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Key Insights

• Butterworth filters are known for their maximally flat response in the passband.
• A two-pole (second-order) high-pass filter provides a roll-off rate of 40 dB/decade in the stopband.
• Designing for unity gain at high frequencies simplifies the circuit and calculations.

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Designing active filters is a fundamental task in electronics, allowing for precise control over frequency responses. The Butterworth filter is a popular choice due to its flat passband characteristic, which minimizes distortion of signals within the desired frequency range. In this response, we will delve into the design of a two-pole high-pass Butterworth active filter with a specified cutoff frequency and unity gain in the high-frequency passband. We will also analyze the filter's gain at a specific frequency to demonstrate its performance.

Understanding Butterworth Filters and High-Pass Characteristics

The Foundation of Filter Design

The Butterworth filter is a type of signal processing filter designed to have a frequency response that is as flat as possible in the passband. This makes it ideal for applications where preserving the amplitude of signals within the passband is crucial. Unlike other filter types such as Chebyshev or elliptic filters, Butterworth filters do not exhibit ripple in the passband or stopband, although they have a slower roll-off rate.

A high-pass filter allows frequencies above a certain cutoff frequency to pass through while attenuating frequencies below it. The cutoff frequency, often defined as the -3dB point, is the frequency at which the filter's gain is reduced by 3 decibels compared to its passband gain. A two-pole, or second-order, high-pass filter incorporates two reactive components (typically capacitors in the case of a high-pass active filter design using the Sallen-Key topology), resulting in a steeper roll-off of 40 dB per decade in the stopband compared to a first-order filter's 20 dB per decade.

The Role of Active Filters

Active filters utilize active components, such as operational amplifiers (op-amps), in addition to passive components (resistors and capacitors). This allows for characteristics that are difficult or impossible to achieve with passive filters alone, including:

• Gain: Active filters can provide gain in the passband, amplifying the signal.
• Buffering: The active component can isolate the filter stages, preventing loading effects.
• Easier design of higher-order filters: Complex filter responses can be achieved by cascading lower-order active filter stages.

For a high-pass filter with unity gain in the high-frequency passband, an active implementation is often preferred for its ability to provide gain equal to 1 (0 dB) and buffer the output.

Designing the Two-Pole High-Pass Butterworth Active Filter

Setting the Specifications

The design requires a two-pole high-pass Butterworth active filter with:

• Cutoff frequency (\(f_{3dB}\)) = 25 kHz
• Unity gain magnitude at high frequency (Passband gain \(A_{\infty}\) = 1)

Choosing the Topology: Sallen-Key Filter

A common and effective topology for realizing second-order active filters is the Sallen-Key configuration. For a high-pass filter, the basic Sallen-Key circuit uses two resistors and two capacitors, along with an operational amplifier. Since we require unity gain in the passband, we can use the unity-gain Sallen-Key topology, which simplifies the component relationships.

A typical Sallen-Key high-pass filter configuration.

Deriving the Component Values

For a unity-gain Sallen-Key high-pass Butterworth filter, the transfer function is given by:

\[ H(s) = \frac{s^2}{s^2 + s\frac{\sqrt{2}}{\omega_c} + \frac{1}{\omega_c^2}} \]

where \(s\) is the complex frequency variable, and \(\omega_c = 2\pi f_{3dB}\) is the angular cutoff frequency.

In the Sallen-Key high-pass configuration with unity gain, the general transfer function is:

\[ H(s) = \frac{s^2 R_1 R_2 C_1 C_2}{s^2 R_1 R_2 C_1 C_2 + s (R_1 C_1 + R_2 C_1 + R_2 C_2(1-K)) + 1} \]

where \(K\) is the gain of the non-inverting amplifier. For unity gain (\(K=1\)), the transfer function simplifies to:

\[ H(s) = \frac{s^2 R_1 R_2 C_1 C_2}{s^2 R_1 R_2 C_1 C_2 + s (R_1 C_1 + R_2 C_1) + 1} \]

By comparing the coefficients of the Butterworth transfer function and the Sallen-Key transfer function with unity gain, we can establish relationships between the component values and the cutoff frequency. For a second-order Butterworth filter with unity gain, the standard normalized transfer function's denominator polynomial is \(s^2 + \sqrt{2}s + 1\).

To match this, we need:

\[ R_1 R_2 C_1 C_2 = \frac{1}{\omega_c^2} \] \[ R_1 C_1 + R_2 C_1 = \frac{\sqrt{2}}{\omega_c} \]

We can simplify the design by choosing \(C_1 = C_2 = C\) and \(R_1 = R_2 = R\). In this case, the equations become:

\[ R^2 C^2 = \frac{1}{\omega_c^2} \implies RC = \frac{1}{\omega_c} \] \[ RC + RC = 2RC = \frac{\sqrt{2}}{\omega_c} \]

This simplifies to \(RC = \frac{\sqrt{2}}{2\omega_c} = \frac{1}{\sqrt{2}\omega_c}\). However, this choice of equal resistors and capacitors does not directly yield the Butterworth response for unity gain. A more common approach for unity-gain Butterworth is to set \(C_1 = C_2 = C\) and choose specific resistor values, or set \(R_1 = R_2 = R\) and choose specific capacitor values based on normalized Butterworth polynomial coefficients.

