# Calculating RC Low-Pass Filter Cut-Off Frequency and Transfer Function

## Learn how to determine the RC low-pass filter's cut-off frequency and transfer function and plot the gain/frequency and phase/frequency response graphs.

An RC low-pass filter circuit allows low-frequency signals to pass through while attenuating high-frequency signals. It consists of a resistor (R) and a capacitor (C) connected in series. The filter operation is based on the time constant of the RC circuit, which determines the rate at which the capacitor charges and discharges.

*Image used courtesy of Adobe Stock*

Figure 1(a) shows a basic RC low-pass filter circuit (R_{1} and C_{1}) together with a signal voltage (v_{i}) and a load resistor (R_{L}). Resistance R_{1} and capacitive reactance X_{C1} constitute a voltage divider, as illustrated in Figure 1(b). When the filter input voltage (v_{i}) has a low frequency (f), the capacitor impedance \(\Bigg[X_{C1}=\frac{1}{2\pi fC_{1}}\Bigg]\) is much larger than the resistance of R_{1}. In this case, there is very little voltage division, and the output voltage (v_{o}) approximately equals the input. When v_{i} has a high frequency, X_{C1} becomes much smaller than R_{1}, and so v_{o} becomes much smaller than v_{i}. It is seen that low-frequency input voltages are passed to the output, and high-frequency inputs are attenuated.

*(a) RC low-pass filter circuit*

*(b) R*_{1} and X_{C1} are voltage dividers

_{1}and X

_{C1}are voltage dividers

**Figure 1.** In a low-pass RC filter, input voltages are divided across the capacitor impedance (X_{C1}) and the resistor (R_{1}). Because X_{C1} is large at low frequencies, low-frequency inputs are passed to the output with little attenuation. High-frequency signals are substantially attenuated as X_{C1} becomes smaller with increasing frequency. Image used courtesy of Amna Ahmad

**Figure 1.**In a low-pass RC filter, input voltages are divided across the capacitor impedance (X

_{C1}) and the resistor (R

_{1}). Because X

_{C1}is large at low frequencies, low-frequency inputs are passed to the output with little attenuation. High-frequency signals are substantially attenuated as X

_{C1}becomes smaller with increasing frequency. Image used courtesy of Amna Ahmad

### Gain/Frequency Response

The typical graph of output-to-input voltage ratio \((\frac{v_{_o}}{v_{_i}}\,in\,decibels)\) plotted versus input frequency (f) for a low-pass filter is shown in Figure 2(a). This is the filter gain/frequency response graph (Bode plot). The output remains substantially equal to the input \((v_{_o}/v_{_i}\approx1\approx0dB)\) for all frequencies up to the cut-off frequency (f_{c}). This is the upper limit of the passband for the filter, and all frequencies above f_{c} are in the attenuation band, as illustrated.

The cut-off frequency is when the output voltage falls 3 dB from its normal (low frequency) level. It is also referred to as the 3 dB and critical frequencies. The cut-off frequency can be expected to occur when X_{C1 }becomes equal to the resistance of R_{1}.

\[X_{C1}=\frac{1}{2\pi_{c}C_{1}}=R_{1}\]

Resulting in

\[f_{c}=\frac{1}{2\pi R_{_1}C_{_1}}(1)\]

Equation 1 is correct only when the resistance of R_{1} is much smaller than the load resistance (R_{1}<_{1} is not smaller than R_{L}, then f_{c} occurs when X_{C1} equals R_{1 }ǁ R_{L}.

An equation for the ratio of output-to-input voltage for the RC low-pass filter is easily derived from the voltage divider in Figure 1(b):

\[\frac{v_{_o}}{v_{_i}}=\frac{X_{_{C1}}}{\sqrt{R^{2}_{1}+X^{2}_{C1}}}(2)\]

When X_{C1} = R_{1}

\[\frac{v_{_o}}{v_{_i}}=\frac{R_{1}}{\sqrt{R^{2}_{1}+R^{2}_{1}}}=\frac{1}{\sqrt{2}}=0.707\]

So, X_{C1}= R_{1 }at f_{c}, and this occurs when v_{o}/v_{i}=0.707 \(\Big(or\frac{1}{\sqrt2}\Big)\), which is also -3 dB below its normal (low frequency) level. The output power is halved at the -3 dB point, giving the half-power point one more name for the cut-off frequency. Ideally, the low-pass filter's gain/frequency response graph should be perfectly flat below the cut-off frequency and perfectly vertical above f_{c} to produce zero signal attenuation with frequencies within the pass band and infinite attenuation of those with frequencies in the attenuation band. This is impossible with practical circuits; however, the steepness of the frequency response graph within the attenuation band can be an important consideration for a filter.

Figure 2(a) shows that for signal frequencies above f_{c}, the output voltage decreases at a rate of 6 dB per octave; a 6 dB reduction each time the frequency is doubled. As illustrated, this fall-off rate (also referred to as the roll-off rate) can also be stated as 20 dB per decade, a 20 dB reduction for ten times increase in signal frequency.

