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Question 1 of 30
1. Question
Consider a scenario where a continuous-time audio signal, containing frequencies up to 15 kHz, is being digitized for processing at Xi’an Technological University Northern College of Information Engineering. If the analog-to-digital converter is configured to sample this signal at a rate of 20 kHz, what specific distortion will manifest in the resulting digital representation, and what is the apparent frequency of the highest original frequency component after sampling?
Correct
The question probes the understanding of fundamental principles in digital signal processing, specifically related to sampling and aliasing, which are core to information engineering. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling frequency is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, a continuous-time signal with a maximum frequency of 15 kHz is being sampled. To avoid aliasing, the sampling frequency must be greater than or equal to twice this maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks about the consequence of sampling at a frequency *below* this minimum requirement. When the sampling frequency (\(f_s\)) is less than the Nyquist rate (\(2f_{max}\)), higher frequency components in the original signal masquerade as lower frequencies in the sampled signal. This phenomenon is called aliasing. Specifically, a frequency \(f\) in the original signal will appear as \(|f – k f_s|\) in the sampled signal, where \(k\) is an integer chosen such that the resulting frequency is within the range \([0, f_s/2]\). If the sampling frequency is 20 kHz, and the signal contains frequencies up to 15 kHz, then the frequency 15 kHz will be aliased. The aliased frequency will be \(|15 \text{ kHz} – 1 \times 20 \text{ kHz}| = |-5 \text{ kHz}| = 5 \text{ kHz}\). This means that the original 15 kHz component will be indistinguishable from a 5 kHz component in the sampled data. This distortion is irreversible without additional information about the original signal’s bandwidth. Understanding this principle is crucial for designing effective digital communication systems and signal processing algorithms, areas of significant focus at Xi’an Technological University Northern College of Information Engineering. The ability to identify and mitigate aliasing is a foundational skill for information engineers.
Incorrect
The question probes the understanding of fundamental principles in digital signal processing, specifically related to sampling and aliasing, which are core to information engineering. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling frequency is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, a continuous-time signal with a maximum frequency of 15 kHz is being sampled. To avoid aliasing, the sampling frequency must be greater than or equal to twice this maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks about the consequence of sampling at a frequency *below* this minimum requirement. When the sampling frequency (\(f_s\)) is less than the Nyquist rate (\(2f_{max}\)), higher frequency components in the original signal masquerade as lower frequencies in the sampled signal. This phenomenon is called aliasing. Specifically, a frequency \(f\) in the original signal will appear as \(|f – k f_s|\) in the sampled signal, where \(k\) is an integer chosen such that the resulting frequency is within the range \([0, f_s/2]\). If the sampling frequency is 20 kHz, and the signal contains frequencies up to 15 kHz, then the frequency 15 kHz will be aliased. The aliased frequency will be \(|15 \text{ kHz} – 1 \times 20 \text{ kHz}| = |-5 \text{ kHz}| = 5 \text{ kHz}\). This means that the original 15 kHz component will be indistinguishable from a 5 kHz component in the sampled data. This distortion is irreversible without additional information about the original signal’s bandwidth. Understanding this principle is crucial for designing effective digital communication systems and signal processing algorithms, areas of significant focus at Xi’an Technological University Northern College of Information Engineering. The ability to identify and mitigate aliasing is a foundational skill for information engineers.
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Question 2 of 30
2. Question
When digitizing an analog audio signal at Xi’an Technological University Northern College of Information Engineering for advanced acoustic analysis, a researcher is working with a signal whose spectrum extends up to 15 kHz. The researcher intends to use a sampling frequency of 25 kHz. What fundamental digital signal processing phenomenon will occur due to this sampling rate, and what is the minimum sampling frequency required to prevent it?
Correct
The question assesses understanding of the foundational principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its practical implications in preventing aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In the given scenario, a signal containing frequencies up to 15 kHz is being sampled. To avoid aliasing, the sampling frequency must be greater than twice the maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If the sampling frequency is set to 25 kHz, which is less than the Nyquist rate, aliasing will occur. Aliasing causes higher frequencies in the original signal to be misrepresented as lower frequencies in the sampled signal, leading to distortion and loss of information. This is a critical concept in the Northern College of Information Engineering’s curriculum, as it underpins the digitization of all forms of information, from audio and video to sensor data. Understanding and applying the Nyquist criterion is essential for designing effective digital systems that accurately capture and process analog information, a core competency for graduates of Xi’an Technological University Northern College of Information Engineering. The ability to identify situations where aliasing will occur and to select appropriate sampling rates is a hallmark of a well-prepared student in this field.
Incorrect
The question assesses understanding of the foundational principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its practical implications in preventing aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In the given scenario, a signal containing frequencies up to 15 kHz is being sampled. To avoid aliasing, the sampling frequency must be greater than twice the maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If the sampling frequency is set to 25 kHz, which is less than the Nyquist rate, aliasing will occur. Aliasing causes higher frequencies in the original signal to be misrepresented as lower frequencies in the sampled signal, leading to distortion and loss of information. This is a critical concept in the Northern College of Information Engineering’s curriculum, as it underpins the digitization of all forms of information, from audio and video to sensor data. Understanding and applying the Nyquist criterion is essential for designing effective digital systems that accurately capture and process analog information, a core competency for graduates of Xi’an Technological University Northern College of Information Engineering. The ability to identify situations where aliasing will occur and to select appropriate sampling rates is a hallmark of a well-prepared student in this field.
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Question 3 of 30
3. Question
During the development of a new communication protocol at Xi’an Technological University Northern College of Information Engineering, researchers are analyzing the fidelity of analog signal transmission. They are considering a scenario where an analog signal, known to contain frequency components up to a maximum of 10 kHz, is digitized. The chosen sampling rate for this digitization process is 15 kHz. What is the most significant consequence of this sampling rate selection on the ability to accurately reconstruct the original analog signal?
Correct
The question probes the understanding of fundamental principles in digital signal processing, specifically concerning the impact of sampling rate on signal reconstruction. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. Consider a scenario where a continuous-time signal contains frequency components up to \(f_{max} = 10\) kHz. According to the Nyquist-Shannon theorem, the minimum sampling frequency required for perfect reconstruction is \(2 \times 10\) kHz = 20 kHz. If the signal is sampled at a rate of 15 kHz, which is below the Nyquist rate, aliasing will occur. Aliasing is the phenomenon where higher frequencies in the original signal are misrepresented as lower frequencies in the sampled signal, leading to distortion and an inability to accurately reconstruct the original waveform. The question asks about the consequence of sampling a signal with a maximum frequency of 10 kHz at 15 kHz. Since 15 kHz is less than the required 20 kHz, frequencies above \(15 \text{ kHz} / 2 = 7.5\) kHz will be aliased. Specifically, a frequency component at \(f\) where \(7.5 \text{ kHz} < f \le 10\) kHz will appear as \(15 \text{ kHz} – f\) in the sampled data. For instance, a 9 kHz component would appear as \(15 \text{ kHz} – 9 \text{ kHz} = 6\) kHz. This distortion means that the original signal cannot be perfectly recovered from the samples. Therefore, the primary consequence is the inability to accurately reconstruct the original signal due to the presence of aliased frequencies. This is a core concept taught in signal processing courses at institutions like Xi'an Technological University Northern College of Information Engineering, emphasizing the critical role of sampling in preserving signal integrity.
Incorrect
The question probes the understanding of fundamental principles in digital signal processing, specifically concerning the impact of sampling rate on signal reconstruction. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. Consider a scenario where a continuous-time signal contains frequency components up to \(f_{max} = 10\) kHz. According to the Nyquist-Shannon theorem, the minimum sampling frequency required for perfect reconstruction is \(2 \times 10\) kHz = 20 kHz. If the signal is sampled at a rate of 15 kHz, which is below the Nyquist rate, aliasing will occur. Aliasing is the phenomenon where higher frequencies in the original signal are misrepresented as lower frequencies in the sampled signal, leading to distortion and an inability to accurately reconstruct the original waveform. The question asks about the consequence of sampling a signal with a maximum frequency of 10 kHz at 15 kHz. Since 15 kHz is less than the required 20 kHz, frequencies above \(15 \text{ kHz} / 2 = 7.5\) kHz will be aliased. Specifically, a frequency component at \(f\) where \(7.5 \text{ kHz} < f \le 10\) kHz will appear as \(15 \text{ kHz} – f\) in the sampled data. For instance, a 9 kHz component would appear as \(15 \text{ kHz} – 9 \text{ kHz} = 6\) kHz. This distortion means that the original signal cannot be perfectly recovered from the samples. Therefore, the primary consequence is the inability to accurately reconstruct the original signal due to the presence of aliased frequencies. This is a core concept taught in signal processing courses at institutions like Xi'an Technological University Northern College of Information Engineering, emphasizing the critical role of sampling in preserving signal integrity.
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Question 4 of 30
4. Question
Consider a scenario within the advanced digital communications curriculum at Xi’an Technological University Northern College of Information Engineering where a researcher is evaluating the impact of signal and noise power adjustments on data transmission quality. If the original signal power is \(S\) and the noise power is \(N\), resulting in an initial signal-to-noise ratio (SNR), and subsequently the signal power is doubled while the noise power is simultaneously halved, how does the new SNR compare to the original SNR in terms of its decibel value?
Correct
The core of this question lies in understanding the principles of signal-to-noise ratio (SNR) and its impact on data integrity within communication systems, a fundamental concept at Xi’an Technological University Northern College of Information Engineering. A higher SNR indicates that the desired signal is significantly stronger than background noise, leading to more accurate data reception and processing. Conversely, a lower SNR means the noise is more prominent, potentially corrupting the signal and causing errors. Consider a scenario where a digital communication system transmits data. The received signal strength is \(S\) and the noise power is \(N\). The signal-to-noise ratio (SNR) is defined as the ratio of signal power to noise power, often expressed in decibels (dB) as \(SNR_{dB} = 10 \log_{10} \left(\frac{S}{N}\right)\). If the signal power is \(S_1\) and the noise power is \(N_1\), the initial SNR is \(SNR_1 = \frac{S_1}{N_1}\). If the signal power is doubled to \(2S_1\) and the noise power is halved to \(N_1/2\), the new SNR becomes \(SNR_2 = \frac{2S_1}{N_1/2} = \frac{4S_1}{N_1} = 4 \times SNR_1\). In decibels, the change in SNR is: \(SNR_{1, dB} = 10 \log_{10} \left(\frac{S_1}{N_1}\right)\) \(SNR_{2, dB} = 10 \log_{10} \left(\frac{2S_1}{N_1/2}\right) = 10 \log_{10} \left(\frac{4S_1}{N_1}\right)\) \(SNR_{2, dB} = 10 \log_{10} (4) + 10 \log_{10} \left(\frac{S_1}{N_1}\right)\) \(SNR_{2, dB} = 10 \log_{10} (4) + SNR_{1, dB}\) Since \(10 \log_{10} (4) \approx 6.02\), the new SNR is approximately \(SNR_{1, dB} + 6.02\) dB. This increase in SNR directly translates to improved data reliability. In the context of information engineering at Xi’an Technological University Northern College of Information Engineering, understanding how to optimize SNR through techniques like increasing transmission power or reducing interference is crucial for designing robust communication protocols and ensuring high-fidelity data transmission, especially in complex environments where signal degradation is a significant concern. The ability to quantify and interpret these changes is a hallmark of advanced engineering practice.
Incorrect
The core of this question lies in understanding the principles of signal-to-noise ratio (SNR) and its impact on data integrity within communication systems, a fundamental concept at Xi’an Technological University Northern College of Information Engineering. A higher SNR indicates that the desired signal is significantly stronger than background noise, leading to more accurate data reception and processing. Conversely, a lower SNR means the noise is more prominent, potentially corrupting the signal and causing errors. Consider a scenario where a digital communication system transmits data. The received signal strength is \(S\) and the noise power is \(N\). The signal-to-noise ratio (SNR) is defined as the ratio of signal power to noise power, often expressed in decibels (dB) as \(SNR_{dB} = 10 \log_{10} \left(\frac{S}{N}\right)\). If the signal power is \(S_1\) and the noise power is \(N_1\), the initial SNR is \(SNR_1 = \frac{S_1}{N_1}\). If the signal power is doubled to \(2S_1\) and the noise power is halved to \(N_1/2\), the new SNR becomes \(SNR_2 = \frac{2S_1}{N_1/2} = \frac{4S_1}{N_1} = 4 \times SNR_1\). In decibels, the change in SNR is: \(SNR_{1, dB} = 10 \log_{10} \left(\frac{S_1}{N_1}\right)\) \(SNR_{2, dB} = 10 \log_{10} \left(\frac{2S_1}{N_1/2}\right) = 10 \log_{10} \left(\frac{4S_1}{N_1}\right)\) \(SNR_{2, dB} = 10 \log_{10} (4) + 10 \log_{10} \left(\frac{S_1}{N_1}\right)\) \(SNR_{2, dB} = 10 \log_{10} (4) + SNR_{1, dB}\) Since \(10 \log_{10} (4) \approx 6.02\), the new SNR is approximately \(SNR_{1, dB} + 6.02\) dB. This increase in SNR directly translates to improved data reliability. In the context of information engineering at Xi’an Technological University Northern College of Information Engineering, understanding how to optimize SNR through techniques like increasing transmission power or reducing interference is crucial for designing robust communication protocols and ensuring high-fidelity data transmission, especially in complex environments where signal degradation is a significant concern. The ability to quantify and interpret these changes is a hallmark of advanced engineering practice.
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Question 5 of 30
5. Question
A research team at Xi’an Technological University Northern College of Information Engineering is developing a new digital communication protocol. They have digitized an audio signal, capturing discrete samples at a specific rate. To verify the integrity of their data transmission and ensure the original audio quality can be restored, they need to understand the fundamental process of converting these digital samples back into a continuous analog waveform. Considering the principles of signal reconstruction in digital signal processing, which method is most appropriate for accurately recreating the original analog signal from its discrete digital samples?
