SNR in FM System MCQ Quiz - Objective Question with Answer for SNR in FM System - Download Free PDF

Last updated on Apr 18, 2025

Latest SNR in FM System MCQ Objective Questions

SNR in FM System Question 1:

Which of the following options is true?

  1. Noise has more effect on lower frequencies in FM
  2. Noise in AM increases as frequency increases
  3. Noise has more effect on higher frequencies in FM
  4. Noise in PM increases exponentially for the entire audio range

Answer (Detailed Solution Below)

Option 3 : Noise has more effect on higher frequencies in FM

SNR in FM System Question 1 Detailed Solution

Explanation:-

The effect of noise in FM signal is determined by the extent to which it changes the frequency of the modulated signal.

Noise in FM

F1 Shubham.B 10-05-21 Savita D7

PSD of the noise at the detector output is directly proportional to the square of the frequency.

F1 Shubham.B 10-05-21 Savita D8

So we can say the effect of noise is more at higher frequencies in FM.

Discussion:-

Figure of Merit (FOM): It is the ratio of output SNR to the input SNR. FOM depends only on the demodulator.

For AM:

In noise performance, the synchronous detector is better than an envelope detector.

In AM noise has nothing to do with frequency, it depends on the amplitude of modulating signal.

FOMAM=13        (for envelope detector).

For PM:

FOMPM=β22

Noise in PM increases exponentially for the entire audio range is a wrong statement.

For FM:

FOMFM=3β22

The figure of merit of FM is far better than the other two. So we can say the noise performance of FM is better than AM and PM.

Important Points

1. To improve noise performance in FM we use the concept of pre-emphasis and de-emphasis.

2. WBFM is preferred over NBFM because the earlier has a better figure of merit.

SNR in FM System Question 2:

A FM system has an SNR of 40 dB at the receiver. The Bandwidth available for the channel is 120 kHz and the message signal bandwidth is 10 kHz. The channel is affected by noise having one-sided power spectral density of No = 10-8 W/Hz. The minimum transmitted power required if the signal is attenuated by 40 dB in transmission through channel is __________.

  1. 267 W
  2. 232 W
  3. 153 W
  4. 198 W

Answer (Detailed Solution Below)

Option 1 : 267 W

SNR in FM System Question 2 Detailed Solution

From Carson’s rule

Bc = 2 (1 + β) fm

120 kHz = 2 (1 + β) (10)

β = 5

For F.M.

(SN)o=32β2(SN)i

(SN)o=40 dB=10log(SN)i

(SN)=104

(SN)i=(23)(104)(1β2)

=8003266.6

≈ 24.26 dB

(SN)=PRNofm

fm = 10, 000

No = 10-8

PR=8300=0.0266

≈  -15.74 dB

Due to channel attenuation, 

PT = -15.74 + 40

= 24.26 dB

≈  266.66 W

SNR in FM System Question 3:

The minimum value of modulation index β for an F.M. system required to produce a noticeable improvement in SNR over a comparable AM system with μ = 1 is ________.(upto 2 decimals)

Answer (Detailed Solution Below) 0.46 - 0.48

SNR in FM System Question 3 Detailed Solution

We know,

FOMFM=32β2

FOMAM=μ22+μ2

FOMFM > FOMAM

32β2>11+2(μ=1)

32β2>13

β>23

Minimum value of β = 0.47

Top SNR in FM System MCQ Objective Questions

Which of the following options is true?

  1. Noise has more effect on lower frequencies in FM
  2. Noise in AM increases as frequency increases
  3. Noise has more effect on higher frequencies in FM
  4. Noise in PM increases exponentially for the entire audio range

Answer (Detailed Solution Below)

Option 3 : Noise has more effect on higher frequencies in FM

SNR in FM System Question 4 Detailed Solution

Download Solution PDF

Explanation:-

The effect of noise in FM signal is determined by the extent to which it changes the frequency of the modulated signal.

Noise in FM

F1 Shubham.B 10-05-21 Savita D7

PSD of the noise at the detector output is directly proportional to the square of the frequency.

F1 Shubham.B 10-05-21 Savita D8

So we can say the effect of noise is more at higher frequencies in FM.

Discussion:-

Figure of Merit (FOM): It is the ratio of output SNR to the input SNR. FOM depends only on the demodulator.

For AM:

In noise performance, the synchronous detector is better than an envelope detector.

In AM noise has nothing to do with frequency, it depends on the amplitude of modulating signal.

FOMAM=13        (for envelope detector).

For PM:

FOMPM=β22

Noise in PM increases exponentially for the entire audio range is a wrong statement.

For FM:

FOMFM=3β22

The figure of merit of FM is far better than the other two. So we can say the noise performance of FM is better than AM and PM.

