Maximum Phase Deviation MCQ Quiz - Objective Question with Answer for Maximum Phase Deviation - Download Free PDF

Explanation:

AM Detection Using PLL Circuit

Definition: Phase-Locked Loop (PLL) circuits are used for a variety of applications in communication systems, including frequency synthesis, demodulation, and signal synchronization. In the context of AM (Amplitude Modulation) detection, PLLs are used to demodulate the AM signal by locking onto the carrier frequency and extracting the modulating signal.

Working Principle: A PLL circuit consists of a phase detector, a low-pass filter, and a voltage-controlled oscillator (VCO). The phase detector compares the phase of the incoming AM signal with the phase of the VCO output. The resulting error signal is filtered and used to adjust the VCO frequency. When the PLL is locked, the VCO frequency matches the carrier frequency of the AM signal, and the error signal corresponds to the modulating signal (audio or data).

Advantages:

• Higher noise immunity compared to conventional peak detector type AM detectors.
• Precise demodulation of the AM signal even in the presence of noise and signal distortion.
• Ability to lock onto the carrier frequency, providing stable and accurate demodulation.

Disadvantages:

• Increased complexity and cost compared to simple peak detector circuits.
• Requires careful design and tuning to ensure proper locking and demodulation.

Applications: PLL-based AM detectors are used in high-fidelity radio receivers, communication systems, and other applications where accurate and reliable demodulation of AM signals is critical.

Correct Option Analysis:

The correct option is:

Option 4: Both S1 and S2

This option correctly identifies both key aspects of AM detection using PLL circuits:

• S1: It has higher noise immunity than the conventional peak detector type AM detector.
  This statement is true because PLL circuits can reject noise and maintain lock on the carrier frequency, leading to better demodulation performance in noisy environments.
• S2: The PLL is locked to the carrier frequency of the AM signal.
  This statement is also true because the fundamental operation of a PLL involves locking onto the carrier frequency of the input signal, ensuring accurate extraction of the modulating signal.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Only S1

While S1 is correct, this option is incomplete because it does not acknowledge the importance of S2, which is also a true statement about PLL-based AM detectors.

Option 2: Only S2

Similar to Option 1, this option is incomplete as it only considers S2 and ignores the validity of S1, which is an important aspect of the advantages of PLL-based AM detectors.

Option 3: Neither S1 nor S2

This option is incorrect because both S1 and S2 are true statements about PLL-based AM detectors. Ignoring both statements would lead to a misunderstanding of the advantages and operation of PLL circuits in AM detection.

Conclusion:

Understanding the role of PLL circuits in AM detection is crucial for recognizing their advantages over conventional peak detector circuits. PLLs offer higher noise immunity and precise locking to the carrier frequency, ensuring accurate demodulation of AM signals. Both S1 and S2 are correct statements that highlight these key benefits, making Option 4 the accurate choice.

The correct answer is: 4) Both S1 and S2

Explanation:

S1: If the input signal frequency changes, the PLL adjusts its frequency output to match the input frequency.

• Correct. This is a fundamental characteristic of a Phase-Locked Loop (PLL).
• In FSK demodulation, when the frequency of the input signal changes (representing binary data), the PLL dynamically adjusts its voltage-controlled oscillator (VCO) output frequency to lock onto the input signal.
   

S2: The phase-locked loop is used to track the changes in the frequency of the modulated signal.

• Correct. A PLL continuously tracks the instantaneous frequency of the input FSK signal and thus can decode the binary information.

Given:

f_c = 10⁵ Hz

x_EM(t) = 10 cos [ω_ct + 5 sin 3000πt + 10 sin 2000πt]

From the given equation:

A_c = 10

The carrier power is given by:

\({P_c} = \frac{{A_c^2}}{2} = \left( {\frac{{{{10}^2}}}{2}} \right) = 50\;W\)

Frequency deviation is given by:

\({\rm{\Delta }}f = \frac{1}{{2π }}\frac{{d\phi }}{{dt}}\)

\( = \frac{1}{{2π }}\frac{d}{{dt}}\left[ {5\sin 3000πt + 10\sin 2000π t} \right]\)

