Double Sideband Suppressed Carrier(DSB-SC) Modulation MCQ Quiz - Objective Question with Answer for Double Sideband Suppressed Carrier(DSB-SC) Modulation - Download Free PDF

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(ω_ct) × m(t)

Here:

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

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

Local Carrier: cos(ω_ct + 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(ω_ct + θ)

Substituting the expression for s(t):

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

The term cos(2ω_ct + 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.

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.

Concept:

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

Since the given modulation is DSB-SC, the carrier is suppressed. ∴ The transmitted power will be:

\({P_{t(DSB-SC)}} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)-P_c\)

\({P_{t(DSB-SC)}} = P_c\frac{\mu ^2}{2}\)

Calculation: 

Given

μ = 0.4

Pc = 80 W

\({P_{t(DSB-SC)}} = 80\times \frac{0.4^2}{2}\)

\({P_{t(DSB-SC)}} = 80\times 0.08\)

Pt(DSB-SC) = 6.40 W

Concept:

Standard expression for AM expression for a message signal AmCosωmt is

S(t) = AC(1 + μCosωmt )cos ωct

Where;

ω_m → Modulating frequency (rad/sec) = 2πf_m

Bandwidth = 2f_m

Calculation:

Given;

Modulating frequency (f_m) = 5 KHz

Bandwidth = 2fm = (2 × 5) = 10 KHz

Correct answer is option (4) : v(t) = (Ec + Em sin(ωmt)) × sin(ωct)

Concept:

The expression of the full-carrier AM (DSB-full-carrier) = (E_c + m(t)) × sin(ωct) 

• ​E_c is peak amplitude of the carrier signal i.e  E_csin(ωct).
• ω_c is the frequency of the carrier signal in rad/sec.
• m(t) is modulating or the message signal.

Calculation:

Given, m(t) = E_msin(ωmt) , where E_m is the peak amplitude of the messege signal and ω_m is the frequency of the message signal in rad/sec.

.Equation for full-carrier AM signal:   

v(t) =  (Ec + m(t)) × sin(ωct) 

⇒ v(t) =  (Ec + Emsin (ωmt) ) × sin(ωct) 

Hence,  the equation of the full-carrier AM (DSB-full carrier) is v(t) =  [Ec + Emsin (ωmt)] × sin(ω_ct) . 

In a balanced modulator, 2 AM-modulators are connected in a way that the resultant signal does not contain the carrier spectrum, i.e. to generate a Double Side-band suppressed carrier (DSB-SC).

The circuit diagram of a balanced modulator is as shown:

\( {s_{AM'}}\left( t \right) = {A_c}\left[ {1 + {k_a}m\left( t \right)} \right]\cos 2\pi {f_c}t\)

\(\\ {s_{AM''}}\left( t \right) = {A_c}\left[ {1 - {k_a}m\left( t \right)} \right]\cos 2\pi {f_c}t\)

The resultant DSB signal is the difference between the two, i.e.

\({s_{DSB}}\left( t \right) = {s_{AM'}}\left( t \right) - {s_{AM''}}\left( t \right) \)

Hence,

\({s_{DSB}}\left( t \right) = \;2{A_c}{k_a}m\left( t \right)\cos 2\pi {f_c}t\)

\( {s_{DSB}}\left( t \right) = \;{A_c}'m\left( t \right)\cos 2\pi {f_c}t\)

1) Square law modulator is used for the generation of conventional AM signal which includes the carrier frequency component as well.

2) An Armstrong modulator is used to generate an FM signal.

3) Envelop detector is used for AM demodulation.

The ring modulator and balanced modulator is generally used for the modulation of DSB SC signal.

SSB/SC is generated using a balanced modulator

DSB -SC:

1) It is a form of Amplitude modulation where only the sidebands are transmitted and the carrier is suppressed.

2) The advantage is 66% power saving compared to conventional AM.

3) The Bandwidth requirement is of 2f_m, with f_m being the maximum frequency component of the message signal.

4) It is used for transmitting stereo information.

Single Side Band (SSB SC):

1) SSB SC is a form of amplitude modulation where only a single sideband is transmitted and 1 sideband and carrier are suppressed.

2) The advantage of SSB is bandwidth and power-saving over both conventional AM and DSB SC, i.e. DSB SC has more power consumption than SSB SC.

3) The Bandwidth requirement is only fm, ∴ It needs half the bandwidth than the DSB-SC transmission.

Comparison:

Concept:

An amplitude modulation spectrum is represented as:

f_c = Carrier Frequency

f_m = Signal frequency

f_c + f_m = Upper Sideband

f_c - f_m = Lower Sideband

Analysis:

Given f_c = 40 kHz and f_m = 5 kHz

The upper and lower sidebands after Amplitude modulation will be:

Upper Sideband = fc + fm = 40 + 5 kHz = 45 kHz

Lower Sideband = fc - fm = 40 - 5 kHz = 35 kHz

In a balanced modulator, 2 AM-modulators are connected in a way that the resultant signal does not contain the carrier spectrum, i.e. to generate a Double Side-band suppressed carrier (DSB-SC).