Correct
The question probes the understanding of signal processing concepts within the context of digital communication, a core area for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital signal that has undergone a process, and the task is to identify the most appropriate method for reconstructing the original analog signal. The core principle here is the Nyquist-Shannon sampling theorem, which states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the original analog signal. This minimum sampling rate is known as the Nyquist rate (\(f_{Nyquist} = 2f_{max}\)). When an analog signal is sampled, information about frequencies above \(f_s/2\) is lost or aliased into lower frequencies. To reconstruct the original analog signal from its discrete samples, a process called digital-to-analog conversion (DAC) is used. A crucial component of the DAC process is the reconstruction filter, often a low-pass filter. This filter is designed to remove any high-frequency components introduced during the sampling process or that might exist in the sampled data, and to interpolate between the sample points to create a smooth analog waveform. The cutoff frequency of this reconstruction filter should ideally be set at \(f_s/2\) to recover the original signal up to its maximum frequency component, assuming the sampling was done at or above the Nyquist rate. If the sampling rate was insufficient (i.e., below the Nyquist rate), aliasing would have occurred, meaning higher frequencies in the original signal would have been misrepresented as lower frequencies. In such a case, even with an ideal reconstruction filter, the original analog signal cannot be perfectly recovered because the necessary high-frequency information is already corrupted or lost. Therefore, the effectiveness of reconstruction hinges on the initial sampling rate relative to the signal’s bandwidth. The question implicitly assumes that the sampling was performed adequately for reconstruction to be a meaningful concept. The process of using a low-pass filter with a cutoff frequency at half the sampling rate is the standard method for analog signal reconstruction from digital samples.
Incorrect
The question probes the understanding of signal processing concepts within the context of digital communication, a core area for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital signal that has undergone a process, and the task is to identify the most appropriate method for reconstructing the original analog signal. The core principle here is the Nyquist-Shannon sampling theorem, which states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the original analog signal. This minimum sampling rate is known as the Nyquist rate (\(f_{Nyquist} = 2f_{max}\)). When an analog signal is sampled, information about frequencies above \(f_s/2\) is lost or aliased into lower frequencies. To reconstruct the original analog signal from its discrete samples, a process called digital-to-analog conversion (DAC) is used. A crucial component of the DAC process is the reconstruction filter, often a low-pass filter. This filter is designed to remove any high-frequency components introduced during the sampling process or that might exist in the sampled data, and to interpolate between the sample points to create a smooth analog waveform. The cutoff frequency of this reconstruction filter should ideally be set at \(f_s/2\) to recover the original signal up to its maximum frequency component, assuming the sampling was done at or above the Nyquist rate. If the sampling rate was insufficient (i.e., below the Nyquist rate), aliasing would have occurred, meaning higher frequencies in the original signal would have been misrepresented as lower frequencies. In such a case, even with an ideal reconstruction filter, the original analog signal cannot be perfectly recovered because the necessary high-frequency information is already corrupted or lost. Therefore, the effectiveness of reconstruction hinges on the initial sampling rate relative to the signal’s bandwidth. The question implicitly assumes that the sampling was performed adequately for reconstruction to be a meaningful concept. The process of using a low-pass filter with a cutoff frequency at half the sampling rate is the standard method for analog signal reconstruction from digital samples.
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Question 6 of 30
6. Question
A research team at Xi’an Technological University Northern College of Information Engineering is developing a system to digitize atmospheric pressure readings, which are known to fluctuate with components up to 15 kHz. They intend to sample these readings at a rate of 20 kHz. If they proceed with sampling without employing an appropriate analog anti-aliasing filter, what is the most accurate description of the outcome regarding the spectral content of the digitized data?
Correct
The question revolves around the fundamental principles of digital signal processing, specifically focusing on the concept of aliasing and its mitigation through anti-aliasing filters. When a continuous-time signal is sampled, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal to avoid aliasing, as stated by the Nyquist-Shannon sampling theorem (\(f_s \ge 2f_{max}\)). Aliasing occurs when frequencies above \(f_s/2\) (the Nyquist frequency) in the original signal are misrepresented as lower frequencies in the sampled signal, leading to distortion. Consider a signal containing frequency components up to \(f_{max} = 15\) kHz. If this signal is sampled at \(f_s = 20\) kHz, the Nyquist frequency is \(f_s/2 = 10\) kHz. Since \(f_{max} > f_s/2\), aliasing will occur. Frequencies between 10 kHz and 15 kHz will be folded back into the range of 0 to 10 kHz. For instance, a 12 kHz component would appear as \(|12 – 20| = 8\) kHz, and a 15 kHz component would appear as \(|15 – 20| = 5\) kHz. To prevent this, an anti-aliasing filter, which is a low-pass filter, is applied to the analog signal *before* sampling. This filter attenuates or removes frequency components above the desired bandwidth, ensuring that the highest frequency component in the signal presented to the sampler is below the Nyquist frequency. If the anti-aliasing filter is designed to pass frequencies up to \(f_{cutoff} = 9\) kHz, then the effective maximum frequency in the signal to be sampled becomes \(f_{max\_filtered} = 9\) kHz. With \(f_{max\_filtered} = 9\) kHz and \(f_s = 20\) kHz, the Nyquist frequency is \(10\) kHz. Since \(f_{max\_filtered} < f_s/2\), the sampling can proceed without aliasing. The question asks about the consequence of *not* using an anti-aliasing filter when the signal's maximum frequency exceeds half the sampling rate. In this scenario, the frequencies in the original signal that are above the Nyquist frequency (\(10\) kHz) will be incorrectly represented as lower frequencies within the sampled spectrum. Specifically, the component at \(15\) kHz will alias to \(|15 – 20| = 5\) kHz. The component at \(12\) kHz will alias to \(|12 – 20| = 8\) kHz. The component at \(10\) kHz (which is exactly the Nyquist frequency) will be represented at \(10\) kHz, but components slightly above it will fold down. The fundamental issue is the introduction of spurious frequency components that were not present in the original signal's intended bandwidth, corrupting the reconstructed signal. This phenomenon is known as spectral folding.
Incorrect
The question revolves around the fundamental principles of digital signal processing, specifically focusing on the concept of aliasing and its mitigation through anti-aliasing filters. When a continuous-time signal is sampled, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal to avoid aliasing, as stated by the Nyquist-Shannon sampling theorem (\(f_s \ge 2f_{max}\)). Aliasing occurs when frequencies above \(f_s/2\) (the Nyquist frequency) in the original signal are misrepresented as lower frequencies in the sampled signal, leading to distortion. Consider a signal containing frequency components up to \(f_{max} = 15\) kHz. If this signal is sampled at \(f_s = 20\) kHz, the Nyquist frequency is \(f_s/2 = 10\) kHz. Since \(f_{max} > f_s/2\), aliasing will occur. Frequencies between 10 kHz and 15 kHz will be folded back into the range of 0 to 10 kHz. For instance, a 12 kHz component would appear as \(|12 – 20| = 8\) kHz, and a 15 kHz component would appear as \(|15 – 20| = 5\) kHz. To prevent this, an anti-aliasing filter, which is a low-pass filter, is applied to the analog signal *before* sampling. This filter attenuates or removes frequency components above the desired bandwidth, ensuring that the highest frequency component in the signal presented to the sampler is below the Nyquist frequency. If the anti-aliasing filter is designed to pass frequencies up to \(f_{cutoff} = 9\) kHz, then the effective maximum frequency in the signal to be sampled becomes \(f_{max\_filtered} = 9\) kHz. With \(f_{max\_filtered} = 9\) kHz and \(f_s = 20\) kHz, the Nyquist frequency is \(10\) kHz. Since \(f_{max\_filtered} < f_s/2\), the sampling can proceed without aliasing. The question asks about the consequence of *not* using an anti-aliasing filter when the signal's maximum frequency exceeds half the sampling rate. In this scenario, the frequencies in the original signal that are above the Nyquist frequency (\(10\) kHz) will be incorrectly represented as lower frequencies within the sampled spectrum. Specifically, the component at \(15\) kHz will alias to \(|15 – 20| = 5\) kHz. The component at \(12\) kHz will alias to \(|12 – 20| = 8\) kHz. The component at \(10\) kHz (which is exactly the Nyquist frequency) will be represented at \(10\) kHz, but components slightly above it will fold down. The fundamental issue is the introduction of spurious frequency components that were not present in the original signal's intended bandwidth, corrupting the reconstructed signal. This phenomenon is known as spectral folding.
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Question 7 of 30
7. Question
Consider a scenario where researchers at Xi’an Technological University Northern College of Information Engineering are developing a new digital communication system. They have an analog audio signal that contains frequency components ranging from \(0 \text{ Hz}\) up to a maximum of \(15 \text{ kHz}\). To accurately convert this analog signal into a digital format without losing critical information or introducing distortion due to misrepresentation of frequencies, what is the absolute minimum sampling frequency that must be employed, adhering strictly to the principles of signal reconstruction?
Correct
The question assesses understanding of signal processing principles, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling rate is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal contains frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing is \(f_s \ge 2 \times f_{max}\). Substituting the given \(f_{max}\), we get \(f_s \ge 2 \times 15 \text{ kHz}\), which means \(f_s \ge 30 \text{ kHz}\). If the sampling frequency is set below this minimum, aliasing will occur. Aliasing is the phenomenon where higher frequencies in the analog signal are incorrectly represented as lower frequencies in the sampled digital signal, leading to distortion and loss of information. For instance, if the sampling frequency were \(25 \text{ kHz}\), a frequency component of \(28 \text{ kHz}\) in the original signal would be aliased to \(25 \text{ kHz} – (28 \text{ kHz} – 25 \text{ kHz}) = 22 \text{ kHz}\), or more generally, the aliased frequency \(f_{alias}\) is given by \(f_{alias} = |f – k \cdot f_s|\) for some integer \(k\), where \(f\) is the original frequency. In this case, \(28 \text{ kHz}\) would alias to \(|28 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = 3 \text{ kHz}\). The question asks for the minimum sampling frequency to prevent aliasing. This directly corresponds to the Nyquist rate. Therefore, the minimum sampling frequency is \(30 \text{ kHz}\). This principle is fundamental in digital signal processing, a core area of study at Xi’an Technological University Northern College of Information Engineering, ensuring accurate conversion of analog information to digital formats for processing and analysis in fields like telecommunications and data acquisition.
Incorrect
The question assesses understanding of signal processing principles, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling rate is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal contains frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing is \(f_s \ge 2 \times f_{max}\). Substituting the given \(f_{max}\), we get \(f_s \ge 2 \times 15 \text{ kHz}\), which means \(f_s \ge 30 \text{ kHz}\). If the sampling frequency is set below this minimum, aliasing will occur. Aliasing is the phenomenon where higher frequencies in the analog signal are incorrectly represented as lower frequencies in the sampled digital signal, leading to distortion and loss of information. For instance, if the sampling frequency were \(25 \text{ kHz}\), a frequency component of \(28 \text{ kHz}\) in the original signal would be aliased to \(25 \text{ kHz} – (28 \text{ kHz} – 25 \text{ kHz}) = 22 \text{ kHz}\), or more generally, the aliased frequency \(f_{alias}\) is given by \(f_{alias} = |f – k \cdot f_s|\) for some integer \(k\), where \(f\) is the original frequency. In this case, \(28 \text{ kHz}\) would alias to \(|28 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = 3 \text{ kHz}\). The question asks for the minimum sampling frequency to prevent aliasing. This directly corresponds to the Nyquist rate. Therefore, the minimum sampling frequency is \(30 \text{ kHz}\). This principle is fundamental in digital signal processing, a core area of study at Xi’an Technological University Northern College of Information Engineering, ensuring accurate conversion of analog information to digital formats for processing and analysis in fields like telecommunications and data acquisition.
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Question 8 of 30
8. Question
Consider a digital communication link established for a research project at Xi’an Technological University Northern College of Information Engineering, where the received signal power is measured at \(10^{-10}\) Watts. The communication channel is characterized by a noise power spectral density of \(10^{-12}\) Watts per Hertz and operates within a bandwidth of 100 kHz. What is the signal-to-noise ratio (SNR) for this transmission link in its linear form?
Correct
The question probes the understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital transmission where the received signal power is \(P_s = 10^{-10}\) Watts and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the channel is \(B = 100\) kHz, which is \(100 \times 10^3\) Hz. The total noise power \(P_n\) in the channel is calculated by multiplying the noise power spectral density by the bandwidth: \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ Watts/Hz}) \times (100 \times 10^3 \text{ Hz})\) \(P_n = 10^{-12} \times 10^5 \text{ Watts}\) \(P_n = 10^{-7} \text{ Watts}\) The signal-to-noise ratio (SNR) is then calculated as the ratio of signal power to noise power: \(SNR = \frac{P_s}{P_n}\) \(SNR = \frac{10^{-10} \text{ Watts}}{10^{-7} \text{ Watts}}\) \(SNR = 10^{-10 – (-7)}\) \(SNR = 10^{-3}\) This value represents the linear ratio. Often, SNR is expressed in decibels (dB). The conversion to decibels is: \(SNR_{dB} = 10 \log_{10}(SNR)\) \(SNR_{dB} = 10 \log_{10}(10^{-3})\) \(SNR_{dB} = 10 \times (-3)\) \(SNR_{dB} = -30 \text{ dB}\) The question asks for the SNR in linear terms, which is \(10^{-3}\). This calculation demonstrates a fundamental understanding of how channel characteristics (noise power spectral density and bandwidth) affect the quality of a digital signal, a critical aspect of information engineering. A low SNR, as calculated here, implies significant noise interference, which would necessitate error correction coding or other signal processing techniques to ensure reliable data transmission, aligning with the practical challenges addressed in the curriculum at Xi’an Technological University Northern College of Information Engineering. Understanding the relationship between signal power, noise power, bandwidth, and SNR is crucial for designing efficient and robust communication systems.