\({\rm{\Delta }}f = \frac{1}{{2π }}[5 \times 3000π\cos 3000πt + 10 \times 2000π \cos 2000π t]\)

\({\rm{\Delta }}{f_{max}} = \frac{1}{{2π }}\left[ {5 \times 3000π + 10 \times 2000π } \right]\)

\( = 17.5\;kHz\)

The maximum phase deviation is given by:

Δϕ_max = max[ϕ]

Δϕ_max = max[5 sin 3000πt + 10 sin 2000πt] = 15 rad

Concept:

A general angle modulated wave is defined as:

\(S\left( t \right) = {A_c}\cos \underbrace {({\omega _c}t + km\left( t \right)}_{\theta \left( t \right)}\)      ---(1)

The instantaneous phase is given as:

ϕ(t) = ω_ct + k m(t)

The phase deviation is:

ϕ(t) - ω_ct = k m(t)

The instantaneous frequency is given as:

\({\omega _i} = \frac{{d\theta \left( t \right)}}{{dt}}\) 

The frequency deviation is given as:

Δω = ω_i - ω_c

Calculation:

Comparing with (1)

Given:

A_c = 5

ω_c = 2π 10⁶

k = 4

m(t) = sin (2π 10³ t)

θ(t) = 2π 10⁶ t + 4 sin (2π 10³ t)

Instantaneous frequency \(= \frac{{d\theta \left( t \right)}}{{dt}}\) 

\( = \frac{d}{{dt}}\left( {2\pi {{10}^6}t + 4\sin \left( {2\pi {{10}^3}t} \right)} \right)\) 

= 2π 10⁶ + 8π × 10³ cos (2π 10³ t)

Instantaneous frequency at t = 0.5 ms

\(= 2\pi {10^6} + 8\pi {10^3}\cos \left( {2\pi {{10}^3} \times \frac{{{{10}^{ - 3}}}}{2}} \right)\) 

≈ 10⁶ Hz

The phase deviation = 4 sin (2π 10³t)

∴ Maximum phase deviation = 4 rad

The frequency deviation \(= \frac{{d\theta \left( t \right)}}{{dt}} - {\omega _c}\)  

= 8π 10³ cos (2π 10³ t)

Maximum frequency deviation = 4000 Hz.

Modulated signal maximum amplitude = 5

∴ Modulated signal power \(= \frac{{{{\left( 5 \right)}^2}}}{2} = 12.5\;W\)  

Concept:

A standard frequency-modulated wave is written in the form:

\({S_{FM}}\left( t \right) = A\;cos\left( {2\pi {f_c}t + 2\pi {k_f}\mathop \smallint \limits_0^t m\left( t \right).dt} \right)\)

where kf is in Hz/V

The phase deviation (PD) for an FM wave will be:

\(P.D.=2\pi k_f\smallint m\left( t \right)dt\)

A standard phase-modulated wave is written in the form:

\({S_{PM}}\left( t \right) = A\;cos\left( {2\pi {f_c}t + k_pm(t)} \right)\)

where kp is in rad/V

The phase deviation of PM wave will be:

PD = kp |m(t)| 

The instantaneous frequency for the PM modulated signal is given by:

\({{{f_{PM}}\left( t \right)}} = {{f_c} + {\frac{1}{2\pi}}.k_p.\frac{d}{{dt}}m\left( t \right)}\)

Also, the instantaneous frequency for the FM modulated signal is given by:

\({{{f_{FM}}\left( t \right)}} = {{{f_c} + k_f{m}\left( t \right)}}\)

Calculation:

Given k_p = 1 rad/V, the instantaneous frequency for PM wave will be:

\({{{f_{PM}}\left( t \right)}} = \left[ {{f_c} + \frac{k_p}{{2\pi }}\frac{d}{{dt}}m\left( t \right)} \right]\)

m(t) = t for 0 ≤ t ≤ 1

\((\frac{d}{{dt}}m\left( t \right))_{max}=1\)

∴ the maximum instantaneous frequency for PM will be:

\({\left( {{f_{PM}}} \right)_{max}} = {f_c} + \frac{k_p}{{2\pi }}\)   --- (1)

Now, the maximum instantaneous frequency for FM will be:

\({\left( {{f_{FM}}\left( t \right)} \right)_{max}} = {\left[ {{f_c} + k_f.{m(t)}} \right]_{max}}\)

With (m(t))_max = 1, the maximum instantaneous frequency for FM will be:

\({\left( {{f_{FM}}} \right)_{max}} = {f_c} + k_f\)     --- (2)

Now, when the maximum phase deviation in both cases is the same, we can write:

\(2\pi {K_f}\smallint m\left( t \right) = {k_p}{\left| {m\left( t \right)} \right|_{max}}\)

Modulus is taken because we are only concerned with the magnitude.

