Pulse Modulation MCQ Quiz - Objective Question with Answer for Pulse Modulation - Download Free PDF

The SNR in a binary PCM system can be expressed as:

SNR = (6 × n) dB

where n is the number of bits used for quantization. This indicates that the SNR increases linearly with the number of bits used in the PCM system.

Effect of Increasing Number of Bits:

When the number of bits in the PCM system is increased from n to n + 2, the SNR also increases. The improvement in SNR can be calculated using the difference in SNR values before and after increasing the number of bits:

Improvement in SNR = SNR (n + 2) - SNR (n)

Substituting the values:

Improvement in SNR = (6 × (n + 2)) - (6 × n)

Improvement in SNR = 6 × n + 12 - 6 × n

Improvement in SNR = 12 dB

Hence the correct answer is 12 dB

Minimum Permissible Bit Rate Calculation:

Given Data:

• Bandwidth (BW) of the audio signal = 15 kHz
• Number of quantization levels = 1024
• No overheads in PCM encoding

Step 1: Determine the number of bits per sample

The number of quantization levels (L) is given as 1024. The number of bits required to represent each quantization level can be calculated using the formula:

Number of bits per sample = log₂(L)

Here, log₂(1024) = 10. Therefore, each sample requires 10 bits to represent the quantized value.

Step 2: Apply Nyquist Theorem to determine the sampling rate

According to Nyquist theorem, the minimum sampling rate (f_s) should be at least twice the bandwidth of the signal:

f_s = 2 × BW

Given BW = 15 kHz:

f_s = 2 × 15 kHz = 30 kHz

Step 3: Calculate the minimum permissible bit rate

The bit rate (R) is given by:

R = Sampling Rate × Number of bits per sample

Substituting the values:

R = 30 kHz × 10 bits/sample = 300 kbps

Therefore, the minimum permissible bit rate for the given communication system is 300 kbps.

Correct Option Analysis:

The correct option is:

Option 2) 300 kbps

This option correctly represents the minimum permissible bit rate for the given communication system using PCM encoding, considering the bandwidth, quantization levels, and sampling rate.

Explanation:

Digital Modulation Technique

Definition: Digital modulation refers to the process of modifying a carrier signal to encode digital information. This is achieved by altering the carrier's properties such as amplitude, frequency, or phase. Digital modulation techniques are fundamental in modern communication systems, as they enable efficient and reliable transmission of digital data over various communication channels.

Phase Shift Keying (PSK):

Definition: Phase Shift Keying (PSK) is a digital modulation technique where the phase of the carrier signal is varied to represent digital data. Each unique phase state corresponds to a specific binary or multi-bit symbol, allowing the transmission of information.

Working Principle: In PSK, the carrier signal's phase is shifted to encode digital information. For example:

• BPSK (Binary Phase Shift Keying): Two distinct phase states (e.g., 0° and 180°) are used to represent binary data (0 and 1).
• QPSK (Quadrature Phase Shift Keying): Four phase states (e.g., 0°, 90°, 180°, and 270°) are used, with each phase state representing two bits of data.
• Higher-Order PSK: More phase states (e.g., 8-PSK, 16-PSK) are used to represent more bits per symbol, increasing the data transmission rate.

Advantages of PSK:

• High spectral efficiency, allowing more data to be transmitted within a given bandwidth.
• Resilience to noise and interference, especially in low signal-to-noise ratio (SNR) environments.
• Widely used in modern digital communication systems, including wireless networks, satellite communications, and optical fiber systems.

Disadvantages of PSK:

• Complexity increases with higher-order PSK, requiring sophisticated demodulation techniques.
• Performance can degrade in the presence of severe phase noise or synchronization issues.

Applications: PSK is extensively used in various digital communication systems, including Wi-Fi, cellular networks, and satellite communications, due to its efficiency and robustness.

Correct Option Analysis:

The correct option is:

Option 3: Phase Shift Keying (PSK)

This option correctly identifies PSK as an example of a digital modulation technique. PSK is a widely used method for encoding digital information by varying the phase of the carrier signal, making it a fundamental concept in digital communication systems.