Incorrect
The question probes the understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital transmission where the received signal power is \(P_s = 10^{-10}\) Watts and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the channel is \(B = 100\) kHz, which is \(100 \times 10^3\) Hz. The total noise power \(P_n\) in the channel is calculated by multiplying the noise power spectral density by the bandwidth: \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ Watts/Hz}) \times (100 \times 10^3 \text{ Hz})\) \(P_n = 10^{-12} \times 10^5 \text{ Watts}\) \(P_n = 10^{-7} \text{ Watts}\) The signal-to-noise ratio (SNR) is then calculated as the ratio of signal power to noise power: \(SNR = \frac{P_s}{P_n}\) \(SNR = \frac{10^{-10} \text{ Watts}}{10^{-7} \text{ Watts}}\) \(SNR = 10^{-10 – (-7)}\) \(SNR = 10^{-3}\) This value represents the linear ratio. Often, SNR is expressed in decibels (dB). The conversion to decibels is: \(SNR_{dB} = 10 \log_{10}(SNR)\) \(SNR_{dB} = 10 \log_{10}(10^{-3})\) \(SNR_{dB} = 10 \times (-3)\) \(SNR_{dB} = -30 \text{ dB}\) The question asks for the SNR in linear terms, which is \(10^{-3}\). This calculation demonstrates a fundamental understanding of how channel characteristics (noise power spectral density and bandwidth) affect the quality of a digital signal, a critical aspect of information engineering. A low SNR, as calculated here, implies significant noise interference, which would necessitate error correction coding or other signal processing techniques to ensure reliable data transmission, aligning with the practical challenges addressed in the curriculum at Xi’an Technological University Northern College of Information Engineering. Understanding the relationship between signal power, noise power, bandwidth, and SNR is crucial for designing efficient and robust communication systems.
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Question 9 of 30
9. Question
Consider a scenario at Xi’an Technological University Northern College of Information Engineering where a research project involves transmitting large datasets over a network link characterized by a fixed packet size of 1000 bytes and a bandwidth of 10 Mbps. The network path exhibits variable latency, ranging from a minimum of 50 milliseconds to a maximum of 150 milliseconds for packet acknowledgments. If the data transfer protocol employed is a simple acknowledgment-based system that can only have one unacknowledged packet in transit at any given time, what is the maximum achievable data throughput in Mbps, assuming the acknowledgment packet transmission time is negligible compared to the data packet transmission and propagation delays?
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The scenario describes a system where data packets are transmitted over a network with varying latency. The core concept being tested is the impact of latency on the effective throughput of a reliable data transfer protocol, specifically one that uses acknowledgments. A common model for this is the Stop-and-Wait protocol, or a similar sliding window protocol where the window size is limited by the round-trip time (RTT). In this problem, we are given the following: – Packet size (\(P\)): 1000 bytes – Bandwidth (\(B\)): 10 Mbps (Megabits per second) – Minimum latency (\(L_{min}\)): 50 ms (milliseconds) – Maximum latency (\(L_{max}\)): 150 ms (milliseconds) First, let’s convert all units to be consistent. Bandwidth in bits per second: \(B = 10 \text{ Mbps} = 10 \times 10^6 \text{ bits/s}\). Packet size in bits: \(P = 1000 \text{ bytes} \times 8 \text{ bits/byte} = 8000 \text{ bits}\). Latency in seconds: \(L_{min} = 50 \text{ ms} = 0.050 \text{ s}\), \(L_{max} = 150 \text{ ms} = 0.150 \text{ s}\). The time it takes to transmit a single packet is the packet size divided by the bandwidth: Transmission time (\(T_{tx}\)) = \(P / B = 8000 \text{ bits} / (10 \times 10^6 \text{ bits/s}) = 0.0008 \text{ s}\). In a system like this, the effective throughput is limited by how quickly data can be sent and acknowledged. The round-trip time (RTT) is the time from sending a packet to receiving its acknowledgment. The minimum RTT is \(2 \times L_{min}\) (assuming acknowledgment transmission time is negligible compared to data packet transmission time and propagation delay) and the maximum RTT is \(2 \times L_{max}\). The maximum amount of data that can be “in flight” at any given time is determined by the bandwidth-delay product. For a simple Stop-and-Wait protocol, the sender can only send one packet at a time and must wait for an acknowledgment before sending the next. The time it takes to send one packet and receive its acknowledgment is approximately the transmission time of the packet plus the RTT. The maximum throughput (\(TP_{max}\)) for a system like this, assuming a window size of 1 (like Stop-and-Wait), is effectively limited by the time it takes to send one packet and receive its acknowledgment. The sender is idle for most of the RTT. The total time for one cycle (send packet, wait for ACK) is \(T_{tx} + RTT\). The amount of data sent in this cycle is \(P\). Therefore, the throughput is \(TP = P / (T_{tx} + RTT)\). To find the maximum throughput, we consider the minimum RTT, which is \(2 \times L_{min}\). Maximum Throughput (\(TP_{max}\)) = \(P / (T_{tx} + 2 \times L_{min})\) \(TP_{max} = 8000 \text{ bits} / (0.0008 \text{ s} + 2 \times 0.050 \text{ s})\) \(TP_{max} = 8000 \text{ bits} / (0.0008 \text{ s} + 0.100 \text{ s})\) \(TP_{max} = 8000 \text{ bits} / 0.1008 \text{ s}\) \(TP_{max} \approx 79365 \text{ bits/s}\) Converting this to Mbps: \(TP_{max} \approx 79365 \text{ bits/s} / (10^6 \text{ bits/s/Mbps}) \approx 0.079365 \text{ Mbps}\). This calculation demonstrates that even with a high bandwidth, significant latency can drastically reduce the effective data transfer rate in protocols that require acknowledgments for each packet or have limited window sizes. The transmission time of the packet (\(T_{tx} = 0.0008\) s) is much smaller than the minimum RTT (\(0.100\) s), meaning the sender spends most of its time waiting. This highlights the importance of minimizing latency and employing more advanced protocols (like sliding window with larger windows) in high-bandwidth, high-latency environments to achieve better utilization of the available bandwidth. The question probes the understanding of how network conditions, specifically latency and bandwidth, interact with protocol design to determine achievable data rates, a crucial concept for network engineers graduating from Xi’an Technological University Northern College of Information Engineering. The effective throughput is not simply the advertised bandwidth but is constrained by the time it takes for the control signals (acknowledgments) to return, a fundamental principle in network performance analysis taught at institutions like Xi’an Technological University Northern College of Information Engineering.
Incorrect
The scenario describes a system where data packets are transmitted over a network with varying latency. The core concept being tested is the impact of latency on the effective throughput of a reliable data transfer protocol, specifically one that uses acknowledgments. A common model for this is the Stop-and-Wait protocol, or a similar sliding window protocol where the window size is limited by the round-trip time (RTT). In this problem, we are given the following: – Packet size (\(P\)): 1000 bytes – Bandwidth (\(B\)): 10 Mbps (Megabits per second) – Minimum latency (\(L_{min}\)): 50 ms (milliseconds) – Maximum latency (\(L_{max}\)): 150 ms (milliseconds) First, let’s convert all units to be consistent. Bandwidth in bits per second: \(B = 10 \text{ Mbps} = 10 \times 10^6 \text{ bits/s}\). Packet size in bits: \(P = 1000 \text{ bytes} \times 8 \text{ bits/byte} = 8000 \text{ bits}\). Latency in seconds: \(L_{min} = 50 \text{ ms} = 0.050 \text{ s}\), \(L_{max} = 150 \text{ ms} = 0.150 \text{ s}\). The time it takes to transmit a single packet is the packet size divided by the bandwidth: Transmission time (\(T_{tx}\)) = \(P / B = 8000 \text{ bits} / (10 \times 10^6 \text{ bits/s}) = 0.0008 \text{ s}\). In a system like this, the effective throughput is limited by how quickly data can be sent and acknowledged. The round-trip time (RTT) is the time from sending a packet to receiving its acknowledgment. The minimum RTT is \(2 \times L_{min}\) (assuming acknowledgment transmission time is negligible compared to data packet transmission time and propagation delay) and the maximum RTT is \(2 \times L_{max}\). The maximum amount of data that can be “in flight” at any given time is determined by the bandwidth-delay product. For a simple Stop-and-Wait protocol, the sender can only send one packet at a time and must wait for an acknowledgment before sending the next. The time it takes to send one packet and receive its acknowledgment is approximately the transmission time of the packet plus the RTT. The maximum throughput (\(TP_{max}\)) for a system like this, assuming a window size of 1 (like Stop-and-Wait), is effectively limited by the time it takes to send one packet and receive its acknowledgment. The sender is idle for most of the RTT. The total time for one cycle (send packet, wait for ACK) is \(T_{tx} + RTT\). The amount of data sent in this cycle is \(P\). Therefore, the throughput is \(TP = P / (T_{tx} + RTT)\). To find the maximum throughput, we consider the minimum RTT, which is \(2 \times L_{min}\). Maximum Throughput (\(TP_{max}\)) = \(P / (T_{tx} + 2 \times L_{min})\) \(TP_{max} = 8000 \text{ bits} / (0.0008 \text{ s} + 2 \times 0.050 \text{ s})\) \(TP_{max} = 8000 \text{ bits} / (0.0008 \text{ s} + 0.100 \text{ s})\) \(TP_{max} = 8000 \text{ bits} / 0.1008 \text{ s}\) \(TP_{max} \approx 79365 \text{ bits/s}\) Converting this to Mbps: \(TP_{max} \approx 79365 \text{ bits/s} / (10^6 \text{ bits/s/Mbps}) \approx 0.079365 \text{ Mbps}\). This calculation demonstrates that even with a high bandwidth, significant latency can drastically reduce the effective data transfer rate in protocols that require acknowledgments for each packet or have limited window sizes. The transmission time of the packet (\(T_{tx} = 0.0008\) s) is much smaller than the minimum RTT (\(0.100\) s), meaning the sender spends most of its time waiting. This highlights the importance of minimizing latency and employing more advanced protocols (like sliding window with larger windows) in high-bandwidth, high-latency environments to achieve better utilization of the available bandwidth. The question probes the understanding of how network conditions, specifically latency and bandwidth, interact with protocol design to determine achievable data rates, a crucial concept for network engineers graduating from Xi’an Technological University Northern College of Information Engineering. The effective throughput is not simply the advertised bandwidth but is constrained by the time it takes for the control signals (acknowledgments) to return, a fundamental principle in network performance analysis taught at institutions like Xi’an Technological University Northern College of Information Engineering.
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Question 10 of 30
10. Question
A network administrator at Xi’an Technological University Northern College of Information Engineering is tasked with ensuring the seamless operation of a high-demand computational fluid dynamics simulation for a faculty research initiative. The simulation generates substantial real-time data streams and is highly susceptible to network latency and packet loss, which can lead to inaccurate results and significant project delays. The administrator is evaluating various network traffic management strategies to guarantee the performance of this critical research traffic. Which network traffic management approach would most effectively address the immediate need to prioritize these sensitive data flows and minimize their exposure to network congestion?
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The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering attempting to optimize data flow for a critical research project involving large datasets and real-time simulations. The core issue is latency and packet loss impacting the project’s efficiency. The administrator is considering implementing Quality of Service (QoS) mechanisms. To address the problem of latency and packet loss for the research project, the administrator needs to prioritize traffic. This involves identifying which types of data are most sensitive to delay and ensuring they receive preferential treatment. The options presented relate to different network management strategies. Option a) focuses on implementing a strict priority queuing (SPQ) mechanism for the research data. SPQ assigns a fixed priority level to traffic classes, and higher priority queues are always serviced before lower priority queues. This directly tackles latency and packet loss for the most critical traffic, aligning with the project’s needs. Option b) suggests implementing weighted fair queuing (WFQ). While WFQ aims to provide fairness among different traffic flows, it might not offer the strict guarantees needed for real-time simulations if the research data is highly sensitive to even minor variations in delay. It aims for fairness rather than absolute priority. Option c) proposes a simple first-come, first-served (FCFS) queuing discipline. This is the default behavior in many networks and would not address the specific problem of prioritizing sensitive research data, likely exacerbating latency and packet loss for that traffic. Option d) suggests increasing the overall bandwidth of the network. While increased bandwidth can help, it doesn’t inherently solve the problem of prioritizing specific traffic types when congestion occurs. Without a prioritization mechanism, even a wider pipe can still experience delays for sensitive data if less critical traffic consumes a significant portion of the available bandwidth. Therefore, implementing a strict priority queuing mechanism for the research data is the most direct and effective approach to mitigate latency and packet loss for the critical research project at Xi’an Technological University Northern College of Information Engineering.
Incorrect
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering attempting to optimize data flow for a critical research project involving large datasets and real-time simulations. The core issue is latency and packet loss impacting the project’s efficiency. The administrator is considering implementing Quality of Service (QoS) mechanisms. To address the problem of latency and packet loss for the research project, the administrator needs to prioritize traffic. This involves identifying which types of data are most sensitive to delay and ensuring they receive preferential treatment. The options presented relate to different network management strategies. Option a) focuses on implementing a strict priority queuing (SPQ) mechanism for the research data. SPQ assigns a fixed priority level to traffic classes, and higher priority queues are always serviced before lower priority queues. This directly tackles latency and packet loss for the most critical traffic, aligning with the project’s needs. Option b) suggests implementing weighted fair queuing (WFQ). While WFQ aims to provide fairness among different traffic flows, it might not offer the strict guarantees needed for real-time simulations if the research data is highly sensitive to even minor variations in delay. It aims for fairness rather than absolute priority. Option c) proposes a simple first-come, first-served (FCFS) queuing discipline. This is the default behavior in many networks and would not address the specific problem of prioritizing sensitive research data, likely exacerbating latency and packet loss for that traffic. Option d) suggests increasing the overall bandwidth of the network. While increased bandwidth can help, it doesn’t inherently solve the problem of prioritizing specific traffic types when congestion occurs. Without a prioritization mechanism, even a wider pipe can still experience delays for sensitive data if less critical traffic consumes a significant portion of the available bandwidth. Therefore, implementing a strict priority queuing mechanism for the research data is the most direct and effective approach to mitigate latency and packet loss for the critical research project at Xi’an Technological University Northern College of Information Engineering.
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Question 11 of 30
11. Question
A research team at Xi’an Technological University Northern College of Information Engineering is developing a new sensor for environmental monitoring. The sensor captures analog data that, after initial processing, contains frequency components ranging up to \(15 \text{ kHz}\). This analog signal is then digitized using a sampling rate of \(25 \text{ kHz}\). What is the most likely consequence for the fidelity of the captured digital data due to this sampling rate, and what specific phenomenon is responsible?