The maximum phase deviation for an FM wave will be 

\(2\pi k_f\times 1= {K_p}\left( 1 \right)\)

2π kf =  kp

Given similar carrier frequency is used in both FM and PM modulation

Instantaneous phase in PM is given by θ_i(t) = ω_ct + k_pm(t)

Instantaneous phase in FM is given by θ_i(t) = ω_ct + k_f

Maximum phase deviation in PM is:

Δθ = K_p x Max (m(t)) 

= K_p x 2

While in FM, it will be:

k_f x 2π x max (\( \mathop \smallint \limits_0^t m\left( t \right)dt\) )

= 2π k_f x 4

Given phase deviation for both modulations are equal, we can write:

8 π k_f = 2 K_p

k_p / k_f = 4π

The correct answer is: 4) Both S1 and S2

Explanation:

S1: If the input signal frequency changes, the PLL adjusts its frequency output to match the input frequency.

• Correct. This is a fundamental characteristic of a Phase-Locked Loop (PLL).
• In FSK demodulation, when the frequency of the input signal changes (representing binary data), the PLL dynamically adjusts its voltage-controlled oscillator (VCO) output frequency to lock onto the input signal.
   

S2: The phase-locked loop is used to track the changes in the frequency of the modulated signal.

• Correct. A PLL continuously tracks the instantaneous frequency of the input FSK signal and thus can decode the binary information.

Explanation:

AM Detection Using PLL Circuit

Definition: Phase-Locked Loop (PLL) circuits are used for a variety of applications in communication systems, including frequency synthesis, demodulation, and signal synchronization. In the context of AM (Amplitude Modulation) detection, PLLs are used to demodulate the AM signal by locking onto the carrier frequency and extracting the modulating signal.

Working Principle: A PLL circuit consists of a phase detector, a low-pass filter, and a voltage-controlled oscillator (VCO). The phase detector compares the phase of the incoming AM signal with the phase of the VCO output. The resulting error signal is filtered and used to adjust the VCO frequency. When the PLL is locked, the VCO frequency matches the carrier frequency of the AM signal, and the error signal corresponds to the modulating signal (audio or data).

Advantages:

• Higher noise immunity compared to conventional peak detector type AM detectors.
• Precise demodulation of the AM signal even in the presence of noise and signal distortion.
• Ability to lock onto the carrier frequency, providing stable and accurate demodulation.

Disadvantages:

• Increased complexity and cost compared to simple peak detector circuits.
• Requires careful design and tuning to ensure proper locking and demodulation.

Applications: PLL-based AM detectors are used in high-fidelity radio receivers, communication systems, and other applications where accurate and reliable demodulation of AM signals is critical.

Correct Option Analysis:

The correct option is:

Option 4: Both S1 and S2

This option correctly identifies both key aspects of AM detection using PLL circuits:

• S1: It has higher noise immunity than the conventional peak detector type AM detector.
  This statement is true because PLL circuits can reject noise and maintain lock on the carrier frequency, leading to better demodulation performance in noisy environments.
• S2: The PLL is locked to the carrier frequency of the AM signal.
  This statement is also true because the fundamental operation of a PLL involves locking onto the carrier frequency of the input signal, ensuring accurate extraction of the modulating signal.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Only S1

While S1 is correct, this option is incomplete because it does not acknowledge the importance of S2, which is also a true statement about PLL-based AM detectors.

Option 2: Only S2

Similar to Option 1, this option is incomplete as it only considers S2 and ignores the validity of S1, which is an important aspect of the advantages of PLL-based AM detectors.

Option 3: Neither S1 nor S2

This option is incorrect because both S1 and S2 are true statements about PLL-based AM detectors. Ignoring both statements would lead to a misunderstanding of the advantages and operation of PLL circuits in AM detection.

