Baseband Transmission MCQ Quiz - Objective Question with Answer for Baseband Transmission - Download Free PDF
Last updated on May 14, 2025
Latest Baseband Transmission MCQ Objective Questions
Baseband Transmission Question 1:
Which of the following components is required to detect (demodulate) Pulse Position Modulation (PPM)?
a) Pulse Generator
b) RS Flip-Flop
c) PWM Demodulator
Answer (Detailed Solution Below)
Baseband Transmission Question 1 Detailed Solution
Explanation:
Pulse Position Modulation (PPM) Demodulation
Pulse Position Modulation (PPM) is a type of signal modulation where the position of a pulse, relative to the position of a reference pulse, is varied according to the sampled value of the modulating signal. In simpler terms, the timing of the pulse is varied to encode the information.
In PPM, the position of each pulse is shifted in time to represent the data being transmitted. The time displacement of each pulse is proportional to the amplitude of the analog signal at the corresponding sampling time. To demodulate this signal, the receiver needs to be able to detect these time shifts accurately.
Components Required for PPM Demodulation:
To demodulate a PPM signal, the following components are typically required:
- Pulse Generator: Used to create reference pulses that help in determining the position of the incoming pulses.
- RS Flip-Flop: Helps in maintaining the state information needed for the timing and synchronization of the demodulated signal.
- PWM Demodulator: While not directly used for PPM demodulation, understanding Pulse Width Modulation (PWM) demodulation can be beneficial as PPM can be derived from PWM signals.
Correct Option Analysis:
The correct option is:
Option 4: Only c
This option correctly identifies that a PWM Demodulator (Option c) is required for the demodulation of PPM. The reasoning is that a PPM signal can be derived from a PWM signal. Therefore, understanding and utilizing a PWM demodulator is crucial for effectively demodulating a PPM signal
Baseband Transmission Question 2:
The impulse response h(t) of the filter matched to the input pulse s(t) is :
Answer (Detailed Solution Below)
Baseband Transmission Question 2 Detailed Solution
Concept:
The shown bock diagram will be considered as the receiver in digital communication.
The impulse response of the filter is calculated by:
h(t) = s*(Tb - t)
Calculation:
In the given signal the duration is T. So, Tb = T
The shifted signal by T is given by:
s(t+T)
The time-reversed signal will be
Option 2 is correct.
Baseband Transmission Question 3:
In the context of digital communications, what is the primary benefit of using a matched filter?
Answer (Detailed Solution Below)
Baseband Transmission Question 3 Detailed Solution
The correct answer is option 1.
Concept
In the context of digital communications, a matched filter is used to maximize the signal-to-noise ratio (SNR) for a given transmitted signal. This is crucial for reliable data transmission and effective communication.
Additional Information
The primary benefit of using a matched filter in digital communications is that it reduces inter-symbol interference. Inter-symbol interference occurs when one symbol interferes with subsequent symbols, leading to errors in the received data.
A matched filter is designed to match the shape of the transmitted pulse, thus maximizing the SNR and minimizing inter-symbol interference.
Baseband Transmission Question 4:
What is the difference between asynchronous and synchronous communication?
Answer (Detailed Solution Below)
Baseband Transmission Question 4 Detailed Solution
The main difference between asynchronous and synchronous communication lies in the timing of the interaction and exchange of information.
Important Points
Asynchronous Communication:
- Asynchronous communication refers to communication that does not happen in real-time or simultaneously.
- In asynchronous communication, there is a time delay between the message sent by one person and the response received from another person.
Examples of asynchronous communication include email, messaging apps (like WhatsApp or Slack), discussion forums, and leaving voicemail messages.
- Participants in asynchronous communication do not need to be online or available at the same time.
- They can read and respond to messages at their convenience.
- Asynchronous communication allows for flexibility in terms of time and location, making it convenient for individuals in different time zones or with varying schedules.
However, since responses are not immediate, it may take longer to have back-and-forth conversations or to resolve urgent matters.
Additional Information
Synchronous Communication:
- Synchronous communication refers to communication that occurs in real-time and requires the participants to be present and engaged simultaneously.
- In synchronous communication, there is an immediate exchange of messages or information between participants.
Examples of synchronous communication include phone calls, video conferences, live chats, and face-to-face conversations.
- Participants in synchronous communication need to be available and actively participating at the same time for effective communication.
