High-Pass RC Filters: Transfer Function & Cutoff Frequency

Hello! Welcome to the next lesson in your radioelectronics course.

In our last session, we thoroughly analyzed the RC low-pass filter, learning how to derive its transfer function and calculate the crucial cutoff frequency that defines its behavior. You saw how taking the output across the capacitor allows low frequencies to pass while attenuating high frequencies.

Introduction

Today's Goal: We will now explore what happens when we swap the positions of the resistor and capacitor. This lesson is dedicated to the RC High-Pass Filter. You will learn how this simple change in topology inverts the filter's characteristic, allowing it to pass high frequencies while blocking low ones. We will derive its unique transfer function and discover its cutoff frequency.

This concept is the natural counterpart to low-pass filtering. Continuing the analogy from our last lesson, if a low-pass filter is like smoothing data to see a trend, a high-pass filter is like detrending data to focus on short-term fluctuations or rapid changes. By the end of this lesson, you will be able to derive and analyze this second fundamental filter type, completing your understanding of first-order RC filters.

Time to complete: Approximately 50 minutes.

1. The High-Pass Filter: A New Configuration and Intuition

Let's start by reconfiguring our circuit. Instead of taking the output across the capacitor, we now take it across the resistor.

This simple change has profound effects. The circuit still acts as a frequency-dependent voltage divider, but the roles are reversed. The following video provides an excellent conceptual introduction.

• Resource: by Phil's Lab
• Watch: 02:00 - 05:13 (Introduction to RC High-Pass Filters)
• Focus on: The explanation of the circuit as a frequency-dependent voltage divider and the intuitive analysis of its behavior at very low and very high frequencies.

Let's summarize the key insights from the video:

• At DC ($\omega \to 0$):

  • The capacitor's impedance $Z_C=\frac{1}{j\omega C}$ goes to infinity. It acts as an open circuit.
  • Since no current can flow through the open circuit, there is no voltage drop across the resistor R ($V_{out}=I\times R=0\times R=0$).
  • The circuit blocks DC and very low frequencies.

• At High Frequencies ($\omega \to \infty$):

  • The capacitor's impedance $Z_C$ approaches zero. It acts like a short circuit.
  • The circuit behaves as if the resistor R is connected directly to the input source.
  • Therefore, ${\mathbf{V}}_{out}\approx {\mathbf{V}}_{in}$. The circuit passes high frequencies.

This behavior—blocking low frequencies and passing high ones—is the definition of a high-pass filter.

2. Deriving the Transfer Function

Now, let's formalize this intuition by deriving the transfer function $H(j\omega )=\frac{{\mathbf{V}}_{out}}{{\mathbf{V}}_{in}}$. The process is identical to the low-pass case: apply the voltage divider rule using complex impedances.

1. Apply Voltage Divider Rule:
   This time, the output is across the resistor, so $Z_R$ is in the numerator.

   $$H(j\omega )=\frac{Z_R}{Z_R+Z_C}=\frac{R}{R+\frac{1}{j\omega C}}$$

2. Simplify Algebraically:
   To clean this up, we multiply the numerator and denominator by $j\omega C$:

   $$H(j\omega )=\frac{R\cdot (j\omega C)}{\left(R+\frac{1}{j\omega C}\right)\cdot (j\omega C)}=\frac{j\omega RC}{j\omega RC+1}$$

This gives us the standard form for the transfer function of a first-order high-pass filter:

Notice how this differs from the low-pass transfer function $H(j\omega )=\frac{1}{1+j\omega RC}$. The presence of the $j\omega$ term in the numerator is what gives this filter its high-pass characteristic.

The next video provides a clear walkthrough of this derivation and the subsequent phase analysis.

• Resource: by ALL ABOUT ELECTRONICS
• Watch: 05:10 - 06:25 (Deriving the Transfer Function and Phase)
• Focus on: The step-by-step derivation of the transfer function and the resulting phase equation. Note that the video arrives at an algebraically equivalent form, $\frac{1}{1-j/(\omega CR)}$.

3. Magnitude and Phase Analysis

Let's break down our derived transfer function, $H(j\omega )=\frac{j\omega RC}{1+j\omega RC}$, into its magnitude and phase components to fully understand its behavior.

• Magnitude (Gain):

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

• Phase Angle:

  $$\angle H(j\omega )=\angle (j\omega RC)-\angle (1+j\omega RC)$$

  The angle of a pure imaginary number like $j\omega RC$ is $90^{\circ }$. The angle of the denominator is $\arctan (\omega RC)$.

  $$\angle H(j\omega )=90^{\circ }-\arctan (\omega RC)$$

  This can also be shown to be equivalent to $\arctan \left(\frac{1}{\omega RC}\right)$, which is a very useful form for analysis.