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The question assesses understanding of the fundamental principles of digital signal processing, specifically focusing on the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In this scenario, the original analog signal contains frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing and ensure perfect reconstruction is \(f_{s,min} = 2 \times f_{max} = 2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The digital system samples this signal at \(25 \text{ kHz}\). Since \(25 \text{ kHz} < 30 \text{ kHz}\), the sampling rate is below the Nyquist rate. When a signal is sampled below its Nyquist rate, higher frequency components in the original signal "fold over" or alias into lower frequencies in the sampled signal. Specifically, frequencies between \(f_s/2\) and \(f_{max}\) will appear as lower frequencies. In this case, \(f_s/2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). The frequency component at \(15 \text{ kHz}\) is greater than \(f_s/2\). The aliased frequency (\(f_{alias}\)) can be calculated using the formula \(f_{alias} = |f – n \cdot f_s|\), where \(f\) is the original frequency and \(n\) is an integer chosen such that \(0 \le f_{alias} < f_s/2\). For \(f = 15 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\), we can find \(n\) such that \(0 \le |15000 – n \cdot 25000| < 12500\). If \(n=1\), \(|15000 – 25000| = |-10000| = 10000\). Since \(10000 < 12500\), the aliased frequency is \(10 \text{ kHz}\). This means the \(15 \text{ kHz}\) component will be indistinguishable from a \(10 \text{ kHz}\) component after sampling at \(25 \text{ kHz}\). This phenomenon, where a higher frequency masquerades as a lower frequency due to undersampling, is known as aliasing. Understanding and mitigating aliasing is a core concept in signal processing education at institutions like Xi'an Technological University Northern College of Information Engineering, particularly for students in information engineering disciplines who will work with digital communication systems and data acquisition.
Incorrect
The question assesses understanding of the fundamental principles of digital signal processing, specifically focusing on the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In this scenario, the original analog signal contains frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing and ensure perfect reconstruction is \(f_{s,min} = 2 \times f_{max} = 2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The digital system samples this signal at \(25 \text{ kHz}\). Since \(25 \text{ kHz} < 30 \text{ kHz}\), the sampling rate is below the Nyquist rate. When a signal is sampled below its Nyquist rate, higher frequency components in the original signal "fold over" or alias into lower frequencies in the sampled signal. Specifically, frequencies between \(f_s/2\) and \(f_{max}\) will appear as lower frequencies. In this case, \(f_s/2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). The frequency component at \(15 \text{ kHz}\) is greater than \(f_s/2\). The aliased frequency (\(f_{alias}\)) can be calculated using the formula \(f_{alias} = |f – n \cdot f_s|\), where \(f\) is the original frequency and \(n\) is an integer chosen such that \(0 \le f_{alias} < f_s/2\). For \(f = 15 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\), we can find \(n\) such that \(0 \le |15000 – n \cdot 25000| < 12500\). If \(n=1\), \(|15000 – 25000| = |-10000| = 10000\). Since \(10000 < 12500\), the aliased frequency is \(10 \text{ kHz}\). This means the \(15 \text{ kHz}\) component will be indistinguishable from a \(10 \text{ kHz}\) component after sampling at \(25 \text{ kHz}\). This phenomenon, where a higher frequency masquerades as a lower frequency due to undersampling, is known as aliasing. Understanding and mitigating aliasing is a core concept in signal processing education at institutions like Xi'an Technological University Northern College of Information Engineering, particularly for students in information engineering disciplines who will work with digital communication systems and data acquisition.
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Question 12 of 30
12. Question
When evaluating the efficacy of various noise reduction techniques for a high-fidelity data transmission system at Xi’an Technological University Northern College of Information Engineering, which fundamental principle most directly underpins the improvement in data integrity?
Correct
The core of this question lies in understanding the principles of signal-to-noise ratio (SNR) and its impact on data integrity in digital communication systems, a fundamental concept within the Information Engineering curriculum at Xi’an Technological University Northern College of Information Engineering. While no explicit calculation is required, the reasoning process involves evaluating how different noise mitigation strategies affect the clarity of the transmitted signal. A higher SNR indicates a stronger signal relative to background noise, leading to more reliable data reception. Techniques that effectively suppress or filter out unwanted interference, thereby increasing the ratio of signal power to noise power, are paramount. Considering the context of advanced information engineering, the focus is on the *fundamental impact* of noise reduction on signal quality. Therefore, strategies that directly enhance the signal’s power relative to the noise floor, or conversely, reduce the noise power without significantly attenuating the signal, are the most impactful. This is often achieved through sophisticated filtering, error correction coding, or modulation schemes designed for robustness. The question probes the candidate’s ability to discern which approach most directly addresses the degradation of information due to noise, a critical skill for designing and analyzing communication systems.
Incorrect
The core of this question lies in understanding the principles of signal-to-noise ratio (SNR) and its impact on data integrity in digital communication systems, a fundamental concept within the Information Engineering curriculum at Xi’an Technological University Northern College of Information Engineering. While no explicit calculation is required, the reasoning process involves evaluating how different noise mitigation strategies affect the clarity of the transmitted signal. A higher SNR indicates a stronger signal relative to background noise, leading to more reliable data reception. Techniques that effectively suppress or filter out unwanted interference, thereby increasing the ratio of signal power to noise power, are paramount. Considering the context of advanced information engineering, the focus is on the *fundamental impact* of noise reduction on signal quality. Therefore, strategies that directly enhance the signal’s power relative to the noise floor, or conversely, reduce the noise power without significantly attenuating the signal, are the most impactful. This is often achieved through sophisticated filtering, error correction coding, or modulation schemes designed for robustness. The question probes the candidate’s ability to discern which approach most directly addresses the degradation of information due to noise, a critical skill for designing and analyzing communication systems.
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Question 13 of 30
13. Question
A digital communication system at Xi’an Technological University Northern College of Information Engineering is experiencing performance issues. The receiver front-end is characterized by a bandwidth of \(5 \times 10^6\) Hz and is subject to thermal noise with a power spectral density of \(10^{-12}\) Watts per Hertz. During a specific transmission interval, the received signal power is measured to be \(10^{-10}\) Watts. What is the signal-to-noise ratio (SNR) at the output of the receiver’s front-end filter, expressed in decibels?
Correct
The question assesses understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept in information engineering. The scenario involves a digital receiver processing a signal. The signal power is given as \(P_s = 10^{-10}\) Watts, and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the receiver’s front-end filter is \(B = 5 \times 10^6\) Hz. First, calculate the total noise power within the bandwidth: Noise Power \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ W/Hz}) \times (5 \times 10^6 \text{ Hz})\) \(P_n = 5 \times 10^{-6}\) Watts Next, calculate the Signal-to-Noise Ratio (SNR) in linear terms: SNR (linear) = \(P_s / P_n\) SNR (linear) = \(10^{-10} \text{ W} / (5 \times 10^{-6} \text{ W})\) SNR (linear) = \(10^{-10} / (5 \times 10^{-6})\) SNR (linear) = \(0.2 \times 10^{-4}\) SNR (linear) = \(2 \times 10^{-5}\) Finally, convert the SNR to decibels (dB): SNR (dB) = \(10 \times \log_{10}(\text{SNR (linear)})\) SNR (dB) = \(10 \times \log_{10}(2 \times 10^{-5})\) SNR (dB) = \(10 \times (\log_{10}(2) + \log_{10}(10^{-5}))\) SNR (dB) = \(10 \times (\log_{10}(2) – 5)\) Using \(\log_{10}(2) \approx 0.301\): SNR (dB) \(\approx 10 \times (0.301 – 5)\) SNR (dB) \(\approx 10 \times (-4.699)\) SNR (dB) \(\approx -46.99\) dB This calculation demonstrates the fundamental relationship between signal power, noise power, bandwidth, and the resulting SNR, a critical metric for evaluating the performance of communication receivers at institutions like Xi’an Technological University Northern College of Information Engineering. Understanding how noise degrades signal quality and how to quantify this degradation is essential for designing efficient and reliable communication systems. The negative decibel value indicates that the noise power is significantly higher than the signal power, which would severely impact the ability to reliably decode transmitted information. This concept is foundational for advanced topics in digital signal processing, error correction coding, and wireless communication protocols taught at the university.
Incorrect
The question assesses understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept in information engineering. The scenario involves a digital receiver processing a signal. The signal power is given as \(P_s = 10^{-10}\) Watts, and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the receiver’s front-end filter is \(B = 5 \times 10^6\) Hz. First, calculate the total noise power within the bandwidth: Noise Power \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ W/Hz}) \times (5 \times 10^6 \text{ Hz})\) \(P_n = 5 \times 10^{-6}\) Watts Next, calculate the Signal-to-Noise Ratio (SNR) in linear terms: SNR (linear) = \(P_s / P_n\) SNR (linear) = \(10^{-10} \text{ W} / (5 \times 10^{-6} \text{ W})\) SNR (linear) = \(10^{-10} / (5 \times 10^{-6})\) SNR (linear) = \(0.2 \times 10^{-4}\) SNR (linear) = \(2 \times 10^{-5}\) Finally, convert the SNR to decibels (dB): SNR (dB) = \(10 \times \log_{10}(\text{SNR (linear)})\) SNR (dB) = \(10 \times \log_{10}(2 \times 10^{-5})\) SNR (dB) = \(10 \times (\log_{10}(2) + \log_{10}(10^{-5}))\) SNR (dB) = \(10 \times (\log_{10}(2) – 5)\) Using \(\log_{10}(2) \approx 0.301\): SNR (dB) \(\approx 10 \times (0.301 – 5)\) SNR (dB) \(\approx 10 \times (-4.699)\) SNR (dB) \(\approx -46.99\) dB This calculation demonstrates the fundamental relationship between signal power, noise power, bandwidth, and the resulting SNR, a critical metric for evaluating the performance of communication receivers at institutions like Xi’an Technological University Northern College of Information Engineering. Understanding how noise degrades signal quality and how to quantify this degradation is essential for designing efficient and reliable communication systems. The negative decibel value indicates that the noise power is significantly higher than the signal power, which would severely impact the ability to reliably decode transmitted information. This concept is foundational for advanced topics in digital signal processing, error correction coding, and wireless communication protocols taught at the university.
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Question 14 of 30
14. Question
A student at Xi’an Technological University Northern College of Information Engineering reports intermittent delays and incomplete data downloads when accessing a research repository hosted on a server located in a different country. Network diagnostics indicate a significant rate of packet loss on the path between the university and the repository. Which protocol’s inherent mechanisms for detecting lost packets and initiating retransmissions is most directly responsible for the observed performance degradation in this scenario?
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The question assesses understanding of network protocol layering and the role of specific protocols within the TCP/IP model, particularly in the context of data transmission and error handling. The scenario describes a situation where a user at Xi’an Technological University Northern College of Information Engineering is experiencing slow data retrieval from a remote server. The core issue is identified as packet loss, which triggers retransmissions. Packet loss at the Transport Layer, specifically within the Transmission Control Protocol (TCP), leads to the sender retransmitting unacknowledged segments. This retransmission process, while crucial for reliability, directly impacts perceived performance by introducing delays. The User Datagram Protocol (UDP), in contrast, does not inherently provide mechanisms for detecting or recovering from packet loss, making it unsuitable for applications requiring guaranteed delivery and thus less likely to be the primary cause of *detected* retransmissions due to loss. The Network Layer (IP) is responsible for routing packets but does not manage reliable delivery or retransmissions; it simply forwards packets. The Data Link Layer handles error detection and correction within a local network segment, but packet loss between distinct networks (as implied by accessing a remote server) is typically managed at higher layers. The Application Layer protocols (like HTTP or FTP) operate on top of the transport layer and rely on its services for reliable data transfer. Therefore, the retransmission mechanism directly addressing packet loss for reliable data streams is a function of TCP.
Incorrect
The question assesses understanding of network protocol layering and the role of specific protocols within the TCP/IP model, particularly in the context of data transmission and error handling. The scenario describes a situation where a user at Xi’an Technological University Northern College of Information Engineering is experiencing slow data retrieval from a remote server. The core issue is identified as packet loss, which triggers retransmissions. Packet loss at the Transport Layer, specifically within the Transmission Control Protocol (TCP), leads to the sender retransmitting unacknowledged segments. This retransmission process, while crucial for reliability, directly impacts perceived performance by introducing delays. The User Datagram Protocol (UDP), in contrast, does not inherently provide mechanisms for detecting or recovering from packet loss, making it unsuitable for applications requiring guaranteed delivery and thus less likely to be the primary cause of *detected* retransmissions due to loss. The Network Layer (IP) is responsible for routing packets but does not manage reliable delivery or retransmissions; it simply forwards packets. The Data Link Layer handles error detection and correction within a local network segment, but packet loss between distinct networks (as implied by accessing a remote server) is typically managed at higher layers. The Application Layer protocols (like HTTP or FTP) operate on top of the transport layer and rely on its services for reliable data transfer. Therefore, the retransmission mechanism directly addressing packet loss for reliable data streams is a function of TCP.
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Question 15 of 30
15. Question
A network administrator at Xi’an Technological University Northern College of Information Engineering is tasked with configuring the network infrastructure to support a critical research initiative focused on real-time environmental monitoring. This initiative involves collecting and processing vast quantities of sensor data from a geographically dispersed array of devices, necessitating a communication protocol that prioritizes rapid data dissemination and minimizes transmission delays for immediate analysis. Considering the stringent latency requirements and the potential for the research application to manage data integrity at the application level, which network transport protocol would be most judiciously selected to facilitate this high-performance data flow?
Correct
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering attempting to optimize data flow for a new research project involving large datasets. The project requires high throughput and low latency for real-time analysis of sensor data from a distributed network of environmental monitoring stations. The administrator is considering different network protocols. The core of the problem lies in understanding the trade-offs between connection-oriented and connectionless protocols in the context of real-time data transmission for research. Connection-oriented protocols, like TCP, establish a dedicated connection before data transfer, ensuring reliable delivery through acknowledgments and retransmissions. This reliability comes at the cost of increased overhead and potential latency, as connection setup and teardown take time, and retransmissions can delay subsequent packets. Connectionless protocols, such as UDP, send data packets without establishing a prior connection. This offers lower overhead and potentially lower latency, making it suitable for real-time applications where occasional packet loss is acceptable or can be handled at the application layer. For a research project demanding real-time analysis of sensor data, where timely delivery of information is paramount and the application layer might implement its own error checking or tolerance for minor data loss, a connectionless protocol is generally more advantageous. The overhead of TCP’s connection management and guaranteed delivery mechanisms would introduce unacceptable delays for real-time processing. While UDP doesn’t guarantee delivery, the research application can be designed to cope with this, perhaps by prioritizing newer data or using application-level techniques to infer missing information. Therefore, UDP is the more appropriate choice for this specific scenario at Xi’an Technological University Northern College of Information Engineering.
