There are 16 quantization levels available for the ADC and the quantization interval for this ADC is 2V.
To find the number of quantization levels and the quantization interval for a 4-bit analog-to-digital converter (ADC) with a maximum detection voltage of 32V, we need to consider the resolution of the ADC.
i) The number of quantization levels (N) can be determined using the formula:
N = 2^B
where B is the number of bits. In this case, B = 4, so the number of quantization levels is:
N = 2^4 = 16
ii) The quantization interval (Q) represents the difference between two adjacent quantization levels and can be calculated by dividing the maximum detection voltage by the number of quantization levels. In this case, the maximum detection voltage is 32V, and the number of quantization levels is 16:
Q = Maximum detection voltage / Number of quantization levels
= 32V / 16
= 2V
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Compute the Fourier Series decomposition of a square waveform with 90% duty cycle
The Fourier series decomposition of the square waveform with a 90% duty cycle is given by: f(t) = (a0/2) + ∑[(an * cos((2πnt)/T)) + (bn * sin((2πnt)/T))]
The Fourier series decomposition for a square waveform with a 90% duty cycle:
Definition of the Square Waveform:
The square waveform with a 90% duty cycle is defined as follows:
For 0 ≤ t < T0.9 (90% of the period), the waveform is equal to +1.
For T0.9 ≤ t < T (10% of the period), the waveform is equal to -1.
Here, T represents the period of the waveform.
Fourier Series Coefficients:
The Fourier series coefficients for this waveform can be computed using the following formulas:
a0 = (1/T) ∫[0 to T] f(t) dt
an = (2/T) ∫[0 to T] f(t) cos((2πnt)/T) dt
bn = (2/T) ∫[0 to T] f(t) sin((2πnt)/T) dt
where a0, an, and bn are the Fourier coefficients.
Computation of Fourier Coefficients:
For the given square waveform with a 90% duty cycle, we have:
a0 = (1/T) ∫[0 to T] f(t) dt = 0 (since the waveform is symmetric around 0)
an = 0 for all n ≠ 0 (since the waveform is symmetric and does not have cosine terms)
bn = (2/T) ∫[0 to T] f(t) sin((2πnt)/T) dt
Computation of bn for n = 1:
We need to compute bn for n = 1 using the formula:
bn = (2/T) ∫[0 to T] f(t) sin((2πt)/T) dt
Breaking the integral into two parts (corresponding to the two regions of the waveform), we have:
bn = (2/T) [∫[0 to T0.9] sin((2πt)/T) dt - ∫[T0.9 to T] sin((2πt)/T) dt]
Evaluating the integrals, we get:
bn = (2/T) [(-T0.9/2π) cos((2πt)/T)] from 0 to T0.9 - (-T0.1/2π) cos((2πt)/T)] from T0.9 to T
bn = (2/T) [(T - T0.9)/2π - (-T0.9)/2π]
bn = (T - T0.9)/π
Fourier Series Decomposition:
The Fourier series decomposition of the square waveform with a 90% duty cycle is given by:
f(t) = (a0/2) + ∑[(an * cos((2πnt)/T)) + (bn * sin((2πnt)/T))]
However, since a0 and an are 0 for this waveform, the decomposition simplifies to:
f(t) = ∑[(bn * sin((2πnt)/T))]
For n = 1, the decomposition becomes:
f(t) = (T - T0.9)/π * sin((2πt)/T)
This represents the Fourier series decomposition of the square waveform with a 90% duty cycle, including the computation of the Fourier coefficients and the final decomposition expression for the waveform.
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The open-loop transfer function of a unit-negative-feedback system has the form of
G(s)H(s) = 1 / s(s+1).
Please determine the following transient specifications when the reference input is a unit step function:
(1) Percentage overshoot σ%;
(2) Peak time tp;
(3) 2% Settling time t.
For the given open-loop transfer function 1 / (s(s+1)), the transient specifications when the reference input is a unit step function can be determined by calculating the percentage overshoot, peak time, and 2% settling time using appropriate formulas for a second-order system.
What is the percentage overshoot?To determine the transient specifications for the given open-loop transfer function G(s)H(s) = 1 / (s(s+1)) with a unit step reference input, we need to analyze the corresponding closed-loop system.
1) Percentage overshoot (σ%):
The percentage overshoot is a measure of how much the response exceeds the final steady-state value. For a second-order system like this, the percentage overshoot can be approximated using the formula: σ% ≈ exp((-ζπ) / √(1-ζ^2)) * 100, where ζ is the damping ratio. In this case, ζ = 1 / (2√2), so substituting this value into the formula will give the percentage overshoot.
2) Peak time (tp):
The peak time is the time it takes for the response to reach its maximum value. For a second-order system, the peak time can be approximated using the formula: tp ≈ π / (ωd√(1-ζ^2)), where ωd is the undamped natural frequency. In this case, ωd = 1, so substituting this value into the formula will give the peak time.
3) 2% settling time (ts):
The settling time is the time it takes for the response to reach and stay within 2% of the final steady-state value. For a second-order system, the settling time can be approximated using the formula: ts ≈ 4 / (ζωn), where ωn is the natural frequency. In this case, ωn = 1, so substituting this value into the formula will give the 2% settling time.
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