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Consider a sine wave of frequency 2 KHz and peak amplitude AM. This signal is applied to a delta modulator of step-size 0.4 Volt and sampling period 3.18 x 10-6 sec. What is the maximum power that
may be transmitted without slopeoverload distortion?
may be transmitted without slopeoverload distortion?
- a)100 W
- b)50 W
- c)150 W
- d)200 W
Correct answer is option 'B'. Can you explain this answer?
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Consider a sine wave of frequency 2 KHz and peak amplitude AM. This si...
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Consider a sine wave of frequency 2 KHz and peak amplitude AM. This si...
Solution:
Given parameters:
Frequency of sine wave, f = 2 kHz
Peak amplitude, A = AM
Step size, Δ = 0.4 V
Sampling period, T = 3.18 x 10^-6 sec
To avoid slope overload distortion, the step size must be less than or equal to the maximum slope of the input signal. The maximum slope of a sine wave is at its zero crossings and is given by:
dV/dt = 2πfA
Substituting the given values, we get:
dV/dt = 2π x 2 x 10^3 x AM
To avoid slope overload distortion, we must have:
Δ ≤ dV/dt x T
Substituting the given values, we get:
0.4 ≤ 2π x 2 x 10^3 x AM x 3.18 x 10^-6
Solving for AM, we get:
AM ≤ 31.83 V
The power of the signal is given by:
P = (AM/√2)^2/R
where R is the load resistance.
The maximum power that may be transmitted without slope overload distortion is obtained by substituting the maximum value of AM:
Pmax = (31.83/√2)^2/R = 50 W
Therefore, the correct answer is option B (50 W).
Given parameters:
Frequency of sine wave, f = 2 kHz
Peak amplitude, A = AM
Step size, Δ = 0.4 V
Sampling period, T = 3.18 x 10^-6 sec
To avoid slope overload distortion, the step size must be less than or equal to the maximum slope of the input signal. The maximum slope of a sine wave is at its zero crossings and is given by:
dV/dt = 2πfA
Substituting the given values, we get:
dV/dt = 2π x 2 x 10^3 x AM
To avoid slope overload distortion, we must have:
Δ ≤ dV/dt x T
Substituting the given values, we get:
0.4 ≤ 2π x 2 x 10^3 x AM x 3.18 x 10^-6
Solving for AM, we get:
AM ≤ 31.83 V
The power of the signal is given by:
P = (AM/√2)^2/R
where R is the load resistance.
The maximum power that may be transmitted without slope overload distortion is obtained by substituting the maximum value of AM:
Pmax = (31.83/√2)^2/R = 50 W
Therefore, the correct answer is option B (50 W).
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Consider a sine wave of frequency 2 KHz and peak amplitude AM. This signal is applied to a delta modulator of step-size 0.4 Volt and sampling period 3.18 x 10-6 sec. What is the maximum power thatmay be transmitted without slopeoverload distortion?a)100 Wb)50 Wc)150 Wd)200 WCorrect answer is option 'B'. Can you explain this answer?
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