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Chapter 30: Problem 45

Design an RC high-pass filter that rejects \(60.0 \mathrm{~Hz}\) line noise from a circuit used in a detector. Your criteria are reduction of the amplitude of the line noise by a factor of 1000 . and total impedance at high frequencies of \(2.00 \mathrm{k} \Omega\) a) What components will you use? b) What is the frequency range of the signals that will be passed with at least \(90.0 \%\) of their amplitude?

Short Answer

Expert verified
Based on the given criteria, we have designed an RC high-pass filter that rejects line noise at 60 Hz. The components required for this filter are a 2.00 kΩ resistor and approximately a 1.333 μF capacitor. The frequency range for signals with at least 90% of their amplitude is approximately 83.355 Hz and above.

Step by step solution

01

Understanding the Criteria

To design an RC high-pass filter, we must first understand the given criteria. We want: 1. The amplitude of the 60 Hz noise to be reduced by a factor of 1000. 2. The total impedance at high frequencies should be 2.00 kΩ. Now let's move on to the calculations.
02

Calculating the Cutoff Frequency

We know that the ratio of the amplitudes (gain) in an RC high-pass filter is given by: Gain = \(\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + \left(\frac{f_c}{f}\right)^2}}\) For a reduction factor of 1000 (0.001), we have: \(\frac{1}{0.001} = \frac{1}{\sqrt{1 + \left(\frac{f_c}{60}\right)^2}}\) Solving for \(f_c\): \(f_c = 60\sqrt{\left(\frac{1}{0.001^2}\right) - 1} \approx 59.983 \mathrm{~Hz}\) This is the cutoff frequency of our high-pass filter.
03

Calculating Required Resistance and Capacitance

The formula for the cutoff frequency in an RC high-pass filter is given by: \(f_c = \frac{1}{2\pi R C}\) Where R is the resistance, C is the capacitance, and Pi is approximately 3.14159. We were also given that the total impedance at high frequencies should be 2.00 kΩ. In a high-pass filter, the impedance is approximately equal to the resistance, since the capacitive reactance is negligible. So, let's use R = 2.00 kΩ. Now, we can solve for C: \(C = \frac{1}{2\pi R f_c} = \frac{1}{2\pi \cdot 2.00 \cdot 10^3 \cdot 59.983 \mathrm{~Hz}} \approx 1.333 \mathrm{~\mu F}\)
04

Conclusion for Given Components

The components we need for our RC high-pass filter are: a) A resistor R with a value of 2.00 kΩ and a capacitor C with a value of approximately 1.333 μF.
05

Calculating Frequency Range with at least 90% Amplitude

We want to find the frequency range at which the output signal amplitude is at least 90% of the input amplitude. We can use the gain equation for this: \(\frac{V_{out}}{V_{in}} = 0.9 = \frac{1}{\sqrt{1 + \left(\frac{f_c}{f}\right)^2}}\) Solving for f: \(\frac{1}{0.9^2} - 1 = \left(\frac{f_c}{f}\right)^2 \Rightarrow f = \frac{f_c}{\sqrt{\frac{1}{0.9^2} - 1}} \approx 1.39 \cdot f_c\) Since \(f_c \approx 59.983 \mathrm{~Hz}\), the frequency range that will be passed with at least 90% of their amplitude is: f \(\geq 1.39 \cdot 59.983 \mathrm{~Hz} \approx 83.355 \mathrm{~Hz}\)

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