Important Points

1. To improve noise performance in FM we use the concept of pre-emphasis and de-emphasis.

2. WBFM is preferred over NBFM because the earlier has a better figure of merit.

SNR in FM System Question 5:

Which of the following options is true?

  1. Noise has more effect on lower frequencies in FM
  2. Noise in AM increases as frequency increases
  3. Noise has more effect on higher frequencies in FM
  4. Noise in PM increases exponentially for the entire audio range

Answer (Detailed Solution Below)

Option 3 : Noise has more effect on higher frequencies in FM

SNR in FM System Question 5 Detailed Solution

Explanation:-

The effect of noise in FM signal is determined by the extent to which it changes the frequency of the modulated signal.

Noise in FM

F1 Shubham.B 10-05-21 Savita D7

PSD of the noise at the detector output is directly proportional to the square of the frequency.

F1 Shubham.B 10-05-21 Savita D8

So we can say the effect of noise is more at higher frequencies in FM.

Discussion:-

Figure of Merit (FOM): It is the ratio of output SNR to the input SNR. FOM depends only on the demodulator.

For AM:

In noise performance, the synchronous detector is better than an envelope detector.

In AM noise has nothing to do with frequency, it depends on the amplitude of modulating signal.

FOMAM=13        (for envelope detector).

For PM:

FOMPM=β22

Noise in PM increases exponentially for the entire audio range is a wrong statement.

For FM:

FOMFM=3β22

The figure of merit of FM is far better than the other two. So we can say the noise performance of FM is better than AM and PM.

Important Points

1. To improve noise performance in FM we use the concept of pre-emphasis and de-emphasis.

2. WBFM is preferred over NBFM because the earlier has a better figure of merit.

SNR in FM System Question 6:

The minimum value of modulation index β for an F.M. system required to produce a noticeable improvement in SNR over a comparable AM system with μ = 1 is ________.(upto 2 decimals)

Answer (Detailed Solution Below) 0.46 - 0.48

SNR in FM System Question 6 Detailed Solution

We know,

FOMFM=32β2

FOMAM=μ22+μ2

FOMFM > FOMAM

32β2>11+2(μ=1)

32β2>13

β>23

Minimum value of β = 0.47

SNR in FM System Question 7:

A FM system has an SNR of 40 dB at the receiver. The Bandwidth available for the channel is 120 kHz and the message signal bandwidth is 10 kHz. The channel is affected by noise having one-sided power spectral density of No = 10-8 W/Hz. The minimum transmitted power required if the signal is attenuated by 40 dB in transmission through channel is __________.

  1. 267 W
  2. 232 W
  3. 153 W
  4. 198 W

Answer (Detailed Solution Below)

Option 1 : 267 W

SNR in FM System Question 7 Detailed Solution

From Carson’s rule

Bc = 2 (1 + β) fm

120 kHz = 2 (1 + β) (10)

β = 5

For F.M.

(SN)o=32β2(SN)i

(SN)o=40 dB=10log(SN)i

(SN)=104

(SN)i=(23)(104)(1β2)

=8003266.6

≈ 24.26 dB

(SN)=PRNofm

fm = 10, 000

No = 10-8

PR=8300=0.0266

≈  -15.74 dB

Due to channel attenuation, 

PT = -15.74 + 40

= 24.26 dB

≈  266.66 W

SNR in FM System Question 8:

For a Gaussian modulating signal with mean zero and variance σ2, the dividing line between narrow band and wideband FM modulation in terms of SNR is at β= _________.

For the purpose of maximum amplitude, assume the maximum amplitude of input Gaussian to be 3σ.

Answer (Detailed Solution Below) 0.5 - 0.55

SNR in FM System Question 8 Detailed Solution

As NBFM is similar to AM, its maximum SNR (putting μ=1 and maximum amplitude A=mp) is m2mp2+m2γ where γ is input SNR. For wideband FM, maximum SNR is 3β2(m2mp2)γ 

Now, the dividing line in terms of SNR will be at that value of β for which SNR of FM is equal to SNR of NBFM.

3β2(m2mp2)γ=m2mp2+m2γβ2=13[11+m2/mp2]

Now, for the given Gaussian, m2=σ2 and mp2=(3σ)2=9σ2 

Thus, β2=13[11+σ29σ2]=13[910] 

β2=3/10β=0.547

SNR in FM System Question 9:

A modulated signal is given as s(t)=100cos[200πt+10tm(τ)dτ], where m(t) is periodic message signal shown below

Gate EC Communication Mock Images-Q17

The SNR of the receive output is _____

Assume effective bandwidth of m(t) to be twice the fundamental frequency and γ=1.

Answer (Detailed Solution Below) 3.7 - 3.8

SNR in FM System Question 9 Detailed Solution

Since, the message signal is integrated to produce phase variation, the signal is FM signal. For FM, SNR=3β2(m2mp2) 

m2=1202m2(t)dt=12(0×1+1×1)=12 and mp2=(1)2=1

Now, BW(m(t))=2f0=1Hz as fo=1T and T=2 sec.