*(a) Gain/frequency response for low-pass filter*

*(b) Insertion loss produced by the filter*

**Figure 2.** Typical gain/frequency response graph for a low-pass filter. At the cut-off frequency (f_{c}), the output level is 3 dB below the normal low-frequency output. At frequencies higher than f_{c}, the gain falls off at 6 dB/octave, the same as 20 dB/decade. The presence of the filter introduces an insertion loss between the input and output. Image used courtesy of Amna Ahmad

**Figure 2.**Typical gain/frequency response graph for a low-pass filter. At the cut-off frequency (f

_{c}), the output level is 3 dB below the normal low-frequency output. At frequencies higher than f

_{c}, the gain falls off at 6 dB/octave, the same as 20 dB/decade. The presence of the filter introduces an insertion loss between the input and output. Image used courtesy of Amna Ahmad

### Insertion Loss

For signal frequencies within the filter passband, there is some attenuation of the input, known as the insertion loss, as illustrated in Figure 2(b). The insertion loss is defined as the loss of signal due to the presence of the filter, and for the filter circuit in Figure 1, it is simply the result of voltage division across R_{1 }and R_{L. }The insertion loss can be determined by measuring the output (load) voltage without the filter in the circuit and measuring it again with the filter connected. The measured quantities are then used to calculate the loss in decibels. The insertion loss can also be determined from the filter component and load resistance values.

### Phase/Frequency Response

Because the low-pass filter output is the capacitor voltage, there is a phase shift between the input and output voltages. The phase/frequency response of the filter is a plot of the phase difference (θ) between the input and output voltages versus signal frequency (f), (see Figure 3). The phase shift is seen to be zero at low signal frequencies where X_{C1}>>R_{1}. As the signal frequency increases, the (lagging) phase shift gradually increases to -45° at f_{c}, and then continues beyond f_{c} to a maximum of -90°. The phase angle (θ) between the input and capacitor (filter output) voltages is (90°- φ), and it can also be determined directly as,

\[\theta=\Big[\frac{-R1}{X_{C1}}\Big](3)\]

At f_{c}, where X_{C1}=R_{1}, θ is calculated from Equation 3 as -45°.

**Figure 3.** Typical phase/frequency response graph for a low-pass filter. At the cut-off frequency (f_{c}), the output voltage is phase shifted by 45° with respect to the input. At frequencies higher and lower than fc, the phase shift changes at a rate of approximately 45°per decade. Image used courtesy of Amna Ahmad

**Figure 3.**Typical phase/frequency response graph for a low-pass filter. At the cut-off frequency (f

_{c}), the output voltage is phase shifted by 45° with respect to the input. At frequencies higher and lower than fc, the phase shift changes at a rate of approximately 45°per decade. Image used courtesy of Amna Ahmad

### Low-Pass Filter Transfer Function

Combining Equations 2 and 3 gives the transfer function for the filter, which is the phasor input-output relationship.

\[\frac{v_{_o}}{v_{_i}}=\frac{X_{C1}}{\sqrt{R^{2}_{1}+X^{2}_{C1}}}\angle tan \,\^{}-1\Big(\frac{-R_{1}}{X_{C1}}\Big)(4)\]

Substituting

\[R_{1}=\frac{1}{2\pi f_{C}C_{1}},andX_{C1}=\frac{1}{(2\pi f_{C1})}\]

Equation 4 can be simplified to

\[\frac{v_{_o}}{v_{_i}}=\Bigg(\frac{1}{\sqrt{(\frac{f}{f_{c}})^{2}+1}}\Bigg)\angle tan\,\^{}-1(\frac{-f}{f_{c}})(5)\]

Expressing \(\frac{v_{_o}}{v_{_i}}\) in decibels

\[\frac{v_{_o}}{v_{_i}}=20log\Bigg(\frac{1}{\sqrt{(\frac{f}{f_{c}})^{2}+1}}\Bigg)\angle tan\,\^{}-1(\frac{-f}{f_{c}})(6)\]

Once f_{c} is calculated for a given RC low-pass filter, Equation 6 can be used to quickly determine the values of v_{o}/v_{i} and θ at various frequencies (multiples of f_{c}) for plotting the gain/frequency and phase/frequency responses.

### RC Low-Pass Filter Takeaways

The RC low-pass filter circuit consists of a resistor and a capacitor, which work together to attenuate high-frequency signals while allowing low-frequency signals to pass through. The filter's cut-off frequency can be determined using the values of the resistor and capacitor. The transfer function is the relationship between the input and output signals of the filter, and the gain/frequency and phase/frequency response graphs provide a visual representation of how the filter affects different frequencies.

1 CommentThank you for these well-documented articles on transfer functions. You may want to consider the fast analytical circuit techniques (FACTs) for higher-order types of filters. I suggest you have a look at this seminar taught in 2016 at an APEC conference: http://powersimtof.com/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf