Double Sideband Suppressed Carrier(DSB-SC) Modulation MCQ Quiz in বাংলা - Objective Question with Answer for Double Sideband Suppressed Carrier(DSB-SC) Modulation - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Apr 13, 2025

পাওয়া Double Sideband Suppressed Carrier(DSB-SC) Modulation उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Double Sideband Suppressed Carrier(DSB-SC) Modulation MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Double Sideband Suppressed Carrier(DSB-SC) Modulation MCQ Objective Questions

Top Double Sideband Suppressed Carrier(DSB-SC) Modulation MCQ Objective Questions

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 1:

A modulator and Demodulator of a DSB-SC system are cascaded as shown in the figure below.

F1 S.B 21.7.20 Pallavi D1.1

The message signal m(t) is band limited to fm and fm ≪ fc. If the average power of the signal M(t) is 80 W, then the average power of the signal y(t) is _______ W. 

Answer (Detailed Solution Below) 5

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 1 Detailed Solution

Concept:

Let the power of a signal m(t) be Pm

Then the power of a signal am(t) will be a2m (a is some constant)

Analysis:

The given is redrawn as:

F1 S.B 21.7.20 Pallavi D1

The output of the first multiplier will be:

s(t) = m(t) sin (2πfct + 30°)

This signal is passed through the second multiplier to give:

x(t) = m(t) sin (2πfc t + 30°)⋅ cos 2πfct

Applying trigonometric identity to the above, we can write:

\(x\left( t \right) = \frac{{m\left( t \right)}}{2}\left[ {\sin \left( {4\pi {f_c}t + 30^\circ } \right) + \sin \left( {30^\circ } \right)} \right]\)

Since the LPF is tuned around fm, and given that fm ≪ fc, the output of the LPF will be:

\(y\left( t \right) = \frac{{m\left( t \right)}}{2}\sin 30^\circ \)

\(y\left( t \right) = \frac{{m\left( t \right)}}{2} \times \frac{1}{2}\)

\(y\left( t \right) = \frac{{m\left( t \right)}}{4}\)

Given power of m(t) = Pm = 80 W,

∴ The power of \(\frac{1}{4}m\left( t \right)\) will be:

\({P_{out}} = {\left( {\frac{1}{4}} \right)^2} \times {P_m}\)

\( = \frac{1}{{16}} \times 80\)

Pout = 5 W

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 2:

Consider an AM signal defined as

s(t) = Ac [1 + 0.4 sin ωmt + 0.2 sin 2ωmt] cos ωc t

Where ωc is the carrier frequency, ωm and 2ωm are two modulating signal frequency components. This signal is transmitted through an antenna. What would be the percentage saving in power transmitted, if instead a suppressed carrier version was sent?

Answer (Detailed Solution Below) 90.5 - 100.5

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 2 Detailed Solution

Concept:

For a multitone modulation with modulation index μ1, μ2, … μn, the net modulation index is given as:

\({\mu _{net}} = \sqrt {\mu _1^2 + \mu _2^2 + \ldots \mu _n^2} \)  

The total power transmitted of an AM signal is defined as:

\({P_t} = {P_c}\left( {1 + \frac{{\mu _{net}^2}}{2}} \right)\) 

Also, the power-saving is defined as the ratio of the amount of power saved to the total transmitted power, i.e.

\(\% \;{P_{saved}} = \frac{{{P_{saved}}}}{{{P_{total}}}} \times 100\) 

Calculation:

Given:

s(t) = Ac [1 + 0.4 sin ωmt + 0.2 sin 2 ωmt] cos ωct

Comparing this with the general expression of multitone AM modulation, we get:

μ1 = 0.4 and μ2 = 0.2

\({\mu _{net}} = \sqrt {\mu _1^2 + \mu _2^2} \) 

\( = \sqrt {{{\left( {0.4} \right)}^2} + {{\left( {0.2} \right)}^2}} \) 

\({\mu _{net}} = \sqrt {0.16 + 0.04} \) 

\({\mu _{net}} = \sqrt {0.2} \) 

Now, the total power transmitted will be:

\({P_t} = {P_c}\left( {1 + \frac{{\mu _{net}^2}}{2}} \right)\) 

\( = \frac{{A_c^2}}{2}\left( {1 + \frac{{0.2}}{2}} \right)\)   

\({P_t} = \frac{{11\;A_c^2}}{{20}}\) 

If the carrier is suppressed, the amount of power saved is:

\({P_{saved}} = \frac{{A_c^2}}{2}\) 

∴ The percentage of power saved will be:

\(\% {P_{saved}} = \frac{{{P_{saved}}}}{{{P_t}}} \times 100\;\% \) 

\( = \frac{{A_c^2/2}}{{11\;A_c^2/20}} \times 100\% \) 

= 90.9%

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 3:

Consider a system shown in fig. Let x(f) and y(f) denote the Fourier transform of x(t) & y(t) respectively. The lowest frequency for which a spectral peak is observed for y(t) is______ kHz.

