Cutoff Frequency of a Low Pass Filter

A low-pass filter permits low-frequency signals while blocking high ones. Its cutoff frequency is the point where output voltage falls to 70.7% of the input, also marked by a 3 dB drop from the 0 Hz level.

Cutoff Frequency of Low Pass Filter — Cutoff Frequency of a Low Pass Filter

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For example, if a low pass capacitive filter has $R=500\Omega$ and $C=7\mu F$ , at what frequency will the output be 70.7%?

A simple capacitive low pass filter with one resistor and one capacitor has a cutoff frequency of $f_{c}=\frac{1}{2\pi RC}$ . Substituting the corresponding R and C values, the cut-off frequency would be 45.473 Hz. So, the output will be 70.7% at 45.473 Hz.

When Bode Plot is plotted for a low pass filter as shown in the image below, the frequency response of the filter seems to be nearly flat for low frequencies.

Until the Cut-off Frequency point is reached, all of the input signals pass directly to the output, which results in a unity gain. This happens when the reactance of the capacitor is large at low frequencies and prevents any current flow through the capacitor. The response of the circuit decreases to zero at a slope of -20dB/ Decade “roll-off” after this cut-off frequency point.

The frequency point at which the capacitive reactance and resistance are equal is known as the cutoff frequency of a low-pass filter. At cutoff frequency, the output signal is attenuated to 70.7% of the input signal value or -3dB of the input.

Consider a first-order low pass filter with a transfer function

$\begin{align*} T(s)=\frac{a_{0}}{s+\omega _{0}} \end{align*}$

Rephrase the above equation by dividing the numerator and denominator by RC

(3) $\begin{align*}T(s)=\frac{1}{1+sRC}\end{align*}$

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(4) $\begin{align*}T(s)=\frac{1/RC}{s+1/RC}\end{align*}$

Hence, $a_{0}=1/RC$ and $\omega_{0}=1/RC$ , where $\omega_{0}$ is the cutoff frequency.

To have a better understanding of cut-off frequency, convert the standard s domain transfer function into equivalent $j\omega$ format.

$\begin{align*}T(s)=\frac{K}{1+s/\omega _{0}}\end{align*}$

$\begin{align*}T(j\omega)=\frac{K}{1+\j{\frac{\omega}{\omega_{0}}}}\end{align*}$

Now, lets evaluate this expression at cutoff frequency

$\begin{align*}T(j\omega=j\omega_{0})=\frac{K}{1+\j{\frac{\omega}{\omega_{0}}}}=\frac{K}{1+j}\end{align*}$

The denominator being a complex number, the magnitude has to be calculated.

$\begin{align*} \left |T(j\omega=j\omega_{0}) \right |=\frac{K}{\sqrt{1^{2}}+1^{2}}=\frac{K}{\sqrt{2}}\end{align*}$

K is the DC gain. When the input frequency increases to the cut-off frequency, the output amplitude will be $\frac{K}{\sqrt{2}}$ . The value $\frac{1}{\sqrt{2}}$ corresponds to -3dB which is nothing but the cutoff frequency.

This transfer-function analysis has shown clearly that the cutoff frequency is just the frequency at which the filter’s amplitude response is decreased by 3 dB corresponding to the very-low-frequency amplitude response.

Cutoff Frequency of a High Pass Filter

A high-pass filter passes signals with a frequency greater than a specified cutoff frequency. It attenuates signals with frequencies lower than that cutoff frequency.

Cutoff Frequency of High Pass Filter — Cut off Frequency of a High Pass Filter

The transfer function is derived in the below equations.

$\begin{align*} Z_{R}=R \; \; and\; \; Z_{C}=\frac{1}{sC} \end{align*}$

The output impedance is given as

$\begin{align*}Z_{out}=Z_{R}\end{align*}$

Input impedance is given as

$\begin{align*}Z_{in}=Z_{R}+Z_{C}\end{align*}$

The transfer function of a high pass filter is defined as the ratio of Output voltage to the input voltage.

$\begin{align*}\frac{V_{out}}{V_{in}}=\frac{Z_{out}}{Z_{in}}\end{align*}$

$\begin{align*} =\frac{Z_{R}}{Z_{R}+Z_{C}}\end{align*}$

$\begin{align*}=\frac{R}{R+\frac{1}{sC}}\end{align*}$

$\begin{align*}=\frac{sCR}{sCR+1}T(S)\end{align*}$

$\begin{align*}=\frac{s}{s+\frac{1}{RC}}\end{align*}$

On comparing the above equation, with the standard form of the transfer function,

$\begin{align*}T(s)=\frac{a_{1}s}{s+\omega _{0}}\end{align*}$

$a_{1}$ is the amplitude of the signal

$\omega _{0}$ is the angular cutoff frequency

The cutoff frequency is known as a frequency creating a boundary between the pass and stop band. If the signal frequency is more than the cutoff frequency for a high pass filter then it will cause the signal to pass. The cutoff frequency equation for the first-order high pass filter is the same as the low pass filter.

$\begin{align*}f_{c}=\frac{1}{2\pi RC}\end{align*}$

Cutoff Frequency of a Low Pass Filter

Cutoff Frequency of a High Pass Filter

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