Let's consider the case where \(R_1 = R_2 = R\). Then the equations become:

\[ R^2 C_1 C_2 = \frac{1}{\omega_c^2} \] \[ R C_1 + R C_1 = R(C_1 + C_2) = \frac{\sqrt{2}}{\omega_c} \]

If we select \(C_1 = C\) and \(C_2 = 2C\), the second equation becomes \(R(C + 2C) = 3RC = \frac{\sqrt{2}}{\omega_c}\), so \(RC = \frac{\sqrt{2}}{3\omega_c}\). The first equation becomes \(R^2 (C)(2C) = 2R^2C^2 = \frac{1}{\omega_c^2}\), so \(RC = \frac{1}{\sqrt{2}\omega_c}\). These do not align.

Let's consider the case where \(C_1 = C\) and \(C_2 = C\). Then the equations become:

\[ R_1 R_2 C^2 = \frac{1}{\omega_c^2} \implies \sqrt{R_1 R_2} C = \frac{1}{\omega_c} \] \[ (R_1 + R_2) C = \frac{\sqrt{2}}{\omega_c} \implies (R_1 + R_2) = \frac{\sqrt{2}}{C \omega_c} \]

For a unity-gain Sallen-Key high-pass Butterworth filter, a common approach is to set \(C_1 = C_2 = C\) and determine the resistor values \(R_1\) and \(R_2\). The required quality factor (Q) for a second-order Butterworth filter is \(Q = \frac{1}{\sqrt{2}}\). The equations relating component values to cutoff frequency and Q for the unity-gain Sallen-Key high-pass filter are:

\[ \omega_c = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}} \] \[ Q = \frac{\sqrt{R_1 R_2 C_1 C_2}}{R_1 C_1 + R_2 C_1} \]

With \(C_1 = C_2 = C\), these simplify to:

\[ \omega_c = \frac{1}{C\sqrt{R_1 R_2}} \] \[ Q = \frac{C\sqrt{R_1 R_2}}{(R_1 + R_2)C} = \frac{\sqrt{R_1 R_2}}{R_1 + R_2} \]

We know \(\omega_c = 2\pi f_{3dB} = 2\pi \times 25 \times 10^3\) rad/s and \(Q = \frac{1}{\sqrt{2}}\). We need to select a convenient capacitor value, for example, \(C = 1 nF\). Then we can solve for \(R_1\) and \(R_2\).

\[ \sqrt{R_1 R_2} = \frac{1}{\omega_c C} = \frac{1}{(2\pi \times 25 \times 10^3)(1 \times 10^{-9})} = \frac{1}{2\pi \times 25 \times 10^{-6}} \approx 6366.2 \] \[ R_1 R_2 \approx (6366.2)^2 \approx 40.53 \times 10^6 \] \[ \frac{\sqrt{R_1 R_2}}{R_1 + R_2} = \frac{1}{\sqrt{2}} \implies R_1 + R_2 = \sqrt{2} \sqrt{R_1 R_2} \approx 1.414 \times 6366.2 \approx 8908.7 \]

We have two equations:

1. \(R_1 R_2 \approx 40.53 \times 10^6\)
2. \(R_1 + R_2 \approx 8908.7\)

We can solve this system of equations. From equation 2, \(R_2 \approx 8908.7 - R_1\). Substitute this into equation 1:

\[ R_1 (8908.7 - R_1) \approx 40.53 \times 10^6 \] \[ 8908.7 R_1 - R_1^2 \approx 40.53 \times 10^6 \] \[ R_1^2 - 8908.7 R_1 + 40.53 \times 10^6 \approx 0 \]

Using the quadratic formula \(R_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-8908.7\), and \(c=40.53 \times 10^6\):

\[ \Delta = b^2 - 4ac = (-8908.7)^2 - 4(1)(40.53 \times 10^6) \approx 79.36 \times 10^6 - 162.12 \times 10^6 < 0 \]

This indicates that with \(C_1 = C_2\) and \(R_1 \neq R_2\), we cannot achieve the exact Butterworth response with unity gain using this direct calculation method without complex resistor values. A common practical approach for unity-gain Sallen-Key Butterworth is to use specific ratios of R and C values based on normalized filter tables or design formulas derived for the unity-gain case. For a second-order unity-gain Butterworth high-pass filter with \(C_1 = C_2 = C\), the resistor values are often chosen such that \(R_1 = R\) and \(R_2 = 2R\), or specific ratios related to the Butterworth polynomial coefficients.