Incorrect
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering attempting to optimize data flow for a new research project involving large datasets. The project requires high throughput and low latency for real-time analysis of sensor data from a distributed network of environmental monitoring stations. The administrator is considering different network protocols. The core of the problem lies in understanding the trade-offs between connection-oriented and connectionless protocols in the context of real-time data transmission for research. Connection-oriented protocols, like TCP, establish a dedicated connection before data transfer, ensuring reliable delivery through acknowledgments and retransmissions. This reliability comes at the cost of increased overhead and potential latency, as connection setup and teardown take time, and retransmissions can delay subsequent packets. Connectionless protocols, such as UDP, send data packets without establishing a prior connection. This offers lower overhead and potentially lower latency, making it suitable for real-time applications where occasional packet loss is acceptable or can be handled at the application layer. For a research project demanding real-time analysis of sensor data, where timely delivery of information is paramount and the application layer might implement its own error checking or tolerance for minor data loss, a connectionless protocol is generally more advantageous. The overhead of TCP’s connection management and guaranteed delivery mechanisms would introduce unacceptable delays for real-time processing. While UDP doesn’t guarantee delivery, the research application can be designed to cope with this, perhaps by prioritizing newer data or using application-level techniques to infer missing information. Therefore, UDP is the more appropriate choice for this specific scenario at Xi’an Technological University Northern College of Information Engineering.
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Question 16 of 30
16. Question
When evaluating the performance of a digital communication link designed for high-speed data transfer, as might be implemented in advanced network infrastructure projects at Xi’an Technological University Northern College of Information Engineering, what is the most direct and fundamental consequence of a substantial increase in the signal-to-noise ratio (SNR) at the receiver, assuming all other system parameters remain constant?
Correct
The question probes the understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept in information engineering. While no direct calculation is required, the underlying principle involves how signal power and noise power interact to determine the clarity of information transmission. A higher SNR indicates a clearer signal relative to background noise. In digital systems, this directly impacts the probability of bit errors. Consider a scenario where a digital signal is transmitted. The received signal consists of the original signal plus additive white Gaussian noise (AWGN). The quality of the received signal is often quantified by the ratio of the signal power (\(P_s\)) to the noise power (\(P_n\)), which is the SNR. A higher SNR means that the signal is significantly stronger than the noise, leading to fewer errors in decoding the transmitted bits. Conversely, a low SNR implies that the noise is comparable to or stronger than the signal, making it difficult to distinguish between transmitted bits (e.g., a ‘0’ and a ‘1’), thus increasing the bit error rate (BER). In the context of Xi’an Technological University Northern College of Information Engineering’s curriculum, understanding the relationship between SNR and BER is crucial for designing efficient and reliable communication systems, such as those used in wireless networks, satellite communications, and data transmission. The ability to interpret how system parameters affect SNR and, consequently, data integrity is a fundamental skill. For instance, increasing transmission power or using directional antennas can boost signal strength, thereby improving SNR. Similarly, employing error-correction coding techniques can mitigate the impact of noise, effectively lowering the BER for a given SNR. Therefore, the most direct consequence of a significantly improved SNR in a digital communication system is a reduction in the likelihood of errors occurring during data reception.
Incorrect
The question probes the understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept in information engineering. While no direct calculation is required, the underlying principle involves how signal power and noise power interact to determine the clarity of information transmission. A higher SNR indicates a clearer signal relative to background noise. In digital systems, this directly impacts the probability of bit errors. Consider a scenario where a digital signal is transmitted. The received signal consists of the original signal plus additive white Gaussian noise (AWGN). The quality of the received signal is often quantified by the ratio of the signal power (\(P_s\)) to the noise power (\(P_n\)), which is the SNR. A higher SNR means that the signal is significantly stronger than the noise, leading to fewer errors in decoding the transmitted bits. Conversely, a low SNR implies that the noise is comparable to or stronger than the signal, making it difficult to distinguish between transmitted bits (e.g., a ‘0’ and a ‘1’), thus increasing the bit error rate (BER). In the context of Xi’an Technological University Northern College of Information Engineering’s curriculum, understanding the relationship between SNR and BER is crucial for designing efficient and reliable communication systems, such as those used in wireless networks, satellite communications, and data transmission. The ability to interpret how system parameters affect SNR and, consequently, data integrity is a fundamental skill. For instance, increasing transmission power or using directional antennas can boost signal strength, thereby improving SNR. Similarly, employing error-correction coding techniques can mitigate the impact of noise, effectively lowering the BER for a given SNR. Therefore, the most direct consequence of a significantly improved SNR in a digital communication system is a reduction in the likelihood of errors occurring during data reception.
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Question 17 of 30
17. Question
Consider a scenario where a critical data stream is being transmitted wirelessly between two nodes within the campus of Xi’an Technological University Northern College of Information Engineering, specifically during a period of high network activity and potential interference from nearby electronic equipment. Which of the following factors, if significantly increased, would most directly and detrimentally impact the fidelity and interpretability of the received data, assuming all other transmission parameters remain constant?
Correct
The core concept tested here is the understanding of signal-to-noise ratio (SNR) in the context of digital communication, a fundamental area within information engineering. While no direct calculation is required, the scenario implies a need to evaluate the impact of noise on signal integrity. A higher SNR indicates a clearer signal relative to background interference, which is crucial for reliable data transmission and processing. In digital systems, this translates to fewer bit errors. The question probes the candidate’s grasp of how environmental factors, such as electromagnetic interference (EMI) or thermal noise, can degrade the quality of an information signal. A robust understanding of signal processing principles and the challenges faced in real-world communication environments, particularly those relevant to advanced information engineering studies at Xi’an Technological University Northern College of Information Engineering, is essential. The ability to discern the primary factor affecting signal clarity in a noisy environment, without resorting to specific numerical values, demonstrates a conceptual mastery of the subject. The emphasis is on identifying the most direct and impactful element that degrades the signal’s intelligibility.
Incorrect
The core concept tested here is the understanding of signal-to-noise ratio (SNR) in the context of digital communication, a fundamental area within information engineering. While no direct calculation is required, the scenario implies a need to evaluate the impact of noise on signal integrity. A higher SNR indicates a clearer signal relative to background interference, which is crucial for reliable data transmission and processing. In digital systems, this translates to fewer bit errors. The question probes the candidate’s grasp of how environmental factors, such as electromagnetic interference (EMI) or thermal noise, can degrade the quality of an information signal. A robust understanding of signal processing principles and the challenges faced in real-world communication environments, particularly those relevant to advanced information engineering studies at Xi’an Technological University Northern College of Information Engineering, is essential. The ability to discern the primary factor affecting signal clarity in a noisy environment, without resorting to specific numerical values, demonstrates a conceptual mastery of the subject. The emphasis is on identifying the most direct and impactful element that degrades the signal’s intelligibility.
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Question 18 of 30
18. Question
Consider a scenario at Xi’an Technological University Northern College of Information Engineering where a research team is developing a new digital audio processing system. They are tasked with digitizing an analog audio signal that contains harmonic components extending up to \(15 \text{ kHz}\). The team decides to implement a sampling process with a sampling frequency of \(25 \text{ kHz}\). What is the fundamental limitation imposed by this sampling rate on the fidelity of the reconstructed analog signal, and why?
Correct
The question probes the understanding of signal processing principles, specifically concerning the impact of sampling rate on signal reconstruction. A fundamental concept in digital signal processing is the Nyquist-Shannon sampling theorem, which states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the original analog signal. This minimum sampling rate is known as the Nyquist rate, where \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal contains frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required for perfect reconstruction is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The digital system samples the signal at \(25 \text{ kHz}\). Since \(25 \text{ kHz} < 30 \text{ kHz}\), the sampling rate is below the Nyquist rate. When sampling occurs below the Nyquist rate, aliasing occurs. Aliasing is a phenomenon where higher frequencies in the analog signal are incorrectly represented as lower frequencies in the sampled digital signal. This distortion makes it impossible to accurately reconstruct the original analog signal from the digital samples. The higher frequencies above \(f_s/2\) (in this case, above \(12.5 \text{ kHz}\)) will fold back into the frequency band below \(12.5 \text{ kHz}\), corrupting the information contained within that band. Consequently, the original analog signal cannot be perfectly reconstructed.
Incorrect
The question probes the understanding of signal processing principles, specifically concerning the impact of sampling rate on signal reconstruction. A fundamental concept in digital signal processing is the Nyquist-Shannon sampling theorem, which states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the original analog signal. This minimum sampling rate is known as the Nyquist rate, where \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal contains frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required for perfect reconstruction is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The digital system samples the signal at \(25 \text{ kHz}\). Since \(25 \text{ kHz} < 30 \text{ kHz}\), the sampling rate is below the Nyquist rate. When sampling occurs below the Nyquist rate, aliasing occurs. Aliasing is a phenomenon where higher frequencies in the analog signal are incorrectly represented as lower frequencies in the sampled digital signal. This distortion makes it impossible to accurately reconstruct the original analog signal from the digital samples. The higher frequencies above \(f_s/2\) (in this case, above \(12.5 \text{ kHz}\)) will fold back into the frequency band below \(12.5 \text{ kHz}\), corrupting the information contained within that band. Consequently, the original analog signal cannot be perfectly reconstructed.
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Question 19 of 30
19. Question
When designing a digital signal processing system at Xi’an Technological University Northern College of Information Engineering, a critical step involves converting an analog signal to a digital format. If the analog signal is known to contain a broad spectrum of frequencies, what fundamental principle must be adhered to during the analog-to-digital conversion process to ensure that high-frequency components do not masquerade as lower-frequency components in the resulting digital representation?
Correct
The question assesses understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. Aliasing occurs when the sampling rate is less than twice the highest frequency component of the analog signal, leading to the misrepresentation of higher frequencies as lower ones. To prevent aliasing, a low-pass filter, known as an anti-aliasing filter, is applied to the analog signal *before* sampling. This filter attenuates frequencies above the Nyquist frequency (\(f_s/2\), where \(f_s\) is the sampling frequency) to ensure that only frequencies below half the sampling rate are present in the signal being sampled. Consider a scenario where an analog signal contains frequency components up to \(f_{max}\). If this signal is sampled at a rate \(f_s\), the Nyquist-Shannon sampling theorem states that \(f_s\) must be greater than \(2f_{max}\) to perfectly reconstruct the signal. If \(f_s \le 2f_{max}\), aliasing will occur. For instance, if a signal has a component at \(f_{signal}\) and the sampling frequency is \(f_s\), a frequency component at \(f_{signal}\) will appear as \(|f_{signal} – k \cdot f_s|\) for some integer \(k\), where \(k \cdot f_s\) is the closest multiple of the sampling frequency to \(f_{signal}\). If \(f_{signal} > f_s/2\), it will alias to a frequency in the range \([0, f_s/2]\). An anti-aliasing filter is a low-pass filter placed before the analog-to-digital converter (ADC). Its cutoff frequency is typically set slightly below \(f_s/2\). This filter removes or significantly attenuates any frequency components in the analog signal that are above \(f_s/2\). By doing so, it ensures that the signal presented to the sampler contains no frequencies that would cause aliasing. Therefore, the primary role of an anti-aliasing filter is to remove spectral content above the Nyquist frequency to prevent the misinterpretation of these frequencies as lower frequencies during the sampling process. This is crucial in digital signal processing applications, including those studied at Xi’an Technological University Northern College of Information Engineering, to maintain signal integrity and accuracy.
Incorrect
The question assesses understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. Aliasing occurs when the sampling rate is less than twice the highest frequency component of the analog signal, leading to the misrepresentation of higher frequencies as lower ones. To prevent aliasing, a low-pass filter, known as an anti-aliasing filter, is applied to the analog signal *before* sampling. This filter attenuates frequencies above the Nyquist frequency (\(f_s/2\), where \(f_s\) is the sampling frequency) to ensure that only frequencies below half the sampling rate are present in the signal being sampled. Consider a scenario where an analog signal contains frequency components up to \(f_{max}\). If this signal is sampled at a rate \(f_s\), the Nyquist-Shannon sampling theorem states that \(f_s\) must be greater than \(2f_{max}\) to perfectly reconstruct the signal. If \(f_s \le 2f_{max}\), aliasing will occur. For instance, if a signal has a component at \(f_{signal}\) and the sampling frequency is \(f_s\), a frequency component at \(f_{signal}\) will appear as \(|f_{signal} – k \cdot f_s|\) for some integer \(k\), where \(k \cdot f_s\) is the closest multiple of the sampling frequency to \(f_{signal}\). If \(f_{signal} > f_s/2\), it will alias to a frequency in the range \([0, f_s/2]\). An anti-aliasing filter is a low-pass filter placed before the analog-to-digital converter (ADC). Its cutoff frequency is typically set slightly below \(f_s/2\). This filter removes or significantly attenuates any frequency components in the analog signal that are above \(f_s/2\). By doing so, it ensures that the signal presented to the sampler contains no frequencies that would cause aliasing. Therefore, the primary role of an anti-aliasing filter is to remove spectral content above the Nyquist frequency to prevent the misinterpretation of these frequencies as lower frequencies during the sampling process. This is crucial in digital signal processing applications, including those studied at Xi’an Technological University Northern College of Information Engineering, to maintain signal integrity and accuracy.
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Question 20 of 30
20. Question
A network administrator at Xi’an Technological University Northern College of Information Engineering is tasked with enhancing the performance of a high-priority research initiative that relies on real-time data streaming and collaborative analysis of large datasets. The current network infrastructure is experiencing intermittent packet loss and significant latency, which are disrupting the research workflow. The administrator is evaluating network management strategies to ensure the smooth and efficient transfer of critical research data and to maintain seamless communication among distributed research team members. Which network management strategy would be most effective in prioritizing the sensitive research traffic and mitigating the impact of network congestion on this specific project?