Let's test these formulas at the frequency extremes:

• At DC ($\omega \to 0$):

  • Gain: $|H(j0)|=\frac{0}{\sqrt{1+0}}=0$. The signal is completely blocked.
  • Phase: $\angle H(j0)\to \arctan (\infty )=90^{\circ }$. The output leads the input by 90 degrees.

• At High Frequencies ($\omega \to \infty$):

  • Gain: $|H(j\omega )|\to \frac{\omega RC}{\sqrt{(\omega RC{)}^2}}=\frac{\omega RC}{\omega RC}=1$. The signal is passed with no attenuation.
  • Phase: $\angle H(j\omega )|\to \arctan (0)=0^{\circ }$. There is no phase shift.

These results perfectly match the intuition we built in the first section.

4. Deriving the Cutoff Frequency

Just like the low-pass filter, the high-pass filter has a cutoff frequency ($f_c$) that marks the boundary between the stopband and the passband. It's defined in exactly the same way: the frequency at which the gain drops to $\frac{1}{\sqrt{2}}$ of its maximum, also known as the -3dB point.

Let's derive the formula for it.

• Resource: by ALL ABOUT ELECTRONICS
• Watch: 02:48 - 05:10 (Deriving Cutoff Frequency)
• Focus on: The algebra used to solve for the frequency where the gain is $1/\sqrt{2}$.

Here is a summary of the derivation steps:

1. Set the magnitude equation equal to $1/\sqrt{2}$: $$|H(j{\omega }_c)|=\frac{{\omega }_{c}RC}{\sqrt{1+({\omega }_{c}RC{)}^2}}=\frac{1}{\sqrt{2}}$$
2. Square both sides: $$\frac{({\omega }_{c}RC{)}^2}{1+({\omega }_{c}RC{)}^2}=\frac{1}{2}$$
3. Cross-multiply and solve for $({\omega }_{c}RC{)}^2$: $$2({\omega }_{c}RC{)}^2=1+({\omega }_{c}RC{)}^2⟹({\omega }_{c}RC{)}^2=1$$
4. Take the square root: $${\omega }_{c}RC=1$$

This gives us the formula for the cutoff frequency in radians per second:

And in Hertz:

This is a remarkable result! The cutoff frequency for a high-pass filter is determined by the exact same formula as for a low-pass filter. The component values $R$ and $C$ define the critical frequency, while the circuit's topology (which component you take the output across) determines whether it's a low-pass or high-pass filter.

5. Visualizing the High-Pass Response

The Bode plot for a high-pass filter visually confirms its characteristics. It is essentially a mirror image of the low-pass filter's plot.

• Magnitude Plot:

  • Stopband: For $\omega \ll {\omega }_c$, the gain is low and increases as frequency rises. This region has a "roll-on" slope.
  • Cutoff Frequency (${\omega }_c$): The gain is exactly $1/\sqrt{2}$ (or -3 dB).
  • Passband: For $\omega \gg {\omega }_c$, the gain is flat and close to 1 (or 0 dB).

• Phase Plot:

  • Starts at +90° for very low frequencies.
  • Passes through +45° exactly at the cutoff frequency ${\omega }_c$.
  • Approaches 0° at very high frequencies.

The video from Phil's Lab provides a great dynamic visualization of these plots.

• Resource: by Phil's Lab
• Watch: 13:17 - 16:08 (Frequency Domain Analysis and Transfer Function)
• Focus on: The shape of the magnitude and phase plots and how they visually represent the high-pass characteristic.

Conclusion

In this lesson, we completed our look at first-order passive filters by analyzing the RC high-pass filter.

Key Takeaways:

• An RC circuit with the output taken across the resistor functions as a high-pass filter.
• Its transfer function is $H(j\omega )=\frac{j\omega RC}{1+j\omega RC}$.
• The cutoff frequency is given by ${\omega }_c=1/RC$ or $f_c=\frac{1}{2\pi RC}$, the same formula as the low-pass filter.
• The circuit's topology dictates its function (low-pass vs. high-pass), while the R and C values set the boundary frequency.

What's Next?

You now understand how to pass low frequencies and how to pass high frequencies. The logical next step is to ask: can we combine these ideas to pass only a specific band of frequencies? In the next lesson, we will explore the concepts of Band-Pass and Band-Stop filters, which do exactly that by strategically combining low-pass and high-pass sections.