Question

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

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

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.

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.

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: \(C = \frac{1}{2\pi R f_c} = \frac{1}{2\pi \cdot 2.00 \cdot 10^3 \cdot 59.983 \mathrm{~Hz}} \approx 1.333 \mathrm{~\mu F}\)

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.

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 \(\geq 1.39 \cdot 59.983 \mathrm{~Hz} \approx 83.355 \mathrm{~Hz}\)

Key concepts

Line Noise Rejection

Line noise, particularly at 60 Hz, is a common issue in many electrical applications. It can interfere with sensitive equipment, like detectors, which need clean signals to function effectively. A key strategy to combat this interference is through using an RC high-pass filter.
The purpose of such a filter is to attenuate, or weaken, unwanted low-frequency signals, allowing higher frequency signals to pass through unimpeded. In this case, we want the filter to reduce the amplitude of 60 Hz line noise by a factor of 1000. This means that the filter's design must ensure that 60 Hz signals are heavily dampened, practically eliminating their impact on the circuit's output.
Achieving such attenuation involves carefully selecting the filter's components to meet this criterion while also keeping the desired signals intact.

Cutoff Frequency Calculation

A crucial factor in designing a high-pass filter is determining the cutoff frequency (\(f_c\)). This frequency defines the threshold below which signals are significantly reduced. For our design, we calculated the cutoff frequency using the attenuation requirement and the formula:

• Gain = \(\frac{1}{\sqrt{1 + \left(\frac{f_c}{f}\right)^2}}\)

Given that the noise reduction required is by a factor of 1000, solving the calculations provides us with \(f_c \approx 59.983 \, \mathrm{Hz}\).
This calculation ensures that signals lower than 59.983 Hz, like the 60 Hz noise, are sharply reduced by the filter, while frequencies above it remain largely unaffected, maintaining their integrity in the circuit's processing.

Impedance in Filter Circuits

Impedance is a critical characteristic in filter circuit design. It combines both resistance and reactance, which is the opposition to changes in current. For high-pass filters, the impedance at high frequencies mostly equals the resistance component because the capacitive reactance decreases as frequency increases.
In this exercise, we need a total impedance of 2.00 kΩ at high frequencies. This requirement significantly influences the choice of resistor. The resistor value directly dictates the impedance, as the capacitive component becomes negligible at these frequencies.
Thus, selecting a resistor of 2.00 kΩ ensures that the filter meets the high-frequency impedance specification, allowing the expected performance of the filter in rejecting unwanted noise.

Frequency Response of Filters

The frequency response of an RC high-pass filter describes how the filter's output varies across different frequencies. It's crucial to understand this behavior to predict how the filter will perform in terms of passing or attenuating signals.
For this particular high-pass filter, our goal is to allow signals with at least 90% amplitude return for frequencies above a certain threshold. This is derived from calculating when the gain equation is 0.9, giving us a frequency of approximately 83.355 Hz.
This means that frequencies beyond 83.355 Hz will be passed through the filter with almost no reduction — a critical zone for ensuring the circuit's operational signals remain unaffected. By setting the cutoff just below this frequency, the filter effectively manages which frequencies are transmitted with minimum distortion, ensuring the desired signals pass while rejecting noise.