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A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δ is 100 mV. The modulator is tested with a this test signal required to avoid slope overload is
- a)2.04 V
- b)1.08 V
- c)4.08 V
- d)2.16 V
Correct answer is option 'B'. Can you explain this answer?
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Linear Delta Modulation
Linear Delta Modulation (LDM) is a type of analog-to-digital conversion technique that is used to convert analog signals into digital signals. It is a form of pulse modulation where the amplitude of the input signal is compared with the amplitude of the previous sample, and the difference is quantized and encoded into a binary signal.
Given Parameters
- Speech signals limited to 3.4 kHz
- Sampling rate is 10 times the Nyquist rate of the speech signal
- Step size is 100 mV
Nyquist Rate
Nyquist rate is the minimum sampling rate required to accurately represent a continuous-time signal in a digital format. The Nyquist rate for a signal with bandwidth B is given by:
Nyquist rate = 2B
In this case, the bandwidth of the speech signal is 3.4 kHz. Therefore, the Nyquist rate is:
Nyquist rate = 2 x 3.4 kHz = 6.8 kHz
Sampling Rate
The sampling rate for the linear delta modulator is given as 10 times the Nyquist rate of the speech signal. Therefore, the sampling rate is:
Sampling rate = 10 x 6.8 kHz = 68 kHz
Step Size
The step size of the linear delta modulator is given as 100 mV.
Test Signal
The test signal required to avoid slope overload is given as:
Test signal = 1.08 V
Calculations
To avoid slope overload in the LDM, the maximum slope of the input signal should be less than the maximum slope that can be tracked by the modulator. The maximum slope that can be tracked by the modulator is given by:
Maximum slope = Step size / Sampling interval
where the sampling interval is the time between two consecutive samples. The sampling interval is given by:
Sampling interval = 1 / Sampling rate
Substituting the given values, we get:
Sampling interval = 1 / 68 kHz = 14.7 μs
Therefore, the maximum slope that can be tracked by the modulator is:
Maximum slope = 100 mV / 14.7 μs = 6.8 V/ms
The test signal is given as 1.08 V. The maximum slope of the test signal is given by:
Maximum slope of test signal = 3.4 kHz x 2π x 1.08 V = 23.14 V/ms
Since the maximum slope of the test signal is less than the maximum slope that can be tracked by the modulator, slope overload is avoided.
Therefore, the correct answer is option B (1.08 V).
Linear Delta Modulation (LDM) is a type of analog-to-digital conversion technique that is used to convert analog signals into digital signals. It is a form of pulse modulation where the amplitude of the input signal is compared with the amplitude of the previous sample, and the difference is quantized and encoded into a binary signal.
Given Parameters
- Speech signals limited to 3.4 kHz
- Sampling rate is 10 times the Nyquist rate of the speech signal
- Step size is 100 mV
Nyquist Rate
Nyquist rate is the minimum sampling rate required to accurately represent a continuous-time signal in a digital format. The Nyquist rate for a signal with bandwidth B is given by:
Nyquist rate = 2B
In this case, the bandwidth of the speech signal is 3.4 kHz. Therefore, the Nyquist rate is:
Nyquist rate = 2 x 3.4 kHz = 6.8 kHz
Sampling Rate
The sampling rate for the linear delta modulator is given as 10 times the Nyquist rate of the speech signal. Therefore, the sampling rate is:
Sampling rate = 10 x 6.8 kHz = 68 kHz
Step Size
The step size of the linear delta modulator is given as 100 mV.
Test Signal
The test signal required to avoid slope overload is given as:
Test signal = 1.08 V
Calculations
To avoid slope overload in the LDM, the maximum slope of the input signal should be less than the maximum slope that can be tracked by the modulator. The maximum slope that can be tracked by the modulator is given by:
Maximum slope = Step size / Sampling interval
where the sampling interval is the time between two consecutive samples. The sampling interval is given by:
Sampling interval = 1 / Sampling rate
Substituting the given values, we get:
Sampling interval = 1 / 68 kHz = 14.7 μs
Therefore, the maximum slope that can be tracked by the modulator is:
Maximum slope = 100 mV / 14.7 μs = 6.8 V/ms
The test signal is given as 1.08 V. The maximum slope of the test signal is given by:
Maximum slope of test signal = 3.4 kHz x 2π x 1.08 V = 23.14 V/ms
Since the maximum slope of the test signal is less than the maximum slope that can be tracked by the modulator, slope overload is avoided.
Therefore, the correct answer is option B (1.08 V).
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A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect answer is option 'B'. Can you explain this answer?
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A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect 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 linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect 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 linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect answer is option 'B'. Can you explain this answer?.
A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect 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 linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect 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 linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δis 100 mV. The modulator is tested with a this test signal required to avoid slope overload isa)2.04 Vb)1.08 Vc)4.08 Vd)2.16 VCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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