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

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 (R₁ and C₁) together with a signal voltage (v_i) and a load resistor (R_L). Resistance R₁ 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 R₁. 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 R₁, 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) R₁ 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 (R₁). 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 R₁.

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

Resulting in

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

Equation 1 is correct only when the resistance of R₁ is much smaller than the load resistance (R₁<L in Figure 1). When R₁ is not smaller than R_L, then f_c occurs when X_C1 equals R₁ 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 = R₁

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

So, X_C1= R₁ 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

 

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 R₁ 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>>R₁. 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=R₁, 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

 

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.