Correct
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering attempting to optimize data flow for a critical research project involving large datasets. The core issue is latency and packet loss impacting real-time collaboration and data transfer speeds. The administrator is considering implementing Quality of Service (QoS) mechanisms. To address the problem of latency and packet loss for the research project, the administrator needs to prioritize traffic. This involves identifying the types of data that are most sensitive to delay and ensuring they receive preferential treatment. For instance, real-time video conferencing for collaborative analysis is highly susceptible to jitter and delay, as is the transmission of critical sensor data that requires immediate processing. Conversely, bulk data transfers for archival purposes, while important, can tolerate higher latency. The most effective approach to manage this is through traffic shaping and policing, often implemented using techniques like Weighted Fair Queuing (WFQ) or DiffServ. WFQ ensures that bandwidth is allocated proportionally to different traffic classes, preventing any single class from monopolizing resources. DiffServ, on the other hand, classifies traffic into different service levels (e.g., Expedited Forwarding for low-latency, assured forwarding for reliable delivery) and applies specific forwarding behaviors. Considering the need for both low latency for interactive sessions and reliable delivery for data integrity, a differentiated services approach is most suitable. This allows for granular control over how different types of traffic are treated based on their sensitivity to delay, jitter, and packet loss. By classifying the research data and collaborative traffic into higher priority classes, the network can ensure these packets are processed and forwarded with minimal delay, thereby improving the overall performance and usability for the research team. The other options, while related to network management, do not directly address the specific problem of prioritizing sensitive research data flow in the face of potential congestion and packet loss as effectively as a well-configured QoS policy based on traffic differentiation.
Incorrect
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering attempting to optimize data flow for a critical research project involving large datasets. The core issue is latency and packet loss impacting real-time collaboration and data transfer speeds. The administrator is considering implementing Quality of Service (QoS) mechanisms. To address the problem of latency and packet loss for the research project, the administrator needs to prioritize traffic. This involves identifying the types of data that are most sensitive to delay and ensuring they receive preferential treatment. For instance, real-time video conferencing for collaborative analysis is highly susceptible to jitter and delay, as is the transmission of critical sensor data that requires immediate processing. Conversely, bulk data transfers for archival purposes, while important, can tolerate higher latency. The most effective approach to manage this is through traffic shaping and policing, often implemented using techniques like Weighted Fair Queuing (WFQ) or DiffServ. WFQ ensures that bandwidth is allocated proportionally to different traffic classes, preventing any single class from monopolizing resources. DiffServ, on the other hand, classifies traffic into different service levels (e.g., Expedited Forwarding for low-latency, assured forwarding for reliable delivery) and applies specific forwarding behaviors. Considering the need for both low latency for interactive sessions and reliable delivery for data integrity, a differentiated services approach is most suitable. This allows for granular control over how different types of traffic are treated based on their sensitivity to delay, jitter, and packet loss. By classifying the research data and collaborative traffic into higher priority classes, the network can ensure these packets are processed and forwarded with minimal delay, thereby improving the overall performance and usability for the research team. The other options, while related to network management, do not directly address the specific problem of prioritizing sensitive research data flow in the face of potential congestion and packet loss as effectively as a well-configured QoS policy based on traffic differentiation.
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Question 21 of 30
21. Question
Consider a scenario where a critical data stream is being transmitted wirelessly between two nodes within the Xi’an Technological University Northern College of Information Engineering campus network. The transmission medium is subject to intermittent electromagnetic interference from nearby experimental equipment. If the signal-to-noise ratio (SNR) at the receiving node drops significantly, what is the most direct and fundamental consequence on the reliability of the received data, assuming no adaptive error correction mechanisms are in place to compensate for the degradation?
Correct
The core principle being tested here is the understanding of signal-to-noise ratio (SNR) and its impact on data transmission reliability, particularly in the context of digital communication systems as studied at institutions like Xi’an Technological University Northern College of Information Engineering. While no direct calculation is required for the final answer, the underlying concept involves how increased noise power relative to signal power degrades the quality of information. A higher SNR indicates a stronger signal compared to background noise, leading to fewer errors in data reception. Conversely, a lower SNR means the noise is more dominant, increasing the probability of bit errors. In digital communication, the bit error rate (BER) is inversely related to the SNR. As the SNR decreases, the BER tends to increase. This is because the receiver has a harder time distinguishing between the intended signal bits (0s and 1s) and the random fluctuations introduced by noise. Techniques like error correction coding are employed to mitigate the effects of low SNR, but they have their limits. Therefore, to maintain a high level of data integrity and minimize the need for retransmissions, engineers strive to maximize the SNR at the receiver. This can be achieved by increasing the transmitted signal power, reducing the bandwidth of the communication channel (which also reduces noise power), or employing more sensitive receiving equipment. The question probes the candidate’s grasp of this fundamental trade-off in communication system design, a crucial area for information engineering.
Incorrect
The core principle being tested here is the understanding of signal-to-noise ratio (SNR) and its impact on data transmission reliability, particularly in the context of digital communication systems as studied at institutions like Xi’an Technological University Northern College of Information Engineering. While no direct calculation is required for the final answer, the underlying concept involves how increased noise power relative to signal power degrades the quality of information. A higher SNR indicates a stronger signal compared to background noise, leading to fewer errors in data reception. Conversely, a lower SNR means the noise is more dominant, increasing the probability of bit errors. In digital communication, the bit error rate (BER) is inversely related to the SNR. As the SNR decreases, the BER tends to increase. This is because the receiver has a harder time distinguishing between the intended signal bits (0s and 1s) and the random fluctuations introduced by noise. Techniques like error correction coding are employed to mitigate the effects of low SNR, but they have their limits. Therefore, to maintain a high level of data integrity and minimize the need for retransmissions, engineers strive to maximize the SNR at the receiver. This can be achieved by increasing the transmitted signal power, reducing the bandwidth of the communication channel (which also reduces noise power), or employing more sensitive receiving equipment. The question probes the candidate’s grasp of this fundamental trade-off in communication system design, a crucial area for information engineering.
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Question 22 of 30
22. Question
Consider a digital communication scenario at Xi’an Technological University Northern College of Information Engineering where a signal is transmitted over a channel with a bandwidth of \(5 \times 10^6\) Hz. The received signal power is measured to be \(10^{-9}\) Watts. The noise in the channel has a power spectral density of \(10^{-12}\) Watts/Hz. What is the signal-to-noise ratio (SNR) in decibels (dB) for this transmission?
Correct
The question probes the understanding of signal-to-noise ratio (SNR) in digital communication systems, a core concept for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital transmission where the received signal power is \(P_s = 10^{-9}\) Watts and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the channel is \(B = 5 \times 10^6\) Hz. The total noise power \(P_n\) in the channel is calculated by multiplying the noise power spectral density by the bandwidth: \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ Watts/Hz}) \times (5 \times 10^6 \text{ Hz})\) \(P_n = 5 \times 10^{-6}\) Watts The signal-to-noise ratio (SNR) is then the ratio of the received signal power to the total noise power: \(SNR = \frac{P_s}{P_n}\) \(SNR = \frac{10^{-9} \text{ Watts}}{5 \times 10^{-6} \text{ Watts}}\) \(SNR = \frac{1}{5000}\) \(SNR = 0.0002\) To express this in decibels (dB), we use the formula: \(SNR_{dB} = 10 \log_{10}(SNR)\) \(SNR_{dB} = 10 \log_{10}(0.0002)\) \(SNR_{dB} = 10 \log_{10}(2 \times 10^{-4})\) \(SNR_{dB} = 10 (\log_{10}(2) + \log_{10}(10^{-4}))\) \(SNR_{dB} = 10 (\log_{10}(2) – 4)\) Using \(\log_{10}(2) \approx 0.301\): \(SNR_{dB} \approx 10 (0.301 – 4)\) \(SNR_{dB} \approx 10 (-3.699)\) \(SNR_{dB} \approx -36.99\) dB The question requires understanding how signal and noise powers interact within a given bandwidth to determine the quality of a communication channel, a fundamental principle in information engineering. A negative SNR in decibels indicates that the noise power is significantly greater than the signal power, which would severely degrade the performance of any digital communication system operating under these conditions. This concept is crucial for designing robust communication protocols and understanding the limitations imposed by channel impairments, directly relevant to the curriculum at Xi’an Technological University Northern College of Information Engineering. The ability to calculate and interpret SNR is essential for analyzing system performance, optimizing data rates, and ensuring reliable information transfer, all key aspects of modern information engineering.
Incorrect
The question probes the understanding of signal-to-noise ratio (SNR) in digital communication systems, a core concept for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital transmission where the received signal power is \(P_s = 10^{-9}\) Watts and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the channel is \(B = 5 \times 10^6\) Hz. The total noise power \(P_n\) in the channel is calculated by multiplying the noise power spectral density by the bandwidth: \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ Watts/Hz}) \times (5 \times 10^6 \text{ Hz})\) \(P_n = 5 \times 10^{-6}\) Watts The signal-to-noise ratio (SNR) is then the ratio of the received signal power to the total noise power: \(SNR = \frac{P_s}{P_n}\) \(SNR = \frac{10^{-9} \text{ Watts}}{5 \times 10^{-6} \text{ Watts}}\) \(SNR = \frac{1}{5000}\) \(SNR = 0.0002\) To express this in decibels (dB), we use the formula: \(SNR_{dB} = 10 \log_{10}(SNR)\) \(SNR_{dB} = 10 \log_{10}(0.0002)\) \(SNR_{dB} = 10 \log_{10}(2 \times 10^{-4})\) \(SNR_{dB} = 10 (\log_{10}(2) + \log_{10}(10^{-4}))\) \(SNR_{dB} = 10 (\log_{10}(2) – 4)\) Using \(\log_{10}(2) \approx 0.301\): \(SNR_{dB} \approx 10 (0.301 – 4)\) \(SNR_{dB} \approx 10 (-3.699)\) \(SNR_{dB} \approx -36.99\) dB The question requires understanding how signal and noise powers interact within a given bandwidth to determine the quality of a communication channel, a fundamental principle in information engineering. A negative SNR in decibels indicates that the noise power is significantly greater than the signal power, which would severely degrade the performance of any digital communication system operating under these conditions. This concept is crucial for designing robust communication protocols and understanding the limitations imposed by channel impairments, directly relevant to the curriculum at Xi’an Technological University Northern College of Information Engineering. The ability to calculate and interpret SNR is essential for analyzing system performance, optimizing data rates, and ensuring reliable information transfer, all key aspects of modern information engineering.
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Question 23 of 30
23. Question
Consider a scenario where a research team at Xi’an Technological University Northern College of Information Engineering is developing a new audio processing module. They are tasked with digitizing a continuous-time audio signal that contains frequency components up to 15 kHz. The team decides to implement a sampling process using a sampling frequency of 20 kHz. What is the most accurate description of the outcome for the frequency component at 15 kHz within the digitized signal?
Correct
The question assesses understanding of signal processing principles, specifically the Nyquist-Shannon sampling theorem and its practical implications in digital signal processing, a core area for students at Xi’an Technological University Northern College of Information Engineering. The theorem states that to perfectly reconstruct a signal, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In this scenario, a continuous-time signal with a maximum frequency of 15 kHz is being digitized. To avoid aliasing, which is the distortion that occurs when the sampling frequency is too low, the sampling frequency must be greater than or equal to twice the maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks about the consequence of sampling at a rate *below* this minimum requirement. If the sampling frequency \(f_s\) is less than \(2f_{max}\), higher frequency components in the original signal will be misrepresented as lower frequencies in the sampled signal. This phenomenon is called aliasing. Aliasing leads to a loss of information and distortion, making accurate reconstruction of the original analog signal impossible. Specifically, frequencies above \(f_s/2\) (the Nyquist frequency) will appear as lower frequencies within the range \(0\) to \(f_s/2\). For instance, a frequency of \(f\) will be aliased to \(|f – k f_s|\) for some integer \(k\) such that the result is in the range \([0, f_s/2]\). In this case, sampling at 20 kHz means the Nyquist frequency is 10 kHz. A 15 kHz signal component would be aliased. The closest frequency to 15 kHz that is less than 10 kHz and can be represented by aliasing from 15 kHz with a 20 kHz sampling rate is \(|15 \text{ kHz} – 1 \times 20 \text{ kHz}| = |-5 \text{ kHz}| = 5 \text{ kHz}\). Thus, the 15 kHz component would incorrectly appear as a 5 kHz component in the digital representation. This demonstrates a fundamental limitation in digital signal processing when sampling is not performed at an adequate rate, directly impacting the fidelity of information representation and processing at institutions like Xi’an Technological University Northern College of Information Engineering.
Incorrect
The question assesses understanding of signal processing principles, specifically the Nyquist-Shannon sampling theorem and its practical implications in digital signal processing, a core area for students at Xi’an Technological University Northern College of Information Engineering. The theorem states that to perfectly reconstruct a signal, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In this scenario, a continuous-time signal with a maximum frequency of 15 kHz is being digitized. To avoid aliasing, which is the distortion that occurs when the sampling frequency is too low, the sampling frequency must be greater than or equal to twice the maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks about the consequence of sampling at a rate *below* this minimum requirement. If the sampling frequency \(f_s\) is less than \(2f_{max}\), higher frequency components in the original signal will be misrepresented as lower frequencies in the sampled signal. This phenomenon is called aliasing. Aliasing leads to a loss of information and distortion, making accurate reconstruction of the original analog signal impossible. Specifically, frequencies above \(f_s/2\) (the Nyquist frequency) will appear as lower frequencies within the range \(0\) to \(f_s/2\). For instance, a frequency of \(f\) will be aliased to \(|f – k f_s|\) for some integer \(k\) such that the result is in the range \([0, f_s/2]\). In this case, sampling at 20 kHz means the Nyquist frequency is 10 kHz. A 15 kHz signal component would be aliased. The closest frequency to 15 kHz that is less than 10 kHz and can be represented by aliasing from 15 kHz with a 20 kHz sampling rate is \(|15 \text{ kHz} – 1 \times 20 \text{ kHz}| = |-5 \text{ kHz}| = 5 \text{ kHz}\). Thus, the 15 kHz component would incorrectly appear as a 5 kHz component in the digital representation. This demonstrates a fundamental limitation in digital signal processing when sampling is not performed at an adequate rate, directly impacting the fidelity of information representation and processing at institutions like Xi’an Technological University Northern College of Information Engineering.
