3 dB frequency of first-order active high-pass filter

Question

Using the KCL law in the red dot in the circuit shown above, I got this equation:

$$\frac{V_\text{in}}{R_1 + \frac{1}{j\omega C}} + \frac{V_\text{out}}{R_2} = 0$$

From this equation the transfer function of this circuit is:

$$T(s) = \frac{V_\text{out}(s)}{V_\text{in}(s)} = -\frac{R_2}{R_1} \times \frac{j\omega}{j\omega + \frac{1}{R_1C}}$$

How can I get 3 dB frequency from the transfer function?

Answer 1

Calculate the magnitude of \$T(s = j\omega)\$ when it is not acting as a filter (which is as \$\omega \to \infty\$ for a high pass filter).

$$|T(j\omega)| = \left|-\frac{R_2}{R_1}\right| \times \frac{|j\omega|}{\left|j\omega + \frac{1}{R_1C}\right|} = \frac{R_2}{R_1} \times \frac{\omega}{\sqrt{\omega^2 + \left(\frac{1}{R_1C}\right)^2}}$$

$$\lim_{\omega \to \infty}|T(j\omega)| = \frac{R_2}{R_1}$$

since the \$\omega\$ term in the denominator dominates the \$1/(R_1C)\$ term and then the fraction with \$\omega\$ simplifies to 1. Alternatively, you can find this by inspection since the capacitor is a short circuit at high frequencies and the circuit becomes just an inverting op amp amplifier.

Next, set the transfer function magnitude equal to \$-3\text{ dB} \approx 1/\sqrt{2}\$ times the previously calculated magnitude.¹ In this case the \$-R_2/R_1\$ terms cancel so you are left with

$$\frac{\omega}{\sqrt{\omega^2 + \left(\frac{1}{R_1C}\right)^2}} = \frac{1}{\sqrt{2}}$$

Solve this equation for \$\omega\$ to obtain the -3 dB frequency, which turns out to be $$\omega = \frac{1}{R_1C}$$

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¹The power gain of a circuit is given in dB as $$10 \log_{10}\left(\frac{P_\text{out}}{P_\text{in}}\right)$$ and the cutoff frequency is typically defined as the frequency at which the output is at half the power of the input: $$10 \log_{10}\left(\frac{1/2 \times P_\text{in}}{P_\text{in}}\right) = 10 \log_{10}(1/2) = -3.0103\text{ dB} \approx -3\text{ dB}$$

The transfer function is given in terms of voltage rather than power so we need to use the fact that \$P = V^2/R\$: $$10 \log_{10}\left(\frac{P_\text{out}}{P_\text{in}}\right) = 10 \log_{10}\left(\frac{V_\text{out}^2 / R}{V_\text{in}^2 / R}\right) = 10 \log_{10}\left(\frac{V_\text{out}^2}{V_\text{in}^2}\right) = 10 \log_{10}\left(\frac{V_\text{out}}{V_\text{in}}\right)^2 = 20 \log_{10}\left(\frac{V_\text{out}}{V_\text{in}}\right)$$

where the last equality follows from the fact that \$\log_b(x^p) = p \log_b(x)\$ and explains why we multiply the logarithm by 10 when working with power but 20 when working with voltage (or current).

Solving for \$V_\text{out}/V_\text{in}\$ at the cutoff frequency:

$$10 \log_{10}\left(\frac{V_\text{out}}{V_\text{in}}\right)^2 = 10 \log_{10}(1/2)$$

$$\left(\frac{V_\text{out}}{V_\text{in}}\right)^2 = 1/2$$

$$\frac{V_\text{out}}{V_\text{in}} = \frac{1}{\sqrt{2}}$$

Answer 2

How can I get 3 dB frequency from transfer function?

It's simpler than that; the half power point (commonly know as the -3 dB point) is when: -

$$R_1 = X_{C}$$.

And we know that \$X_{C} = \dfrac{1}{\omega C}\$ therefore: -

$$\omega = \dfrac{1}{R_1 C}\hspace{1cm}\text{or}\hspace{1cm} f = \dfrac{1}{2\pi R_1 C}$$

Answer 3

From the transfer function DC gain is 1 so