfc t + kp m(t) 13000t = 10, 000t + kp m(t) 3000t = kp m(t)
m(t) = t31000
t3000=
03. Ans: (c) Sol: Given
fc = 100 × 103 Hz
kf = 10×103Hz m(t) / max = +1 , m(t) / min =-1 fi = fc ± ∆f = fc ± kf Am = 100×103 ± 10×103 (m(t)) = 110 kHz & 90 kHz
04. Ans (c) Sol: S(t) = Ac cos (2πfct+kpm(t)) θi(t)
)t(dtd
21f ii θπ
=
= π21 (2πfct + kpm(t))
= fc + )t(mdtdk
21
pπ
π+=
−
410
12k
ff3
pcmax
3pc 104
2k
f ××π
+=
31042
kHz100 ××π
π+=
=102 kHz
−=
−
410
1kff3pcmin
= fc - 2 kHz fmin = 98kHz 05. Ans: (c) Sol: Given, S(t) = Ac cos (θi(t)) = Ac cos (ωct +φ(t) ) m(t) = cos (ωmt) fi(t) = fc+2πk(fm)2 cos ωmt
dt
)t(d21f i
iθ
π=
θi(t) = ∫ 2π fi(t)dt
dt]tcos)f(k2f[2)t( m2
mci ωπ+π= θ ∫
θi(t) = 2πfct + (2πfm)2 k t
tcos
m
m
ωω
θi(t)= ωct+ωmk sin ωmt 06. Ans (d) Sol: If m(t) is other then cos (ωmt) then s(t) = Ac cos (ωct + kf ∫ m(t) dt) (or) Ac cos (ωct + kp m(t)dt)
+1
-1
T=10-3sec
T/4
LEVEL – 2 (Solutions)
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01. Ans: (c) Sol: Given data: IF = 455 kHz fs = 1200 kHz Image frequency fsi = fs + 2×IF
= 1200 + 2 × 455 = 2110 kHz 02. Ans (b) Sol: Given fsi = 2500kHz fs = 1600kHz IF =?
kHz450
Hz102
16002500IF 3
=
×
−
=
03. Ans: (b) Sol: Image frequency (fsi) = fs + 2IF = 500 + 2 × 465 = 1430 kHz 04. Ans: (a) 05. Ans: (d) 06. Ans: (a)
Sol: The super heterodyne receiver for AM signal shown in the figure.
07. Ans: (b) Sol: The maximum frequency deviation for the
FM broad cast system is 75 KHz. 08. Ans: (b) 09. Ans: (d)
10. Ans: (c) 11. Ans: (a) 12. Ans: (a) Sol: Noise performance of FM is better than
that of AM. 13. Ans: (d) Sol: If amplifier in a radio receiver is always
tuned to a fixed frequency, called Intermediate different frequencies. A → 2
Mixer is used to generate different frequencies. B → 4
Detector detects the Modulating signals from the Modulated carrier. C → 1
In AGC, the gain of the amplifying stages of the receiver, should be automatically controllable, depending on the strength of the incoming signal. D → 3
14. Ans: (c) 15. Sol: Ring modulator - DSBSC Generator
VCO - FM Generator Foster seely - demodulation of FM Discriminator Mixer - Frequency conversion
16. Ans (a) 17. Ans (b) Sol: Given
fs = 500 kHz IF = 46.5 kHz Q = 50 α(in dB) = ?
fsi = fs + 2IF
Mixer IF Amp
AM detector
AF Amp LS
Local oscillator
RFAmp
Power Amp
Receivers 1
6 LEVEL – 1 (Solutions)
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22 Electronics and Communication Engg. ACE = 1.43 × 106
si
s
s
si
ff
ff
−=ρ = 2.51
22Q1 ρ+=α = 125.52
20 log (α) = 42dB 18. Ans: (a) Sol: As Per FCC regulations, the B.W of each
AM Broad casting channel is 10 kHz A → 1
Telephone Channel B.W is 4 kHz, as the speech signal is band limited to 4 kHz. B → 2
The B.W assigned for an FM Channel is 200 kHz C → 3
Each TV Channel is of B.W 7 MHz D → 4
01. Ans (d) Sol: Given fs = 4 to 10 MHz IF = 108 MHz fsi = ? fsi = fs + 2IF = 7.6 MHz to 13.6 MHz 02. Ans: (b) 03. Ans: (a) Sol: Image frequency fsi = fs + 2IF
= 700 × 103 + 2( 450) = 1600 kHz Local oscillator frequency, fl = fs + IF (fl)max = (fs)max + IF = 1650 +450
= 2100 kHz (fl)min = (fs)min + IF = 550 + 450
= 1000 kHz
R = 41.410002100
ff
CC 22
min
max
min
max =
=
=
l
l
04. Ans (a) Sol: fs(range) = 88-108MHz Given condition fIf < fLO, fsi>108 MHz fsi = fs + 2IF fsi > 108 MHz fs + 2IF > 108 MHz 88MHz+2IF > 108 MHz IF > 10MHz
Among the given options IF = 10.7 MHz 05. Ans (a) Sol: Range of variation in local oscillator
frequency is fL = fsmin + IF = 88 + 10.7 fL =98.7 MHz fLmax = fsmax + IF =108 +10.7 fLmax = 118.7 MHz
LEVEL – 2 (Solutions)
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1
7
01. Sol: n = 5 Rb = 50Mbps Maximum B.W of a message signal is Rb = fb = n.fs
50×106 = 5×fs fs = 10MHz ∴ maximum B.W of message signal
= MHz52
102fs ==
02. Ans (d) Sol: Given L = 64 2n = 64 ⇒ n = 6
2
)Q( maxe∆
=
L
VVWhere minmax −=∆
642VV
2minmax
×−
=∆
128
VV)Q( minmax
maxe−
=
03. Ans (b) Sol: n =10 Rb = 100 kbps Rb = nfs = 100×103 10 fs =100×103 fs = 10×103
fs = 2fm ⇒ fm = kHz52fs =
04. Ans (b)
05. Ans (d) Sol: Given
fs = 44.1kHz n = 16 Bit Rate = nfs
=16×44.1kHz bits/sec Bit Rate = 16×44.1kHz×60.bits/minute
For 50 minutes, No of bits in a piece of music is 50 × Bit rate = 2.1168 G.bits
06. Refer above Ans: 07. Ans: (c) Sol: PWM is the method of transmitting a
continuous and analog signal, using pulse train as a carrier.
08. Ans (c) Sol: Given fm = 5kHz fs = 2fm = 10kHz s/sec L = 256 ⇒ 2n = 28 n = 8 Sampling rate in minute = fs = 600kHz s/minute For 10 minutes, Total No of samples taken = 6×106 Samples 09. Ans (c) Sol: Total No of bits generator
= (6×106)×8 = 48 mega bits.
Pulse Code Modulation LEVEL – 1 (Solutions)
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24 Electronics and Communication Engg. ACE 01.
Sol: nminmax
2VV −
=∆
n21α∆ ;
1
2
n
n
2
1
22
=∆∆
n
3n
2 221.0 +
=∆
811.02 ×=∆
0125.02 =∆ None of the option correct
02. Ans: (3)
Sol: (BW)PCM = 2f sn
Where ‘n’ is the number of bits to encode the signal and L = 2n, where ‘L’ is the number of quantization levels.
L1 = 4 ⇒ n1 = 2 L2 = 64 ⇒ n2 = 6
326
nn
(BW)(BW)
1
2
1
2 ===
⇒ (BW)2 = 3 (BW)1 03. Ans: (b) Sol: The o/p of the quantizer will be in error if
the channel noise Magnitude exceeds Half of the step size.
