Laplace Domain RC Circuit

For an RC circuit in the Laplace domain:

Our transfer function is:

$$\text{TF}=\frac{1}{1+sRC}=A(s)=\frac{V_{out}(s)}{V_{in}(s)}[\text{V}/\text{V}]$$

We can replace $s=j\omega$ to see the frequency behavior.

$$\begin{array}{rl}\text{Magnitude}:|A(j\omega )|& =\frac{|1|}{|1+j\omega RC|}\\ & =\frac{1}{\sqrt{1+(\omega RC{)}^2}}\\ \text{Phase}:\angle A(j\omega )& =\angle \text{Numerator}-\angle \text{Denominator}\\ & =\angle 1-\angle (1+j\omega RC)\\ & =0-\angle (1+j\omega RC)\end{array}$$

What is the frequency-dependent behavior for this circuit if $V_{in}$ is AC? We have $A(s)=\frac{1}{1+sRC}$, but what does this mean?

• Low frequency: The capacitor looks like an open circuit, so $V_{out}=V_{in},A=1$
• High frequency: The capacitor looks like a short circuit, so $V_{out}=0,A=0$
• In between: $V_{out}$ is somewhere between $0$ and $V_{in}$, with $A$ decreasing as $f$ increases.
• Therefore, this is a passive Low-pass Filter.

We can see this LPF behavior by focusing on the magnitude, $|A(j\omega )|$:

$$\begin{array}{rl}|A(j\omega )|& =\frac{1}{\sqrt{1+(\omega RC{)}^2}}\\ \omega & \to 0,|A(j\omega )|\to \frac{1}{\sqrt{1+0}}=1\\ \omega & \to \infty ,|A(j\omega )|\to \frac{1}{\omega RC}=0\end{array}$$

When does it transition from $A\approx 1$ to $A\approx 0$?

There is a special frequency where $\text{Re}=\text{Im}$ in the denominator.

$$\begin{array}{rl}\text{Re}& =\text{Im}\\ 1& =\omega RC\\ & \therefore {\omega }_c=\frac{1}{RC}=\frac{1}{\tau }\end{array}$$

This special frequency is called the:

• Cut-off frequency
• Corner frequency
• -3dB frequency, since $20{\log }_{10}(1/\sqrt{1+1}=20{\log }_{10}(1/\sqrt{2})=-3$
• Half power frequency.

We can make a plot of $|A(j\omega )|$ vs $\omega =2\pi f$:

As $\omega$ goes above ${\omega }_c$, we have

$$|A(j\omega )|\approx \frac{1}{\sqrt{(\omega RC{)}^2}}=\frac{1}{\omega RC}=\frac{1}{\omega /{\omega }_c}=\frac{{\omega }_c}{\omega }$$

(Recall ${\omega }_c=\frac{1}{RC}$).