Conclusion:

Understanding the role of PLL circuits in AM detection is crucial for recognizing their advantages over conventional peak detector circuits. PLLs offer higher noise immunity and precise locking to the carrier frequency, ensuring accurate demodulation of AM signals. Both S1 and S2 are correct statements that highlight these key benefits, making Option 4 the accurate choice.

Concept:

A general angle modulated wave is defined as:

S(t) = A_c cos (ω_c t + km(t))   ---(1)

The instantaneous phase is given as:

θ_i(t) = ωc t + km(t)

The phase deviation is:

θ(t) = km(t)

and the frequency deviation is:

\({\rm{\Delta }}f = \frac{1}{{2\pi }}k\frac{d}{{dt}}\left( {m\left( t \right)} \right)\)

Application:

Given carrier frequency = 1 kHz, i.e.

ω_c = 2π × 1000 rad/sec

The output signal is given as:

x(t) = cos (2π 1100t)

This can be written as:  

x(t) = cos (2π 1000t + 2π 100t)

Comparing this with the standard angle modulated wave of Equation (1), we get the phase deviation as:

Δθ = θ(t) = 200 πt

And the frequency deviation will be:

\({\rm{\Delta }}f = \frac{1}{{2\pi }} \cdot \frac{{dθ \left( t \right)}}{{dt}}\)

\( = \frac{1}{{2\pi }} \times 200\pi \)

Concept:

The general expression of an FM wave is given by:

\(S_{FM}(t)={A_C}\cos \left( {{\omega _c}t +2π k_f\smallint m\left( t \right)dt} \right)\)

The instantaneous phase of the FM wave is given by:

\(ϕ_i(t)={\omega _c}t +2π k_f\smallint m\left( t \right)dt\)

Now, the phase deviation of a frequency modulated wave will be given by:

\(\Delta \phi _i =2π k_f\smallint m\left( t \right)dt\)

The maximum phase deviation will be:

\(|\Delta \phi _i|_{max} =|2π k_f\smallint m\left( t \right)dt|_{max}\)

Calculation:

Given:

kf = 0.5 Hz/V, 

Since m(t) has only positive values,

\({\left| {\mathop \smallint \nolimits_{ - \infty }^t m\left( t \right)dt} \right|_{max}}\) = Area under m(t)

= 4 × 2 = 8 V-sec

From equation (1)

\({\rm{\Delta }}{\phi _{max}} = 2\pi {k_f}{\left| {\mathop \smallint \nolimits_{ - \infty }^t m\left( t \right)dt} \right|_{max}}\)

= 2π (0.5 Hz/V)(8 V-sec) rad

= 2π (0.5) (8) rad

= 8π rad

Concept:

The general expression of an FM wave is given by:

\(S_{FM}(t)={A_C}\cos \left( {{\omega _c}t +2π k_f\smallint m\left( t \right)dt} \right)\)

The instantaneous phase of the FM wave is given by:

\(ϕ_i(t)={\omega _c}t +2π k_f\smallint m\left( t \right)dt\)

Now, the phase deviation of a frequency modulated wave will be given by:

\(\Delta \phi _i =2π k_f\smallint m\left( t \right)dt\)

The maximum phase deviation will be:

\(|\Delta \phi _i|_{max} =|2π k_f\smallint m\left( t \right)dt|_{max}\)

Also, the general expression for a PM wave is given by:

\(S_{PM}(t)={A_c}\cos \left( {{\omega _c}t + {K_p}m\left( t \right)} \right)\)

The instantaneous phase is given by:

\(ϕ_i(t)={\omega _c}t + {K_p}m( t)\)

Now, the phase deviation of a phase-modulated wave will be given by:

\(\Delta \phi _i =k_pm(t)\)

The maximum phase deviation will be:

\(|\Delta \phi _i|_{max} =|k_p m(t)|_{max}\)

Calculation:

The maximum phase deviation in both cases is the same, i.e.

\(|k_p m(t)|_{max}=|2π k_f\smallint m\left( t \right)dt|_{max}\)   ---(1)

From the given signal m(t), the maximum value is 1 and the maximum value of the integration of m(t) will be the area from 0 to 2 (∵ After t = 2, the negative area will be added), i.e.

