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Filters - Low Pass and High Pass

The type of filter being used.
V out V in 47nF C 1 5kΩ R 1
Capacitance value of $C_1$
Resistance cap for the pot $R_1$.
The type of potentiometer being used.
This knob controls the potentiometer (represented as $R_1$) in the circuit, altering its value as it turns. The value shown on the knob is the percent rotation (from 0 to 100%). Try rotating it!
Watch what happens to the frequencies of a guitar chord as the filter changes. Notes labeled according to Helmholtz pitch notation.
You can also manually enter a frequency here instead.

This calculator can be used to design either low-pass filters or high-pass filters. Choose your filter type, enter a value for the capacitor, enter a value for the potentiometer, and then select the taper for the potentiometer. Click and hold to rotate the knob and vary the resistance. As the resistance changes with the sweep of the pot, the cutoff frequency ($f_c$) will change; this value is displayed on the Bode plot directly below the knob. A Bode plot is a graph of the frequency response of a system.

Below the Bode plot is another graph displaying the selected guitar chord. Altering the values of the low/high pass filter will show the effects on the multiple frequencies of the chord in this graph.

The frequencies of these guitar chords are filtered based on the high/low pass filter above. On a high pass filter, values lower than the frequency cutoff ($f_c$) point will be filtered out - you will see the magnitude of their waveforms decrease as they pass the frequency cutoff. In a low pass filter, frequency values higher than the frequency cutoff ($f_c$) point will be filtered out. The amount of gain in the frequency waveform (the magnitude of the wave) will be reduced as the frequency is filtered. Unfiltered frequencies will show the full gain (1).

Passive low-pass and high-pass filters are found in a multitude of circuits - including the tone knob on a guitar, the tone stack in amplifiers, and tone controls in pedals. Even voltage controlled OTA low-pass filters found in synthesizers are derived from these simple circuits. Low frequencies are allowed to pass in a low-pass filter whereas high frequencies are allowed to pass in a high-pass filter. The cutoff sets the point where the frequencies are reduced, resulting in attenuation. Everything below the cutoff point in a low-pass filter is considered within the pass band and everything above it is within the stop band. With a high-pass filter it’s just the opposite. Everything above the cutoff point is considered within the pass band and everything below it is within the stop band.

The most common versions of these circuits are RC networks comprised of a single resistor and a single capacitor. A potentiometer used as a variable resistor is often used in place of the resistor to vary the cutoff frequency.

RC Low-Pass Filter with Variable CutoffRC High-Pass Filter with Variable Cutoff

These can be combined in different ways as well. The Big Muff Pi’s tone control famously uses a low-pass filter and a high-pass filter with a potentiometer mixing between the two.

High-pass / Low-pass Mix

When designing a filter for audio, we’ll want to know the frequency of the cutoff point. This is calculated using the same formula for both low-pass filters and high-pass filters:

$$f_c = \frac{1}{2\pi RC}$$

$f_c$ is the cutoff frequency in hertz. $R$ is the value of the resistor in ohms. $C$ is the value of the capacitor in farads.

Suppose we have a circuit in which we want to filter out frequencies above 5,000Hz. Let’s also say that we have a 500kΩ resistor. We need to find the capacitor value to achieve the cutoff point of 5kHz. So $f_c$ = 5,000 and $R$ = 500,000. Solving the equation for $C$ we find:

$$C = \frac{1}{2\pi R f_c}$$$$C = \frac{1}{2 \pi \times 500{,}000\text{Ω} \times 5{,}000\text{Hz}}$$$$C \approx 63.66 \times 10^{-12}\text{F}$$

After converting our answer to picofarads we find that we will need a 63.66pF capacitor to use with the 500k resistor to get a cutoff frequency of 5kHz. This isn’t a common capacitor value so we can see what happens when we use the more common 62pF value with the formula.

