Question

A particular RC low-pass filter has a breakpoint frequency of \(200 .\) Hz. At what frequency will the output voltage divided by the input voltage be \(0.100 ?\)

Short answer

Answer: The frequency is approximately 1414.21 Hz.

Step-by-step solution

Recall the Formula for RC Low-Pass Filter

The transfer function (the ratio of the output voltage to the input voltage) for an RC low-pass filter is given by: $$\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + (\omega RC)^2}}$$ where: - \(V_{out}\) is the output voltage - \(V_{in}\) is the input voltage - \(\omega\) is the angular frequency in radians per second (\(\omega = 2\pi f\), where \(f\) is the frequency in Hertz) - \(R\) is the resistance in ohms - \(C\) is the capacitance in farads - \(f\) is the frequency in Hertz

Find the RC Value from Breakpoint Frequency

The breakpoint frequency is given as 200 Hz. At the breakpoint frequency, the transfer function is \(\frac{1}{\sqrt{2}}\). So, we can write: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{1 + (2\pi\cdot200\cdot RC)^2}}$$ Now, we solve for the RC value: $$1 + (2\pi\cdot200\cdot RC)^2 = 2$$ $$(2\pi\cdot200\cdot RC)^2 = 1$$ $$2\pi\cdot200\cdot RC = 1$$ $$RC = \frac{1}{2\pi\cdot200}$$

Use the Transfer Function Formula

Now that we have the RC value, we can use the given relation between \(V_{out}\) and \(V_{in}\): $$\frac{V_{out}}{V_{in}} = 0.1$$ Substitute the transfer function formula, and set it equal to 0.1: $$0.1 = \frac{1}{\sqrt{1+(\omega RC)^2}}$$

Solve for Angular Frequency

Now, we solve for the angular frequency: $$1+(\omega RC)^2 = \frac{1}{0.1^2}$$ $$(\omega RC)^2 = \frac{1}{0.01} - 1$$ $$\omega RC = \sqrt{\frac{1}{0.01}- 1}$$ Substitute the RC value: $$\omega =\frac{\sqrt{\frac{1}{0.01}- 1}}{\frac{1}{2\pi\cdot200}}$$

Convert Angular Frequency to Hertz

Now, convert the angular frequency to Hertz using the relation \(\omega = 2\pi f\), where \(f\) is the frequency in Hertz: $$f = \frac{\omega}{2\pi}$$ $$f = \frac{\sqrt{\frac{1}{0.01} - 1}}{\frac{1}{2\pi\cdot200}\cdot2\pi}$$ Calculate the result: $$f \approx 1414.21 \text{ Hz}$$

Conclusion

The frequency at which the output voltage divided by the input voltage for this particular RC low-pass filter is 0.1 is approximately 1414.21 Hz.

Key concepts

Breakpoint Frequency

The breakpoint frequency, also known as the cutoff or corner frequency, is a crucial concept in the study of filters, including the RC low-pass filter. It is the frequency at which the output signal's power drops to half its value at low frequencies, or equivalently, when the output voltage is reduced to approximately 70.7% (or \(1/\sqrt{2}\)) of the input voltage.

For an RC low-pass filter, the breakpoint frequency is directly related to the resistance (R) and capacitance (C) of the circuit: \(f_c = 1/(2\pi RC)\). When the frequency of the input signal is below the breakpoint frequency, the filter passes signals with little attenuation. In contrast, signals with frequencies higher than the breakpoint frequency are significantly attenuated.

Understanding the breakpoint frequency is essential for predicting how the filter will behave in response to different frequency signals, making it a key design parameter in electronic circuits.

Transfer Function

The transfer function is an expression that determines how the output of a system relates to its input. In the context of RC low-pass filters, the transfer function tells us the magnitude by which the input signal's amplitude is reduced at a particular frequency.

The transfer function is defined as the ratio of the output voltage (\(V_{out}\)) to the input voltage (\(V_{in}\)), given as a function of angular frequency \(\omega\): \(\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + (\omega RC)^2}}\). As the frequency increases, the denominator of the transfer function increases, which results in a decreased output voltage.

The transfer function facilitates the understanding of the response of the filter at different frequencies, and it can be graphically represented as a Bode plot. For students, analyzing the transfer function can reveal how effectively the filter will smooth out or eliminate unwanted high-frequency noise.

Angular Frequency

Angular frequency, usually denoted as \(\omega\), is the rate of change of the phase of a sinusoidal waveform, or in simpler terms, it's the speed at which a wave oscillates in radians per second. It is connected with the linear frequency (f), which is measured in Hertz (Hz), by the relation \(\omega = 2\pi f\).

Since the response of an RC low-pass filter is frequency-dependent, the concept of angular frequency becomes crucial in its analysis. In calculations and filter design, especially when using the transfer function, angular frequency provides a direct and effective way to account for the filter's behavior across different frequencies.

Understanding angular frequency helps students to convert between familiar Hertz and the radians per second needed for analyzing the filter's response in terms of phase shifts and the effect these have on the signal passing through the filter.