Question

Design an RC high-pass filter that rejects 60.0 - 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

Question: Design an RC high-pass filter that rejects 60 Hz line noise with a reduction factor of 1000 and a total impedance of 2 kΩ at high frequencies. Find the components to be used and the frequency range of the signals that will be passed with at least 90% of their amplitude. Answer: A suitable RC high-pass filter can be constructed using a \(2\,\mathrm{k}\Omega\) resistor and a \(1.33\,\text{nF}\) capacitor. Signals with a frequency of approximately \(66,667\,\text{Hz}\) and higher will be passed with at least 90% of their amplitude.

Step-by-step solution

Calculate the resistance and capacitance of the RC circuit

We can begin by calculating the cut-off frequency (\(f_c\)) of the RC high-pass filter. The cut-off frequency is related to the reduction factor (\(k\)) and the line noise frequency (\(f_n\)) by the following equation: \(f_c = k \cdot f_n\) Now, plug in the given values of the reduction factor and the line noise frequency to find the cut-off frequency. \(f_c = 1000 \cdot 60 \,\text{Hz}\) \(f_c = 60,000 \,\text{Hz}\) The cut-off frequency is related to the resistance (\(R\)) and capacitance (\(C\)) of the RC circuit by the following equation: \(f_c = \frac{1}{2\pi R C}\) We are given the total impedance at high frequencies, which is the resistance in our circuit. So, \(R = 2 \,\mathrm{k}\Omega = 2 \times 10^3\,\Omega\) Now, rearrange the equation for the cut-off frequency to find the capacitance: \(C = \frac{1}{2\pi R f_c}\) Plug in the values of resistance and cut-off frequency: \(C = \frac{1}{2\pi \times 2 \times 10^3\,\Omega \times 60,000\,\text{Hz}}\) \(C = 1.33 \times 10^{-9}\,\text{F}\) We can use a \(2\,\mathrm{k}\Omega\) resistor and a \(1.33\,\text{nF}\) capacitor for the RC high-pass filter.

Calculate the frequency range for signals passed with at least 90% amplitude

To find the frequency range for signals passed with at least 90% amplitude, we will use the filter's cut-off frequency and the amplitude transfer function of the RC filter. The amplitude transfer function for a high-pass filter is: \(H_A = \frac{V_\text{out}}{V_\text{in}} = \sqrt{\frac{f^2}{f_c^2 + f^2}}\) Here \(V_\text{out}\) and \(V_\text{in}\) are output and input voltages, respectively, and \(f\) is the frequency. We are asked to find the frequency range at which the amplitude transfer function is at least 90% or 0.9. \(0.9 = \sqrt{\frac{f^2}{f_c^2 + f^2}}\) Square both sides and rearrange the equation to solve for \(f\): \(f^2 = 0.9^2(f_c^2 + f^2)\) \(f^2(1 - 0.9^2) = 0.9^2 f_c^2\) \( f = \sqrt{\frac{0.9^2 f_c^2}{1 - 0.9^2}}\) Now plug in the cut-off frequency we calculated previously: \(f = \sqrt{\frac{0.9^2 \times (60,000\,\text{Hz})^2}{1 - 0.9^2}}\) \(f \approx 66,667\,\text{Hz}\) Therefore, the frequency range of the signals that will be passed with at least 90% of their amplitude is approximately \(66,667\,\text{Hz}\) and higher.