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A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec and the samples are applied to an idealrectangular LPF with cat-off frequency of 1100 Hz, then the output of the filter contains
- a)only 800 Hz component
- b)800 and 900 Hz components
- c)800 Hz and 1000 Hz components
- d)800 Hz, 900 and 1000 Hz components
Correct answer is option 'B'. Can you explain this answer?
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A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec an...
Since the sampling rate is 1800 samples/sec the highest frequency that can be recovered is 900 Hz.
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A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec an...
Given information:
- Signal frequency = 1.0 kHz
- Sampling rate = 180 samples/sec
- Cut-off frequency of LPF = 1100 Hz
Sampling:
- The signal is sampled at the rate of 180 samples/sec.
- Sampling rate is less than twice the signal frequency, but since the signal is not band-limited, aliasing will occur.
Aliasing:
- Due to aliasing, the signal will appear to have a lower frequency.
- Let's find the alias frequency using the formula: fa = |fs - n*f|
- fs = sampling frequency = 180 samples/sec
- f = signal frequency = 1.0 kHz
- n = integer such that |fs - n*f| <=>
- We can find n as follows:
- n = floor(fs/f) = floor(180/1000) = 0
- Therefore, the alias frequency is given by:
- fa = |fs - n*f| = |180 - 0*1000| = 180 Hz
LPF filtering:
- The sampled signal is passed through an ideal rectangular LPF with a cut-off frequency of 1100 Hz.
- Frequencies below 1100 Hz are passed through unchanged, while frequencies above 1100 Hz are completely attenuated.
- Therefore, the output of the LPF will contain frequencies up to 1100 Hz.
Frequency components in the output:
- The output will contain the alias frequency and any frequency components up to 1100 Hz that were present in the original signal.
- The frequency components up to 1100 Hz can be calculated using the formula: kf = m*fs ± f, where m is an integer.
- Let's calculate the frequency components:
- kf = ±fs ± f = ±180 ± 1000 = 820 Hz and 1160 Hz.
- Therefore, the output will contain components at 820 Hz (alias frequency) and at 800 Hz and 900 Hz (original signal components).
- The correct answer is option 'B': 800 and 900 Hz components.
- Signal frequency = 1.0 kHz
- Sampling rate = 180 samples/sec
- Cut-off frequency of LPF = 1100 Hz
Sampling:
- The signal is sampled at the rate of 180 samples/sec.
- Sampling rate is less than twice the signal frequency, but since the signal is not band-limited, aliasing will occur.
Aliasing:
- Due to aliasing, the signal will appear to have a lower frequency.
- Let's find the alias frequency using the formula: fa = |fs - n*f|
- fs = sampling frequency = 180 samples/sec
- f = signal frequency = 1.0 kHz
- n = integer such that |fs - n*f| <=>
- We can find n as follows:
- n = floor(fs/f) = floor(180/1000) = 0
- Therefore, the alias frequency is given by:
- fa = |fs - n*f| = |180 - 0*1000| = 180 Hz
LPF filtering:
- The sampled signal is passed through an ideal rectangular LPF with a cut-off frequency of 1100 Hz.
- Frequencies below 1100 Hz are passed through unchanged, while frequencies above 1100 Hz are completely attenuated.
- Therefore, the output of the LPF will contain frequencies up to 1100 Hz.
Frequency components in the output:
- The output will contain the alias frequency and any frequency components up to 1100 Hz that were present in the original signal.
- The frequency components up to 1100 Hz can be calculated using the formula: kf = m*fs ± f, where m is an integer.
- Let's calculate the frequency components:
- kf = ±fs ± f = ±180 ± 1000 = 820 Hz and 1160 Hz.
- Therefore, the output will contain components at 820 Hz (alias frequency) and at 800 Hz and 900 Hz (original signal components).
- The correct answer is option 'B': 800 and 900 Hz components.
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A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec and the samples are applied to an idealrectangular LPF with cat-off frequency of 1100 Hz, then the output of the filter containsa)only 800 Hz componentb)800 and 900 Hz componentsc)800 Hz and 1000 Hz componentsd)800 Hz, 900 and 1000 Hz componentsCorrect answer is option 'B'. Can you explain this answer?
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A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec and the samples are applied to an idealrectangular LPF with cat-off frequency of 1100 Hz, then the output of the filter containsa)only 800 Hz componentb)800 and 900 Hz componentsc)800 Hz and 1000 Hz componentsd)800 Hz, 900 and 1000 Hz componentsCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec and the samples are applied to an idealrectangular LPF with cat-off frequency of 1100 Hz, then the output of the filter containsa)only 800 Hz componentb)800 and 900 Hz componentsc)800 Hz and 1000 Hz componentsd)800 Hz, 900 and 1000 Hz componentsCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec and the samples are applied to an idealrectangular LPF with cat-off frequency of 1100 Hz, then the output of the filter containsa)only 800 Hz componentb)800 and 900 Hz componentsc)800 Hz and 1000 Hz componentsd)800 Hz, 900 and 1000 Hz componentsCorrect answer is option 'B'. Can you explain this answer?.
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