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Consider a linear DM system designed to accommodate analog message signals limited to bandwidth of 3.5 kHz. A sinusoidal test signals of amplitude Amax = 1 V and frequency fm = 800 Hz is applied to system. The sampling rate of the system is 64 kHz.
Que: The minimum value of the step size to avoid overload is
- a)240 mV
- b)120 mV
- c)670 mV
- d)78.5 mV
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Consider a linear DM system designed to accommodate analog message sig...
Given:
B = 3.5 kHz,
Am = 1 V,
fm = 800 Hz and fs = 64 kHz
Since fs = 64 kHz, the time duration of the sample will be:
Let the step-size of the delta modulated signal be δ.
δ ≥ (1.56 × 10-5) × 1 × (2π × 800)
δ ≥ 7.84 × 10-2 V
∴ The minimum step size to avoid slope overload will be:
δ = 78.5 mV
Since fs = 64 kHz, the time duration of the sample will be:
Let the step-size of the delta modulated signal be δ.
δ ≥ (1.56 × 10-5) × 1 × (2π × 800)
δ ≥ 7.84 × 10-2 V
∴ The minimum step size to avoid slope overload will be:
δ = 78.5 mV
Most Upvoted Answer
Consider a linear DM system designed to accommodate analog message sig...
Given information:
- Bandwidth of the analog message signals: 3.5 kHz
- Amplitude of the sinusoidal test signal (Amax): 1 V
- Frequency of the sinusoidal test signal (fm): 800 Hz
- Sampling rate of the system: 64 kHz
Explanation:
The minimum value of the step size is determined by the Nyquist criterion, which states that the sampling rate should be at least twice the bandwidth of the signal to avoid aliasing.
Step 1: Calculate the maximum frequency component of the analog message signal:
The maximum frequency component of the analog message signal can be calculated using the formula:
Maximum frequency component (fmax) = fm + bandwidth/2
Given: fm = 800 Hz, bandwidth = 3.5 kHz
Substituting the values, we get:
fmax = 800 Hz + 3.5 kHz/2
= 800 Hz + 1.75 kHz
= 2550 Hz
Step 2: Calculate the minimum sampling rate:
According to the Nyquist criterion, the minimum sampling rate should be at least twice the maximum frequency component of the analog message signal. Therefore,
Minimum sampling rate = 2 * fmax
Substituting the value of fmax, we get:
Minimum sampling rate = 2 * 2550 Hz
= 5100 Hz
Step 3: Calculate the maximum step size:
The maximum step size is determined by the quantization range of the analog-to-digital converter (ADC). In this case, the maximum amplitude of the sinusoidal test signal is given as Amax = 1 V.
The maximum step size (Δ) can be calculated using the formula:
Δ = (2 * Amax) / (2^N)
Where N is the number of bits used by the ADC.
Step 4: Calculate the number of bits required:
To determine the number of bits required, we need to calculate the quantization levels using the formula:
Quantization levels (L) = 2^N
The number of quantization levels is determined by the quantization range, which is given by:
Quantization range = 2 * Amax
Substituting the value of Amax, we get:
Quantization range = 2 * 1 V
= 2 V
Since the quantization range is divided into quantization levels, we have:
Quantization levels = 2^N
Substituting the value of the quantization range, we get:
2 V = 2^N
Taking the logarithm of both sides, we get:
log2(2 V) = log2(2^N)
Simplifying the equation, we get:
1 = N
Therefore, the number of bits required (N) is 1.
Step 5: Calculate the maximum step size:
Substituting the values of Amax and N into the formula, we get:
Δ = (2 * Amax) / (2^N)
= (2 * 1 V) / (2^1)
= 2 V / 2
= 1 V
Step 6: Calculate the minimum step size:
The minimum step size is equal to half
- Bandwidth of the analog message signals: 3.5 kHz
- Amplitude of the sinusoidal test signal (Amax): 1 V
- Frequency of the sinusoidal test signal (fm): 800 Hz
- Sampling rate of the system: 64 kHz
Explanation:
The minimum value of the step size is determined by the Nyquist criterion, which states that the sampling rate should be at least twice the bandwidth of the signal to avoid aliasing.
Step 1: Calculate the maximum frequency component of the analog message signal:
The maximum frequency component of the analog message signal can be calculated using the formula:
Maximum frequency component (fmax) = fm + bandwidth/2
Given: fm = 800 Hz, bandwidth = 3.5 kHz
Substituting the values, we get:
fmax = 800 Hz + 3.5 kHz/2
= 800 Hz + 1.75 kHz
= 2550 Hz
Step 2: Calculate the minimum sampling rate:
According to the Nyquist criterion, the minimum sampling rate should be at least twice the maximum frequency component of the analog message signal. Therefore,
Minimum sampling rate = 2 * fmax
Substituting the value of fmax, we get:
Minimum sampling rate = 2 * 2550 Hz
= 5100 Hz
Step 3: Calculate the maximum step size:
The maximum step size is determined by the quantization range of the analog-to-digital converter (ADC). In this case, the maximum amplitude of the sinusoidal test signal is given as Amax = 1 V.
The maximum step size (Δ) can be calculated using the formula:
Δ = (2 * Amax) / (2^N)
Where N is the number of bits used by the ADC.
Step 4: Calculate the number of bits required:
To determine the number of bits required, we need to calculate the quantization levels using the formula:
Quantization levels (L) = 2^N
The number of quantization levels is determined by the quantization range, which is given by:
Quantization range = 2 * Amax
Substituting the value of Amax, we get:
Quantization range = 2 * 1 V
= 2 V
Since the quantization range is divided into quantization levels, we have:
Quantization levels = 2^N
Substituting the value of the quantization range, we get:
2 V = 2^N
Taking the logarithm of both sides, we get:
log2(2 V) = log2(2^N)
Simplifying the equation, we get:
1 = N
Therefore, the number of bits required (N) is 1.
Step 5: Calculate the maximum step size:
Substituting the values of Amax and N into the formula, we get:
Δ = (2 * Amax) / (2^N)
= (2 * 1 V) / (2^1)
= 2 V / 2
= 1 V
Step 6: Calculate the minimum step size:
The minimum step size is equal to half
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Consider a linear DM system designed to accommodate analog message signals limited to bandwidth of 3.5 kHz. A sinusoidal test signals of amplitude Amax = 1 V and frequency fm = 800 Hz is applied to system. The sampling rate of the system is 64 kHz.Que: The minimum value of the step size to avoid overload isa)240 mVb)120 mVc)670 mVd)78.5 mVCorrect answer is option 'D'. Can you explain this answer?
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Solutions for Consider a linear DM system designed to accommodate analog message signals limited to bandwidth of 3.5 kHz. A sinusoidal test signals of amplitude Amax = 1 V and frequency fm = 800 Hz is applied to system. The sampling rate of the system is 64 kHz.Que: The minimum value of the step size to avoid overload isa)240 mVb)120 mVc)670 mVd)78.5 mVCorrect answer is option 'D'. 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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