Communication system file 1 images Q11

Answer (Detailed Solution Below) 0

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 3 Detailed Solution

Output of 1st modulator

Communication system file 1 images Q11a

Output of HPF

Communication system file 1 images Q11b

Output of 2nd balanced modulator (only positive frequency)

Communication system file 1 images Q11c

The lowest frequency for which a spectral peak is observed for y(t) is at 0 kHz, and then at 2 khz and so on as shown in the figure 

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 4:

An AM broadcast station is found to be transmitting modulating frequencies up to 5 kHz and the station is allowed to transmit on a frequency of 950 kHz, then what will be the bandwidth (in kHz) allotted to the station?

  1. 5
  2. 10
  3. 20
  4. 25

Answer (Detailed Solution Below)

Option 2 : 10

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 4 Detailed Solution

Concept:

Standard expression for AM expression for a message signal AmCosωmt is

S(t) = AC(1 + μCosωmt )cos ωct

Where;

ωm → Modulating frequency (rad/sec) = 2πfm

Bandwidth = 2fm

Calculation:

Given;

Modulating frequency (fm) = 5 KHz

Bandwidth = 2fm = (2 × 5) = 10 KHz

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 5:

Quadrature null effect is present in

a) DSB-SC

b) AM

c) SSB

d) FM

Which of the following option is correct

  1. (a) only
  2. (a) & (b) only
  3. (b) & (c) only
  4. all (a), (b), (c), (d)

Answer (Detailed Solution Below)

Option 2 : (a) & (b) only

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 5 Detailed Solution

Quadrature Null Effect (QNE) in AM demodulator circuit is observed if the carrier signal generated at the local oscillator is not synchronized in phase with the incoming input signal. 

For AM and DSB-SC modulation schemes, the demodulated signal when there is a phase difference of ϕ is given by:

\({s_0(t)} = \frac{{A_c^2}}{2}m\left( t \right)\cos ϕ\)

When ϕ = 90°, the demodulated signal is 0.

However, in SSB, the demodulated signal after passing through a low pass filter is given by:

\({s_0}\left( t \right) = \frac{{A_c^2}}{4}\left[ {m\left( t \right)cosϕ \pm \hat m\left( t \right)sinϕ } \right]\)

If ϕ = 0°, we get \(\frac{{A_c^2}}{4}m\left( t \right)\)

If ϕ = 90°, we get \(\frac{{A_c^2}}{4}\hat m\left( t \right)\)

In both of the above cases, the output signal is never zero and reconstruction of the message signal is easy.

So, no QNE is observed for the SSB modulation scheme.

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 6:

A multiplier shown in the figure is used as a balanced modulator. The inputs to terminals X and Y are sinusoidal signals:

m(t) = 6 sin 2 π (1200) t

c(t) = 6 sin 2 π (12000) t

F1 S.B Madhu 19.12.19 D1

Which of the following frequency component is present at the output s(t)?

  1. 10.8 kHz
  2. 13.2 kHz
  3. 12 kHz
  4. Both (1) and (2)

Answer (Detailed Solution Below)

Option 4 : Both (1) and (2)

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 6 Detailed Solution

Given the signals at terminals X and Y as:

m(t) = 6 sin 2 π (1200)t

c(t) = 6 sin 2 π (12000)t

The output of the multiplier will be,

s(t) = k (m(t) c(t)), for some constant k.

= 36 k sin 2π (1200)t sin 2π (12000)t

=\(\frac{{36k}}{2}[\cos 2\pi \left( {12000 - 1200} \right)t - \cos 2\pi \left( {12000 + 1200} \right)t]\;\)

= 18k [cos 2π (10800)t – cos 2π (13200)t]

Thus, the frequency components present at the output are:

f1 = 10800 Hz = 10.8 kHz

f2 = 13200 Hz = 13.2 kHz

So, Option (4) is correct.

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 7:

The carrier used in the detection of a DSB-SC signal has a constant phase error. As a result of this, which one of the following is correct?

  1. The output gets phase distorted
  2. The output gets reduced
  3. The signal cannot be detected
  4. There is no effect on the output

Answer (Detailed Solution Below)

Option 2 : The output gets reduced

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 7 Detailed Solution

In coherent detection of a DSB-SC the modulated signal is multiplied by local carrier signal and the higher frequencies are filtered off.

let ϕ be the constant phase error in load oscillator. It is multiplied by message signal.

\(\begin{array}{l} S\left( t \right) = \left[ {m\left( t \right)\cos {\omega _c}t} \right]\cos \left( {{\omega _c}t + \phi } \right)\\ = \frac{{m\left( t \right)}}{2}\left[ {\cos \left( {{\omega _c}t + \phi - {\omega _c}t} \right) + \cos \left( {{\omega _c}t + \phi + {\omega _c}t} \right)} \right]\\ = \frac{{m\left( t \right)}}{2}\cos \phi + \frac{{m\left( t \right)}}{2}\cos \left( {2{\omega _c}t + \phi } \right) \end{array}\)

The higher frequency signal is filtered off and we get diminished output by a factor cos ϕ 

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 8:

A balanced modulator is used in the generation of

  1. DSB – SC signal
  2. FM signal
  3. PM signal
  4. PAM signal

Answer (Detailed Solution Below)

Option 1 : DSB – SC signal

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 8 Detailed Solution

Balanced modulators and ring modulators are popularly used to generate DSB-SC signals.

