1 Amplitude Modulation - gschyd.files.wordpress.com · 07. Refer Solution no: 4 . 08. Ans: (c) Sol:...

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

I100115I =

2

μ1II2

Ct 2+=

2

μ1II2

tt 12+=

2

μ1100115 22

=−

µ = 0.803 06. Ans: (b)

Sol: (t)mK1

(t)m̂K)(ηEfficiency

22a

22a

+=

Where, (t)m2 → mean square value of m(t)

Ka = Amplitude sensitivity m(t) → square wave 2m

2 A(t)m =

100AK1

AKη% 2

m2a

2m

2a ×

+=

]AKμ[100μ1

μη% ma22

=×+

=

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 = 0.3

2 + 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 = cf1001

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 1021010

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

22

2 µ+µ

=η

22

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


ACE Communications Postal Coaching Solutions. 3

17. Ans: (a)

Sol: given Pc = 1000R2A2c =

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 minmaxC

+=

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 2t

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

∴ 2221t 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 4

2µcA 4

3µcA 42µcA

1.


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 22222C =

×=

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: 2322

21t µ+µ+µ=µ

222t 9.08.05.0 ++=µ 3.1=

459.02 2t

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

ACE Engineering Publications (A unit of ACE Engg. Academy – Hyderabad, Vijayawada, Visakhapatnam, Tirupati, Delhi, Bhubaneswar, Bangalore, Pune &Chennai)

(Copyrights Reserved)


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


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


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 = 2221 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)

*



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


01. Ans: (b) Sol: sinc2 (1000t) cos 2π fct 2× 1000 = 2000 = 2kHz 02. Ans: (a)

Sol: 2fm = cf1002

2 × 10k = cf1002

⇒ 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 22 2m2

1m2c +=

+= 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 f1c t + m(t)]

3

= a0 [Ac 1 cos 2πf 1c t + m(t)]

+a1 [ (Ac1)3cos32π fc 1t + m3(t) + 3Ac1m2(t).cos 2π fc1t

+ 3(Ac1.cos2π fc t )2.m(t)]

The AM – DSB – SC signal lies in a1.3m(t) . ( Ac1. cos2 π fc t)2

= 3a1(Ac1)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 fc1 = 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)


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 = aV12-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



10 Electronics and Communication Engg. ACE

Power = W4004

116004AA 2m

2C =

×=

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)t}2]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


)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 cc =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 22s22

d µ=µ

21

AA

2s

2d =

707.0AA

s

d =


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 = ( ) 10

25.2410

=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)


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


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 ctotal 100502

1001002

2

=××

==

30. Ans: (b) Sol: % of total power at 10 kHz? Jo(4) = -0.4

)(JR2

AP 2o

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


16 Electronics and Communication Engg. ACE = 50 × 1 = 50

β = 51050

fmf ==

∆

∴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)



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 2c

β+β+β+β+β=± )(

24J)(

23J)(

22J)(

21J)(2

2oJ

R

2cA

mc f4fP

β+β+β+β=

±)(J)(J)(J)(

2J

RA

P 242

22

1

2o

2c

f4f mc

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

)(22JR2

2CA2)(21JR2

2CA2)(20JR2

2CAP β+β+β=

( )β+β+ 24JR1

2CA223JR2

2CA2

++

+−+−=

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 = R2A2C (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



% 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 3e| i ××= = 128×105

65c

| 109.1010128f ×−×=

= 19×105

MHz

fpc

2.91

481019 5

=

××=

∆fi = 25 ∆fi = 25×64 ∆fN = 25×64×48

= 76.8kHz


01. Ans: (d) Sol: [ ]t150cos40t15030sint102π2(t)ψ 6i ++×= ∴ ∆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)


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 1042k

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 ii

θπ

=

θi(t) = ∫ 2π fi(t)dt

dt]tcos)f(k2f[2)t( m2

mci ωπ+π= θ ∫

θi(t) = 2πfct + (2πfm)2 k ttcos

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



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



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 2

2

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



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( minmaxmaxe

−=

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)


24 Electronics and Communication Engg. ACE 01.

Sol: nminmax

2VV −

=∆

n21α∆ ;

1

2

n

n

2

1

22

=∆∆

n3n

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



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 23W41t

W41t −

=→→

−

−=

23

W1613Wπ64Wπ4

1)(Wπ4

0.5Wπ8Wπ4

=−−

=


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 ffff 21 ++=

= 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

+ (3s

f )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)



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


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)



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

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)


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 = fn∆

Let n = 1 Tb = 12

1

CC ff − =

K151

= 66.6 sec

Let n = 2 Tb = K15

2 = 133.3 µsec

n = 3 Tb = K153 = 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



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



n/2

Ideal AWGN

n/2

−B B


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

===


P{X = 4} = 0

P{X = 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


[ ]202 ))tcos()t(x(E)]t(z[E θ+ω= )]t([cosE)]t(x[E 0

22 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(2

121d

21.

21

0)2(21

21

+π××π

=

21

=

21.m)]t(z{E 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(4

1

∫ ∫π π

θτππ

+θθ+τ+ππ

=2

0

2

0cc d)f2cos(4

1d))t(f2(2cos41

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) = 1.01011

101

==×

Ryx(τ) Sy (β) FT

1/10

1 -9 0


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 2s

2c + 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



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 3π

=×

)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)



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 =

222

12

Noω+α

α−=

ω+ααα

−= 22y2.

21

2No)f(S

Applying Inverse Fourier transform

α−τδ=τ τα−e

2)(

2N

)(R 0y

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


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


fA(A)

1/2 0 1

SX(ω)

400δ(ω−104)

6

0 9 10 11 ω(103)


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)

⇒0n

R (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



-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) = 2N 0 00 BfB ≤≤−

= 0 else where 10. Ans: (b) Sol: We know that,

Taking Inverse Fourier Transform

[ ] dfe)t(S)t(SF f2jyy1 τπ∞

∞−

− ∫=

∫=τ−

τπ0

0

B

B

f2j0y dfe2

N)(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


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 = Ac2Ka2P

= 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




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

=

40

10NS

=

9102

−=η

9102 −×=η⇒ Power loss = 40dB (PL) = 104 ( ) ( ) dBLdBtdB )P(PP pi −=

=

L

t

PP

Pp

i

03. Sol:

04. Sol: For SSB modulation

⇒ 4i

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 NN

S3S ×

=

kHz10102103 94 ××××= − = 0.6 Lit PSP ×=∴

4106.0 ×= KW6= 06. Ans: (b) Sol: ∆f = 75 kHz fm = 15kHz

40

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



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


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 2

V

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



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[21A 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

2n21N2

2

1

22

)SQNR()SQNR(

=

8252

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




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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