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Question 24 of 30
24. Question
A network engineer at Xi’an Technological University Northern College of Information Engineering is designing a network infrastructure to support a new high-performance computing cluster for advanced materials science research. This cluster generates massive amounts of simulation data that require rapid transfer and low latency for real-time visualization and analysis. To ensure the research data’s timely delivery and prevent degradation due to other network activities, the engineer is considering implementing a comprehensive Quality of Service (QoS) framework. Considering the critical nature of this research traffic, which fundamental QoS component is primarily responsible for actively prioritizing and expediting the transmission of these high-priority data packets over less time-sensitive traffic when network links become saturated?
Correct
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering tasked with optimizing data flow for a new research project involving large datasets. The project requires high bandwidth and low latency for real-time data analysis. The administrator is considering implementing a Quality of Service (QoS) policy. The core of QoS is to prioritize certain types of network traffic over others to ensure performance for critical applications. In this context, the research data analysis is the critical application. When evaluating QoS mechanisms, several techniques are employed. Classification and marking are the initial steps, identifying and labeling traffic based on predefined criteria (e.g., application type, source/destination IP, port numbers). Congestion management then uses queuing algorithms (like Weighted Fair Queuing or Strict Priority Queuing) to order packets for transmission when the network is overloaded. Congestion avoidance mechanisms (like Random Early Detection – RED) attempt to prevent congestion before it occurs by dropping packets probabilistically. Shaping and policing are used to control the rate of traffic, ensuring it adheres to bandwidth limits. The question asks which QoS component is *most directly* responsible for ensuring that the high-priority research data packets are processed and transmitted ahead of less critical traffic during periods of network congestion. While all components play a role in overall QoS, the mechanism that actively manages the order of packet transmission based on priority is congestion management, specifically through queuing strategies. Congestion avoidance aims to prevent queues from becoming too long, shaping and policing control traffic rates, and classification/marking are preparatory steps. Congestion management, through its queuing algorithms, is the direct enforcer of priority during congestion. Therefore, the most appropriate answer is congestion management.
Incorrect
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering tasked with optimizing data flow for a new research project involving large datasets. The project requires high bandwidth and low latency for real-time data analysis. The administrator is considering implementing a Quality of Service (QoS) policy. The core of QoS is to prioritize certain types of network traffic over others to ensure performance for critical applications. In this context, the research data analysis is the critical application. When evaluating QoS mechanisms, several techniques are employed. Classification and marking are the initial steps, identifying and labeling traffic based on predefined criteria (e.g., application type, source/destination IP, port numbers). Congestion management then uses queuing algorithms (like Weighted Fair Queuing or Strict Priority Queuing) to order packets for transmission when the network is overloaded. Congestion avoidance mechanisms (like Random Early Detection – RED) attempt to prevent congestion before it occurs by dropping packets probabilistically. Shaping and policing are used to control the rate of traffic, ensuring it adheres to bandwidth limits. The question asks which QoS component is *most directly* responsible for ensuring that the high-priority research data packets are processed and transmitted ahead of less critical traffic during periods of network congestion. While all components play a role in overall QoS, the mechanism that actively manages the order of packet transmission based on priority is congestion management, specifically through queuing strategies. Congestion avoidance aims to prevent queues from becoming too long, shaping and policing control traffic rates, and classification/marking are preparatory steps. Congestion management, through its queuing algorithms, is the direct enforcer of priority during congestion. Therefore, the most appropriate answer is congestion management.
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Question 25 of 30
25. Question
Consider a scenario where an analog signal, containing frequency components up to 15 kHz, is sampled by a system at Xi’an Technological University Northern College of Information Engineering Entrance Exam. The sampling process utilizes a frequency of 25 kHz. If the original signal contains a specific frequency component at 14 kHz, what will be the apparent frequency of this component after sampling, assuming no anti-aliasing filter is employed?
Correct
The question assesses understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In this scenario, the analog signal contains frequency components up to 15 kHz. Therefore, the minimum sampling frequency required to avoid aliasing is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The actual sampling frequency used is 25 kHz. Since 25 kHz is less than the required Nyquist rate of 30 kHz, aliasing will occur. Aliasing causes higher frequencies in the original signal to be misrepresented as lower frequencies in the sampled signal. Specifically, frequencies above \(f_s/2\) (the Nyquist frequency, which is \(25 \text{ kHz} / 2 = 12.5 \text{ kHz}\)) will fold back into the frequency range below \(12.5 \text{ kHz}\). A frequency component at 14 kHz, which is above the Nyquist frequency of 12.5 kHz, will be aliased. The aliased frequency (\(f_{alias}\)) can be calculated using the formula \(f_{alias} = |f – n \cdot f_s|\), where \(f\) is the original frequency and \(n\) is an integer chosen such that \(0 \le f_{alias} < f_s/2\). For \(f = 14 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\), we test values of \(n\): – If \(n=0\), \(f_{alias} = |14 – 0 \cdot 25| = 14 \text{ kHz}\). This is greater than \(f_s/2 = 12.5 \text{ kHz}\), so this is not the correct aliased frequency within the baseband. – If \(n=1\), \(f_{alias} = |14 – 1 \cdot 25| = |-11| = 11 \text{ kHz}\). This frequency is within the range \(0 \le 11 \text{ kHz} < 12.5 \text{ kHz}\). Therefore, the 14 kHz component will be incorrectly represented as 11 kHz in the sampled data. This phenomenon is a critical concern in digital signal processing, particularly in areas like telecommunications and audio processing, where accurate representation of signals is paramount. Students at Xi'an Technological University Northern College of Information Engineering Entrance Exam are expected to grasp these fundamental concepts to design and analyze digital systems effectively, ensuring signal integrity and preventing distortion. Understanding aliasing is crucial for selecting appropriate sampling rates and implementing anti-aliasing filters, which are standard practices in the field.
Incorrect
The question assesses understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In this scenario, the analog signal contains frequency components up to 15 kHz. Therefore, the minimum sampling frequency required to avoid aliasing is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The actual sampling frequency used is 25 kHz. Since 25 kHz is less than the required Nyquist rate of 30 kHz, aliasing will occur. Aliasing causes higher frequencies in the original signal to be misrepresented as lower frequencies in the sampled signal. Specifically, frequencies above \(f_s/2\) (the Nyquist frequency, which is \(25 \text{ kHz} / 2 = 12.5 \text{ kHz}\)) will fold back into the frequency range below \(12.5 \text{ kHz}\). A frequency component at 14 kHz, which is above the Nyquist frequency of 12.5 kHz, will be aliased. The aliased frequency (\(f_{alias}\)) can be calculated using the formula \(f_{alias} = |f – n \cdot f_s|\), where \(f\) is the original frequency and \(n\) is an integer chosen such that \(0 \le f_{alias} < f_s/2\). For \(f = 14 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\), we test values of \(n\): – If \(n=0\), \(f_{alias} = |14 – 0 \cdot 25| = 14 \text{ kHz}\). This is greater than \(f_s/2 = 12.5 \text{ kHz}\), so this is not the correct aliased frequency within the baseband. – If \(n=1\), \(f_{alias} = |14 – 1 \cdot 25| = |-11| = 11 \text{ kHz}\). This frequency is within the range \(0 \le 11 \text{ kHz} < 12.5 \text{ kHz}\). Therefore, the 14 kHz component will be incorrectly represented as 11 kHz in the sampled data. This phenomenon is a critical concern in digital signal processing, particularly in areas like telecommunications and audio processing, where accurate representation of signals is paramount. Students at Xi'an Technological University Northern College of Information Engineering Entrance Exam are expected to grasp these fundamental concepts to design and analyze digital systems effectively, ensuring signal integrity and preventing distortion. Understanding aliasing is crucial for selecting appropriate sampling rates and implementing anti-aliasing filters, which are standard practices in the field.
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Question 26 of 30
26. Question
Consider a scenario where a critical data packet is being transmitted wirelessly from a remote sensor network to a central processing unit at Xi’an Technological University Northern College of Information Engineering. Due to unexpected atmospheric interference, the signal-to-noise ratio (SNR) at the receiver drops from a healthy \(30 \text{ dB}\) to a mere \(5 \text{ dB}\). What is the most probable and direct consequence of this drastic reduction in SNR on the integrity of the transmitted data packet?
Correct
The core concept here revolves around understanding the principles of signal-to-noise ratio (SNR) and its impact on data transmission quality, particularly in the context of digital communication systems as studied at institutions like Xi’an Technological University Northern College of Information Engineering. While no direct calculation is required, the question probes the understanding of how noise affects the fidelity of a transmitted signal. A higher SNR indicates that the signal power is significantly greater than the noise power, leading to a clearer and more reliable reception. Conversely, a lower SNR means the noise is more dominant, corrupting the signal and potentially causing errors in data interpretation. The question asks to identify the primary consequence of a drastically reduced SNR. A drastically reduced SNR means the noise level has increased substantially relative to the signal. This increased noise interferes with the original signal’s waveform, making it harder for the receiving equipment to distinguish the intended information from the unwanted fluctuations. In digital systems, this can lead to bit errors, where a transmitted ‘0’ is misinterpreted as a ‘1’, or vice versa. Such errors degrade the overall quality of the received data, impacting the accuracy and reliability of communication. Therefore, the most direct and significant consequence of a drastically reduced SNR is the increased likelihood of data corruption and the degradation of information fidelity. This understanding is fundamental for students in information engineering, as it directly relates to the design and performance of communication channels, error detection and correction codes, and the overall robustness of information systems. The ability to interpret the impact of noise on signal quality is a critical skill for analyzing and improving communication protocols and system designs, aligning with the rigorous academic standards at Xi’an Technological University Northern College of Information Engineering.
Incorrect
The core concept here revolves around understanding the principles of signal-to-noise ratio (SNR) and its impact on data transmission quality, particularly in the context of digital communication systems as studied at institutions like Xi’an Technological University Northern College of Information Engineering. While no direct calculation is required, the question probes the understanding of how noise affects the fidelity of a transmitted signal. A higher SNR indicates that the signal power is significantly greater than the noise power, leading to a clearer and more reliable reception. Conversely, a lower SNR means the noise is more dominant, corrupting the signal and potentially causing errors in data interpretation. The question asks to identify the primary consequence of a drastically reduced SNR. A drastically reduced SNR means the noise level has increased substantially relative to the signal. This increased noise interferes with the original signal’s waveform, making it harder for the receiving equipment to distinguish the intended information from the unwanted fluctuations. In digital systems, this can lead to bit errors, where a transmitted ‘0’ is misinterpreted as a ‘1’, or vice versa. Such errors degrade the overall quality of the received data, impacting the accuracy and reliability of communication. Therefore, the most direct and significant consequence of a drastically reduced SNR is the increased likelihood of data corruption and the degradation of information fidelity. This understanding is fundamental for students in information engineering, as it directly relates to the design and performance of communication channels, error detection and correction codes, and the overall robustness of information systems. The ability to interpret the impact of noise on signal quality is a critical skill for analyzing and improving communication protocols and system designs, aligning with the rigorous academic standards at Xi’an Technological University Northern College of Information Engineering.
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Question 27 of 30
27. Question
Consider a digital communication link established for a research project at Xi’an Technological University Northern College of Information Engineering, transmitting data across a noisy channel. The received signal power is measured at \(10^{-9}\) Watts. The channel exhibits a noise power spectral density of \(10^{-12}\) Watts per Hertz, and the operational bandwidth of the transmission is \(10^5\) Hertz. What is the signal-to-noise ratio (SNR) of this communication link, expressed in decibels?
Correct
The question probes the understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital transmission where the received signal power is \(P_s = 10^{-9}\) Watts and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the channel is \(B = 10^5\) Hz. The noise power \(P_n\) in the channel is calculated by multiplying the noise power spectral density by the bandwidth: \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ Watts/Hz}) \times (10^5 \text{ Hz})\) \(P_n = 10^{-7}\) Watts The signal-to-noise ratio (SNR) is then calculated as the ratio of signal power to noise power: \(SNR = \frac{P_s}{P_n}\) \(SNR = \frac{10^{-9} \text{ Watts}}{10^{-7} \text{ Watts}}\) \(SNR = 10^{-2}\) To express this in decibels (dB), we use the formula: \(SNR_{dB} = 10 \log_{10}(SNR)\) \(SNR_{dB} = 10 \log_{10}(10^{-2})\) \(SNR_{dB} = 10 \times (-2)\) \(SNR_{dB} = -20\) dB A negative SNR in decibels indicates that the noise power is greater than the signal power, which is a critical consideration for reliable data transmission. Understanding how to calculate and interpret SNR is fundamental for analyzing the performance of communication links and designing robust systems, aligning with the rigorous curriculum at Xi’an Technological University Northern College of Information Engineering, particularly in areas like digital signal processing and wireless communications. This calculation demonstrates a practical application of fundamental communication theory, emphasizing the impact of noise on signal quality.
Incorrect
The question probes the understanding of signal-to-noise ratio (SNR) in the context of digital communication systems, a core concept for students at Xi’an Technological University Northern College of Information Engineering. The scenario describes a digital transmission where the received signal power is \(P_s = 10^{-9}\) Watts and the noise power spectral density is \(N_0 = 10^{-12}\) Watts/Hz. The bandwidth of the channel is \(B = 10^5\) Hz. The noise power \(P_n\) in the channel is calculated by multiplying the noise power spectral density by the bandwidth: \(P_n = N_0 \times B\) \(P_n = (10^{-12} \text{ Watts/Hz}) \times (10^5 \text{ Hz})\) \(P_n = 10^{-7}\) Watts The signal-to-noise ratio (SNR) is then calculated as the ratio of signal power to noise power: \(SNR = \frac{P_s}{P_n}\) \(SNR = \frac{10^{-9} \text{ Watts}}{10^{-7} \text{ Watts}}\) \(SNR = 10^{-2}\) To express this in decibels (dB), we use the formula: \(SNR_{dB} = 10 \log_{10}(SNR)\) \(SNR_{dB} = 10 \log_{10}(10^{-2})\) \(SNR_{dB} = 10 \times (-2)\) \(SNR_{dB} = -20\) dB A negative SNR in decibels indicates that the noise power is greater than the signal power, which is a critical consideration for reliable data transmission. Understanding how to calculate and interpret SNR is fundamental for analyzing the performance of communication links and designing robust systems, aligning with the rigorous curriculum at Xi’an Technological University Northern College of Information Engineering, particularly in areas like digital signal processing and wireless communications. This calculation demonstrates a practical application of fundamental communication theory, emphasizing the impact of noise on signal quality.
