# The RC Filter Transfer Function

Most people are familiar with the simple first-order RC low pass filter:

RC Low Pass Filter

Also well known is the equation for calculating the -3dB (aka, half-power) cutoff frequency of the RC low pass filter:

$$\mathbf{f}_{c} = \frac{1}{2 \pi RC}\\$$

It’s an easy equation to memorize, but if you’re interested in where this equation comes from, read on; if you’re familiar with resistor voltage dividers, this will be a piece of cake!

First, we need to find the transfer function of this circuit, which is simply the ratio between the input and output voltages.

The RC low pass filter is really just a resistor divider circuit where the lower resistor has been replaced with a capacitor. A capacitor’s impedance is, of course, frequency dependent:

$$\mathbf{X}_{c} = \frac{1}{j \omega C} \tag{1}\\$$

$$j \omega = \sqrt{\text{-1}} \times 2 \pi f \tag{2}\\$$

So, the transfer function for the RC circuit is the same as for a voltage divider:

$$\mathbf{V}_{out} = \mathbf{V}_{in} \times \frac{\mathbf{R}_{2}}{\mathbf{R}_{1} + \mathbf{R}_{2}} \tag{3}\\$$

$$\frac{\mathbf{V}_{out}}{\mathbf{V}_{in}} = \frac{\mathbf{R}_{2}}{\mathbf{R}_{1} + \mathbf{R}_{2}} \tag{4}\\$$

…just with the lower resistance replaced with the capacitor’s impedance:

$$\frac{\mathbf{V}_{out}}{\mathbf{V}_{in}} = \frac{\frac{1}{j \omega C}}{R + \frac{1}{j \omega C}} \tag{5}\\$$

The right hand side of the equation contains a compound fraction, which can be simplified by multiplying both the numerator and denominator by the least-common-denominator (jwC). In doing so, we find that:

$$\frac{\frac{1}{j \omega C}}{R + \frac{1}{j \omega C}} \times \frac{j \omega C}{j \omega C} = \frac{\frac{j \omega C}{j \omega C}}{j \omega CR + \frac{j \omega C}{j \omega C}} = \frac{1}{j \omega R C + 1} \tag{6}\\$$

And thus:

$$\frac{\mathbf{V}_{out}}{\mathbf{V}_{in}} = \frac{1}{j \omega R C + 1} \tag{7}\\$$

This is the transfer function for a first-order low-pass RC filter. Next, we need to use this equation to find the frequency at which the output power drops by -3dB.

The half power point (aka, -3dB point) is the frequency at which the output power is one half of the input power; in other words, we’re interested in the magnitude (aka, absolute value) of the circuit’s output, and more specifically, the frequency at which that output drops to one half of the input power.

Let’s start by finding the magnitude of our transfer function:

$$\bigg|\frac{\mathbf{V}_{out}}{\mathbf{V}_{in}}\bigg| = \bigg|\frac{1}{j \omega R C + 1}\bigg| = \frac{| 1 |}{\big| j \omega R C + 1 \big|} \tag{8}\\$$

Note that the denominator of our transfer function is a complex number, that is, it contains the sum of a real component (1) and an imaginary component (jwRC). Real and imaginary numbers lie on different axes in the complex plane:

Real vs imaginary axes

Thus, if we wish to find the magnitude of a complex number, we have to find the sum of the real and imaginary components, which are at right angles to each other in the complex plane:

Magnitude = sqrt(8^2 + 6^2) = 10

Graphically, we can see that this forms a triangle with the magnitude as the hypotenuse, which necessitates the use of the pythagorean theorem in the denominator of our transfer function:

$$\bigg|\frac{\mathbf{V}_{out}}{\mathbf{V}_{in}}\bigg| = \frac{| 1 |}{\big| j \omega R C + 1 \big|} = \frac{1}{\sqrt{(\omega R C)^2 + 1^2}} \tag{9}\\$$

As with calculating the sum of any sequence of numbers, we aren’t concerned about the individual parts that make up the total value, only with the total sum itself. Thus we don’t care how much of the magnitude was “real” and how much was “imaginary”, we’re just concerned with finding how big their total sum is. So, in taking the magnitude of the transfer function (or any complex number), only real numbers remain.

We now have an equation that describes the output magnitude of the RC low pass filter. From this, we can apply some algebraic manipulation to solve for the -3dB cutoff frequency.

So far, our transfer equation has been specified in terms of voltage gain, but we are actually interested in the half-power (-3dB) point. By definition, when the output power is one half of the input power, the voltage gain will be one divided by the square root of two:

$$P_{dB} = 20\log_{10}\bigg(\frac{\mathbf{V}_{out}}{\mathbf{V}_{in}}\bigg) \tag{10}\\$$

$$\frac{\mathbf{V}_{out}}{\mathbf{V}_{in}} = \frac{1}{\sqrt{2}} \tag{11}\\$$

$$P_{dB} = 20\log_{10}\bigg(\frac{1}{\sqrt{2}}\bigg) = -3dB \tag{12}\\$$

So at the half-power point, the following equation must be satisfied:

$$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{(\omega R C)^2 + 1}} \tag{13}\\$$

Thus we can conclude that:

$$(\omega R C)^2 + 1 = 2 \tag{14}\\$$

And then:

$$(\omega R C)^2 = 1 \tag{15}\\$$

Which is the same as:

$$\omega^2 R^2 C^2 = 1 \tag{16}\\$$

Rearranging to get w by itself, and simplifying to eliminate the squares:

$$\omega^2 = \frac{1}{R^2 C^2} \tag{17}\\$$

$$\sqrt{\omega^2} = \sqrt{\frac{1}{R^2 C^2}} \tag{18}\\$$

$$\sqrt{\omega^2} = \frac{\sqrt{1}}{\sqrt{R^2 C^2}} \tag{19}\\$$

$$\omega = \frac{1}{R C} \tag{20}\\$$

Remembering that w is really two times pi times the frequency, we can rearrange to solve for frequency:

$$2\pi f = \frac{1}{R C} \tag{21}\\$$

$$f = \frac{\frac{1}{R C}}{2\pi} \tag{22}\\$$

So, finally:

$$f = \frac{1}{2\pi R C} \tag{23}\\$$

This tells us the frequency, in terms of our R and C components, at which the output power drops to one half of the input power.

## 2 thoughts on “The RC Filter Transfer Function”

1. Eric says:

Awesome and easy explanation, thank’s a lot!

2. Milan says:

To the point explanation.