2
\$\begingroup\$

enter image description here

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?

\$\endgroup\$
5
  • \$\begingroup\$ What have you tried so far? \$\endgroup\$
    – winny
    Commented Sep 29, 2022 at 15:16
  • \$\begingroup\$ Do you know how to convert -3 dB to a voltage ratio? Or, there is a shortcut for first order filters if you remember the relationship between real and imaginary at the cutoff frequency. \$\endgroup\$
    – Mattman944
    Commented Sep 29, 2022 at 15:33
  • \$\begingroup\$ If you write \$T(s)\$ then you have to follow with \$s\$, not \$j\omega\$, otherwise you need to write \$T(j\omega)\$ (this is also a hint). \$\endgroup\$ Commented Sep 29, 2022 at 15:46
  • \$\begingroup\$ What is difference between s and jw ?? I thought they are same. \$\endgroup\$
    – moonjy1120
    Commented Sep 29, 2022 at 16:17
  • \$\begingroup\$ @moonjy1120 they are just different notations. s comes from the Laplace transform while \$jw\$ is the Fourier representation. While they are equivalent you are mixing domains \$\endgroup\$ Commented Sep 29, 2022 at 17:08

3 Answers 3

3
\$\begingroup\$

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 $$\omega = \frac{1}{R_1C}$$


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:

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

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

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

\$\endgroup\$
2
\$\begingroup\$

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: -

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

\$\endgroup\$
1
\$\begingroup\$

From the transfer function enter image description here DC gain is 1 so
enter image description here

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Your answer for the -3 dB frequency is correct but the DC gain is not 1, it is 0. The high frequency gain is \$-R_2/R_1\$. \$\endgroup\$
    – Null
    Commented Sep 29, 2022 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.