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Question 28 of 30
28. Question
In the context of digital communication systems, particularly those developed and researched at institutions like Xi’an Technological University Northern College of Information Engineering, what is the primary implication of a significantly elevated signal-to-noise ratio (SNR) on data transmission reliability and achievable throughput?
Correct
The core of this question lies in understanding the principles of signal-to-noise ratio (SNR) and its impact on data integrity in digital communication systems, a fundamental concept at Xi’an Technological University Northern College of Information Engineering. A higher SNR indicates a stronger signal relative to background noise, leading to more reliable data transmission and fewer errors. Conversely, a lower SNR degrades the signal quality, increasing the probability of bit errors. Consider a scenario where a digital communication system is designed to transmit data at a rate of \(R\) bits per second. The channel has a bandwidth of \(B\) Hertz. The received signal power is \(S\) Watts, and the noise power spectral density is \(N_0\) Watts per Hertz. The total noise power within the bandwidth \(B\) is approximately \(N = N_0 \times B\). The signal-to-noise ratio (SNR) is then calculated as \(SNR = \frac{S}{N} = \frac{S}{N_0 B}\). Shannon’s channel capacity theorem states that the maximum achievable data rate \(C\) for a channel with bandwidth \(B\) and SNR is given by \(C = B \log_2(1 + SNR)\). This theorem is crucial for understanding the theoretical limits of communication systems. If a system operates with a high SNR, it means the signal is significantly stronger than the noise. This allows for a higher data rate with a lower probability of error, as the receiver can more easily distinguish between signal bits and noise fluctuations. For instance, if the SNR is very high, \(1 + SNR \approx SNR\), and the capacity approaches \(C \approx B \log_2(SNR)\). This implies that with a strong signal, the channel can support more information per unit of time. Conversely, a low SNR means the noise is comparable to or even greater than the signal. This severely limits the achievable data rate and increases the bit error rate (BER). To maintain a certain data rate with a low SNR, more robust error correction coding schemes are required, which often come at the cost of increased overhead and reduced effective data throughput. Therefore, maintaining a high SNR is paramount for efficient and reliable digital communication, a key area of study in information engineering at Xi’an Technological University Northern College of Information Engineering. The question probes the understanding of how signal quality directly influences the system’s ability to convey information accurately and efficiently.
Incorrect
The core of this question lies in understanding the principles of signal-to-noise ratio (SNR) and its impact on data integrity in digital communication systems, a fundamental concept at Xi’an Technological University Northern College of Information Engineering. A higher SNR indicates a stronger signal relative to background noise, leading to more reliable data transmission and fewer errors. Conversely, a lower SNR degrades the signal quality, increasing the probability of bit errors. Consider a scenario where a digital communication system is designed to transmit data at a rate of \(R\) bits per second. The channel has a bandwidth of \(B\) Hertz. The received signal power is \(S\) Watts, and the noise power spectral density is \(N_0\) Watts per Hertz. The total noise power within the bandwidth \(B\) is approximately \(N = N_0 \times B\). The signal-to-noise ratio (SNR) is then calculated as \(SNR = \frac{S}{N} = \frac{S}{N_0 B}\). Shannon’s channel capacity theorem states that the maximum achievable data rate \(C\) for a channel with bandwidth \(B\) and SNR is given by \(C = B \log_2(1 + SNR)\). This theorem is crucial for understanding the theoretical limits of communication systems. If a system operates with a high SNR, it means the signal is significantly stronger than the noise. This allows for a higher data rate with a lower probability of error, as the receiver can more easily distinguish between signal bits and noise fluctuations. For instance, if the SNR is very high, \(1 + SNR \approx SNR\), and the capacity approaches \(C \approx B \log_2(SNR)\). This implies that with a strong signal, the channel can support more information per unit of time. Conversely, a low SNR means the noise is comparable to or even greater than the signal. This severely limits the achievable data rate and increases the bit error rate (BER). To maintain a certain data rate with a low SNR, more robust error correction coding schemes are required, which often come at the cost of increased overhead and reduced effective data throughput. Therefore, maintaining a high SNR is paramount for efficient and reliable digital communication, a key area of study in information engineering at Xi’an Technological University Northern College of Information Engineering. The question probes the understanding of how signal quality directly influences the system’s ability to convey information accurately and efficiently.
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Question 29 of 30
29. Question
A network administrator at Xi’an Technological University Northern College of Information Engineering is tasked with enhancing the security posture of a newly established high-performance computing cluster used for advanced computational fluid dynamics simulations. The cluster’s primary data repository resides on a dedicated server with the IP address \(10.0.0.20\), and the cluster nodes are assigned IP addresses within the \(192.168.10.0/24\) subnet. To prevent unauthorized data exfiltration and ensure the integrity of simulation results, the administrator must configure the network firewall. The requirement is to permit only essential data transfer from the cluster nodes to the data repository on the standard MySQL port (\(3306\)), while strictly prohibiting any other outbound traffic originating from the cluster subnet to any external destination. Which firewall rule configuration best satisfies these stringent security mandates for the university’s research infrastructure?
Correct
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering implementing a new firewall policy to segment a research network. The goal is to isolate sensitive data from general student access while allowing specific, controlled communication. The administrator needs to configure rules that permit traffic from a designated research server (IP address \(192.168.10.5\)) to a specific database server (IP address \(10.0.0.20\)) only on port \(3306\) (standard for MySQL). Additionally, all other traffic originating from the research network segment (IP range \(192.168.10.0/24\)) to any destination outside this segment should be blocked. To achieve this, the firewall needs a rule that explicitly allows the desired traffic and a subsequent rule that denies all other traffic from the source segment. Rule 1: Allow traffic from \(192.168.10.5\) to \(10.0.0.20\) on TCP port \(3306\). Source IP: \(192.168.10.5\) Destination IP: \(10.0.0.20\) Protocol: TCP Destination Port: \(3306\) Action: Allow Rule 2: Deny all traffic from the research network segment \(192.168.10.0/24\) to any destination. Source IP: \(192.168.10.0/24\) Destination IP: Any Protocol: Any Action: Deny The question asks for the most effective firewall rule configuration to achieve the stated objectives. The correct approach involves a specific allow rule followed by a broader deny rule. This is a fundamental principle in firewall management: be explicit about what is allowed, and then deny everything else. This layered security approach ensures that only intended communication pathways are open, minimizing the attack surface. The specific allowance for the database connection on port 3306 is crucial for the research activities, while the general block prevents unauthorized access from the research segment to other parts of the university network or the internet. This aligns with the university’s commitment to data security and responsible network usage, particularly for specialized research environments.
Incorrect
The scenario describes a network administrator at Xi’an Technological University Northern College of Information Engineering implementing a new firewall policy to segment a research network. The goal is to isolate sensitive data from general student access while allowing specific, controlled communication. The administrator needs to configure rules that permit traffic from a designated research server (IP address \(192.168.10.5\)) to a specific database server (IP address \(10.0.0.20\)) only on port \(3306\) (standard for MySQL). Additionally, all other traffic originating from the research network segment (IP range \(192.168.10.0/24\)) to any destination outside this segment should be blocked. To achieve this, the firewall needs a rule that explicitly allows the desired traffic and a subsequent rule that denies all other traffic from the source segment. Rule 1: Allow traffic from \(192.168.10.5\) to \(10.0.0.20\) on TCP port \(3306\). Source IP: \(192.168.10.5\) Destination IP: \(10.0.0.20\) Protocol: TCP Destination Port: \(3306\) Action: Allow Rule 2: Deny all traffic from the research network segment \(192.168.10.0/24\) to any destination. Source IP: \(192.168.10.0/24\) Destination IP: Any Protocol: Any Action: Deny The question asks for the most effective firewall rule configuration to achieve the stated objectives. The correct approach involves a specific allow rule followed by a broader deny rule. This is a fundamental principle in firewall management: be explicit about what is allowed, and then deny everything else. This layered security approach ensures that only intended communication pathways are open, minimizing the attack surface. The specific allowance for the database connection on port 3306 is crucial for the research activities, while the general block prevents unauthorized access from the research segment to other parts of the university network or the internet. This aligns with the university’s commitment to data security and responsible network usage, particularly for specialized research environments.
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Question 30 of 30
30. Question
Consider a continuous-time signal \(x(t) = \cos(200\pi t) + \sin(500\pi t)\) that is to be sampled for digital processing at Xi’an Technological University Northern College of Information Engineering. If the sampling frequency \(f_s\) is set to 400 Hz, what is the most accurate description of the resulting sampled signal and its implications for signal reconstruction?
Correct
The question probes the understanding of signal processing principles, specifically concerning the impact of sampling rate on aliasing and the Nyquist-Shannon sampling theorem. The scenario describes a continuous-time signal \(x(t) = \cos(200\pi t) + \sin(500\pi t)\). The highest frequency component in this signal is determined by the arguments of the cosine and sine functions. For \(\cos(200\pi t)\), the angular frequency is \(\omega_1 = 200\pi\) radians per second. The corresponding frequency \(f_1\) is \(\omega_1 / (2\pi) = 200\pi / (2\pi) = 100\) Hz. For \(\sin(500\pi t)\), the angular frequency is \(\omega_2 = 500\pi\) radians per second. The corresponding frequency \(f_2\) is \(\omega_2 / (2\pi) = 500\pi / (2\pi) = 250\) Hz. Therefore, the maximum frequency present in the signal \(x(t)\) is \(f_{max} = 250\) Hz. According to the Nyquist-Shannon sampling theorem, to perfectly reconstruct a continuous-time signal from its samples, the sampling frequency \(f_s\) must be strictly greater than twice the maximum frequency component of the signal. This minimum required sampling frequency is known as the Nyquist rate, which is \(2f_{max}\). In this case, the Nyquist rate is \(2 \times 250 \text{ Hz} = 500\) Hz. The question states that the signal is sampled at a rate of \(f_s = 400\) Hz. Since \(400 \text{ Hz} < 500 \text{ Hz}\), the sampling rate is below the Nyquist rate. When the sampling rate is less than the Nyquist rate, aliasing occurs. Aliasing is the phenomenon where higher frequencies in the original signal are misrepresented as lower frequencies in the sampled signal, leading to distortion and an inability to reconstruct the original signal accurately. Specifically, frequencies above \(f_s/2\) (the folding frequency) will be aliased. The folding frequency here is \(400 \text{ Hz} / 2 = 200\) Hz. The frequency component at 250 Hz is greater than the folding frequency of 200 Hz. This component will be aliased. The aliased frequency \(f_{alias}\) can be calculated using the formula \(f_{alias} = |f – k f_s|\), where \(f\) is the original frequency and \(k\) is an integer chosen such that \(0 \le f_{alias} \le f_s/2\). For \(f = 250\) Hz and \(f_s = 400\) Hz, we can choose \(k=1\). Then \(f_{alias} = |250 \text{ Hz} – 1 \times 400 \text{ Hz}| = |-150 \text{ Hz}| = 150\) Hz. This 150 Hz component will be present in the sampled signal, masquerading as a genuine 150 Hz component. The 100 Hz component is below the folding frequency, so it will be sampled correctly and will appear as 100 Hz in the sampled signal. Therefore, the sampled signal will contain a 100 Hz component and an aliased 150 Hz component, making accurate reconstruction of the original 250 Hz component impossible. The correct answer is that aliasing will occur, and the 250 Hz component will appear as 150 Hz.
Incorrect
The question probes the understanding of signal processing principles, specifically concerning the impact of sampling rate on aliasing and the Nyquist-Shannon sampling theorem. The scenario describes a continuous-time signal \(x(t) = \cos(200\pi t) + \sin(500\pi t)\). The highest frequency component in this signal is determined by the arguments of the cosine and sine functions. For \(\cos(200\pi t)\), the angular frequency is \(\omega_1 = 200\pi\) radians per second. The corresponding frequency \(f_1\) is \(\omega_1 / (2\pi) = 200\pi / (2\pi) = 100\) Hz. For \(\sin(500\pi t)\), the angular frequency is \(\omega_2 = 500\pi\) radians per second. The corresponding frequency \(f_2\) is \(\omega_2 / (2\pi) = 500\pi / (2\pi) = 250\) Hz. Therefore, the maximum frequency present in the signal \(x(t)\) is \(f_{max} = 250\) Hz. According to the Nyquist-Shannon sampling theorem, to perfectly reconstruct a continuous-time signal from its samples, the sampling frequency \(f_s\) must be strictly greater than twice the maximum frequency component of the signal. This minimum required sampling frequency is known as the Nyquist rate, which is \(2f_{max}\). In this case, the Nyquist rate is \(2 \times 250 \text{ Hz} = 500\) Hz. The question states that the signal is sampled at a rate of \(f_s = 400\) Hz. Since \(400 \text{ Hz} < 500 \text{ Hz}\), the sampling rate is below the Nyquist rate. When the sampling rate is less than the Nyquist rate, aliasing occurs. Aliasing is the phenomenon where higher frequencies in the original signal are misrepresented as lower frequencies in the sampled signal, leading to distortion and an inability to reconstruct the original signal accurately. Specifically, frequencies above \(f_s/2\) (the folding frequency) will be aliased. The folding frequency here is \(400 \text{ Hz} / 2 = 200\) Hz. The frequency component at 250 Hz is greater than the folding frequency of 200 Hz. This component will be aliased. The aliased frequency \(f_{alias}\) can be calculated using the formula \(f_{alias} = |f – k f_s|\), where \(f\) is the original frequency and \(k\) is an integer chosen such that \(0 \le f_{alias} \le f_s/2\). For \(f = 250\) Hz and \(f_s = 400\) Hz, we can choose \(k=1\). Then \(f_{alias} = |250 \text{ Hz} – 1 \times 400 \text{ Hz}| = |-150 \text{ Hz}| = 150\) Hz. This 150 Hz component will be present in the sampled signal, masquerading as a genuine 150 Hz component. The 100 Hz component is below the folding frequency, so it will be sampled correctly and will appear as 100 Hz in the sampled signal. Therefore, the sampled signal will contain a 100 Hz component and an aliased 150 Hz component, making accurate reconstruction of the original 250 Hz component impossible. The correct answer is that aliasing will occur, and the 250 Hz component will appear as 150 Hz.