04. Ans (c) Sol: Given Vmax = + 1, Vmin = -1 Quantization bits = n fm = 5kHz nmin = ? (Qe)max < 5mV
(Qe)max = 2∆
m52<
∆
∆ < 10 m
m102
VVn
minmax <−
m1022
n <
2n-1 >100 n > 7.64 05. Ans: (d) Sol: Bit rate = nfs = 8×(2×5k) = 80kbps 06. Ans (c) Sol: Given,
Two signals are sampled with fs = 44100s/sec & each sample contains ‘16’ bits Due to additional bit there is a 100% overhead. Out put bit rate =?
|s
|b fnR =
|s|
s f2f = = 2 [44100]
)eouslytansimulsampledsignalstwo( n| = 2n ( )bitsadditionalbyoverheadtodue Rb = 4 (nfs)
= 2.822Mbps 07. Ans (c) Sol: No, of bits recorded over an hour
= Rb × 3600 = 10.16G bits 08. Ans (b) Sol: Given Variation of Input = (0-4) volts No, of quantization levels = 2 L = 2 2n = 2 ⇒ n = 1
We know that,
2
Q2 e
∆≤≤
∆−
In the step 0−2 error varies from -1 to 1
LEVEL – 2 (Solutions)
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ACE Communications Postal Coaching Solutions. 25 In the step 2−3.5 error varies from -1.5 to
+ 0.5 So minimum error is -1.5V and maximum error is 0.5V 09. Ans: (c)
Sol: )tW16(1tWπ4
t)Wπ(4sinp(t) 22−=
At W41t = ;
00
W41P =
Use L-Hospital Rule
)t(3Wπ64Wπ4
t)Wπ(4cosWπ4Ltp(t)Lt 23
W41t
W41t −
=→→
−
−=
23
W1613Wπ64Wπ4
1)(Wπ4
0.5Wπ8Wπ4
=−−
=
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01. Ans: (c) Sol: ‘fS’ from the TDM system is = 2400 + 1200 + 1200 = 4800 samples/sec. n = 12 bits ∴ Bit rate = n fS = 12 × 4800 = 57.6 kbps 02. Ans (b) Sol: Given kHz2.7fkHz6.3f
11 sm =⇒=
kHz4.2ffkHz2.1ff3232 ssmm ==⇒==
3sss ffff21++=
= 12kHz No. of Levels used = 1024
⇒ n = 10bits ∴ Bit rate = nfs =10 × 12 kHz =120 kbps None of the option is true If there is another signal m4(t) = 12 kHz Then Rb = 144 kbps 03. Ans (c) 04. Ans: (c) Sol: (fs)min for each signal = 10 kHz Existing fs = 2 ×10 kHz = 20 kHz. No. of samples/sec from the TDM system
= 4 × 20 = 80 kHz.
(BW)min = 21 (sampling rate)
= 40 KHz. 05. Ans (b) Sol: Given m = 20
n = 7 fs = 8ks/s Each sample contains 7 bits + synchronization bit = 8 bits ∴ Bit rate Rb = 20×8×8000 = 1280 kbps
06. Ans (b) 07. Ans (b) Sol: m = 96 fs = 8kHz n = 8 1 frame = [n×m+1] Rb = (96×8+1)8×103 = 6.152Mbps 01. Ans: (a) Sol: (fs)min = (
1sf )min+ (2sf )min
+ (3sf )min + (
4sf )min
= 200 + 200 + 400 + 800 = 1600 Hz 02. Ans: (d) Sol: (BW)min = W + W +2W + 3W = 7 W 03. Ans: (a) Sol: Peak amplitude → Am Peak to peak amplitude Am
2∆− ≤ qe ≤
2∆
PCM maximum tolerable 2∆ = 0.2% Am
∆ = L
peaktoPeak ⇒ L2
m/A2 = 100
2.0 Am
(∆ = LAm2 )
⇒ L = 500
Time Division Multiplexing 8 LEVEL – 1 (Solutions)
LEVEL – 2 (Solutions)
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ACE Communication Postal Coaching Solutions. 27 2n = 500 n = 9 rb = n(fS)TDM + 9 fS = RN + 20%RN = RN + 0.2RN fS = 1.2RN = 1.2 ×2×ω fS = 2.4 K samples/sec (fS)TDM = 5(fS) = 5 × 2.4 K = 12 K sample/sec rb = (nfS) + 0.5%(nfS)
= (9 ×12k) + 100
5.0 (9×12k)
= 108540 bPS 04. Sol: No. of patients = 10 ECG signal B.W = 100Hz (Qe)max ≤ (0.25) %Vmax
maxnmax V
10025.0
22V2
≤×
2n ≥ 400 n ≥ 8.64 n = 9
Bit rate of transmitted data = 10×9×200 = 18kbps 05. Ans (c) Sol:
Minimum B.W of
TDM is ∑=
ωN
1ii
06. Sol: Given
L = 256 ⇒ n = 8 kHz5f
1m =
kHz10f1m =
kHz5f3m =
fs = 2 [5kHz + 10kHz+5kHz] = 40 kHz n = 8 bits
Bit rate = nfs
= 8×40 kHz = 320kbps
i) Bit duration = sec125.3R1
b
µ=
ii) Minimum channel B.W = 2
R b
kHz320= iii) Commutator speed = fs revolution/sec
= 40K×60 RPM = 240K RPM iv) No. of Levels used = 512 ⇒ n = 9 360kHzB.W =⇒
Increase in channel B.W = 40 kHz 07. Sol: n = 8, B.W = 2 kHz
Am%12=
∆ (∆ = 2A2 m )
L2
A2 m = 100
1 Am
⇒ L = 100 ⇒ n = 7 fS = RN + 25%RN fS = 1.25 RN = 1.25 ×2 ×2K = 5 K (fS)TDM = nfS
= 8 ×(5K) = 40 K samples/sec rb = n(RS)TDM = 7 ×(40K) = 280
k.bits/sec rb = 280 kbits/sec
BT = )1(2rb α+ (α = 0.2)
BT = ( )2.01K2
280+
BT = 168 kHz 08. Sol: IES Conventional Refer any standard Text
book
C1 C2……….CN
ωi
T
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01. Ans (a) Sol: Maximum slope that a stair case can track
= 50V/sec 02. Ans (c) Sol: Rb = 5kbps
sec2001051
R1TT 3
bbs µ=
×===
03. Ans (d) Sol: m(t) = cos2π (800)t
∆ = 0.1V To avoid slope overload error
)(tmdtd
Ts
≥∆
0.1(fs) ≥ 2 π 800 fs ≥ 2 π(8000) fs ≥ 50.265kHz 04. Ans (b) Sol: Granular noise occur when step size is
larger than the slope of message signal 05. Ans: (d) Sol: To reduce slope over loading step size is
to be increased 06. Sol: m(t) = 0.01 t fs = 20Hz Optimum value of step size = slope of
message signal
)t(mdtdfs =∆⇒
∆ (20) = 0.01 ∆ = 500×10-6V 07. Ans (d) Sol: m(t) = cos2π(800)t ∆ = 0.1V fs = ?
To avoid distortion ∆fs = )t(mdtd
1.08002fs
×π=
= 2 π×8000 08. Ans (c) Sol: Granular noise occurs when
)t(mdd
Ts
>∆
∆ > (10) Ts )givent10)t(m( = 09. Ans: (b) Sol: To avoid slope over loading, rate of rise
of the o/p of the Integrator and rate of rise of the Base band signal should be the same.