- Synchronous communication allows for immediate feedback, real-time interaction, and quick decision-making.
- However, it may require coordination and scheduling to ensure all participants can be present at the same time, especially when dealing with individuals in different time zones or with busy schedules.
In summary, asynchronous communication occurs without real-time interaction and allows participants to communicate at their convenience, while synchronous communication happens in real-time and requires simultaneous engagement for immediate interaction. Both types of communication have their benefits and are used in different contexts depending on the nature of the communication and the preferences and availability of the participants
Baseband Transmission Question 5:
Each of the following sentences has been divided into four parts – (a), (b), (c), (d) – one of which has an error. Choose the part which contains the error.
I talked (a) / to my neighbourer (b) / to settle the issue (c) / that had been hanging for a long time.(d)
Answer (Detailed Solution Below)
Baseband Transmission Question 5 Detailed Solution
The correct answer is '(b)'.
Key Points
- There is an error in part (b).
- 'neighbourer' should be replaced by 'neighbour'.
- neighbour means a person living next door to or very near to the speaker or person referred to.
- Example:- "our garden was the envy of the neighbours"
- The correct sentence will be - I talked to my neighbour to settle the issue that had been hanging for a long time.
Hence, the correct answer is Option 2.
Top Baseband Transmission MCQ Objective Questions
Each of the following sentences has been divided into four parts – (a), (b), (c), (d) – one of which has an error. Choose the part which contains the error.
I talked (a) / to my neighbourer (b) / to settle the issue (c) / that had been hanging for a long time.(d)
Answer (Detailed Solution Below)
Baseband Transmission Question 6 Detailed Solution
Download Solution PDFThe correct answer is '(b)'.
Key Points
- There is an error in part (b).
- 'neighbourer' should be replaced by 'neighbour'.
- neighbour means a person living next door to or very near to the speaker or person referred to.
- Example:- "our garden was the envy of the neighbours"
- The correct sentence will be - I talked to my neighbour to settle the issue that had been hanging for a long time.
Hence, the correct answer is Option 2.
A random variable X takes values −1 and +1 with probabilities 0.2 and 0.8, respectively. It is transmitted across a channel which adds noise N, so that the random variable at the channel output is Y = X + N. The noise N is independent of X, and is uniformly distributed over the interval [−2 , 2]. The receiver makes a decision
\(x = \left\{ {\begin{array}{*{20}{c}} { - 1,~if~Y \le \theta }\\ { + 1,if~Y > \theta } \end{array}} \right.\)
where the threshold θ ∈ [−1,1] is chosen so as to minimize the probability of error Pr[X̂ ≠ X]. The minimum probability of error, rounded off to 1 decimal place, is_______.
Answer (Detailed Solution Below) 0.1
Baseband Transmission Question 7 Detailed Solution
Download Solution PDFNoise is distributed uniformly
PDF of noise
When – 1 is transmitted
PDF of Received signa;
Let x = threshold voltage
\({P_{{e_1}}} = \left( {0.2} \right)\left( {\frac{1}{4}} \right)\left( {1 - x} \right)\)
When +1 is transmitted
Received pdf
\({P_{{e_2}}} = \left( {0.8} \right)\left( {\frac{1}{4}} \right)\left( {1 + x} \right)\)
Total probability of error
\({P_e} = {P_{{e_1}}} + {P_{{e_2}}}\)
\(= \left( {0.2} \right)\left( {\frac{1}{4}} \right)\left( {1 - x} \right) + \left( {0.8} \right)\left( {\frac{1}{4}} \right)\left( {1 + x} \right)\)
\(\Rightarrow \frac{{\left( {0.2} \right)\left( {1 - x} \right) + \left( {0.8} \right)\left( {1 + x} \right)}}{4}\)
\({P_e} = \frac{{1 + 0.6x}}{4}\)
-1 ≤ x ≤ 1
Minimum probability is obtained when x = -1
\({P_e} = \frac{{1 + \left( {0.6} \right)\left( { - 1} \right)}}{4}\)
\(\Rightarrow \frac{{0.4}}{4} = 0.1\)
Error detection and correction is done
Answer (Detailed Solution Below)
Baseband Transmission Question 8 Detailed Solution
Download Solution PDF- Networks must be able to transfer data from one device to another with complete accuracy.
