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A low-pass first-order instrument has a time constant of 20ms. Find the frequency,in hertz, of the input at which the output will be 93% of the dc out put. Find the phase angle at this frequency? i need full detailed answer

User JaminSore
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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.

User Peter Pei Guo
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