∴∆fs = slope of base band signal ∆× 32 × 103 = 125 ∆ = 2−8 Volts. 10. Ans (d) 01. Ans: (b) Sol: x(t) = Emsin2πsint
ST∆ <
dttdm )( → slope overload distortion
takes place
W
H
Delta Modulation 9 LEVEL – 1 (Solutions)
LEVEL – 2 (Solutions)
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ACE Communication Postal Coaching Solutions. 29 ∆fS < Em2πfm
⇒ π2
Sf∆ < EmRm (∆ = 0.628)
⇒ π×
2K40628.0 < Emfm
fS = 40 kHz ⇒ 4 kHz < EmRm Check for options (a) Em × fm = 0.3 × 8 K = 2.4 kHz (4K</ 2.4 K) (b) Em × fm = 1.5 × 4K = 6 kHz (4K < 6 K) correct (c) Em × fm = 1.5 × 2 K =3 kHz (4K </ 3K) (d) Em × fm = 30 × 1 K = 3 kHz (4K </ 3K)
02. Sol: Given m(t) = 6 sin (2π×103) + 4 sin (4π×103t) ∆ = 0.314 V Maximum slope of m(t)
= 2
t/))t(m(dtd π
=
= 2π×103(6)+4π×103[4] = 28π×103
03. Sol: Pulse rate which avoid distortion
)t(mdtdfs =∆
314.0
1028f5×π
=
fs = 280×103 pulses/sec
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bk
Tb
PSK Modulator
DPSK dk Logic function
01. Ans: (c) Sol: (BW)BPSK = 2fb = 20 kHz (BW)QPSK = fb = 10 kHz 02. Ans:(c) Sol: In DPSK,
0 → is represented with a carrier has phase ‘π’ 1 → is represented with a carrier has phase ‘0’
dk = 0 → output carrier phase of DPSK modulation is ‘π’
dk = 1 → output carrier phase of DPSK modulation is ‘0’
Logic function may EX-OR or EX-NOR. We have to find out logic function from the given information
dk = bk ⊕ dk − 1 (or) dk = bk ⊙ dk − 1 bk = 1, dk = 0, dk − 1 = 0 0 = 1 ⊕ 0 it is wrong
0 = 1 ⊙ 0 it is correct Logic function is EX-NOR operation
The remaining carrier phase 0 π π π 03. Ans: Quadrature Phase Shift Keying 04. Ans: (a) & (c) 05. Ans (c)
Sol: Baud rate = mlog
R
2
b
Mbps174log
34
2
==
06. Ans (b) Sol: Given
Bit stream 110 111001 Reference bit = 1 07. Ans (b)
1 1 0 1 1 1 0 0 1 1
1 1 0 0 0 0 1 0 0
0 0 π π π π 0 π π
b|(t) = b(t) Q(t)
b(t) b|(t)
Q(t)
Message bits bk 1 1 0 0 1 1
0 0
π π dk
First two reception bits
Logic function
1 1 0 0 1 1
0 0
π π
⊙
1 0 0 0
⊙ ⊙ ⊙
π π π 0
Band Pass Data Transmission 10 LEVEL – 1 (Solutions)
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ACE Communication Postal Coaching Solutions. 31 01. Sol: fC1 = 25 kHz, fC2 = 10 kHz To take non cohenent ∆f = rb
fC2 −fc1 = bT
1
⇒ Tb = f
n∆
Let n = 1 Tb = 12
1
CC ff − =
K151
= 66.6 sec
Let n = 2 Tb = K15
2 = 133.3 µsec
n = 3 Tb = K15
3 = 200 µsec
02. Ans: (b)
Sol: fH = 25 kHz ; fL = 10 kHz
∴ Center frequency
=
+
21025 kHz = 17.5 kHz
∴ Frequency offset, Ω = 2π (25 − 17.5) × 103
= 2π (7.5) × 103
= 15 × 103π rad/sec.
The two possible FSK signals are
orthogonal, if 2ΩT = nπ
⇒ 2(15π) × 103 × T = nπ
⇒ 30 × 103 × T = n (integer)
This is satisfied for, T = 200µsec.
03. Ans (a) Sol: rb = 8 kbps Cohenent detection
∆f = 2
nrb
Best possible n = 1
∆f = 2K8 = 4K
To verify the options ∆f = 4k i.e. fC2 −fC1 = 4K (a) 20 K - 16 K = 4 K (b) 32 K – 20 K = 12 K (c) 40 K – 20 K = 20 K (d) 40 K – 32 K = 8 K
LEVEL – 2 (Solutions)
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0125.015.00025.005.015.0
01. Ans: (d)
Sol: Since all the 4 levels are equi-probable, Entropy H = 2 bits/sample. Since, two quantized samples are transmitted per sec, message rate r = 2. Thus, the information rate R = r.H = 4 bps.
02. Ans: (b) Sol: Huffman encoder is the most efficient
source encoder
L = 1×0.5+2×0.25+2×0.25 = 1.5 bits/symbol
Average bit rate = 3000 × 1.5 = 4500 bps 03. Ans: (a) Sol:
max)x(H of a discrete source is nlog2
bits/message. Thus H(x) increases as log n
04. Sol: Given symbols = 64 Maximum entropy = log2M = log264 = 6 bits/symbol 05. Sol: Given,
4 symbols with probabilities: 0.1, 0.2, 0.3 & 0.4
⇒ Average code word length = − (0.1log20.1+0.2log20.2+0.3log20.3+0.4log20.4)
= 1.8464 06. Ans (d)
07. Sol: Maximum entropy of a binary source: Mlog/)x(H 2max =
2log/)x(H 2max =
= 1 bit/symbol 08. Ans: (c) Sol: Assuming all the 64 levels are
equiprobable, H= 64log2 = 6 bits/pixel
Total No. of pixels = 625 × 400 × 400 = 100 M pixels /sec Data rate = 6 bits/pixel×100×106 pixel/sec
= 600 Mbps 09. Sol: Average rate of information = 512×512×log2256×30 = 66.15Mbps 01. Sol:
C = Blog2(1+NS )
+×=
→∝→∝ nBS1logB
Sn
nSLimCLim 2BB
elognSCLim 2B
=→∝
Information Theory 11
LEVEL – 1 (Solutions)
LEVEL – 2 (Solutions)
n/2
Ideal AWGN
n/2
−B B
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ACE Communication Postal Coaching Solutions. 33
(→∝n
Lim xlog
+
Q11 = loge)
nS44.1CLim
B=
→∝
02. Ans: (b) Sol: Max. entropy = 512×512 × 8log2
= 786432 bits 03. Ans: (b)
Sol: C = B log (1+NS )
Since NS >>1. 1+
NS ≅
NS
∴ C1 = B log NS
C2 = B log (2.NS )
= B log2 + B log (NS )
= C1 +B
04. Sol: Given B. W = 3 kHz SNR = 10dB
⇒ 10 log10 (SNR) = 10 SNR = 10| = 10 No of characters = 128
Channel capacity = B log2
+
NS1
= 3 × 103 log2(1 + 10) = 10378bps
05. Sol: No of characteristics can be sent without
any error cps.14827c
Mlogc
2
===
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PX = 4 = 0
PX = xi = 0
4 xi x
PX(x)
01. Ans: (a)
Sol: p(x) = π2
1 2x 2
e−
is the density of
standardized Gaussian random variable. 02. Ans: (c) Sol: A continuous Random variable X takes
every value in a certain range, the probability that X = x, is zero for every x in that range.
Given 184)(x
X
2
eπ23
1(x)P−
−= is a
continuous Random variable therefore probability of the event X = 4 is zero.
03. Ans: (a) 04. Ans: (d) Sol: Var(−kx) = E[(−kx)2] − [E(−kx)]2 = k2.E(x2) − [−k.E(x)]2 = k2. Var(x) 05. Ans: (c)
Sol: p(v) =
4k v ; 0 ≤ v ≤ 4
= 0 ; else where
1/2k1dvv.4k4
0
=⇒=
∫
E(V2) = ∫∫ =4
0
34
0
2 .dvv81.p(v)dvv = 8
06. Ans (c) Sol: Given f(x) = 0.5 e- x By observing figure Mean = 0 Standard
deviation2
x )mean(valuesquaremean −=σ
valuesquaremean=
∫∞
∞−
= dx)x(fxvaluesquaremean 2
∫∞
∞−
−= )e5.0(x x2 dx
∫∞
−×=0
x2ex25.0 dx
∫∞
−=0
x2ex dx
=∞
−−
−−
− ∫
0
xx2
1ex2
xex
∞−−
−
−
−−
+−= ∫0
xxx2
1e2
1ex2ex
[ ]∞−−− −+−= 0xxx2 e2xe2ex
= 0 + 2 = 2
2x =σ∴ 07. Ans: (d) Sol: Z (t) = x(t) cos ( ω0t + θ) ;
x (t) is a Random process with mean = 0 Variance = m2
‘θ’ is a random variable
π<θ<π
=θ 2021)(f x
fx (θ)
0 2π
π21
Random Variables and Noise 12 LEVEL – 1 (Solutions)
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ACE Communication Postal Coaching Solutions. 35
[ ]20
2 ))tcos()t(x(E)]t(z[E θ+ω=
)]t([cosE)]t(x[E 022 θ+ω×=
[ ] θπ
θ+ω=θ+ω ∫π
d21).t(cos)t(cosE 0
2
0
20
2
∫π
θ+ω+
π=
2
0
0
2)t(2cos1
21
∫ ∫π π
θθ+ωπ
+θπ
=2
0
2
00 d)t(2cos(
21
21d
21.