Important Information

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

Option 1: Single Sideband Modulation (SSB)

SSB is a type of analog modulation technique used in amplitude modulation (AM) systems. It involves transmitting only one sideband (either the upper or lower sideband) of the modulated signal while suppressing the carrier and the other sideband. SSB is primarily used in analog communication systems, such as amateur radio and long-distance voice communication, and is not an example of a digital modulation technique.

Option 2: Narrow Band FM

Narrow Band FM is another analog modulation technique where the frequency of the carrier signal is varied in proportion to the modulating signal. It is used in applications such as two-way radio communication and public safety communication. Similar to SSB, it is not a digital modulation technique.

Option 4: Double Sideband Amplitude Modulation

This is a classical analog modulation method where both the upper and lower sidebands are transmitted along with the carrier signal. It is commonly used in AM radio broadcasting but is not related to digital modulation techniques.

Conclusion:

Understanding the distinction between analog and digital modulation techniques is crucial for identifying the correct option. Among the options provided, only Phase Shift Keying (PSK) is a digital modulation technique, as it involves encoding digital information by varying the phase of the carrier signal. The other options, including Single Sideband Modulation (SSB), Narrow Band FM, and Double Sideband Amplitude Modulation, are examples of analog modulation techniques, which are fundamentally different from digital modulation.

Concept:

The quantization noise power P_q for a uniform quantizer is given by:

P_q = $\frac{{\mathrm{\Delta }}^2}{12}$

where $\mathrm{\Delta }$ is the quantization step size.

Given:

Full-scale input voltage = 2B

Number of quantization levels = 2^N

Calculation:

Quantization step size:

$\mathrm{\Delta }=\frac{2B}{2^N}=\frac{B}{2^{N-1}}$

Quantization noise power:

$\frac{{\mathrm{\Delta }}^2}{12}=\frac{1}{12}\cdot {\left(\frac{B}{2^{N-1}}\right)}^2=\frac{B^2}{12\cdot 2^{2(N-1)}}=\frac{B^2}{3\cdot 2^{2N}}$

Hence, the correct answer is option 1) $\frac{{\mathrm{B}}^2}{3\left(2^{2\mathrm{N}}\right)}$

Explanation:

Signaling Rate of Time Division Multiplexing (TDM) Signal

Definition: Time Division Multiplexing (TDM) is a digital multiplexing technique used to combine multiple data streams or signals into one stream by assigning each signal a different time slot in the sequence. The signaling rate of a TDM signal refers to the rate at which the multiplexed data is transmitted over the communication channel. This rate depends on the number of input channels (M), the sampling frequency (fs), and the bandwidth of the input signals (W).

Explanation of the Correct Option:

The correct option is:

Option 1: r = Mfs ≥ 2MW

In a TDM system, each input channel has a specific sampling rate (fs). According to the Nyquist sampling theorem, the minimum sampling rate for any signal must be at least twice its highest frequency component (2W) to ensure accurate signal reconstruction. Thus, for M input channels, the total signaling rate (r) must satisfy the condition:

r = M × fs, where fs ≥ 2W

Therefore:

r = M × fs ≥ 2M × W

This equation ensures that the TDM system operates effectively without aliasing and that all the input signals are sampled and transmitted correctly. The signaling rate must be at least 2M × W to accommodate all M channels, as each channel requires a bandwidth of W and a sampling rate of at least 2W.

Key Points:

• M: Number of input channels in the TDM system.
• fs: Sampling frequency of each channel, which must be at least twice the bandwidth (2W) to satisfy the Nyquist criterion.
• W: Bandwidth of each input channel.
• The total signaling rate (r) ensures that all channels are multiplexed and transmitted without loss of information.

Conclusion: The signaling rate of a TDM signal for M input channels is given by r = Mfs ≥ 2MW, as stated in Option 1. This ensures that the system adheres to the Nyquist sampling theorem and can effectively transmit all input signals.