A simplified design approach for the unity-gain Sallen-Key high-pass Butterworth filter with \(C_1 = C_2 = C\) yields the following relationships for \(R_1\) and \(R_2\):

\[ R_1 = \frac{\sqrt{2}}{2 \omega_c C} \] \[ R_2 = \frac{\sqrt{2}}{ \omega_c C} \]

Let's use \(C = 1 nF\).

\[ \omega_c = 2\pi \times 25 \times 10^3 = 50\pi \times 10^3 \text{ rad/s} \] \[ R_1 = \frac{\sqrt{2}}{2 \times (50\pi \times 10^3) \times (1 \times 10^{-9})} = \frac{\sqrt{2}}{100\pi \times 10^{-6}} \approx \frac{1.414}{314.16 \times 10^{-6}} \approx 4500 \Omega \] \[ R_2 = \frac{\sqrt{2}}{ (50\pi \times 10^3) \times (1 \times 10^{-9})} = \frac{\sqrt{2}}{50\pi \times 10^{-6}} \approx \frac{1.414}{157.08 \times 10^{-6}} \approx 9000 \Omega \]

So, for a practical design with \(C_1 = C_2 = 1 nF\), we can use standard resistor values close to \(R_1 = 4.5 k\Omega\) and \(R_2 = 9 k\Omega\). Common standard values would be \(R_1 = 4.3 k\Omega\) or \(4.7 k\Omega\) and \(R_2 = 8.2 k\Omega\) or \(9.1 k\Omega\).

Alternatively, if we set \(R_1 = R_2 = R\), we would need to use unequal capacitor values \(C_1\) and \(C_2\). For a unity-gain Sallen-Key high-pass Butterworth filter with \(R_1 = R_2 = R\), the relationships are:

\[ R^2 C_1 C_2 = \frac{1}{\omega_c^2} \] \[ R (C_1 + C_2) = \frac{\sqrt{2}}{\omega_c} \]

Let's choose \(R = 10 k\Omega\). Then:

\[ (10 \times 10^3)^2 C_1 C_2 = \frac{1}{(50\pi \times 10^3)^2} \] \[ 100 \times 10^6 C_1 C_2 = \frac{1}{2500\pi^2 \times 10^6} \] \[ C_1 C_2 = \frac{1}{2500\pi^2 \times 10^{12}} \approx \frac{1}{24674 \times 10^{12}} \approx 40.5 \times 10^{-18} \] \[ 10 \times 10^3 (C_1 + C_2) = \frac{\sqrt{2}}{50\pi \times 10^3} \] \[ C_1 + C_2 = \frac{\sqrt{2}}{50\pi \times 10^6} \approx \frac{1.414}{157.08 \times 10^6} \approx 9 \times 10^{-9} \]

We have:

1. \(C_1 C_2 \approx 40.5 \times 10^{-18}\)
2. \(C_1 + C_2 \approx 9 \times 10^{-9}\)

This is similar to the resistor calculation. Solving for \(C_1\) and \(C_2\) would yield complex numbers. Therefore, for unity gain Sallen-Key Butterworth, using equal capacitors and unequal resistors based on the \(R_1 = \frac{\sqrt{2}}{2 \omega_c C}\) and \(R_2 = \frac{\sqrt{2}}{ \omega_c C}\) relationship (or equivalent forms) is the standard approach.

Component Selection Summary (Example with C = 1 nF):

• Capacitors: \(C_1 = C_2 = 1 nF\)
• Resistors: \(R_1 \approx 4.5 k\Omega\), \(R_2 \approx 9 k\Omega\)
• Operational Amplifier: A general-purpose op-amp with sufficient bandwidth for 25 kHz and low noise.

Using standard E12 series resistor values, we could choose \(R_1 = 4.7 k\Omega\) and \(R_2 = 8.2 k\Omega\). This will slightly shift the cutoff frequency from the ideal 25 kHz, but provides a practical implementation.

Determining the Gain at a Specific Frequency

Applying the Transfer Function

To determine the magnitude of the gain at a specific frequency \(f\), we use the magnitude of the transfer function \(|H(j\omega)|\), where \(\omega = 2\pi f\). For a Butterworth filter, the magnitude response is given by:

\[ |H(j\omega)| = \frac{1}{\sqrt{1 + \left(\frac{\omega_{c}}{\omega}\right)^{2n}}} \]

where \(n\) is the order of the filter (in our case, \(n=2\)) and \(\omega_c\) is the angular cutoff frequency.