21
0)2(21
21
+π××π
=
21
=
21.m)]t(zE 22 =∴
= 0.5 m2 Common Data Solutions for 08 & 09
Given, y(t) = x (t) cos (2πfct + θ)
π≤θ≤π
=θ 2021)(f
08. Ans: (a) Sol: ACF [Y(t)] = ?
ACF [Y(t) ] = ACF [x(t) cos(2πfct +θ)] = ACF[x(t)]×ACF[cos(2πfct+θ)]
∫π
θ θθθ+πθ+π=τ2
0cc d)(f)]tf2cos()tf2(cos()(R
∫π
θτπ+θ+τπ+ππ
=2
0ccc d)]f2cos()2f2tf4[cos(
41
∫ ∫π π
θτππ
+θθ+τ+ππ
=2
0
2
0cc d)f2cos(
41d))t(f2(2cos
41
π×τππ
+=τθ 2)f2(cos410)(R c
τπ=τθ cf2cos21)(R
τπτ=τ cxy f2cos)(R5.0)(R
09. Ans: (a) Sol: Ry (τ) = Rx 0.5 cos ωcτ
We know that
Applying Fourier Transform [ ])ft()ft(5.0)f(S)f(S ccxy +δ+−δ∗=
)]ff(S)ff(S[5.0 cxcx ++−= 10. Ans: (b) Sol: Given, X & Y are two Random Variable Y = cosπx
f(x) =1 21x
21
<<−
= 0 else where fy
2 = ?
dydx)x(f)y(f =
)y(cos1x 1−
π=
dyy1
11dx2−
−×
π=
2y1
1dydx
−π
−=
2y1
1)y(f−π
=
222y ]y[E]y[E −=σ
11. Ans: (a) Sol: Given, R.V is uniformly distributed between -9
to 1
P(z > 0) = 1.01011
101
==×
Ryx(τ) Sy (β) FT
1/10
1 -9 0
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36 Electronics and Communication Engg. ACE 12. Ans: (d) 13. Ans: (c) Sol: A Gaussian pulse in time domain is also
Gaussian in frequency domain 14. Ans: (c) 15. Ans: (d) Sol: Narrow band representation of noise is
n(t) = nc(t)cosωct − ns(t)sinωct. Its envelope is R(t) = ,)t(n)t(n 2
s2c + where
nc(t) and ns(t) are two independent, zero mean Gaussian processes, with same variance. The resulting envelope is Rayleigh Random Variable.
16. Ans: (c)
Sol: H(f) = fRC2j1
1π+
= cf/f.j1
1+
|H(f)|2 = 2c
2
2c
fff+
o/p PSD = |H(f)|2 . i/p PSD
= 2c
2
2c
fff+
. K
o/p Noise Power = ∫∞
∞−
dfPSD)(o/p
= K ∫∞
∞− + 2c
2
2c
fff .df = K π fc
(By substitution f = fc tan θ) 17. Ans: (d) 18. Ans: (a) Sol: Power of a signal, g(t) is the time average
of Energy
∴ Pg = ∫−
→∞
2/T
2/T
2
Tdt)]t(g[
T1Lt
If signal is a.g(t), its Power
= ∫−
∞→=
2/T
2/Tg
22
TP.adt)]t(g.a[
T1Lt
PSD = Power /unit BW
If PSD of g(t) is Sg(ω) = Pg/BW, PSD of ‘a g(t)’ , is
a2 Pg/ BW = a2 Sg (ω) 19. Ans: (b) Sol: Output PSD = |H(ω)|2 × i/p PSD
= 2tj de2 ω− × N0
= 4 N0 o/p Noise Power = o/p PSD × B.W
= 4N0B 20. Ans: (a) Sol: Given, Differential equation of a system is
)t(x)t(dtd)t(y)t(y
dtd
−×=+
Applying Fourier transform, )1sf)(f()jf1)(f(y −×=+⇒
jf1jf1
)f(x)f(y
++−
=
→ The transform function of system is a All pass filter
∴Sy(f) = Sx(f) 21. Ans: (a) Sol: Given,
PSD of Noise =2
0η
T = 270 C ⇒ 300K Pη = K.T.B η0 = KT = 1.38×10-23300
1501038.12
PSD 230 ××=η
=
2310207
=
η0/2
f(Hz)
SN(f)
PSD of Noise
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ACE Communication Postal Coaching Solutions. 37
3dB
Sn(f)
0.5×103
f(kHz) -2 0 2
22. Sol: Pn = K.T.B
21023001036.121 623 ×××
×××= −
= 8.28×10-15W 23. 24. Ans: (a) Sol: S(ω) = |H(ω)|2 × i/p PSD = |H(ω)|2
∴ |H(ω)|2 = 21616ω+
H(ω) = ω+ j4
4
The above can be the Transfer function of
an R-L LPF given by LjR
Rω+
25. Ans: (a) Sol: R = 4Ω & L = 1H 26. Ans: (b)
Sol: Sx(f) = 2δ(f) + 0.5
−
10f
1 ;f≤10Hz
= 0 else where DC. Power is present when f = 0
∴ DC power = 2 W
AC power =
−∫
− 10f
15.010
10
= 5 W
27. Ans: (a) Sol:
Given,
Hz/W102
90 −=η
Hz/W102 90
−×=η
RC21f0 π
=
RC21108 3
π=×
)108(2
1RC 3×π=
)108(24102
RC4NoP 3
9
×π××
==−
= 25.2µW 28. Ans: (a) Sol: PSD of Narrow Band Noise is as shown
in figure When fc = 10kHz the power of in phase component = ?
Power of in phase component 33 105.0102
212 −+ ××××× = 1 W
29. Ans: (a) Sol: f0 f(Hz)
20η
PSD of Noise
Sn(f)
f(Hz)
SN(f+10) 0.5×103
f(kHz) 2 0 -12 20 SN (f-10)
0.5×103
f(kHz) 22 20 -2 0
0.5×10-3
f(kHz) 1 -1 -23 -21
SN (f+11)
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38 Electronics and Communication Engg. ACE
Power of inphase component = 0.5×10-3×2×103 = 1 W 30. Ans: (a) Sol: Given,
h (t) = δ(t)-∝e-∝t u(t)
)f(2j11)f(Hπ+α
α−=
ω+α
α−=
j1
y(f) = s(t) * h(t)
by)f(H)f(S)f(S 2Ny =
22
2
12
Noω+α
α−=
ω+ααα
−= 22y2.
21
2No)f(S
Applying Inverse Fourier transform
α
−τδ=τ τα−e2
)(2
N)(R 0
y
31. Ans: (a) Sol: If X(t1) and X(t2) are two samples
obtained from a random process X(t) at t1 and t2 instances. Then 1. E[Xk(t1) Xj(t2)] = E[Xk(t1)] E[Xj(t2)]
X(t1), X(t2) are statistically independent, for all values of k and j where k and j are positive integers
2. RXX(t1, t2) = E[X(t1) X(t2)] = 0 X(t1), X(t2) are orthogonal 3. Cov(t1, t2) = RXX(t1, t2)−E[X(t1)] E[X(t2)] = 0
i.e., E[X(t1) X(t2)] = E[X(t1)] E[X(t2)]
X(t1) X(t2) are uncorrelated Assume that the Random process at the input of the LPF is X(t), which is given zero mean white Gaussian Noise and output of the LPF is Y(t). The spectral densities relation as shown in Fig. below
According to Wiener-Khinchin theorem
S(f))R(τ F→←
ACF and PSD form a F.T pair ∴ ACF of Y(t)
RY(τ) = N0B Sinc(2B0τ) The mean value of Random process Y(t) µY(t) = µX(t) . H(0) [µY(t) = E[Y(t)], µX(t) = E[X(t)]] µY(t) = 0 . (1) = 0
Y(t) is also zero mean Gaussian Random process.