Additional Information:

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

Option 2: r = Mfs ≤ 2MW

This option is incorrect because the signaling rate cannot be less than or equal to 2MW. The condition r = Mfs ≥ 2MW ensures that the Nyquist criterion is satisfied and that all input signals are accurately sampled and transmitted. If r ≤ 2MW, the system would not meet the minimum requirements for sampling and could result in aliasing or loss of information.

Option 3: r = Mfs ≤ MW

This option is incorrect as it violates the Nyquist sampling theorem. If the signaling rate is less than or equal to MW, the sampling frequency (fs) for each channel would be less than 2W, leading to aliasing and inaccurate signal reconstruction. Such a scenario is not feasible for a TDM system.

Option 4: r = Mfs ≥ MW

While this condition might seem plausible at first glance, it is insufficient to ensure accurate signal reconstruction for all channels. The Nyquist theorem requires the sampling frequency to be at least twice the bandwidth (fs ≥ 2W). Therefore, the signaling rate must satisfy r = Mfs ≥ 2MW, not just MW, to avoid aliasing and ensure proper operation.

Conclusion:

Option 1 (r = Mfs ≥ 2MW) is the correct choice as it ensures that the TDM system adheres to the Nyquist criterion, allowing accurate sampling and transmission of all input signals. The other options fail to meet this critical requirement and would result in improper operation of the TDM system.

In TV transmission:

• A video is Vestigial sideband modulated, which is a type of amplitude modulated waveform
• The Video signals are thus encoded in amplitude variations of the carriers
• The Audio signal is encoded in FM waveform
• Thus, the audio signals are encoded as frequency variations of the carrier

Noise is the signal that affects amplitude majorly. Thus, the video signal is distorted from amplitude variations.

Quantization: 

• It is the process through which a range of continuous analog values are quantized or rounded off to a single value, thereby forming samples of a discrete digital signal.
• Quantization Error occurs when there is a difference between an input value and it’s quantized value. 
• Quantization occurs when an analog signal is converted into it’s digital form, thus it occurs in Pulse Code modulation (PCM).

PCM:

• PCM stands for Pulse Code Modulation.
• It is a technique by which an analog signal gets converted into digital form to have signal transmission through a digital network.

• The major steps involved in PCM are sampling, quantizing, and encoding.

• With PCM, the amplitude of the analog signal is sampled at regular intervals and translated into a binary number.
• The difference between the original signal and the translated digital signal is called the quantizing error.

Some Advantages associated with PCM are:

• Immunity to transmission noise and interference.
• It is possible to regenerate the coded signal along the transmission path.
• The Quantization Noise depends on the number of quantization levels and not on the number of samples produced per second.
• The storage of Coded signals is easy.

The disadvantages of PCM includes:

• Requires larger Bandwidth.
• Need synchronization
• Not compatible with analog systems.

PCM:

• PCM stands for Pulse Code Modulation.
• With PCM, the amplitude of the analog signal is sampled at regular intervals and translated into a binary number.
• The difference between the original signal and the translated digital signal is called the quantizing error.

Some Advantages associated with PCM are:

• Immunity to transmission noise and interference.
• It is possible to regenerate the coded signal along the transmission path.
• The Quantization Noise depends on the number of quantization levels and not on the number of samples produced per second.
• The storage of Coded signals is easy.

The disadvantages of PCM includes:

• Requires larger Bandwidth.
• Need synchronization
• Not compatible with analog systems.

Advantages of FM over AM are:

• Improved signal to noise ratio.
• Smaller geographical interference between neighbouring stations.
• Less radiated power.
• Well defined service areas for given transmitter power.

Disadvantages of FM:

• Much more Bandwidth
• More complicated receiver and transmitter.

