Baseband Modulation MCQ Quiz - Objective Question with Answer for Baseband Modulation - Download Free PDF
Last updated on May 30, 2025
Latest Baseband Modulation MCQ Objective Questions
Baseband Modulation Question 1:
The roll-off factor (β:beta) in raised cosine filter represents the ratio of:
Answer (Detailed Solution Below)
Baseband Modulation Question 1 Detailed Solution
Concept:
The roll-off factor (β) in a raised cosine filter represents the ratio of:
Correct Option Explanation:
The correct answer is option 4: excess bandwidth to the Nyquist bandwidth.
The raised cosine filter is a type of filter used in digital communication systems to shape the transmitted signal's spectrum, ensuring that intersymbol interference (ISI) is minimized. The roll-off factor, β, is a critical parameter in the design of the raised cosine filter. It determines the excess bandwidth used by the filter beyond the minimum required Nyquist bandwidth.
The Nyquist bandwidth is the minimum bandwidth required to transmit data without ISI, given by the symbol rate (Rs). However, practical filters cannot achieve a perfect rectangular frequency response, so additional bandwidth is introduced. This additional bandwidth is characterized by the roll-off factor (β).
Mathematically, the total bandwidth (B) of a raised cosine filter is given by:
B = Rs × (1 + β)
Where:
- Rs is the symbol rate (also known as the Nyquist rate).
- β is the roll-off factor, which ranges from 0 to 1.
The roll-off factor (β) represents the ratio of the excess bandwidth to the Nyquist bandwidth. In other words, it indicates how much additional bandwidth is introduced by the filter relative to the minimum required Nyquist bandwidth.
Additional Information
Option 1: Excess power of system to the minimum Nyquist bandwidth
This option is incorrect because the roll-off factor (β) does not represent a ratio involving power. The roll-off factor is strictly related to bandwidth and not power. It describes the additional bandwidth used by the filter beyond the Nyquist bandwidth, not the excess power of the system.
Option 2: Additional bandwidth to total system bandwidth
This option is incorrect because the roll-off factor (β) specifically represents the ratio of excess bandwidth to the Nyquist bandwidth, not the total system bandwidth. The total system bandwidth includes the Nyquist bandwidth and the excess bandwidth. Therefore, this option does not accurately describe the roll-off factor.
Option 3: Total available bandwidth to the Nyquist bandwidth
This option is incorrect because the roll-off factor (β) represents the ratio of excess bandwidth to the Nyquist bandwidth, not the total available bandwidth. The total available bandwidth is the sum of the Nyquist bandwidth and the excess bandwidth, so this option does not accurately describe the roll-off factor.
Baseband Modulation Question 2:
In the communication of the PPM method, the information is encoded in the:
Answer (Detailed Solution Below)
Baseband Modulation Question 2 Detailed Solution
Concept:
Pulse Position Modulation (PPM)
- Pulse Position Modulation (PPM) is a form of signal modulation where the message information is encoded in the timing of a series of signal pulses.
- In PPM, the position of each pulse, relative to the position of a clock pulse, is varied according to the sampled value of the message signal.
- PPM is used in applications where the presence of the signal is more important than the amplitude or width, such as in optical communication systems.
Explanation of the Correct Option:
The correct option is Option 1: position of the pulse.
In Pulse Position Modulation (PPM), the information is encoded in the position of the pulse. This means that the time at which the pulse occurs is varied to represent the information being sent. Here’s a detailed explanation:
1. PPM Basics: PPM involves the transmission of pulses at different positions within a given time frame. The exact position of each pulse within the frame represents the data being communicated. This method is particularly useful in environments where signal timing can be precisely controlled and detected.
2. How It Works: In PPM, a reference clock or synchronization pulse is used as a baseline. The position of each pulse relative to this reference is adjusted to convey the information. For example, if we are encoding digital data, a pulse might be shifted to one position for a binary '0' and another position for a binary '1'. In analog applications, the pulse position might vary continuously to represent the amplitude of the analog signal.