Let Y(ts), Y(2ts) are two consecutive samples obtained from Y(t) at ts and 2ts instances (uniform sampling) E[Y(ts)]=E[Y(2ts)] = 0 (E[Y(t)] = 0)
RYY(ts) = RYY(t, 2ts) = E[Y(ts) Y(2ts)] (RYY(τ) = E[Y(t) Y(t + T)] ) E[Y(ts) Y(2ts)] = RYY(ts)
= N0 B Sinc(2Bts) Given B = 10 KHz, fs = 0.03 ms RYY(ts) = N0×10×103 Sinc(2×10×103
×0.03 ×10−3)
h(t) S(t) Y(t)
0.5×10-3
f(kHz) 1 -1 21
SN(f-11)
23
0.5×10-3
1 -1 f(kHz)
SX(f)
f
H(f)
f
1
−B B
SY(f)
f
N0/2
−B B
Ideal LPF
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ACE Communication Postal Coaching Solutions. 39 = 104 N0 Sinc(0.6)
=
xπxπSin(x)Sinc
≠ 0 → Not orthogonal Cov (ts, 2ts) = RYY(ts)–E(Y(ts)] E(Y(2ts)
= RYY(ts) – 0.(0) ≠ 0 → correlated 01. Ans: (b)
Sol: E(X) = ∫− −
=
3
1
3
1
2
2x
41dx)x(p.x = 1
E(X2) = 3
1
33
1
2
3x
41dx)x(px
−−
=∫ = 7/3
Var(X) = E(X2) – [E(X)]2 =341
37
=−
02. Sol: RXX(t1, t2) = E[X(t1)X(t2)] = E[Acosωt1Acosωt2) = cosωt1cosωt2E[A2] [ E [A2] = 1/3]
= 31 cosωt1cosωt2
σ2 = ( )121 2
→ variance
E[A2] = σ2 + [E[A]]2
= 41
121+
E [A2] = 124 =
31
03. Sol: RXY (t1, t2) = E[X(t1)Y(t2)] Let t2 −t1 = τ E[(Acosωt1 + Bsinωt1)(Bcosωt2
−Asinωt2)] E[AB] = E[A] E[B] E[AB] = 0 E [BA] = 0 E[A2] = σ2
E[B2] = σ2 [= cosωt1.cosω2E[AB]−sinωt1sinωt2
E[BA] – E[A2] cosωt1sinωt2+E[B2] sinωt1 cosωt2]
= 0-0 −σ2cosωt1sinωt2 +σ2sinωt1cosωt2 = −σ2(cosωt1sinωt2 + σ2sinωt1cosωt2) = -σ2sinω(t2−t1) (τ = (t2−t1) = −σ2sinτ 04. Sol: X(t) = +ve req E[X2(t)] and E[X(t)]
E[X2(t)] = ∫∝
∝−
ωωπ
dS XX )(21
= ( )
×+
π62000
214001
= π
6400
E[X(t)] = 0 [ The given function is periodic
function]
Ans : π
6400 , 0
LEVEL – 2 (Solutions)
fA(A)
1/2 0 1
SX(ω)
400δ(ω−104)
6
0 9 10 11 ω(103)
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40 Electronics and Communication Engg. ACE 05. Sol:
RX(τ) = 2πτ−e
Y(t) = X(t)*h(t) |H(f)|2 = (4π2f2+1) RXX(T) →←FT SXX(f)
22 fFT ee ππτ −− →←
Normalised Gaussian function SYY(f) = |H(f)|2SXX(f)
= (4π2f2 +1)2fe π−
06. Sol:
Y(t) = ( )dtt(X)t(Xdtd
−+
Y(f) = j2πf ( )dftie π21 −+ X(f)
H(f) = )()(
fXfY = j2πf( dftje π21 −+ )
2)( fH = 4cos2πftd SYY(f) = |H(f)|2SXX(f) = 4π2f2(2cos(πftd))2SXX(f) At SYY(f) = 0
πftd = (2n+1)dt2
1
f = ( ) 3105.0211n2 −××
+
f = (2n+1)103 f = (2n+1)R0 f0 = 1 kHz 07. Ans: (b) Sol:
Uncorrelated ⇒ cov(τ) ⇒ RXX(τ) −µ2×(τ) cov(τ) = RXX(τ)
⇒0nR (τ) = 0
⇒Nω0sin(2wτ) = 0, sinCx = 0; x is an integer
2wτ = m
τ = w
m2
, integer m = 1, 2, 3 …….
Common Data Solutions for Q. 08 & Q.09
−= −
88
10
f110)f(S 810f ≤
0= 810f >
×−= −
= 8
68
MHz49f 101049110)f(S
= 0.51×10-8
8
8
68
MHz51f
1049.0
10105110)f(S
−
−=
×=
×−=
08. The verify the limits – 108 ≤ f ≤ 108
H(f)=j2πf
H(f)=j2πf−1)
Y(t)
No/2 H(f)
−ω
ω −ω
ω
No/2
x(t)
Delay 0.5ms
+ Y1(t)
dtd
Y(t) +
+
S(f)
f
f(MHz) +49 51 -51 -49
10-8
f(MHz) 49 51 -49 -51
f(MHz) 51 49 49 -51
Sy(x)
.0.51×10-4 .0.49×10-8
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ACE Communication Postal Coaching Solutions. 41
-1
f
09. Ans: (b) Sol: In phase component power =?
Power (In phase) = 1×10-8×2×106 = 2×10-2 Common Data Solutions for 10 & 11 Sy(f) = sx(f)H(f)2
Sy(f) = 2
N 0 00 BfB ≤≤−
= 0 else where 10. Ans: (b) Sol: We know that,
Taking Inverse Fourier Transform
[ ] dfe)t(S)t(SF f2jyy
1 τπ∞
∞−
− ∫=
∫=τ−
τπ0
0
B
B
f2j0y dfe
2N
)(R
0
0
B
B
t2j0
f2je
2N
−
π
π
=
−=
τπ−τπ
πτ j2ee
2N 00 B2jB2j
)(
0
)B2sin(2N
00 τππτ
=
τπτπ
=0
000 B2
)B2sin(BN
)B2(csinBN)(R 000y τ=τ 11. Ans: (b) Sol:
B21ofmultiplett 21 =−
Sy(f-50)
f(MHz) 101 -1 99 1
0
Sy(f+10)+Sy(f-10)
1×10-8
-1 1 f(MHz)
1
-B0 B0 f(mHz)
N0/2 ?
ACF Sx(f) F.T
0B24−
0B23−
0B22−
0B21−
0B21
0B22
0B23
0B24
Rx(τ)
N0B0
t1 t2
f(MHz) 0 -1
Sy(f+50)
-109 -101 10
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01 Ans: (b) Sol: cos2(20) = 0.883 02. Ans: (c) Sol: XAM(t)=10[1+0.5sin2πfmt]cos (2πfct)
Average side band power = 2
P 2Cµ
PC = 2)10( 2
= 50 W
∴ PSB = W25.62
)5.0(50 2
=×
03. Ans: (b)
Sol: Noise Power = Area under PSD curve
= 4
××
2NB
21 0 = N0. B
∴ BN4
25BN
25.6PowerNoise
P
00
SB ==
04. Ans: (d) Sol: The output of signal to Noise Ratio
FM0
0
NS
= 3(mf)2
AM0
0
NS
For improvement to be noticeable, 3(mf)2
= 1 (or) mf = 3
1
05. Ans: (b)
Sol: The loss of message at low prediction NS
is called threshold effect. The name comes about because, there is
some value of inputNS , above which
signal distortion due to noise is negligible and below which the system performance deteriorates rapidly.
06. Ans: (d) Sol: Given, fm = 5 kHz
Ac = 2V µ = 0.5
8102
−=η
η = 2×10-8 S0 = Ac
2Ka2P
= 222
m2
2 Ac21
2AKaAc µ=
××××
××=
− 38
22
1
db0
0
1051022)5.0(4log10
NS
= 33.97dB ≈ 34dB 07. Ans: (d) Sol:
×××
+=
− )105102
)25.01(
22
log10NS
38
22
dBi
i
= 43.51 dB
01. Ans: (d) Sol: Output of the multiplier
= m(t). cosωot cos(ωot + θ)
= [ ]θ+θ+ω cos)t2cos(2
)t(mo
Output of LPF V0(t) = θcos2
)t(m
m(t)θcos21
=
Power of o/p signal = ∫><
∞→T
20T
dt(t)vT1Lt
dtm(t)θcos21
T1Lt
2
TT ∫
><∞→
=
= ∫
><∞→
T
2
T
2 dt(t)mT1Ltθcos
41
m2 Pθcos
41
=
Noise Analog Communications 13
LEVEL – 1 (Solutions)
LEVEL – 2 (Solutions)
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ACE Communication Postal Coaching Solutions. 43 02. Ans: (a) Sol:
ni = n0 ni = n0 ×W = 10−20 × 100 × 106
Si = 4L
t
10mw1
PP
= = 1 × 10−7
ni = 10−20 ×100 × 106
i
i
nS = 12
7
1010
−
−
= 105 = 50 dB
i
i
nS = 50 dB
Solutions for Common Data Questions 03 to 05
Given,
B.W of an audio signal = 10 kHz
dB40NS
0
=
4
0
10NS
=
9102
−=η
9102 −×=η⇒ Power loss = 40dB (PL) = 104 ( ) ( ) dBLdBtdB
)P(PPp
i −=
=
L
t
PP
Pp
i
03. Sol:
04. Sol: For SSB modulation
⇒ 4
i
i
0
0 10NS
NS
==
(Only SSB modulation in one sided 2n )
Pt = ?