Concept:

The number of levels for an n-bit PCM system is given by:

L = 2n

We can also state that the number of bits for a given quantization level will be:

n = log2 L

Calculation:

The number of levels given = 4096, i.e. L = 4096

The number of bits used will be:

n = log2 (4096)

= log2 (2¹²)

= 12 log2 (2)

n = 1

The bandwidth of PCM is given by:

$BW=n{f}_s$

n = number of bits to encode

fs = sampling frequency

Concept:

The number of levels for an n-bit PCM system is given by:

L = 2n

We can also state that the number of bits for a given quantization level will be:

n = log₂ L

Also, the bandwidth of PCM is given by:

$BW=n{f}_s$

n = number of bits to encode

fs = sampling frequency

Calculation:

For L = 4 quantization levels, the number of bits n = log2 4 = 2 bits. The bandwidth is, therefore:

B.W. = 2 fs

Similarly, For L = 64 quantization levels, the number of bits n = log2 64 = 6 bits. The bandwidth is, therefore:

B.W. = 6 fs

Clearly, the Bandwidth is increased by 3 times.

The AM, FM and PM output waveforms are as shown:

In Frequency modulation, the frequency of the carrier is varied according to the amplitude of the message signal and the amplitude of the carrier remains constant.

Concept:

Two types of distortions (Errors) occurs in Delta modulation system i.e.

1) Slope overload error

2) Granular Error

This is explained with the help of the given figure:

To avoid the slope overload error, the optimum or desired condition is:

$\frac{\mathrm{\Delta }}{T_s}=\frac{dm\left(t\right)}{dt}$

i.e. the step rise of the quantized output must follow the input

Slope overload occurs when:

$\frac{\mathrm{\Delta }}{T_s}<\frac{dm\left(t\right)}{dt}$

To prevent/avoid slope overload error, the condition that shall be satisfied is:

$\frac{dm\left(t\right)}{dt}\le \frac{\mathrm{\Delta }}{T_s}$

m(t) is a sinusoidal waveform given as:

m(t) = A_m sin ω_m t

The condition to avoid slope overload, therefore, becomes:

$\frac{d}{dt}(A_m\sin {\omega }_{m}t)\le \frac{\mathrm{\Delta }}{T_s}$

$A_m\cos {\omega }_{m}t\left({\omega }_m\right)\le \frac{\mathrm{\Delta }}{T_s}$

A_m cos (2π (mt)).2πf_m ≤ Δ⋅f_s

$A_m\le \frac{\mathrm{\Delta }{f}_s}{2\pi f_m\cdot \cos \left(2\pi f_{m}f\right)}$

For maximum Amplitude, the above condition becomes:

$A_m\le \frac{\mathrm{\Delta }\cdot f_s}{2\pi f_m}$

• PCM (Pulse Code Modulation) is a digital scheme for transmitting analog data.
• The amplitude of an analog signal can take any value over a continuous range, i.e. it can take on infinite values.
• But, digital signal amplitude can take on finite values.
• Analog signals can be converted into digital by sampling and quantizing.

Advantages of PCM:

• Encoding is possible in PCM.
• Very high noise immunity, i.e. better performance in the presence of noise.
• Convenient for long-distance communication.
• Good signal to noise ratio.

Disadvantage of PCM:

• The circuitry is complex.
• It requires a large bandwidth.
• Synchronization is required between transmitter and receiver.

Concept:

Slope overload distortion: When the maximum slope of the message is more than Δ f_s then there is slope overload distortion in the delta modulator.

condition to prevent slope overload distortion 

$|{\displaystyle \frac{dm(t)}{dt}}{|}_{MAX}\le |\Delta \times f_s|$ ,

where m(t)→ message signal, f_s→ sampling frequency

Calculation:

Given f_m = 10 kHz

Input is 1 V_pp so the maximum  input voltage is 0.5 volt

let applied input is sinusoidal than

m(t) = A_msin(2πf_mt)

where 

$|\frac{dm(t)}{dt}|=2\pi f_m{A}_{m}cos(2\pi f_{m}t)$

$|\frac{dm(t)}{dt}{|}_{MAX}=2\pi f_m{A}_m$

f_s = 10× (Nyquist rate)

f_s =10×(2f_m)= 20(10k)=200kHZ

 By the condition to prevent slope overload distortion 

$|{\displaystyle \frac{dm(t)}{dt}}{|}_{MAX}\le |\Delta \times f_s|$

2πA_mf_m ≤ Δ (200k)

2π(0.5)(10k) ≤ Δ (200k)

Δ ≥ 0.157

Δ _minimum = 0.157 V