Question

In the RC high-pass filter shown in the figure, \(R=10.0 \mathrm{k} \Omega\) and \(C=0.0470 \mu \mathrm{F}\) What is the 3.00 -dB frequency of this circuit (where \(\mathrm{dB}\) means basically the same for electric current as it did for sound in Chapter 16 )? That is, at what frequency does the ratio of output voltage to input voltage satisfy \(20 \log \left(V_{\text {out }} / V_{\text {in }}\right)=-3.00 ?\)

Short answer

Answer: The 3.00-dB frequency of this RC high-pass filter circuit is approximately 723.6 Hz.

Step-by-step solution

Find the voltage transfer function

The voltage transfer function for an RC high-pass filter is given by the formula: \(H(f) = \frac {V_{out}}{V_{in}} = \frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}}\). We will plug in the given values of R and C into this formula to find the transfer function.

Set \(20 \log \left(V_{\text {out }} / V_{\text{in }}\right)\) equal to -3.00

Given the equation, \(20 \log \left(V_{\text {out }} / V_{\text{in }}\right)=-3.00\), we will substitute \(V_{\text{out}}/V_{\text{in}}\) with the transfer function we found in Step 1: \(20 \log \left( \frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}} \right) = -3.00\).

Solve for frequency f

To solve for the frequency, f, follow these steps: 1. Divide both sides by 20: \(\log \left( \frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}} \right) = -0.15\). 2. Use the properties of logarithms to remove the log and square root: \(\frac {1}{\sqrt{1 + (\frac{1}{2\pi fRC})^2}} = 10^{-0.15}\) and then \({(10^{-0.15})^2 = \frac{1}{1 + (\frac{1}{2\pi fRC})^2}\) 3. Rearrange the equation to solve for \(f\): \((\frac{1}{2\pi fRC})^2 = \frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}\). 4. Take the square root of both sides: \(\frac{1}{2\pi fRC} = \sqrt{\frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}}\). 5. Rearrange to solve for f: \(f = \frac{1}{2\pi RC \sqrt{\frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}}}\).

Calculate the 3.00-dB frequency

Now, plug in the given values of R and C into the equation to find the 3.00-dB frequency: \(f = \frac{1}{2\pi (10.0 \times 10^3 \Omega)(0.0470 \times 10^{-6} F) \sqrt{\frac{1 - (10^{-0.15})^2}{(10^{-0.15})^2}}}\) After calculating the above expression, we get: \(f \approx 723.6 Hz\) Therefore, the 3.00-dB frequency of this RC high-pass filter circuit is approximately 723.6 Hz.

Key concepts

3.00-dB Frequency

The 3.00-dB frequency, often called the cutoff frequency, is an essential concept for understanding filter circuits. It defines the point where the output signal's power is reduced to half its input power, resulting in a drop of 3.00 decibels. In practice, this is where the filter begins to significantly attenuate the signal. For an RC high-pass filter, which allows high-frequency signals to pass while attenuating lower frequencies, the formula used to find this frequency includes both the resistance \(R\) and the capacitance \(C\).

Mathematically, the step-by-step calculation provides the actual frequency where this attenuation happens. Using the computation:

• The logarithmic expression \(20 \log(V_{out} / V_{in}) = -3.00\) helps identify the critical frequency range.
• Solving directly for \(f\) gives the specific cutoff frequency: approximately 723.6 Hz for the given values of \(R\) and \(C\).

Voltage Transfer Function

The voltage transfer function of a filter describes how the output voltage relates to the input voltage. For an RC high-pass filter, it's defined as:

\[ H(f) = \frac {1}{\sqrt{1 + \left(\frac{1}{2\pi fRC}\right)^2}} \]

This function is crucial as it characterizes the frequency response of the circuit. As frequency increases, the role of the capacitor decreases, allowing more of the input signal to pass through.

Key aspects include:

• The function tells us how much of the signal's amplitude is preserved at a given frequency.
• Understanding this interaction helps in predicting the filter's behavior across different frequency ranges.
• This concept ties back to the 3.00-dB frequency where attenuation reaches that specific point.

Electrical Circuits

An electrical circuit designed to perform specific signal processing tasks, like an RC high-pass filter, consists of various interconnected components like resistors (R) and capacitors (C).

These components create a filter that affects how signals of different frequencies pass through. The resistor controls the flow of current, while the capacitor stores and releases electrical energy, affecting signal phase and frequency response.

Understanding how these components interact allows engineers to design filters with intended behaviors:

• The formula \(f = \frac{1}{2\pi RC}\) is derived from the fundamental characteristics of these components.
• Altering \(R\) or \(C\) will modify the filter's response, key in customizing circuits for specific applications.

Signal Processing

Signal processing in the context of electrical engineering involves manipulating signals to enhance or modify their properties. Filters like RC high-pass filters play a pivotal role in this by allowing certain frequencies to pass while attenuating others.

In practical terms, they are used in applications such as:

• Communications, to remove low-frequency noise from audio signals.
• Audio devices, to handle bass and treble separately, enhancing sound quality.

By understanding how these filters work, one can tailor the signal processing chain to meet the desired outcomes, ensuring signals retain quality and clarity while unnecessary elements are attenuated.