0
0
i
i
nS
nS
= = 104
Si = 104 × 10 × 103 × 2 × 10−9 w/Hz Si = 20 × 10−2 (Si)dB = (Pt)dB −(Pt)dB (Pt)dB = (Si)dB +(PL)dB Pt = SiPL = 20 ×10−2×104 PL = 2 kW 05. Ans: (c) Sol: For AM
FOM = )1if(31
=µ
i
i
0
0
NS
31
NS
=⇒
i0
0i N
NS
3S ×
=
kHz10102103 94 ××××= − = 0.6 Lit PSP ×=∴
4106.0 ×= KW6= 06. Ans: (b) Sol: ∆f = 75 kHz fm = 15kHz
4
0
10dB40NS
==
TX RX
Video signal ω=100 MHz
PL= 40 dB=104
TX RX
BW=10kHz PL= 40 dB=104
DSB Pt = ? Audio
0
0
nS
=104 Rx0
0
nS
=0
FoM=1
η2
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44 Electronics and Communication Engg. ACE
m
2
ff;
23FOM ∆
=ββ=
2
i
i
0
0
23
NSNS
β=
( ) 20
i
132
NS
NS
β××
==
( ) dB24iNS
dB =
07.
Sol: iN
S
= 10 dB
FOM = 31
0N
S
= 10
31× = 3.33
0N
S
= 3.3
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1
l
l
t −T
l
t
T
01. Ans: (c) 02. Ans: (c)
Sol: The impulse response of the filter matched to S (t) is h(t) = s (T−t)
s(−t) =
h(t) = s(T−t) = 03. Ans: (c) Sol: For every 1 bit increase in data word
length, S/Nq ratio improves by a factor of 4. For an increase of 2 bits , the improvement factor is 16.
04. Ans: (c) Sol: ∫
−
=V
Vx
22 dx(x)fx)E(Xpower,Signal
3
V3
xV21 2V
V
3
=
=
−
In uniform quantization,
power,NoisenuantizatioQ12ΔN
2
q =
LevelsofNumbervaluePeaktoPeaksize)(StepΔwhere =
LV2Δ =
L → Number of Quantization levels
2
2
qL3
VN,powerNoisenuantizatioQ =
2
2
2
2
q
L
L3V3
V
NS
==
05. Ans: (b) Sol: For tone modulation
qNS = 1.8 + 6n
= 1.8 + 48 = 49.8 dB 06. 07. Ans: (b) Sol: Bit rate R = n fS
= 8 × 8 kHz = 64 kbps
10 log qN
S = 10 log 23 L2
= 10 log ( )28223 = 49.9 dB
08. Ans (c) 09. Ans (a) Sol: In a PCM,
npp
2
2V
;12
PowerNoise =∆∆
=
n21
∝∆
)n(2
)1n(2
P
P
22
NN
2
1+
= ⇒ 4NN
2
1
P
P =
dB6)N(N dBPP 12−==
Noise Power decrease by 6dB
fx(x)
V −V
V21
n
Quantization Noise 14
LEVEL – 1 (Solutions)
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46 Electronics and Communication Engg. ACE 10. Ans (a) Sol: 44forAe)x(f x ≤≤−= − x = 0 elsewhere
∫∞
∞−
− == 1Ae x
∫ ∫−
−+ =+=0
4
4
0
xx 1AeAe
11
eA1eA
4
0
x0
4
x
=
−
+
−
−
[ ] 1.0]1e[Ae1A 44 =−+− −−
[ ] 1e1A 4 =−= −
5093.0]e1[2
1A 4 =−
= −
4x4e5093.0)x(f x ≤≤−= −
By observing, Quantized values are -3,-1,1,3
∫−
θ−=4
4
2q dx)x(f))x(x(N
∫ −+∫ −= −−4
2
x2x2
0
2 ]dxe5093.0)3x(dxe5096.0)1x(2
= 0.3793 11. Ans (b) Sol: Vmax = 2 : Vmin = -2V
No of Levels used = 4 ⇒ 2n = 4 ⇒ n = 2
RMS value of quantization error = 12
2∆
12∆
=
14
)2(2=
−−=∆
2886.0121
12==
∆⇒
≈ 0.29 12. Ans (d) Sol: Given No. of bits increased from 5 to 8
⇒ (SQNR) α 22n
2n2
1N2
2
1
22
)SQNR()SQNR(
=
82
52
2
1
22
)SQNR()SQNR(
×
×
=
(SQNR)2 = 26 (SQNR)1
(SQNR)2 = 64 (SQNR)1 01. Ans: (a) Sol: For Bipolar pulses,
PSD =b
2
T|)P(ω| . sin2
2Tω b
The zero magnitude occurs for f = n/Tb. ∴The width of the major lobe = 1/Tb
= fb
∴(B W) min = fb
Here, Data rate = nfs = 8(8 kHz) = 64 kbps ∴(B.W)min = 64 kHz 02. Ans: (c) Sol: Since the signal is uniformly distributed,
f(x) = 101 for −5≤ x ≤5
= 0 : else where.
Signal Power = ∫−
5
5
2x f(x)dx
= 2volts325
o/p
i/p -1
-3
0 -1 -4
3
1
2 4
LEVEL – 2 (Solutions)
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ACE Communication Postal Coaching Solutions. 47
Step size = V039.0210
LV
8pp ==−
Nq = mW126.012
2
=∆
Signal to noise ratio, SNR in dB is
=
powerNoisepowersignallog10SNR
dB48100.126
25/3log10 3 =
×
= −
03. Ans: (b) Sol: For every one bit increase in data word
length, quantization Noise Power
becomes 41 th of the original. Hence, Data
word length n = 9 bits ∴L = 2n = 29 = 512
[
04. Sol: VP – P = −5V to 5V 20logL = 43.5 L = 102.175 = 149.6
⇒ ∆ = L
VV LH − = 175.210)5(5 −−
∆ = 0.06683 05. Ans: (c) Sol:
∫−
=
5
5
22 dx101x]E[XpowerSignal
)250(301
3101
5
5
3
=
=
−
x W325
=
Quantization Noise power
= E[[X- Q(X)]2]
dx(x)fq(x)][x5
5X
2∫ −=−
∫−
−
−−=4.9
5
2 dx1014.95)]([x
times).....(50dx101]4.85)([x
4.8
4.9
2 +−−+ ∫−
−
∫ −+05.0
0
2
101)025.0( dxx
times).....(100dx101])075.0[(x
1.0
05.0
2 +−+ ∫
∫ ∫−
−
−++=4.9
5
0.05
0
22 dx1010.025)(x100dx
1014.95)(x50
0.05
0
34.9
5
3
30.025)(x10
34.95)(x5
−+
+=
−
−
](0.025)[(0.025)3
10](0.05)[(0.05)35 3333 +++=
](0.025)[(0.025)3
10)1012510(12535 3366 ++×+×= −−
)10(3.1253
10)1012510(12535 566 −−− ×+×+×=
66 103
5.312103
1250 −− ×+×=
= 520.83333 × 10−6
××
= −4dB 105.2325log10(SNR)
dB42.04= ≈ 42 dB
−5 −4.9 −4.8
−4.95 −4.85
0.05
0.025
0.1V
0.05V
fx(x)
101
x 0
50 levels
100 levels Rx(x)
X
20∆
2
3∆
1 0
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48 Electronics and Communication Engg. ACE 06. Sol: E[X-Q(x)2]
= ∫ ∫ −−3.0
0
1
3.0
22 dx)1()71.0(x(dx)1()0x(
= 1
3.0
33.0
0
3
3)7.0x(
3x
−+
= ( )3
)4.0(3
)3.0(33.0 333
++
= 0.198 07. Ans: (d) Sol: Given,
m(t) = 16 t -1≤ t ≤ 1 = 0 else where No. of encoding bits = 3 Quantization Noise power =?
n2
2minmax
2
q 212)VV(
12N
×−
=∆
=
32
2
212)32(
××= W33.1=
08. Ans: (b) . Sol: Since, all the quantization levels are
equiprobable,
∫−
=⇒=a
a 32a
31dx
41
09. Ans: (a)
Sol: 814dx.x
41dx).x(fx
3/2
3/2
23/2
3/2
2 == ∫∫−−
10. Ans: (a) Sol:
It is a 6-level quantizer and given three consecutive decision boundaries are −1, 0, 1 and remaining decision boundaries are selected by the condition so as to maximize the entropy. For maximum Entropy, all quantization regions are equi-probable. So that every region has same
probability that is equal to
61
( 6 levels) Probability = Area under the curve
of PDF The Area under the curve of Region 5
and Region 6 are equal (x1 − 1) b = b (5 − x1) ⇒ 2 x1 = 6 ⇒ x1 = 3 Similarly Area under the curve of
Region (1) and Region (2) are equal (−1 + x2) b = (−x2 + 5) b ⇒ 2 x2 = 6 ⇒ x2 = 3
Given reconstruction levels are the mid-
points of the decision boundaries. This is
shown in the above fig.