Area covered by m(t) from 0 to 2 will be:

\(A=\frac{1}{2}\times 1\times 1+1\times 1=\frac{3}{2}\)

Substituting this in equation (1), we get:

\(k_p(1)=2\pi k_f(\frac{3}{2})\)

\(k_p=3\pi k_f\)

Given similar carrier frequency is used in both FM and PM modulation

Instantaneous phase in PM is given by θ_i(t) = ω_ct + k_pm(t)

Instantaneous phase in FM is given by θ_i(t) = ω_ct + k_f

Maximum phase deviation in PM is:

Δθ = K_p x Max (m(t)) 

= K_p x 2

While in FM, it will be:

k_f x 2π x max (\( \mathop \smallint \limits_0^t m\left( t \right)dt\) )

= 2π k_f x 4

Given phase deviation for both modulations are equal, we can write:

8 π k_f = 2 K_p

k_p / k_f = 4π

We have 

\({\rm{s}}\left( {\rm{t}} \right){\rm{}} = {\rm{cos}}\left[ {2{\rm{\pi }}\left( {2 \times {{10}^6}{\rm{t}} + 30{\rm{sin}}\left( {50{\rm{t}}} \right){\rm{}} + {\rm{}}40{\rm{cos}}\left( {50{\rm{t}}} \right)} \right)} \right]\)

we have,

\(\theta (t)= {4\pi \times {{10}^6}{\rm{t}} + 60\pi{\rm{sin}}\left( {50{\rm{t}}} \right){\rm{}} + {\rm{}}80\pi{\rm{cos}}\left( {50{\rm{t}}} \right)}\)

comparing with \(\theta (t)=\omega t+\phi(t)\) , we have

\(\phi(t)= 60\pi{\rm{sin}}\left( {50{\rm{t}}} \right){\rm{}} + {\rm{}}80\pi{\rm{cos}}\left( {50{\rm{t}}} \right)\)

\(\Rightarrow\phi(t)= 100\pi\left({{60\pi\over100\pi}{\rm{sin}}\left( {50{\rm{t}}} \right){\rm{}} + {\rm{}}{80\pi\over100\pi}{\rm{cos}}\left( {50{\rm{t}}} \right)}\right)\)

\(\Rightarrow\phi(t)= 100\pi\left({\sin{(\varphi)}{\rm{sin}}\left( {50{\rm{t}}} \right){\rm{}} + {\rm{}}{\cos(\varphi)}{\rm{cos}}\left( {50{\rm{t}}} \right)}\right)\)

where \(\sin{\varphi}={60\pi\over100\pi}\) and \(\cos{\varphi}={80\pi\over100\pi}\)

Thus, \( \phi(t)= 100\pi\cdot\cos\left(50t-\varphi\right)\)

Hence, maximum phase deviation \(\phi(t)|_{max}\) is given by

\(\phi(t)|_{max}=100\pi\)

The correct answer is: 4) Both S1 and S2

Explanation:

S1: If the input signal frequency changes, the PLL adjusts its frequency output to match the input frequency.

• Correct. This is a fundamental characteristic of a Phase-Locked Loop (PLL).
• In FSK demodulation, when the frequency of the input signal changes (representing binary data), the PLL dynamically adjusts its voltage-controlled oscillator (VCO) output frequency to lock onto the input signal.
   

S2: The phase-locked loop is used to track the changes in the frequency of the modulated signal.

• Correct. A PLL continuously tracks the instantaneous frequency of the input FSK signal and thus can decode the binary information.

The general form of a phase modulated signal is, \(\rm{x_{PM} (t) = A cos(ω_ct + \beta m(t))}\) 

Comparing it with, \(\rm{x_{PM}(t) = 10 cos(ω_ct + 3 sinω_mt)}\) we have

\(∴ β = 3\)

when \(\rm{f_m}\) is doubled then \(\rm{f_m=2\ kHz}\).

Then, bandwidth \(\rm{f_B =2\left(\beta+1\right)f_m}\)

\(\rm{\Rightarrow f_B= 2 (3 + 1)2 = 16 KHz}\)