The circuit diagram of a balanced modulator is as shown:

F1 S.B 8.5.20 Pallavi D4

 

Balanced Modulator:-

1) It is used to generate a DSB-SC wave.

2) It consists of two AM modulators in a balanced configuration to suppress the carrier signal.

3) The message signal m(t) is applied at one modulator and – m(t) is applied at another modulator.

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 9:

A DSB-SC signal is being detected synchronously. The phase error in the locally generated carrier will _________.

  1. cause phase distortion and reduce the output also
  2. increase the amplitude of the demodulated signal 
  3. completely block the demodulated signal always
  4. have no effect on the demodulated signal

Answer (Detailed Solution Below)

Option 1 : cause phase distortion and reduce the output also

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 9 Detailed Solution

Explanation:

5Double Sideband Suppressed Carrier (DSB-SC) is a type of amplitude modulation where the carrier is suppressed, and only the sidebands carry the information. In synchronous detection, a locally generated carrier is used at the receiver to demodulate the signal. The phase and frequency of this locally generated carrier must align precisely with the transmitted carrier for accurate demodulation.

Phase Error in Synchronous Detection: A phase error occurs when there is a phase difference between the locally generated carrier at the receiver and the carrier used during transmission. This phase mismatch can significantly affect the demodulated output.

Impact of Phase Error: In the case of a DSB-SC signal, the demodulated output is given by the product of the transmitted signal and the locally generated carrier. When there is a phase error, the output signal gets distorted. Mathematically, the transmitted DSB-SC signal can be expressed as:

Transmitted Signal: s(t) = A × cos(ωct) × m(t)

Here:

  • A = Amplitude of the signal
  • ωc = Angular frequency of the carrier
  • m(t) = Modulating signal

The locally generated carrier at the receiver can be expressed as:

Local Carrier: cos(ωct + θ)

Where θ is the phase error. The demodulated output is obtained by multiplying the transmitted signal with the local carrier:

Demodulated Output:

y(t) = s(t) × cos(ωct + θ)

Substituting the expression for s(t):

y(t) = [A × cos(ωct) × m(t)] × cos(ωct + θ)

Using the trigonometric identity:

cos(x) × cos(y) = 0.5 × [cos(x-y) + cos(x+y)]

y(t) = 0.5 × A × m(t) × [cos(θ) + cos(2ωct + θ)]

The term cos(2ωct + θ) represents a high-frequency component that is filtered out during demodulation, leaving:

y(t) = 0.5 × A × m(t) × cos(θ)

This shows that the amplitude of the demodulated signal is scaled by cos(θ). If there is a phase error θ, the output amplitude is reduced by the factor cos(θ). Additionally, the phase error introduces phase distortion in the demodulated signal.

Correct Option Analysis:

The correct option is:

Option 1: Cause phase distortion and reduce the output also.

This option is correct because a phase error in the locally generated carrier introduces distortion in the output and reduces its amplitude by the factor cos(θ). The demodulated signal's quality depends on the alignment of the local carrier with the transmitted carrier, and any phase mismatch degrades the output.

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 10:

What is the main advantage of Double-sideband suppressed-carrier (DSB-SC) modulation over conventional AM?

  1. Lower power consumption 
  2. Smaller bandwidth efficiency
  3. Simpler demodulation process
  4. Higher fidelity in signal reproduction

Answer (Detailed Solution Below)

Option 1 : Lower power consumption 

Double Sideband Suppressed Carrier(DSB-SC) Modulation Question 10 Detailed Solution

Concept

Double-sideband suppressed-carrier (DSB-SC) modulation is a type of amplitude modulation (AM) where the carrier signal is suppressed and only the sidebands containing the information are transmitted. This technique is often used in communication systems to improve efficiency and reduce power consumption.

In conventional AM, the carrier is transmitted along with the sidebands, which consumes a significant amount of power without carrying any information. In DSB-SC, the carrier is suppressed, meaning that only the sidebands are transmitted. This results in lower power consumption because the power that would have been used to transmit the carrier is saved. 

Option 2: Smaller bandwidth efficiency

DSB-SC does not necessarily have a smaller bandwidth efficiency compared to conventional AM. Both modulation techniques use the same amount of bandwidth because they both transmit the upper and lower sidebands. However, the key advantage of DSB-SC lies in its power efficiency, not bandwidth efficiency.

Option 3: Simpler demodulation process

DSB-SC modulation actually requires a more complex demodulation process compared to conventional AM. Demodulating a DSB-SC signal usually requires a coherent detector or a synchronous demodulator, which is more complex than the envelope detector used for conventional AM.

Option 4: Higher fidelity in signal reproduction

Higher fidelity in signal reproduction is not a specific advantage of DSB-SC modulation over conventional AM. Both modulation techniques can achieve high fidelity if implemented correctly. The main distinction between them is the power efficiency, not the fidelity of the reproduced signal.

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