3. Advantages of PPM: One of the main advantages of PPM is its robustness to amplitude variations and noise. Since the information is conveyed through the timing of the pulses rather than their amplitude or width, PPM signals are less affected by amplitude noise and can be more easily detected in noisy environments. This makes PPM particularly suitable for optical communication systems, where signal strength can vary due to atmospheric conditions.
4. Applications: PPM is widely used in optical communication systems, such as fiber-optic communication and free-space optical communication, where the precise timing of light pulses can be accurately controlled and detected. It is also used in some radio communication systems and in telemetry for encoding sensor data.
5. Implementation: Implementing PPM requires precise timing control and detection mechanisms. The transmitter must be able to generate pulses at exact positions within the time frame, and the receiver must be able to accurately measure the arrival time of each pulse to decode the information. This often involves the use of high-speed clocks and synchronization techniques to ensure accurate timing.
Overall, the position of the pulse is the key parameter in PPM that carries the information, making Option 1 the correct choice.
Important Information
To analyze why the other options are incorrect, let’s look at each one in detail:
Option 2: power of the pulse
- In PPM, the power of the pulse does not convey the information. The power of the pulse remains constant, and it is the timing or position of the pulse that changes to encode the data.
- Power modulation is not a characteristic of PPM; rather, it is used in other modulation schemes such as Pulse Amplitude Modulation (PAM), where the amplitude (and therefore power) of the pulse is varied.
Option 3: amplitude of the pulse
- In PPM, the amplitude of the pulse is not used to encode information. The amplitude of the pulses remains constant, and it is their position within the time frame that varies to represent the data.
- Amplitude modulation is characteristic of schemes like PAM, where the amplitude of the pulses is varied according to the signal being transmitted.
Option 4: width of the pulse
- In PPM, the width of the pulse does not carry the information. The width of the pulses remains constant, and it is their position in time that is modulated.
- Pulse Width Modulation (PWM) is a different modulation scheme where the width of the pulses is varied to represent the signal. In PPM, however, the pulse width is not relevant to the encoding of information.
In conclusion, Pulse Position Modulation (PPM) encodes information in the position of the pulses, making Option 1 the correct choice. The other options—power, amplitude, and width of the pulse—are characteristics of different modulation schemes and do not apply to PPM.
Baseband Modulation Question 3:
The bandwidth of a raised cosine filter with roll-off factor β (0 to 1) and symbol rate RS is given by:
Answer (Detailed Solution Below)
Baseband Modulation Question 3 Detailed Solution
Concept:
A Raised Cosine Filter is a type of filter used in digital communication systems to shape the transmitted signal spectrum and reduce intersymbol interference (ISI).
The filter is characterized by a roll-off factor \( \beta \) and a symbol rate Rs.
Definition:
The roll-off factor \( \beta \) defines how much excess bandwidth is used beyond the Nyquist bandwidth (Rs/2)
The roll-off factor lies between \( 0 \leq \beta \leq 1 \), where:
- β = 0: Ideal brick-wall filter (practically unrealizable)
- β = 1: Maximum bandwidth usage (100% excess)
Formula for Bandwidth:
The total bandwidth (BW) of a raised cosine filter is given by:
\( BW = \frac{R_s}{2}(1 + \beta) \)
Where:
- \( R_s \) = symbol rate
- \( \beta \) = roll-off factor
Baseband Modulation Question 4:
If the number bits per sample in a PCM system is increased from 16 to 17, the improvement in signal to quantization noise ratio will be :
Answer (Detailed Solution Below)
Baseband Modulation Question 4 Detailed Solution
Explanation:
Improvement in Signal to Quantization Noise Ratio in PCM System
Definition: In a Pulse Code Modulation (PCM) system, the signal to quantization noise ratio (SQNR) is a measure of the quality of the digitized signal. It compares the power of the original signal to the power of the quantization noise, which is introduced during the digitization process.