- An error occurs when a it is altered between transmission and reception (1 is transmitted and 0 is received, and vice versa)
- For reliable communication, errors must be detected and corrected.
- Error detection means deciding whether the received data is correct or not without having a copy of the original message.
- Error detection and correction uses the concept of redundancy, which means adding extra bits for detecting errors at the destination.
- These redundant bits are added by the sender and removed by the receiver.
The error detection methods include:
- VRC (Vertical Redundancy Check): It is an error-detecting code commonly used in digital networks and storage devices to detect an error in data sent by the sender.
- LRC (Longitudinal Redundancy Check): It is an error-checking method used on an eight-bit ASCII character
- CRC (Cyclic Redundancy Check)
- Checksum
Let \(\rm y\left( t \right) = {e^{ - \pi {t^2}}}\) and \(\rm h(t)\) is a filter matched to \(\rm y(t)\). If \(\rm y(t)\) is applied as input to \(\rm h(t)\) then the Fourier transform of the output is:
Answer (Detailed Solution Below)
Baseband Transmission Question 9 Detailed Solution
Download Solution PDF\(\rm y\left( t \right) \leftrightarrow G\left( f \right) = {e^{ - \pi {f^2}}} \)
Since \(\rm h(t)\) is a matched filter.
\(\rm h\left( t \right) = y\left( { - t} \right)\)
\(\rm H\left( f \right) = G\left( { - f} \right) = G\left( f \right) = {e^{ - \pi {f^2}}}\)
Now output \(\rm {Y_o}\left( f \right)\) is given by:
\(\rm \eqalign{ & \rm \;{Y_o}\left( f \right) = Y\left( f \right)\;H\left( f \right) \cr & \rm {Y_o}\left( f \right) = G\left( f \right) \times H\left( f \right) = {e^{ - \pi {f^2}}} \times {e^{ - \pi {f^2}}} \cr & \rm \Rightarrow {Y_o}\left( f \right) = {e^{ - 2\pi {f^2}}} \cr} \)
A binary source generates symbols 𝑋 ∈ {−1, 1} which are transmitted over a noisy channel. The probability of transmitting 𝑋 = 1 is 0.5. Input to the threshold detector is 𝑅 = 𝑋 + 𝑁. The probability density function fN(𝑛) of the noise 𝑁 is shown below:
If the detection threshold is zero, then the probability of error (correct to two decimal places) is ________.
Answer (Detailed Solution Below) 0.12 - 0.14
Baseband Transmission Question 10 Detailed Solution
Download Solution PDFWhen 1 is transmitted received signal is:
R = 1 + N
Probability of error \(= \frac{1}{2}\left( 1 \right)\left( {\frac{1}{4}} \right)\)
[Area of shaded region] = 1/8
When -1 is transmitted, the received signal is:
Probability of error = Area of the shaded region = 1/8
Total probability of error is:
P[E] = P(x = -1)PE1 + P(x = 1) PE2
\(= \frac{1}{2}\left( {\frac{1}{8}} \right) + \frac{1}{2}\left( {\frac{1}{8}} \right)\)
\(= \frac{1}{8} = 0.125\)
Modulation technique used in low speed modem is ______.
Answer (Detailed Solution Below)
Baseband Transmission Question 11 Detailed Solution
Download Solution PDFMODEM:
- A modem is a short form of Modulator/Demodulator.
- The modem is a hardware component/device which can connect computer and other devices such as router and switch to the internet.
- Modems convert or modulate the analog signals coming from telephone wire into digital form i.e in form of 0 s and 1 s.
- Modems of the present time can transfer data at rates of 300-2400 bps (bits per second).
- Generally, two types of modulation techniques are used:
- Frequency-shift keying(FSK) for low-speed modems.
- Phase shift keying (PSK) for high-speed modems.
ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space. In these schemes, bit-by-bit transmission through free space occurs.
FSK (Frequency Shift Keying):
In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.
For binary ‘1’ → S1 (A) = Acos 2π fHt
For binary ‘0’ → S2 (t) = A cos 2π fLt . The constellation diagram is as shown:
ASK(Amplitude Shift Keying):
In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:
For binary ‘1’ → S1 (t) = Acos 2π fct
For binary ‘0’ → S2 (t) = 0
The Constellation Diagram Representation is as shown:
where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.
