Question

Design an RC high-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 components to be used are a resistor R with a value of 1 kΩ and a capacitor C with a value of approximately 18.26 nF. The phase of the output voltage relative to the input voltage at 5 kHz is approximately 60°.

Step-by-step solution

Determine the Resistance R and Capacitance C values

To design the RC high-pass filter, we must first determine the value of the resistor R and capacitor C. We are given that the impedance at very high frequencies is 1 kΩ, which is the same value as R. So, R = 1 kΩ. The transfer function for an RC high-pass filter is given by: $$ \frac{V_{out}}{V_{in}} = \frac{j\omega RC}{1 + j\omega RC} $$ where \(\omega = 2\pi f\). We are given that at \(f = 5.00 \mathrm{kHz}\), the voltage ratio is 0.500. Plug in the given values, and we get: $$ 0.500 = \frac{j2\pi(5\times10^3)(1000)C}{1 + j2\pi(5\times10^3)(1000)C} $$ Solving for C, we get: $$ C = \frac{1}{2\pi (5\times10^3)(1000) \sqrt{3}}\approx 18.26 \mathrm{nF} $$

Determine the Phase Relationship

To find the phase of \(V_{out}\) relative to \(V_{in}\) at the frequency of \(5.00 \mathrm{kHz}\), we need to calculate the phase angle of the transfer function. The phase angle \(\phi\) is given by: $$ \phi = \arctan(\frac{\text{Imaginary Part}}{\text{Real Part}}) = \arctan(\frac{j\omega RC}{1 + j\omega RC}) $$ At the given frequency of 5 kHz, the phase angle \(\phi\) is: $$ \phi = \arctan(\frac{2\pi(5\times10^3)(1000)(18.26\times10^{-9})}{1 + 2\pi(5\times10^3)(1000)(18.26\times10^{-9})}) \approx 60^\circ $$ a) The components to be used are a resistor R with a value of 1 kΩ and a capacitor C with a value of approximately 18.26 nF. b) The phase of \(V_{out}\) relative to \(V_{in}\) at the frequency of 5 kHz is approximately 60°.

Key concepts

Impedance Calculation

Understanding impedance in an RC high-pass filter is crucial for designing circuits that cater to specific frequencies. Impedance, represented by the symbol `Z`, is analogous to resistance but it differs due to its frequency-dependent nature. It combines the resistance `R` and the reactance `X`, and is calculated using complex numbers due to the phase difference between voltage and current.

For a capacitor, the reactance `X_C` is defined as \(\frac{1}{\(\omega C\)}\), where \(\omega = 2\pi f\) is the angular frequency and `C` is the capacitance. Reactance decreases with increasing frequency, which is key to how high-pass filters work, allowing high frequencies to pass while attenuating low frequencies.

In our example, at very high frequencies, the capacitor's reactance becomes negligible compared to the resistor's resistance, which is why the given impedance \(1.00 \mathrm{k}\Omega\) is essentially just the resistance `R` of the resistor. This makes the calculation straightforward for this scenario.

Phase Relationship

The phase relationship in an RC circuit refers to the difference in phase between the input voltage \(V_{\text{in}}\) and output voltage \(V_{\text{out}}\). This difference arises because the capacitor in the circuit introduces a time delay for the voltage to build up across it, leading to the output voltage lagging behind the input voltage.

In our RC high-pass filter design example, we use the transfer function's argument to calculate the phase angle. By taking the arctangent of the ratio between the imaginary and real parts of the complex transfer function, we find the phase angle \(\phi\). For the given frequency of 5 kHz, we established a phase shift of approximately 60°, which indicates that the output voltage lags the input voltage by that angle at this specific frequency. Phase considerations are essential when designing circuits for applications requiring precise timing or coordination of signals.

Transfer Function

The transfer function is a mathematical representation that describes the relationship between the input and output of a system—in our case, an RC high-pass filter. It is defined as the ratio of the output voltage \(V_{\text{out}}\) to the input voltage \(V_{\text{in}}\) and is expressed in terms of complex frequencies to encapsulate both magnitude and phase information of the signal.

For the high-pass filter, the transfer function \(\frac{V_{\text{out}}}{V_{\text{in}}} = \frac{j\omega RC}{1 + j\omega RC}\) shows that the output signal's amplitude and phase are frequency-dependent. This function helps us determine the behavior of the filter at different frequencies. By setting the desired voltage ratio and solving for `C`, as we did in the provided example, we can design our filter to meet specified criteria.

Understanding the transfer function is key to predicting how the filter will respond to different input signals and is an essential tool for electrical engineers when manipulating signal frequencies for various practical applications.