Quantization Noise: Quantization noise is the error introduced when an analog signal is quantized to a finite number of levels. This noise is inherent in the process of converting an analog signal to a digital signal and is a function of the number of bits used in the quantization process.
Key Formula: The SQNR in a PCM system can be approximated using the following formula:
SQNR (dB) ≈ 6.02N + 1.76
where N is the number of bits per sample.
Analysis:
Given the initial number of bits per sample (N1) is 16, the initial SQNR (SQNR1) can be calculated as follows:
SQNR1 (dB) ≈ 6.02 × 16 + 1.76 = 96.32 dB
When the number of bits per sample is increased to 17 (N2), the new SQNR (SQNR2) can be calculated as:
SQNR2 (dB) ≈ 6.02 × 17 + 1.76 = 102.34 dB
The improvement in SQNR is the difference between the new SQNR and the initial SQNR:
Improvement in SQNR (dB) = SQNR2 - SQNR1
Improvement in SQNR (dB) = 102.34 dB - 96.32 dB = 6.02 dB
However, since we are looking for an approximate improvement, we can round this value to the nearest whole number, which gives us:
Improvement in SQNR ≈ 6 dB
Correct Option:
The correct option is:
Option 3: 6 dB
This option reflects the fact that increasing the number of bits per sample by 1 in a PCM system results in an improvement of approximately 6 dB in the signal to quantization noise ratio.
Baseband Modulation Question 5:
In differential encoding the __________ difference between two waveforms is measured.
Answer (Detailed Solution Below)
Baseband Modulation Question 5 Detailed Solution
The correct answer is: 3) Phase
Explanation:
In differential encoding, the phase difference between successive signal elements is used to encode the data. This technique is often used in digital modulation schemes like Differential Phase Shift Keying (DPSK), where information is represented by changes in phase rather than absolute phase values.
-
It does not depend on magnitude, frequency, or period, but on how the phase changes from one symbol to the next.
Hence, the blank should be filled with phase
Top Baseband Modulation MCQ Objective Questions
In digital transmission, the modulation technique that requires the minimum bandwidth is:
Answer (Detailed Solution Below)
Baseband Modulation Question 6 Detailed Solution
Download Solution PDF- In PCM an analog signal is sampled and encoded into different levels before transmission
- The bandwidth of PCM depends on the number of levels
- If each sample is encoded into n bits, then the bandwidth of PCM is nfs
- The bandwidth of DPCM is almost the same as that of PCM signal, the only difference between PCM and DPCM is that the dynamic range is reduced in the DPCM signal
- However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ
- Hence there is bandwidth saving in Delta modulation
A comparison of different modulation schemes is as shown in the table below:
Parameter |
PCM |
DM |
DPCM |
Number of bits |
It can use 4, 8 or 16 bits per sample |
It uses only one bit for one sample |
Bits can be more than one but are less than PCM |
Level/Step size |
Step size is fixed |
Step size is fixed and cannot be varied |
A fixed number of levels are used. |
Quantiz-ation error or Distortion |
Quanti-zation error depends on the number of levels used |
Slope overload distortion and granular noise is present |
Slope overload distortion and quantization noise is present |
Bandwidth of the transmi-ssion channel |
Highest bandwidth is required since the number of bits is high |
The lowest band-width is required |
The bandwidth required is lower than PCM |
Signal to Noise ratio |
Good |
Poor |
Fair |
Area of Applic-ation |
Audio and Video Telephony |
Speech and images |
Speech and video |
An analog voltage is in the range of 0 to 8 V is divided in eight equal intervals for conversion to 3-bit digital output. The maximum quantization error is
Answer (Detailed Solution Below)
Baseband Modulation Question 7 Detailed Solution
Download Solution PDFConcept:
The concept of converting an analog signal to its digital counterpart is explained with the help of the following diagram:
∴ The maximum quantization is given as:
\({Q_{e\left( {max} \right)}} = \frac{{\rm{\Delta }}}{2}\)
Δ = step size given by:
\({\rm{\Delta }} = \frac{{{V_{max}} - {V_{min}}}}{L}\)
L = Number of levels
Calculation:
With n = 3, the number of levels will be:
L = 23 = 8
With analog input in the range 0 to 8 V and L = 8, the step size will be:
\({\rm{\Delta }} = \frac{{8 - 0}}{{8}} = 1\)
Now, the maximum quantization error will be:
\({Q_{e\left( {max} \right)}} = \frac{{1}}{2} = 0.5\)
In a 10-bit PCM system, a message signal having maximum frequency of 4 KHz is to be transmitted. If the bit rate of this PCM system is 60 Kbits / sec, the appropriate sampling frequency is
Answer (Detailed Solution Below)
Baseband Modulation Question 8 Detailed Solution
Download Solution PDFConcept:
The bandwidth of a PCM system for an encoded signal sampled at a frequency of fs is given by:
Bit rate = n fS
fS = Sampling frequency
n = number of bits used for encoding.