The area under Region (2) = (1/6)
61a
61(1)(a) =⇒=
The area under Region (5) = (1/6)
121b
61b2 =⇒=
−5 5
1 2 3 4 5 6
a
fX(x)
x1 −1
−1+x2
−x2
−x2+5
1
x1-1 5 −x1
x
b
−5 5
1 2 3 4 5 6
a
fX(x)
3 −1 −3 1 x
b
−4 −2 −0.5 0.5 2 4
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ACE Communication Postal Coaching Solutions. 49 11. Ans: (d) Sol: The signal power
∫−
=5
5X
22 dx(x)fx][XE
∫∫ ∫−
−
−
−
−
++=0
1
23
5
1
3
22 dxbxdxbxdxbx
∫∫ ∫ +++5
3
21
0
3
1
22 dxbxdxbxdxax Substi
tute ‘a’ and ‘b’ values
0
1
31
3
33
5
3
3x
61
3x
121
3x
121
−
−
−
−
−
+
+
=
5
3
33
1
31
0
3
3x
121
3x
121
3x
61
+
+
+
E[X2] = 7
The Quantization Noise power
∫−
−
−=−5
5X
22 dx(x)f(x)]q[x]Q(X)]E[[X
∫ ∫−
−
−
−
−−+−−=3
5
1
3
22 dx1212)]([xdx
1214)]([x
∫ ∫−
−+−−+0
1
1
0
22 dx610.5][xdx
610.5)]([x
∫ ∫ −+−+3
1
5
3
22 dx1214][xdx
121]2[x
[ ] [ ]0
1
313
335
3
21x
181)2x(
361)4x(
361
−
−
−
−
−
+++++=
[ ] [ ]53
313
3
1
0
3
4)(x3612)(x
361
21x
181
−+−+
−+
41
362
362
721
721
362
362
=+++++=
28
417
][X]Q[XE]E[X(SNR) 2
2
Q ==−
=
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l
l
t
l
t T
t − 2
1
−1
−1
t 2 1
1
−1
t 2 1
1
−1
s(t)
o/p 1
−1 1
2 3 t
(a)
o/p +1
−1
1 2
3 t
(b)
t s(τ) h(−τ)
1
τ −2 −1
0 τ
+1
−1
2 0
1
01. Ans: (c) Sol: The impulse response of the filter
matched to S (t) is h(t) = s (T−t) s(−t) = h(t) = s(T−t) = 02. Ans: (c) Sol: The i/p signal is The impulse response of the Matched
filter is h(t) = s(T−t), where T = 2 s(−t) =
s(−t +2) = h(t) =
The o/p of the matched filter is
y(t) = s(t) * h(t) = ∫∞
∞−=τ
ττ−τ d)t(h)(s
Using graphical interpretation,
For t = 1
The o/p y(t) at t = +1 is = −1 The given Answers:
The o/p should be −1 at t = 1 Hence, ‘a’ is not the Answer
h(− τ + 1) s(τ).h(1 − τ) Area
−1 τ −1
−1
1
1 0
0 1
−1
τ
t
1
Matched Filter 15
LEVEL – 1 (Solutions)
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ACE Communication Postal Coaching Solutions. 51
The convolution should extend from t = 0 to t = 4 sec Hence, ‘b’ is not the Answer
Therefore ‘c’ is the Answer
‘d’ is not the Answer 03. Ans: (d)
Sol:
∫
∫∞
∞−
∞
∞−≤
df|H(f)|N
df|H(f)||S(f)|
NS
20
22
∫
∫∞
∞
∞−≤
0
20
2
df|H(f)|N
df|H(f)|E
(N0 is one sided)
∫
∫∞
∞
≤
0
20
0
2
df|H(f)|N
df|H(f)|2E
NS
0N
E2NS
≤
Matched filter
2010
102NE2
NS
6
5
0
0 =×
==
−
−
dB13(20)log10NS
dB
==
04. Ans: (d) Sol: The time domain representation of the o/p
of a Matched filter is proportional to Auto correlation function of the i/p signal, except for a time delay
Rss (τ) = ∫−
τ+410
0
dt)t(S).t(S
∫−
+××=410
0
66 dtτ)](t10(2π10sint).1010sin(2π
[ ] dtτ)10π2t10cos(4π)τ10cos(2π50410
0
666∫−
×+×−×=
= 50 × 10−4 cos(2π × 106) τ
∴ The Peak is 5mV 05. Ans: (b) Sol: x (t) = 20cos(2π × 106t) 0.1ms < t < 0.2ms = 0 otherwise
We know that S0 (t) = Rx (t-τ)
⇒ So (t) ∫ τ+ms2.0
ms1.0
dt)t(s).t(s
∫ τ+×π×π=ms2.0
ms1.0
66 dt)t(102cos20)t102cos(20
∫
π+τ×π−×π=
ms2.0
ms1.0
662 dt)t2cos()102t104(cos(21)20(
∫ +τ×π
=
ms2.0
ms1.0
62
0dt).102cos(220
200 cos (2π×106τ).[0.2ms-01ms] 20×10-3cos (2π×106τ) 06. Sol: Given, S (t) = 10 cos 2 π ×106t 0<t<10-6sec S0 (t) =? We know that So (t) = R (t-τ)
∫−
τ+ππ=
610
0
660 dt)t(102cos10)t(102cos10)t(S
( ) [ ]010102cos2
10 662
−τπ= −
= 50×10-6cos2π106τ
(d)
1 2 3 t
o/p
1 2 3 t
(c) o/p
− 1
+2
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52 Electronics and Communication Engg. ACE
2 4 t
1
1
−1
−2 −4 t
+1
−1
2 4 t 0
0
1
−1
2 4 t
1
0 1 t
p(t)
g(t) =
1 2 t
1
07. Ans: (b) Sol: The matched filter has maximum value of
output at t = T is energy of the signal
∫ ∫+=⇒1
0
3
2
22s dt)1(AdtAE
222 A2AA =+= 01. Ans: (c) Sol: δ(t−2) × g(t) =
s(t) = g(t) − δ(t−2) × g(t) =
The impulse response of the corresponding Matched filter is h(t) = s(4−t)
s(−t) = s(−t + 4) = = − s(t)
Common Data Solutions for 02 & 03 Given: s1 (t) = A 2
Ttt2Tt 00 +≤≤−
= 0 elsewhere
2Ttt2
Tt)tt(2
cosB)f(S 000
2 +≤≤−
τ−π
=
= 0 elsewhere
02.
Sol: (SNR)0 = 0
S
NE
= N
T.2
B2
= ND2
TB2
03. Ans: (b) Sol: Given,
NE2
NE2
N)t(S
N)t(S 2s0102 1 =⇒=
T2
BTA2
2 = ⇒ 2
BA =
04. Ans: (d) Sol: Out put of the matched is maximum
which is equal to the energy in the signal
∫ ∫+=1
0
2
1
2 dt)1(dtt.1E
[ ]121
0
3
t13t
+
=
341
31
=+=
f(Hz)
SN(f)
N
PSD of a noise signal
1 2
1
LEVEL – 2 (Solutions)
N0
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ACE Communication Postal Coaching Solutions. 53 The time instant which occurs the
maximum value is its time period T = 2 05. Ans: (c) Sol: Given,
ω
−=
ω−
je1)f(H
tj
ω
−ω
=ω−
je
j1)f(H
tj
Applying I.F.T
h(t) = 0.5(sgn(t)- sgn (t - T0))
ω
=j2))t(sgn(F
= 0.5[2 u(t)-1-[2u(t-T0)-1]]
= [u(t)-u(t-T0)]
We know that
h (t) = s∗(t-T)
∴Si (t)
06. Ans: (d) Sol: The maximum value in the output is
energy inside the signal
( ) ∫=⇒T
2max0
2T
dt.2tS
∫=T
2T
dt.14
]2TT[4 −=
= 2T
S(f)
T/2 T f
-2
0 T
1
t
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01. Ans: (a) Sol: P( x = 0) = 1/2 ;
P(x = 1) = 1/2 Let z be the o/p.