PSK(Phase Shift Keying):
In PSK (phase shift keying) binary 1 is represented with a carrier signal and binary O is represented with 180° phase shift of a carrier
For binary ‘1’ → S1 (A) = Acos 2π fct
For binary ‘0’ → S2 (t) = A cos (2πfct + 180°) = - A cos 2π fct
The Constellation Diagram Representation is as shown:
Which of the following gives maximum probability of error
Answer (Detailed Solution Below)
Baseband Transmission Question 12 Detailed Solution
Download Solution PDFAnalysis:
The probability of error for ASK, PSK, and FSK is given as
\(P_{e_{ASK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}})} \space \)
\(P_{e_{PSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{N_0}}})} \space \)
\(P_{e_{FSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}})} \space \)
Q(x) is a decreasing function therefore as x increases the value of Q(x) decreases
\( \sqrt {\dfrac {A^2_c T_b \space }{N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{2N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{4N_0}} \space \)
Therefore,
\( Q (\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}) \space > \space Q(\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}) \space > \space Q( \sqrt {\dfrac {A^2_c T_b \space }{N_0}}) \space \)
Pe ASK > PeFSK > Pe PSK
ASK modulation scheme gives Maximum probability of error.
Consider a binary, digital communication system which used pulses \({\rm{g}}\left( {\rm{t}} \right)\) and \(- {\rm{g}}\left( {\rm{t}} \right)\) for transmitting bits over an AWGN channel. If the receiver uses a matched filter, which one of the following pulses will give the minimum probability of bit error?
Answer (Detailed Solution Below)
Baseband Transmission Question 13 Detailed Solution
Download Solution PDFConcept:
Probability of error for PSK is given as:
\({P_e} = Q\left( {\sqrt {\frac{{2E}}{{{N_0}}}} \;} \right)\)
Complementary error function Q(x) is defined as:
\(Q\left( x \right) = \mathop \smallint \nolimits_x^\infty \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{t^2}}}{2}}}dt\)
Q(x) decreases as x increases.
Calculation:
Since:
\({P_e} = Q\left( {\sqrt {\frac{{2E}}{{{N_0}}}} \;} \right)\)
\(Q\left( {\sqrt {\frac{{2E}}{{{N_0}}}} \;} \right)\) decreases as \(\sqrt {\frac{{2E}}{{{N_0}}}} \) increases.
Therefore, to minimize the probability of error, maximize the energy of the signal E.
Energy of the signal for option A = EA = 1 × 1 = 1
Energy of the signal for option B:
\({E_B} = 2 \times \mathop \smallint \nolimits_0^{1/2\;} {\left( {2t} \right)^2}dt = 2 \times \frac{{4{t^3}}}{3}|\begin{array}{*{20}{c}} {1/2}\\ 0 \end{array} = \frac{8}{3} \times \frac{1}{8} = \frac{1}{3}\)
The energy of the signal for option C:
\({E_C} = \mathop \smallint \nolimits_0^{1\;} {\left( {1 - t} \right)^2}dt = \mathop \smallint \nolimits_0^{1\;} \left( {1 - 2t + {t^2}} \right)dt\)
\( = \left( {t - {t^2} + \frac{{{t^3}}}{3}} \right)\begin{array}{*{20}{c}} 1\\ 0 \end{array}\)
\({E_C} = 1 - 1 + \frac{1}{3}\)
The energy of the signal for option D:
\({E_D} = \mathop \smallint \nolimits_0^{1\;} {\left( t \right)^2}dt = \mathop \smallint \nolimits_0^{1\;} {t^2}dt\)
\( = \left( {\frac{{{t^3}}}{3}} \right)\begin{array}{*{20}{c}} 1\\ 0 \end{array} = \frac{1}{3}\)
Since EA > EB = EC = ED
Therefore, the probability of error will be minimum for option A.
Which of the following statements about the matched filter in a communication receiver are correct ?
(A) It may produce phase error if synchronization is improper
(B) Its impulse response depends on the singal shape
(C) The characteristics of the matched filter is matched with the transmitted data
(D) It produces inter symbol interference
(E) It measures the correlation between incoming received message and its impulse response
Choose the correct answer from the options given below :
Answer (Detailed Solution Below)
Baseband Transmission Question 14 Detailed Solution
Download Solution PDFMatched Filter:
- Matched Filter is used to produce an output in such a way that it maximizes the ratio of the output peak power to the mean noise power in its frequency response.