$$f_c = \frac{1}{2\pi RC}$$$$f_c = \frac{1}{2 \pi \times 500{,}000\text{Ω} \times (62 \times 10^{-12}\text{F})}$$$$f_c = 5{,}134.03\text{Hz}$$

In this case, the closest common value gets the cutoff frequency fairly close to the desired 5kHz. If it needed to be exactly at 5kHz we could always solve for the resistor value and use a trimmer to dial in the exact resistance.

$$R = \frac{1}{2 \pi C f_c}$$$$R = \frac{1}{2 \pi \times (62 \times 10^{-12}\text{F}) \times 5{,}000\text{Hz}}$$$$R = 513{,}403.04\text{ Ω}$$

We could either use a 1 Meg trimmer to get this resistance or use resistors in series/parallel.

RC filters have some key characteristics that you may want to consider before choosing them for your design. They are first-order filters because they have one pole; this is due to the fact that they only have one reactive component, the capacitor. With a single pole, the filter will always have a -6dB/octave or -20dB/decade slope. If the amount of poles increases, the slope will also increase. This can be seen with the famous Moog filter, which has 4 poles and a -24dB/octave or -80dB/decade slope. While our RC filter and the Moog filter both function very similarly, the sound is very different.

Low-pass Bode Plot

High-pass Bode Plot

Another interesting aspect of the RC filters is their effect on the phase angle of different frequencies. At the cutoff frequency the phase is 45° out of phase. For a low-pass filter the phase shift is -45° and for a high-pass filter the phase shift is +45°.

Low-pass Frequency Waveform (-45° phase shift)

High-pass Frequency Waveform (+45° phase shift)

Using the following we can find the phase angle of a set frequency in a low-pass filter. Phase Shift $\Phi$ is the phase shift in radians. $ƒ$ is the frequency in hertz. $R$ is the resistor value in ohms. $C$ is the capacitor value in farads.

$$ \Phi_{\text{Phase Shift}} = -\arctan{(2 \pi f R C)}$$

If we use the component values we solved for above ($R = \text{513,403.04Ω}$, $C = 62 \times 10^{-12}$) and use the cutoff frequency in the formula ($f_c = \text{5,000Hz}$) we get the following:

$$\Phi_{\text{Phase Shift}} = -\arctan{(2 \pi \times 5{,}000 \times 513{,}403.04 \times (62 \times 10^{-12}))}$$$$\Phi_{\text{Phase Shift}} = -\arctan{(2 \pi \times 0.1592)}$$$$\Phi_{\text{Phase Shift}} = -\arctan{(0.9999)}$$$$\Phi_{\text{Phase Shift}} = -0.7853 \text{ radians}$$

We’ll use the following to convert the radians to degrees

$$\text{degrees} = \text{radians} \times \frac{180}{\pi}$$$$\text{degrees} = -0.7853 \times \frac{180}{\pi}$$$$\text{degrees} = -45$$

Suppose we used the same components in a high pass filter and wanted to check the phase angle at our cutoff point. We’ll use the following to find the phase shift in high-pass filters:

$$ \Phi_{\text{Phase Shift}} = \arctan{(2 \pi f R C)}$$

Using the same values as above

$$\Phi_{\text{Phase Shift}} = \arctan{(2 \pi \times 5{,}000 \times 513{}403.04 \times (62 \times 10^{-12}))}$$$$\Phi_{\text{Phase Shift}} = \arctan{(2 \pi \times 0.1592)}$$$$\Phi_{\text{Phase Shift}} = \arctan{(0.9999)}$$$$\Phi_{\text{Phase Shift}} = 0.7853 \text{ radians}$$$$\text{degrees} = \text{radians} \times \frac{180}{\pi}$$$$\text{degrees} = 0.7853 \times \frac{180}{\pi}$$$$\text{degrees} = 45$$

These equations give us the expected -45° and +45° results for the phase shift at the cutoff frequency for the low-pass and high-pass filters. You can use these equations to check the phase shift of any frequency in your circuit.

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