F(z/0) = 5.0
1 for − 0.25 ≤ z ≤ 0.25
= 0 else where F(z/1) = 1 for 0 ≤ z ≤ 1 = 0 else where
P(1/0) = ∫ =250
20
1050
1.
.
.dz.
P(0/1) = ∫ =20
0
201.
.dz
Average bit error probability is P(0). P(1/0) + P(1). P(0/1)
= 21 (0.1 + 0.2) = 0.15
02. Ans: (a) Sol: Given,
HzW10N 2
0−=
Rb = 105 bpe
( )∫ ∫ −+=⇒b b
b
T
0
T2
T
22d dtAdtAE
]T[A]T[A b2
b2 +=
b2TA2=
Q(x) = 10-6 when x = 4.75
0
d
N2E
x =
0
b2
N2TA2
75.4 =⇒
b022 RN)75.4(A ××=
A = 106.21V
03. Ans: (d) Sol: Given,
10102
−=η
10102 −×=η Rb = ? Pe=10-4 when x = 3.7
7.3N
E2
0=⇒
7.3N
TA.2
0
b2
=
( )2
No7.3T102
b6 =×−
kbps730N)7.3(
2T1
02
b
==
04. Ans: (a) Sol: As No of bits increases ⇒ Band Width decreases ⇒ Probability of error increases 05. Ans: (b) Sol: o/p Noise Power = o/p PSD × B.W = 10-20 × 2 × 106 = 2 × 10-14 W Since mean square value = Power
2
2α
= 2 ×10-14 ⇒ α = 107
06. Ans: (d)
Sol: When a 1 is transmitted:
Yk = a + Nk
Threshold Z = 2a = 10-6
⇒ a = 2×10-6 For error to occur, Yk < 10-6
Probability of Error 16
LEVEL – 1 (Solutions)
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ACE Communication Postal Coaching Solutions. 55 2×10-6 +Nk < 10-6 Nk < − 10-6
∴P(0/1) = ∫−−
∞−
610
:dn)n(P
710
10,..)5.0(6
=∫−−
∞−
− αα α withdne n
= 0.5 × e-10
When a ‘0’ is Transmitted:
Yk = Nk
For error to occur, Yk > 10-6
∴ P(1/0) = ∫∞
−610
ndP(n)
10e0.5 −×= Since, both bits are equiprobable, the Probability of bit error
= 21 [P(0/1) + P(1/0)]
= 0.5 × e-10 07. Ans: (b)
Sol: Probability of error in coherent
BPSK =
ONE2Q
Phase difference 450 decreases the signal
energy by a factor of cos2 450 = 21
=
0e N
EQP
08. Ans: (d)
Sol: 2
1
max0
0mine N
S81erfc2
1)P(
=
∫ ωη
=
=
T
0
22
max0
0 dt.ctcosA2NS
ASK
η
=tA2
∫ ω−−ωη
=
=
T
0
2cc
max0
0 dt)tcos)A(tcosA(2NSPSK
η
=tA4 2
25.041
PSKASK
==
09. Ans: (b)
Sol:
−
π+ω= )1i(
m2t(cos
TE2)t(S cb
−
πω−
−ω= )1i(
m2sin.tsin)1i(
mT2cos..tcos
TE2
ccb
)1i(m2sinEtsin
T2)1i(
m2cosEcos
T2
cb
tb
−π
ω−
−π
ω=
Given binary digital communication m = 2
πω cosEtcosT2
cb
tcos2functionbasic cω=
21Tb =⇒
( ) )1i(sinE]tsin2[)1f(cosEtcos2 cc −πω−−πω
Distance between two points is:
0)EE( 2 ++
E2E4 = Energy of the signal:
∫ ωbT
0
2c )tcosA(
2TA2
=
A2
TA2
2TA
2d b2
b2
=×
==⇒
( )21Tb =
∴ d = A
( 0,E ) (- 0,E )
x x
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56 Electronics and Communication Engg. ACE 10. Ans: (a) Sol: Given, Bipolar Binary signal
HzW10
25−=
η
Hzw102 5−×=η⇒
kbps26.7T1R
bb ==
sec1037.1T 4b
−×=⇒ A= ? ; Pe = 10-4 (Q(x) = 10-4 when x = 3.7)
[ ] 4
0
d 107.3QN2E
Q)x(Q −==
=
)1(71.3N2E
0
d −−−−=
( )∫ −=T
0
221d dt)t(S)t(SE
( )∫=T
0
2d dt1)A2(E
= 4A2T
2
0
b2
)71.3(N2
TA4=
2A2 = (3.71)2×2×10-5×7.36×103 A = 0.999 A ≈ 1
Option: a
01. Ans: (d) Sol:
1r2d = 2r
(d/2)8πsin =
d0.7072
dr1 ==⇒ d1.307
8πsin2
dr2 ==⇒
02. Ans: (d)
Sol: 4-PSK, 8-PSK both have same error probability when both signals have same minimum distance between pairs of signal points.
=
0
2min
e N2d
QP
=
Mπsin
NE2
Q2P 2
0
se
Where Es is the average symbol energy
Given both constellation dmin is same
i.e., ‘d’
Average Symbol Energy:
4
EEEE)(E 4321 ssss
PSK4s
+++=
Where ksE is the symbol ‘Sk’ Energy
= (distance from the origin to the symbol ‘Sk’)2
21
21
21
21
21
PSK4s r4
rrrr)(E =+++
=
T
A
-A
r1
I
d
Q
r1
r1 S1
S4
r2 I
d
Q d/2
r2
r2 8π
S2
S3
LEVEL – 2 (Solutions)
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ACE Communication Postal Coaching Solutions. 57 Similarly, For 8 PSK
22PSK8s r)(E =
22
1
2
PSK4s
PSK8s
0.707d1.307d
rr
)(E)(E
=
=
In dB,
2
(dB)PSK4s(dB)PSK8s 0.7071.307log10)(E)(E
=−
= 5.33 dB
dB5.33)(E)(E PSK4sPSK8s +=
8 PSK required additional 5.33 dB
03. Ans: (b)
Sol: Constellation 1:
s1(t) = 0 ;
s2(t) = − 2 a φ1 + 2 a φ2
s3(t) = −2 2 a.φ1 ;
s4(t) = − 2 a φ1 − 2 a φ2
Energy of S1(t) = ES1 = 0 ; ES2 = 4a2 ; ES3 = 8a2; ES4 = 4a2
Average Energy of constellation 1
= 4
EEEE 4S3S2S1S +++ = 4a2
Constellation 2:
s1(t) = aφ1 ⇒ ES1 = a2
s2(t) = a.φ2 ⇒ ES2 = a2 s3(t) = −a.φ1 ⇒ ES3 = a2
s4(t) = −a.φ2 ⇒ ES4 = a2 Average Energy of constellation 2
= 24S3S2S1S a4
EEEE=
+++
The required Ratio is 4
04. Ans: (a) Sol: The distance between the two closest
points in constellation 1 is d1 = 2a. The same in constellation 2,
d2 = 2 a Since d1 > d2, Probability of symbol error in constellation 1 is lower
05. Ans: (b) Sol: Binomial distribution can be applied here.
Let getting an error be success. It is given that P(success) = p
P(failure) = 1 − p ∴ P(X = at most 1) = P(X =0) + P(X =1)
= 8C0.(p)0. (1 − p)8−0 + 8C1.(p)1 (1 − p)8−1 = (1 − p)8 + 8p(1 − p)7
06. Ans: (a) Sol: P(0/1) = P(1/0) = p
⇒ P(1/1) = P(0/0) = 1−p. Reception with error means getting at the
most one 1. ∴ P(reception with error) = P(X = 0) + P(X = 1)
= 0C3 (1−p)0 p3 +
1C3 (1−p)1p2
= p3 + 3p2(1−p)
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