Analysis:

The circuit diagram of a balanced modulator is as shown:

\( {s_{AM'}}\left( t \right) = {A_c}\left[ {1 + {k_a}m\left( t \right)} \right]\cos 2\pi {f_c}t\)

\(\\ {s_{AM''}}\left( t \right) = {A_c}\left[ {1 - {k_a}m\left( t \right)} \right]\cos 2\pi {f_c}t\)

The resultant DSB signal is the difference of the two, i.e.

\({s_{DSB}}\left( t \right) = {s_{AM'}}\left( t \right) - {s_{AM''}}\left( t \right) \)

Hence,

\({s_{DSB}}\left( t \right) = \;2{A_c}{k_a}m\left( t \right)\cos 2\pi {f_c}t\)

\( {s_{DSB}}\left( t \right) = \;{A_c}'m\left( t \right)\cos 2\pi {f_c}t\)

The frequency spectrum of the DSB-SC signal:

The standard DSB-SC equation is given as:

\({S_{DSB}}\left( t \right) = \;{A_c}m\left( t \right)\cos 2\pi {f_c}t\)

Assuming a single tone message signal as:

\(m\left( t \right) = \;{A_m}\cos 2\pi {f_m}t\)

The DSB-SC signal can be written as:

\({S_{DSB}}\left( t \right) = \frac{{{A_c}{A_m}}}{2}\cos 2\pi ({f_c} + {f_m})t + \;\frac{{{A_c}{A_m}}}{2}\cos 2\pi ({f_c} - {f_m})t\)

∴ The output of balanced modulator contains the modulation frequency, LSB and USB.

Concept:

The generalized AM expression is represented as:

s(t) = Ac [1 + μa mn (t)] cos ωc t

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

The above expression can be expanded to get:

\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)

The total power is the sum of the carrier power and the sideband power, i.e.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

The power in a single sideband will be:

\({P_s} = \frac{1}{2}\times P_c\frac{{{μ^2}}}{2}\)

\(As \ P_c=\frac{A_c^2}{2}\),

the above can be written as:

\({P_s} = \frac{1}{2}\times \frac{A_c^2}{2}\frac{{{μ^2}}}{2}\)

\({P_s} = \frac{{{A_c^2μ^2}}}{8}\)   ---(1)

The ratio of total sideband power to total input power gives the maximum transmission power efficiency.

Calculation:

Total sideband power is:

Multiplying equation (1) with 2, we will get total sideband power as:

\({P_{sb}} = \frac{{{A_c^2μ^2}}}{4}\)

When μ = 1,

\({P_{sb}} = \frac{{{A_c^2}}}{4}\)

Total efficiency (η) is given by:

\(η=\frac{P_{sb}}{P_T} \ \times \ 100\)  

P_T = Transmitted power

As P_sb = P_T  in DSB-SC as the carrier is suppressed so:

η = 100%

Hence option (4) is the correct answer.

Important Points

DSB -SC:

1) It is a form of Amplitude modulation where only the sidebands are transmitted and the carrier is suppressed.

2) The advantage is 66% power saving compared to conventional AM.

3) The Bandwidth requirement is of 2fm, with fm being the maximum frequency component of the message signal.

4) It is used for transmitting stereo information.

Comparison:

Concept: Low pass filter allow the low frequency components ( up to Bandwidth of filter) to pass through it.

Calculation:

x(t) = m(t) cos (2πf_ct)

m(t) = 4 cos (1000 πt)

The output of the multiplier(s(t)) will be:

s(t) = x(t) cos (2π(f_c + 40)t)

s(t) = m(t) cos (2πf_ct) cos (2π(f_c + 40)t)

\(\cos \left( A \right)\cos \left( B \right) = \frac{1}{2}(\cos \left( {A - B} \right) + \cos \left( {A + B} \right))\)

\(s\left( t \right) = \frac{{m\left( t \right)}}{2}\left[ {\cos \left( {2\pi ({f_c} - {f_c} - 40} \right)t + \cos \left( {2\pi \left( {{f_c} + {f_c} + 40} \right)t} \right)} \right]\)

\(s\left( t \right) = \frac{{m\left( t \right)}}{2}\left( {\cos \left( {2\pi \left( {40} \right)t + \cos (2\pi (2{f_c} + 40} \right)t} \right)\)

