# RC filter transfer function and gain

For a standard basic RC Filter, the transfer function is as follows: $${ V_{out} \over V_{in} } = H(j\omega) = { 1 \over 1 + j\omega RC }$$

However, when simulating the circuit, I find the output voltage to equal: $${ V_{out} \over V_{in} } = |H(j\omega)| = \left|{ 1 \over 1 + j \omega RC }\right| = { 1 \over \sqrt{ 1 + (\omega RC)^2 }}$$

Why is the actual output equal to the modulus of the transfer function, not the transfer function its self? What does the original transfer function tell me, if anything?

• You're missing the "j" (imaginary unit) in the transfer function. – John D Oct 31 '14 at 18:06
• so I am, my mistake. – ACarter Oct 31 '14 at 18:08

## 3 Answers

The transfer function is not $H(\omega)$, it is $H(j\omega)$ (note the $j$, which makes it complex):

$$H(j\omega) = \frac{1}{1+j\omega RC}$$

This is important because the transfer function captures the phase in addition to the amplitude.

The amplitude is

$$|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}$$

by the definition of complex magnitude. This is the gain you measure from input to output and it may be all you care about, especially for a first order system. However, the phase $$\angle H(j\omega) = -\arctan(\omega RC)$$ can be quite important (e.g. for ensuring stability), especially for higher order transfer functions.

The main, direct use of the transfer function is to capture both gain and phase in one expression, but can also be used for time-domain analysis, as the transfer function is the Laplace transform of the impulse response. It is also useful for characterizing a multiple stage system, since the transfer function $H(j\omega)$ of a system consisting of stage $H_1(j\omega)$ followed by $H_2(j\omega)$ is simply $H(j\omega) = H_1(j\omega)H_2(j\omega)$ whereas in the time-domain you need to convolve the impulse responses $h_1(t)$ and $h_2(t)$ to find the overall system impulse response of $h(t)$.

• I see. Is the transfer function only useful for deriving expressions for the gain and phase then? – ACarter Oct 31 '14 at 18:13
• @ACarter I added some information in my answer about other uses of the transfer function. – Null Oct 31 '14 at 18:23
• Acarter- in short: The transfer function is a complex function (as indicated by the "j" in the denominator). For such a function, you always must discriminate between themagnitude and a phase response. And that`s what each simulator also does: Separate display of magnitude and phase as a function of frequency - unless you require a Nyquist diagram which shows the Im part as a function of the Re part - both as a function of frequency. – LvW Oct 31 '14 at 20:21

If you had written (or had had knowledge to write) the dependencies of $V_{out}$ and $V_{in}$ voltages in your expression, you could have seen the reason while writing your question.

The transfer function you wrote is in the frequency domain, therefore the $V_{out}$ and $V_{in}$ voltages are dependent on the $w$ (angular frequency) variable.

$${ V_{out}(jw) \over V_{in}(jw) } = H(j\omega) = { 1 \over 1 + j\omega RC }$$

However, your observations are in the time domain, so they are dependent on the $t$ (time) variable.

$${ V_{out,average}(t) \over V_{in,average}(t) } = { V_{out,peak}(t) \over V_{in,peak}(t) } = { V_{out,RMS}(t) \over V_{in,RMS}(t) } = { V_{out,envelope}(t) \over V_{in,envelope}(t) } = |H(j\omega)| = \left|{ 1 \over 1 + j \omega RC }\right| = { 1 \over \sqrt{ 1 + (\omega RC)^2 }}$$

Keep in mind that the ratio of instantaneous values of $V_{out}(t)$ and $V_{in}(t)$ are not necessarily equal to $|H(j\omega)|$ because of any possible phase difference between the input and output.

Why is the actual output equal to the modulus of the transfer function,

Recall that, when doing an AC analysis, one is working with phasor voltages and currents.

A phasor has a magnitude and phase but no time dependence. We keep track of the amplitude and phase of a sinusoidal time function with the magnitude and phase of the associated phasor.

When you plot 'the' output voltage, you're actually plotting the magnitude of the output voltage phasor.

For concreteness, let's look at the transfer function you have:

$$\frac{V_{out}}{V_{in}} = \frac{1}{1 + j\omega RC}$$

This is clearly a complex valued function of frequency. Let's write this function in polar form (magnitude and phase):

$$\frac{V_{out}}{V_{in}} = \left|\frac{V_{out}}{V_{in}}\right|e^{j\phi}=\frac{1}{\sqrt{1 + (\omega RC)^2}}e^{j\phi}$$

where

$$\tan\phi = -\omega RC$$

Now note that the magnitude of the transfer function matches your simulation output.

So, remember, when in AC analysis, a plot of the voltage or current (versus frequency) is a plot of the magnitude of the phasor voltage or current. One can also plot the phase as shown below: Also, if you haven't already, take a look at Bode plot.