For a high-pass filter, the gain is 1 at high frequencies and rolls off as frequency decreases. The formula for the magnitude response of an n-th order high-pass Butterworth filter with a high-frequency gain of \(A_{\infty}\) is:

\[ |H(j\omega)| = \frac{A_{\infty}}{\sqrt{1 + \left(\frac{\omega_{c}}{\omega}\right)^{2n}}} \]

In our case, \(A_{\infty} = 1\) and \(n = 2\), and \(\omega_c = 2\pi \times 25 \text{ kHz}\). We want to find the gain at \(f = 30 \text{ kHz}\). So, \(\omega = 2\pi \times 30 \text{ kHz}\).

\[ \frac{\omega_c}{\omega} = \frac{2\pi \times 25 \times 10^3}{2\pi \times 30 \times 10^3} = \frac{25}{30} = \frac{5}{6} \] \[ |H(j\omega)| = \frac{1}{\sqrt{1 + \left(\frac{5}{6}\right)^{2 \times 2}}} = \frac{1}{\sqrt{1 + \left(\frac{5}{6}\right)^{4}}} = \frac{1}{\sqrt{1 + \frac{625}{1296}}} \] \[ |H(j\omega)| = \frac{1}{\sqrt{\frac{1296 + 625}{1296}}} = \frac{1}{\sqrt{\frac{1921}{1296}}} = \sqrt{\frac{1296}{1921}} \approx \sqrt{0.6746} \approx 0.8213 \]

Converting Gain to Decibels

To express the gain in decibels (dB), we use the formula:

\[ \text{Gain (dB)} = 20 \log_{10} (|H(j\omega)|) \] \[ \text{Gain (dB)} = 20 \log_{10} (0.8213) \approx 20 \times (-0.0855) \approx -1.71 \text{ dB} \]

Therefore, the magnitude of the gain at 30 kHz is approximately -1.71 dB.

Visualizing the Filter's Performance

Frequency Response Characteristics

A Butterworth filter's key characteristic is its flat response in the passband and monotonic roll-off in the stopband. The steepness of the roll-off depends on the order of the filter. A second-order filter has a roll-off of -40 dB per decade in the stopband.

Comparison of frequency responses for different filter types, including Butterworth.

Radar Chart Analysis of Filter Attributes

We can represent the characteristics of this Butterworth filter using a radar chart, highlighting different performance aspects based on our design and the nature of Butterworth filters.

The radar chart provides a visual summary of the filter's performance characteristics. A 2nd order Butterworth filter excels in passband flatness and has reasonable phase linearity compared to other filter types like Chebyshev. Its stopband roll-off is moderate (40 dB/decade). Component sensitivity and ease of design are also considered reasonably good for this order and topology.

Practical Considerations and Implementation

Choosing Components and Op-Amp

When building this filter, select components with appropriate tolerances. Resistors with 1% tolerance and capacitors with 5% or 10% tolerance are generally suitable for many applications. The choice of operational amplifier is critical. The op-amp should have a gain-bandwidth product significantly higher than the cutoff frequency (25 kHz) to ensure proper operation. It should also have a slew rate sufficient to handle the maximum expected signal amplitude at the highest frequencies in the passband.

For example, if the maximum output voltage swing is \(\pm 5V\) at 25 kHz, the required slew rate should be at least \(2\pi \times 25 \text{ kHz} \times 5 \text{ V} \approx 0.785 \text{ V}/\mu s\). A higher slew rate provides more headroom.

Testing and Verification

After building the circuit, its frequency response should be tested using a signal generator and an oscilloscope or a spectrum analyzer. By applying sine waves of varying frequencies and measuring the output voltage, the gain at different frequencies can be determined and plotted. This allows for verification of the cutoff frequency and the gain in the passband and stopband.

Using an oscilloscope to visualize the effect of a high-pass filter on a signal.

Comparing Filter Characteristics

Butterworth vs. Other Filter Types

It's useful to compare the Butterworth filter with other common filter types to understand its trade-offs:

Filter Type	Passband Response	Stopband Roll-off	Phase Response
Butterworth	Maximally Flat	Moderate (-20n dB/decade)	Reasonably Linear
Chebyshev (Type I)	Equal Ripple	Steeper than Butterworth	Non-linear
Chebyshev (Type II)	Flat in Passband, Equal Ripple in Stopband	Steeper than Butterworth	Non-linear
Elliptic (Cauer)	Equal Ripple in both Passband and Stopband	Steepest	Highly Non-linear
Bessel	Gradual Roll-off	Slowest	Most Linear Phase

The choice of filter type depends on the specific application requirements. Butterworth filters are suitable for applications where a flat passband and good phase linearity are important, even at the cost of a less steep roll-off compared to Chebyshev or elliptic filters.

Frequently Asked Questions

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Recommended Further Exploration

• Designing higher order Butterworth filters
• Sallen-Key filter design with gain
• Frequency response analysis of active filters
• Butterworth filter applications

References