n is related to the number of quantization levels (L) as:
L = 2n
n = log2 L
Calculation:
the maximum frequency for the signal will be 4 kHz.
n = 10 bits, Rb = 60 Kbits/sec
Bit rate = n fS
fs = 6 kHz
And fm is given as 4 kHz i.e
fs ≥ 2fm
fs ≥ 8 kHz
So, fs = 6kHz leads to undersampling.
Hence, option 1 and 2 can't be correct.
Option 4: fs = 9 kHz leads to oversampling
Hence, option 3 is correct fs = 8 kHz
∴ The appropriate value of the sampling frequency (fs) will be 8 kHz
Following type of multiplexing cannot be used for analog signalling?
Answer (Detailed Solution Below)
Baseband Modulation Question 9 Detailed Solution
Download Solution PDFMultiplexing is the process of combining multiple signals into one signal, over a shared medium.
If the analog signals are multiplexed, then it is called as analog multiplexing. Similarly, if the digital signals are multiplexed, then it is called as digital multiplexing.
Types of Multiplexing:
(1) Analog Multiplexing
- Frequency Division Multiplexing
- Wavelength Division Multiplexing
(2) Digital Multiplexing
- Time Division Multiplexing
- Synchronous TDM
- Asynchronous TDM
Analog Multiplexing:
The signals used in analog multiplexing techniques are analog in nature. The analog signals are multiplexed according to their frequency (FDM) or wavelength (WDM).
Frequency Division Multiplexing
- In analog multiplexing, the most used technique is Frequency Division Multiplexing (FDM).
- This technique uses various frequencies to combine streams of data, for sending them on a communication medium, as a single signal.
- Example − A traditional television transmitter, which sends a number of channels through a single cable uses FDM.
Wavelength Division Multiplexing
- Wavelength Division multiplexing (WDM) is an analog technique, in which many data streams of different wavelengths are transmitted in the light spectrum.
- If the wavelength increases, the frequency of the signal decreases.
- A prism, which can turn different wavelengths into a single line, can be used at the output of MUX and input of DEMUX.
- Example − Optical fiber communications use WDM technique, to merge different wavelengths into a single light for communication.
Digital Multiplexing:
The term digital represents the discrete bits of information. Hence, the available data is in the form of frames or packets, which are discrete.
Time Division Multiplexing
- In Time Division Multiplexing (TDM), the time frame is divided into slots. This technique is used to transmit a signal over a single communication channel, by allotting one slot for each message.
- Time Division Multiplexing (TDM) can be classified into Synchronous TDM and Asynchronous TDM.
Synchronous TDM
- In Synchronous TDM, the input is connected to a frame. If there are ‘n’ number of connections, then the frame is divided into ‘n’ time slots. One slot is allocated for each input line.
- In this technique, the sampling rate is common for all signals and hence the same clock input is given. The MUX allocates the same slot to each device at all times.
Asynchronous TDM
- In Asynchronous TDM, the sampling rate is different for each of the signals and a common clock is not required.