- The frequency response of Magnitude and phase angle of the Matched Filter varies uniformly with frequency.
- The impulse response of the matched filter depends upon the signal shape. It is the mirror image of the received signal about a time instant.
- The Characteristics of the matched filter are matched with the transmitted data.
- The matched filter reduces INTERSYMBOL INTERFERENCE by attenuating the beginning and end of each symbol period.
- Matched Filter measures the correlation between incoming received message and its impulse response.
Important Points:
- A time delay is necessary for the specification of the filter for reasons of physical realizability since there can be no output from the filter until the signal is applied.
- The frequency-response function of the matched filter is the conjugate of the spectrum of the received waveform except for the phase shift exp (- j2Πft).
- This phase shift varies uniformly with frequency. Its effect is to cause a constant time delay.
A zero mean white Gaussian noise having power spectral density \(\frac{{{{\rm{N}}_0}}}{2}\) is passed through an LTI filter whose impulse response \({\rm{h}}\left( {\rm{t}} \right)\) is shown in the figure. The variance of the filtered noise at \({\rm{t}} = 4\) is
Answer (Detailed Solution Below)
Baseband Transmission Question 15 Detailed Solution
Download Solution PDFConcept:
Convolution of a signal x(t) with unit impulse δ(t) is the signal itself. i.e. x(t) ⊕ δ(t) = x(t)
Fourier transform of auto-correlation function of a power signal x(t) is power spectral density Sx(f). i.e. \({R_X}\left( \tau \right)\mathop \leftrightarrow \limits^{FT} {S_X}\left( f \right)\)
And E(x2 (t)) = RX (0)
The variance of the signal x(t) is defined as:
\(var\left( {x\left( t \right)} \right) = E\left( {{x^2}\left( t \right)} \right) - (E{\left( {x\left( t \right)} \right)^2}\)
Fourier transform of unit impulse is 1.
\(\delta \left( t \right)\mathop \leftrightarrow \limits^{FT} 1\)
Calculation:
Let n(t) be the input white noise with zero mean and \(\frac{{{N_0}}}{2}\) power spectral density.
Mean of the white noise = E(n(t)) = 0
Power spectral density is:
\({S_n}\left( f \right) = \frac{{{N_0}}}{2}\) ;
And the auto-correlation function is:
\({R_n}\left( \tau \right)\mathop \leftrightarrow \limits^{FT} {S_n}\left( f \right)\)
\(\frac{{{N_0}}}{2}\mathop \to \limits^{IFT} \frac{{{N_0}}}{2}\delta \left( t \right)\)
\({R_n}\left( \tau \right) = \frac{{{N_0}}}{2}\delta \left( t \right)\)
Let yn(t) is the output noise.
Mean of the output noise:
\( = E\left( {{y_n}\left( t \right)} \right) = E\left( {n\left( t \right) \times \mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right)dt} \right)\)
\( = E\left( {n\left( t \right)} \right) \times \mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right)dt\)
\( = 0 \times \mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right)dt = 0\)
The variance of the output noise is:
\(Var\left( {{y_n}\left( t \right)} \right) = E\left( {y_n^2\left( t \right)} \right) - (E{\left( {{y_n}\left( t \right)} \right)^2}\)
\( = E\left( {y_n^2\left( t \right)} \right)\)
\(E\left( {y_n^2\left( t \right)} \right) = {R_{{y_n}}}\left( 0 \right)\)
\({R_{{y_n}}}\left( \tau \right) = h\left( \tau \right)*{h^*}\left( { - \tau } \right)*{R_n}\left( \tau \right)\)
\( = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right).h\left( {t + \tau } \right)dt} \right)*\frac{{{N_0}}}{2}{\rm{\delta }}\left( {\rm{\tau }} \right)\)
\( = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right).h\left( {t + \tau } \right)dt} \right) \times \frac{{{N_0}}}{2}\;\)
\({R_{{y_n}}}\left( 0 \right) = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right).h\left( t \right)dt} \right)*\frac{{{N_0}}}{2} = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty {h^2}\left( t \right)dt} \right)*\frac{{{N_0}}}{2} = 3{A^2} \times \frac{{{N_0}}}{2}\)
\(Var\left( {{y_n}\left( t \right)} \right) = \frac{3}{2}{A^2}.{N_0}\)