Question

Design an RC band-pass filter that passes a signal with frequency \(5.00 \mathrm{kHz},\) has a ratio \(V_{\text {out }} / V_{\text {in }}=0.500,\) and has an impedance of \(1.00 \mathrm{k} \Omega\) at very high frequencies. a) What components will you use? b) What is the phase of \(V_{\text {out }}\) relative to \(V_{\text {in }}\) at the frequency of \(5.00 \mathrm{kHz} ?\)

Short answer

Answer: The required components for the filter are a 1.00 kΩ resistor and a 10.5 nF capacitor. The phase difference between the input and output voltages at 5.00 kHz is approximately 60°.

Step-by-step solution

Analyze the RC band-pass filter characteristics

An RC band-pass filter consists of two filters: a low-pass filter followed by a high-pass filter. The transfer function depends on the values of the resistors and capacitors used in the design. The ratio \(V_{out}/V_{in}\) can be expressed as: $$\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + (RC\cdot 2\pi \cdot f)^2}}$$ Where \(R\) is the resistance in ohms, \(C\) is the capacitance in farads, and \(f\) is the frequency in hertz.

Find the values of the components using the specifications

We know that at 5.00 kHz, the ratio \(V_{out}/V_{in}\) is given to be 0.500. Plugging the given values into the formula above, we get: $$0.500 = \frac{1}{\sqrt{1 + (RC \cdot 2\pi \cdot 5000)^2}}$$ To find the values of the components, let's start by simplifying the equation: $$0.500^2 = 1 - (RC \cdot 2\pi \cdot 5000)^2$$ $$(RC \cdot 2\pi \cdot 5000)^2 = 1 - 0.500^2$$ $$RC = \frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000}$$ The impedance at very high frequencies is given to be 1.00 kΩ, which means the resistance \(R\) is also 1.00 kΩ. $$1000 \cdot C = \frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000}$$ Now, solve for \(C\): $$C = \frac{\frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000}}{1000}$$ $$C \approx 1.05 \times 10^{-8} F$$ The required components for the filter are a 1.00 kΩ resistor and a 10.5 nF capacitor. Answer to part a): A 1.00 kΩ resistor and a 10.5 nF capacitor

Calculate the phase difference between input and output voltages at 5.00 kHz

The phase difference between the input and output voltages can be calculated using the formula: $$\phi = \arctan(RC \cdot 2\pi \cdot f)$$ Using the values calculated for \(RC\): $$\phi = \arctan(\frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000} \cdot 2\pi \cdot 5000)$$ $$\phi = \arctan(\sqrt{1 - 0.500^2})$$ $$\phi \approx 60^\circ$$ The phase difference between the input and output voltages at 5.00 kHz is approximately 60°. Answer to part b): The phase difference is approximately 60° at 5.00 kHz.

Key concepts

Resistor and Capacitor Values

In an RC band-pass filter, selecting the appropriate resistor and capacitor values is essential for achieving the desired performance. These values directly impact the filter's frequency characteristics, including its ability to allow or reject certain frequencies. The resistance (\(R\)) and capacitance (\(C\)) together determine the filter's cut-off frequencies and the bandwidth of the pass-band.
When designing an RC band-pass filter for a specific frequency, you need to ensure that the product of the resistor and capacitor values (\(RC\)) meets the desired specifications. In this exercise, the requirement is to allow a signal with a frequency of 5.00 kHz. The target voltage ratio (\(V_{out}/V_{in}\) equal to 0.500) is achieved by solving the equation:

• \(\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + (RC \cdot 2\pi \cdot f)^2}}\)

By manipulating this equation, and given certain constraints (such as an impedance of 1.00 k\(\Omega\) at high frequencies), you solve for \(R\) and \(C\).
In practice, this meant choosing a resistor of 1.00 k\(\Omega\) and calculating the necessary capacitance, which in this case is approximately 10.5 nF. These components are foundational as they set the stage for the filter's overall functionality.

Frequency Response

The frequency response of an RC band-pass filter describes how the filter behaves as different frequencies pass through it. This characteristic dictates which frequencies are allowed through and which are attenuated. For the filter noted in this example, it centers around 5.00 kHz, meaning it passes signals close to this frequency while attenuating others.
The frequency response is a critical concept as it ensures the filter performs its intended function, allowing a specific range of frequencies to pass and blocking others. The filter's ability to produce a desired \(V_{out}/V_{in}\) ratio is a direct consequence of its frequency response. The designed filter achieves a \(V_{out}/V_{in}\) ratio of 0.500 at 5.00 kHz, perfectly illustrating this concept.
A band-pass filter like this combines the features of a low-pass filter and a high-pass filter. By doing so, it creates a "window" of frequencies it transmits. Properly tuning \(R\) and \(C\) ensures this window is placed correctly to pass the desired frequency of 5.00 kHz, providing the expected output signal level while minimizing distortion.

Phase Difference

Phase difference is an essential concept to understand when dealing with AC signals and their alterations in filters. For a band-pass filter, there's often a phase shift between the input and output signals. This phase shift arises due to the reactive nature of capacitors and resistors at different frequencies.
In the given exercise, the phase difference between the output and input voltages at the target frequency of 5.00 kHz was calculated to be 60°. This shift is induced by the signal's passage through the filter circuitry, specifically the interaction between the resistance (\(R\)) and the capacitance (\(C\)).
The phase difference can be determined using the formula:

• \(\phi = \arctan(RC \cdot 2\pi \cdot f)\)

This provides insight into how the signal's phase is altered relative to its input. At certain frequencies, this shift is more pronounced, and understanding it can be critical for applications requiring precise timing between signals. For many electronic devices, maintaining the correct phase relationship is crucial, thus highlighting the significance of this parameter in filter design.