- If the allotted device for a time slot transmits nothing and sits idle, then that slot can be allotted to another device, unlike synchronous
- This type of TDM is used in Asynchronous transfer mode networks.
If the number of bits per sample in a PCM system is increased from 8 to 16, then the bandwidth will be increased by
Answer (Detailed Solution Below)
Baseband Modulation Question 10 Detailed Solution
Download Solution PDFConcept:
The number of levels for an n-bit PCM system is given by:
L = 2n
Also, the bandwidth of PCM is given by:
\(BW=n{f_s}\)
n = number of bits to encode
fs = sampling frequency
Calculation:
For n = 8, the bandwidth will be:
B.W. = 8 fs
Similarly, For n = 16, the bandwidth will be:
B.W. = 16 fs
We observe that the Bandwidth is increased by 2 times.
To transmit N signals each band limited to fm Hz by time division multiplexing will require a minimum bandwidth of
Answer (Detailed Solution Below)
Baseband Modulation Question 11 Detailed Solution
Download Solution PDFFormula: For Time division multiplexing (TDM) minimum Band width is given by-
\(\left(B.W\right)_{min}=\dfrac{R_b}{2}\)
Where, Rb = Nnfs
N = No. of signals
n = no. of Bit
fs = sampling frequency = 2 × fm
fm = msg signal Bandwidth
\(\left(B.W\right)_{min}=\dfrac{Nn2f_m}{2}=Nnf_m\)
Solution:
Given:
N signal with fm Bandwidth.
TDM minimum B.W = Nnfm
Assume n = 1 Bit
BWmin = NfmWhat is the frequency and type of modulation used in an IR Remote controls for TV?
Answer (Detailed Solution Below)
Baseband Modulation Question 12 Detailed Solution
Download Solution PDFIR Remote controls for TV:
- Today’s modern remote controls work by modulating the output from an infra-red LED.
- The circuit board consists of circuitry to sense the connections or detect the button being pressed and produces the signal in Morse code form which is amplified by the transistors and then given to IR LED.
- Pulse code modulation technique is used.
- The reason for modulation is to separate the remote IR range from the IR light emitted by other bodies in the vicinity.
- A series of pulses of varying width is sent to a gate that turns on or off, the modulator which is usually 38 kHz.
For a 10 bit PCM system the signal to quantization noise ratio is 62 dB. If the number of bits is increased by 2, then the signal to quantization noise ratio will
Answer (Detailed Solution Below)
Baseband Modulation Question 13 Detailed Solution
Download Solution PDFConcept:
The Signal to Noise (SNR) Ratio for a PCM system is given by:
SNR = (1.8 + 6n) dB
Where ‘n’ is the number of bits per sample.
Calculation:
For n number of bits, the Signal to Noise Ratio will be:
SNR1 = (1.8 + 6n) dB
With the increase in two-bit ( "n+2" bits), the SNR becomes:
SNR2 = (1.8 + 6(n+2)) dB
SNR2 - SNR1 = 12 dB
i.e if the number of bits increased from 10 bits to 12 bits, then the SNR increases by 12 dB.
Some salient features of a PCM system 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 number of quantization levels is 16 and the maximum signal frequency is 4kHz in the PCM system. Find the bit transmission rate.
Answer (Detailed Solution Below)
Baseband Modulation Question 14 Detailed Solution
Download Solution PDFConcept:
The bit transmission rate in the PCM system is given by:
Rb = m × n × fs
where Rb = Bit transmission rate
fs = Sampling frequency
m = No. of message signal
n = No. of bits required for encoding
n is related to no. of quantization level by:
L = 2n
Calculation:
Given, L = 16
16 = 2n
n = 4
fm = 4 kHz
fs = 2fm
fs = 8 kHz
m = 1
Rb = 1 × 4 × 8
Rb = 32 kbs
The main advantage of PCM system is lower
Answer (Detailed Solution Below)
Baseband Modulation Question 15 Detailed Solution
Download Solution PDFPCM (Pulse Code Modulation):
- 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.