Question

Use a $25\mathrm{mH}$ inductor to design a low-pass, $RL$, passive filter with a cutoff frequency of $2.5\mathrm{kHz}$ a) Specify the value of the resistor. b) $\mathrm{A}$ load having a resistance of $750\mathrm{S}$ is connected across the output terminals of the filter. What is the corner, or cutoff, frequency of the loaded filter in hertz?

Short answer

The resistor value is approximately 392.7 ohms, and the loaded cutoff frequency is approximately 1.66 kHz.

Step-by-step solution

Understand the cutoff frequency formula for an RL filter

The cutoff frequency $f_c$ for a low-pass RL filter is given by the formula $f_c=\frac{R}{2\pi L}$, where $R$ is the resistance in ohms and $L$ is the inductance in henries.

Convert inductance units

The inductance is given as $25$ mH, which needs to be converted to henries. Thus, $L=25\times {10}^{-3}$ H.

Solve for the resistor value for given cutoff frequency

To find the resistor $R$ that achieves a cutoff frequency $f_c=2.5$ kHz, plug the values into the formula to get $2.5\times {10}^3=\frac{R}{2\pi \times 25\times {10}^{-3}}$. Solve for $R$ to find $R=2\pi \times 25\times {10}^{-3}\times 2.5\times {10}^3\approx 392.7$ ohms.

Determine new cutoff frequency with load

When a load is connected, the total resistance becomes $R_{total}$ which is the parallel combination of the original $R$ (392.7 ohms) and the load resistance (750 ohms). Use the parallel resistor formula: $\frac{1}{R_{total}}=\frac{1}{392.7}+\frac{1}{750}$ to find $R_{total}\approx 260.9$ ohms.

Calculate the loaded cutoff frequency

Substitute $R_{total}=260.9$ ohms and $L=25\times {10}^{-3}$ H into the cutoff frequency formula: $f_c=\frac{260.9}{2\pi \times 25\times {10}^{-3}}\approx 1.66$ kHz.

Key concepts

Cutoff Frequency

The cutoff frequency is a crucial parameter in an RL low-pass filter. This frequency marks the point where the output signal's amplitude decreases to 70.7% of the input signal's amplitude. In technical terms, it is where the power drops to half, often referred to as the -3 dB point.

When designing a low-pass filter, the cutoff frequency determines which frequencies are allowed to pass and which are attenuated. For the RL filter, the cutoff frequency can be calculated using the formula:

• $f_c=\frac{R}{2\pi L}$

where $f_c$ is the cutoff frequency in hertz, $R$ is the resistance in ohms, and $L$ is the inductance in henries. Understanding this concept allows you to select appropriate resistor and inductor values to fit your filtering needs.

In our example, the RL filter with a cutoff frequency of 2.5 kHz effectively limits frequencies higher than this, ensuring a desired low-pass behavior.

Inductor Value

Inductors play a vital role in RL low-pass filters by opposing changes in current. This property makes them suitable for filtering high-frequency signals. The value of the inductor directly influences the filter's cutoff frequency.

In the given problem, an inductor of 25 mH is utilized. To apply this in calculations, make sure to convert millihenries (mH) to henries (H):

• 25 mH = 25 $\times {10}^{-3}$ H

This conversion is important for ensuring all units are compatible, allowing accurate calculations of the filter's cutoff frequency based on the formula $f_c=\frac{R}{2\pi L}$.

By choosing the right inductor value, you can tailor the filter to suit specific applications, whether that involves audio processing or radio frequency tuning.

Resistor Calculation

Accurate resistor calculation is essential for setting the correct cutoff frequency in an RL filter. The resistor and inductor work jointly to determine frequency response.

To calculate the necessary resistor value for a specific cutoff frequency, rearrange the cutoff frequency formula:

• $R=2\pi L\times f_c$

In the exercise, a cutoff frequency of 2.5 kHz is targeted using a 25 mH inductor. Substituting these values into the formula gives:

• $R=2\pi \times 25\times {10}^{-3}\times 2.5\times {10}^3\approx 392.7$ ohms

Understanding how to calculate the resistor value allows for precision in designing filters that meet your intended specifications, ensuring optimal performance for the desired application.

Low-Pass Filter

A low-pass filter, like the RL filter, is designed to allow low-frequency signals through while attenuating high-frequency signals. This ability is pivotal in various applications, from audio processing to reducing noise in electronic signals.

The RL low-pass filter is a simple yet effective design that involves a resistor (R) and inductor (L) in series. The filter's effectiveness hinges on the correct calculation of its components to achieve the desired cutoff frequency. This frequency is where the filter begins to significantly attenuate higher frequencies.

When additional components, like a load resistance, are connected, the filter's properties may change. For example, adding a 750-ohm load in parallel affects the cutoff frequency, which in turn requires recalculation using the parallel resistor formula:

• $\frac{1}{R_{total}}=\frac{1}{R}+\frac{1}{R_{load}}$

This results in an adjusted cutoff frequency, ensuring the filter remains effective, even with varying conditions. This adaptability is part of what makes low-pass filters so invaluable in controlling signal frequencies.