Answer:
To find the frequency at which the output of a low-pass first-order instrument will be 93% of the DC output, we can use the formula for the transfer function of a first-order low-pass filter:
H(s) = 1 / (1 + s * T)
Where:
H(s) is the transfer function
s is the complex frequency variable
T is the time constant of the filter
To find the frequency at which the output is 93% of the DC output, we need to find the frequency at which the magnitude of the transfer function is 0.93. Let's solve for this frequency.
Magnitude of the transfer function |H(s)| = sqrt(Re^2 + Im^2)
At the desired frequency, |H(s)| = 0.93
0.93 = 1 / sqrt(1 + (2πfT)^2)
Squaring both sides:
0.93^2 = 1 / (1 + (2πfT)^2)
0.93^2 + 0.93^2(2πfT)^2 = 1
0.8649 + 0.8649(2πfT)^2 = 1
0.8649(1 + (2πfT)^2) = 1
1 + (2πfT)^2 = 1 / 0.8649
(2πfT)^2 = 1 / 0.8649 - 1
(2πfT)^2 = 0.158
Taking the square root:
2πfT = sqrt(0.158)
f = sqrt(0.158) / (2πT)
Given T = 20ms, which is 20 * 10^-3 seconds:
f = sqrt(0.158) / (2π * 20 * 10^-3)
f ≈ 0.398 / 0.1257
f ≈ 3.17 Hz
Therefore, the frequency at which the output of the low-pass first-order instrument will be 93% of the DC output is approximately 3.17 Hz.
Now, let's calculate the phase angle at this frequency.
The phase angle of a first-order low-pass filter can be calculated using the formula:
θ = -arctan(2πfT)
Using the same values for T and f as before:
θ = -arctan(2π * 3.17 * 20 * 10^-3)
θ ≈ -arctan(0.4006)
θ ≈ -21.88 degrees
Therefore, the phase angle at the frequency of 3.17 Hz is approximately -21.88 degrees.