Now, from definition of s(t),

kf=10,Δfmax=kp2πm(t)|max=102πΔfmax=5π 

Now, β=ΔfmaxBW=5/π1β=5/π 

Thus, (SNR)=3(5π)2(1/21)=75/2π2 

SNR=3.799

SNR in FM System Question 10:

For tone modulation, the dividing line between narrow band and wideband FM modulation in terms of SNR is at β = _______.

Answer (Detailed Solution Below) 0.4 - 0.5

SNR in FM System Question 10 Detailed Solution

Since, NBFM is similar to AM, the SNR of AM is  its maximum SNR is

 m2mp2+m2γ     (for SNRmax  μ=1,A=mp) 

For wideband FM, maximum SNR is 3β2(m2mp2)γ

Now, the dividing line in terms of SNR performance will be that value of β for which SNR of FM is equal to maximum SNR of NBFM. Thus,

3β2(m2mp2)γ=m2mp2+m2γβ2=13[11+m2mp2]

For tone modulation,

m2=A22 and mp2=A2 

β2=13[11+12]=29β=0.47

SNR in FM System Question 11:

A zero mean Gaussian random process m(t) having variance σm is fed as a baseband signal to an FM communication system. The channel is affected by white noise. The output SNR is

For the purpose of maximum amplitude, assume the maximum amplitude of input Gaussian to be 3σm.

  1. 32β2γ
  2. 13β2γ
  3. 3β2γ
  4. 23β2γ

Answer (Detailed Solution Below)

Option 2 : 13β2γ

SNR in FM System Question 11 Detailed Solution

We have, (S0N0)FM=3β2(m2mp2)γ

m2=σm2 and maximum amplitude mp=3σm

Thus, (S0N0)FM=3β2(σm29σm2)γ

(S0N0)FM=13β2γ

SNR in FM System Question 12:

Message signal, m(t)=a1cosω1t+a2cosω2t, (ω2>ω1).The condition for the output SNR of PM to be better than output SNR of FM is

x=a1/a2 and y=ω1/ω2

  1. (1+x)2<3(1+xy)2
  2. (1+x)2>3(1+xy)2
  3. 3(1+x)2>(1+xy)2
  4. 3(1+x)2<(1+xy)2

Answer (Detailed Solution Below)

Option 2 : (1+x)2>3(1+xy)2

SNR in FM System Question 12 Detailed Solution

We have, (SNR)PM=(Δω)2(m2mp2)γ, where mp=dm(t)dt|max 

And, (SNR)FM=3(ΔfB)2(m2mp2)γ, where mp=m(t)|max 

Now, (SNR)PM(SNR)FM=(2πΔf)2mp23(ΔfB)2mp2 

(SNR)PM(SNR)FM=(2πB)2mp23mp2 

Now m(t)=a1cosω1t+a2cosω2t

m=dm(t)dt=a1ω1sinω1ta2ω2sinω2t 

mp=dm(t)dt|max=(a1ω1+a2ω2)  (∵ at some time, both sinω1t and sinω2t will be 1 together)

Similarly, mp=m(t)|max=(a1+a2) 

Substituting, we have, (SNR)PM(SNR)FM=(2πB)2(a1+a2)23(a1ω1+a2ω2)2 

Now bandwidth of m(t)=ω2rad/sec    (ω2>ω1)

2πB=ω2 

Thus, (SNR)PM(SNR)FM=ω22(a1+a2)23(a1ω1+a2ω2)2=ω22(a1a2+1)2a223(a2ω2)2(a1ω1a2ω2+1)2 

(SNR)PM(SNR)FM=(a1a2+1)23(a1ω1a2ω2+1)2(SNR)PM(SNR)FM=(x+1)23(xy+1)2 

Thus, for (SNR)PM>(SNR)FM, it is required that (1+x)2>3(1+xy)2

SNR in FM System Question 13:

Consider an FM system with Δf=80 kHz and maximum signal frequency f=20 KHz. Assume that average power of input baseband signal, SX=12 . If the signal were to be sent without any modulation over a baseband system, it would have a certain SNR. However, if it is sent using FM, then the improvement in SNR over the baseband system is_______dB.

Answer (Detailed Solution Below) 13.8 - 14

SNR in FM System Question 13 Detailed Solution

We have output SNR of an FM system is:

(SN)o=3(Δff)2.SX.γ

where γ=SiηB is SNR of baseband system or channel signal to noise ration. As for baseband So=Si(SoηB=SiηB).

Thus (SN)oγ=3(Δff)2.SX=3×(80×10320×103)2×12

(SN)oγ=24

Thus improvement in SNR(in dB) is

10log((SN)oγ)=10log(SN)o10logγ=10log24=13.802.

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