\(s\left( t \right) = \frac{{4\cos \left( {1000\pi t} \right)}}{2}(\cos \left( {2\pi \cdot 40t} \right) + \cos \left( {2\pi \left( {2{f_c} + 40} \right)t} \right)\)

s(t) = 2 cos (100 πt)⋅cos (2π 40t) + 2 cos (1000 πt) cos (2π(2f_c + 40)t

Again using the same trigonometric identify, the above equation can be written as:

s(t) = cos (1000πt – 80πt) + cos (1000πt + 80πt) + cos (1000πt – 2π(2f_c + 40)t) + cos (1000πt + 2π (2f_c + 40)t)

s(t) = cos (920πt) + cos (1080πt) + cos (1000πt – 2π (2f_c - 40)t + cos(1000πt + 2π (2f_c + 40)t)

The frequency specimen of s(t) can be drawn as:

The output of the low pass filter will contain frequencies of 460 Hz only as shown:

∴ y(t) = cos (920 πt)

Correct answer is option (4) : v(t) = (Ec + Em sin(ωmt)) × sin(ωct)

Concept:

The expression of the full-carrier AM (DSB-full-carrier) = (E_c + m(t)) × sin(ωct) 

• ​E_c is peak amplitude of the carrier signal i.e  E_csin(ωct).
• ω_c is the frequency of the carrier signal in rad/sec.
• m(t) is modulating or the message signal.

Calculation:

Given, m(t) = E_msin(ωmt) , where E_m is the peak amplitude of the messege signal and ω_m is the frequency of the message signal in rad/sec.

.Equation for full-carrier AM signal:   

v(t) =  (Ec + m(t)) × sin(ωct) 

⇒ v(t) =  (Ec + Emsin (ωmt) ) × sin(ωct) 

Hence,  the equation of the full-carrier AM (DSB-full carrier) is v(t) =  [Ec + Emsin (ωmt)] × sin(ω_ct) . 

Concept:

A single sideband (SSB) AM Signal is represented as

\(s\left( t \right) = \frac{{Ac\;m\left( t \right)}}{2}\cos 2\pi fct \pm \frac{{Ac}}{2}\hat m\left( t \right)\sin 2\pi fct\)

m̂(t) → represent Hilbert transform of message signal

(-) → sign represent USB

(+) → sign represent LSB

Hilbert transform:

It is a specific Linear operator that takes a function u(t) of a real variable and produces another function of real variable H(u)(t)

→ It imparts a phase shift of ± 90° (π/2 radians) to every frequency component of a function.

\(\sin \left( {\omega t} \right)\mathop \to \limits^{H.T} - \cos \left( {\omega t} \right) = \sin \left( {\omega t + \frac{\pi }{2}} \right)\)

\(\cos \left( {\omega t} \right)\mathop \to \limits^{H.T} \sin \left( {\omega t} \right) = \cos \left( {\omega t + \frac{\pi }{2}} \right)\)

Explanation:

From the given Block diagram.

I₁ = (Am sin ω_mt)(Ac cos ω_ct)

I₂ = (Am cos ω_mt)(Ac cos ω_ct)

V_out = I₁ + I₃

From I₁  

\({I_1} = \frac{{AcAm}}{2}\left[ {\sin \left( {{\omega _m} + {\omega _c}} \right)t - \sin ({\omega _c} - {\omega _m})t} \right]\)

Then

\(\because \left( {\begin{array}{*{20}{c}} {2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)}\\ {2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)} \end{array}} \right)\)

\({I_2} = \frac{{AcAm}}{2}\left[ {\cos \left( {{\omega _c} + {\omega _m}} \right)t + \cos ({\omega _c} - {\omega _m})t} \right]\)

\({I_3} = \frac{{AcAm}}{2}\left[ {\cos \left( {{\omega _c} + {\omega _m}} \right)t - \frac{\pi }{2} + \cos ({\omega _c} - {\omega _m})t - \frac{\pi }{2}} \right]\)

\( = \frac{{AcAm}}{2}\left[ {\sin \left( {{\omega _c} + {\omega _m}} \right)t + \sin ({\omega _c} - {\omega _m})t} \right]\)

\({V_{out}} = {I_1} + {I_3} = \frac{{AcAm}}{2}\left[ {\underbrace {2\sin \left( {{\omega _c} + {\omega _m}} \right)t}_{upper\;side\;band}} \right]\)

Option (A) correct choice.

Short trick:

Observe the V_out in block diagram i.e. V_out = I₁ + I₃

But in I₃ there is –π/2 phase change so the SSB becomes in the Form of

S(t) = m(t) cos ω_ct – m̂(t) sin ω_ct

And (-) sign represent USB only.

So the student can tick the correct answer.

s₁(t) = A_c [1 + k_am(t) ] cosω_ct

s₂(t) = A_c [1 - k_am(t) ] cosω_ct

s(t) =  Ac [1 + kam(t) ] cosωct -  Ac [ 1 - kam(t) ] cosωct

= 2 A_c k_a m(t) cosωct

s(t) ∝ m(t) cosωct

Hence, Envelope of s(t) = |m(t)|