I'm trying to calculate the transfer function of this high pass filter : Buttleworth High Pass Filter

So basically I'm trying to find how I can find \$V_{\text{out}}\$.

Due to the amplifier we know that $$V_{in} = V_{\text{out}}$$ I know how to calculate \$V_{\text{in}}\$ but I can't seem to find \$V_{\text{out}}\$ because I am supposed to find this transfer function (since it is a Butterworth high pass filter):

$$|H_{ph}(j \omega)| = \frac{1}{\sqrt{1+ \left(\frac{\omega_c}{\omega}\right)^{2n}}} = \frac{\left(\frac{\omega}{\omega_c}\right)^{n}}{\sqrt{1+ \left(\frac{\omega}{\omega_c}\right)^{2n}}}$$

here \$n=1\$ because it's a first order filter

So I was wondering if any of you could help me and find this transfer function knowing that:

$$\underline{H}(j\omega) = \frac{\underline{V}_{\text{out}}}{\underline{V}_{\text{in}}}$$

  • \$\begingroup\$ Do you want the Laplace TF \$\endgroup\$ – Chu Jun 2 '16 at 17:15
  • \$\begingroup\$ Use 1/(Cs) as the impedance of the capacitor, where s=jomega. Then solve the voltage divider equation for Vout. Once you do that, take the magnitude of the result and you get the equation you are looking for with n=1. \$\endgroup\$ – John D Jun 2 '16 at 17:18
  • \$\begingroup\$ In these simple circuit, you should derive the equation by yourself. \$\endgroup\$ – MathieuL Jun 2 '16 at 17:40
  • \$\begingroup\$ @JohnD True ! I was trying to use Millman here but it's easier with the voltage divider, thank you ! \$\endgroup\$ – David Rigaux Jun 3 '16 at 11:48

Most of you reasonning concerning the Vout and Vin is correct. To find the transfer function, you need to do the Voltage divider.

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

if we manipulate the equation:

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

the cuf off frequency in this problem is:

$$\omega_{c}= \frac{1}{CR} $$

so the equation become:

$$H(j\omega) =\frac{V_{out}}{V_{in}}= \frac{j\omega}{j\omega + \omega_{c}} $$

So there you have the general transfer equation. The op-amp is just a follower with unity Gain.

You can rearrange the equation to fit your format

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

  • \$\begingroup\$ To be a bit pedantic....The cut-off frequency is DEFINED as the frequency where the transfer function is reduced by a factor of 1/SQRT(2) if compared with the magnitude at f=0. In the present example, this is the frequency where the imaginary part of the denominator equals the real part. Therefore, this definition RESULTS in wc=1/RC (for the present example only!). \$\endgroup\$ – LvW Jun 2 '16 at 19:26
  • \$\begingroup\$ @LvW good point, I will precise it in the answer. \$\endgroup\$ – MathieuL Jun 2 '16 at 19:28
  • \$\begingroup\$ Thank you very much for your answer, but how do I know that \$ \omega_c = \frac{1}{CR}\$ ? \$\endgroup\$ – David Rigaux Jun 3 '16 at 11:46
  • \$\begingroup\$ @DavidRigaux omega_c represent where you have 3 dB of attenuation by comparison to f = 0. You can follow LvW's instruction to find omega_c. However, it depends of your course, I guess, personnaly, I rarely write omega_{c} because I find it less intuitive to look it at than with the component's coefficient. \$\endgroup\$ – MathieuL Jun 3 '16 at 12:20
  • \$\begingroup\$ @MathieuL Thank you I will then try to understand what LvW is saying ! I will come back if I do not fully understand. \$\endgroup\$ – David Rigaux Jun 3 '16 at 13:45

John's method will certainly work, and works without you needing to know some of the formulas that you've given.

However, if you want to use your equation for \$|H(j\omega)|\$, then you need \$\omega_c\$.


$$\omega=2\pi 1000$$

$$|H(j\omega)|=\frac{1}{\sqrt{1+\left(\frac{1000}{2\pi 1000}\right)^2}}$$

Either will give you about 0.9876.

  • 1
    \$\begingroup\$ FYI, EE uses \$ instead of just $ to start and end mathjax. (keeps things from getting mixed up when new users want to talk about how much something costs) \$\endgroup\$ – The Photon Jun 2 '16 at 18:18
  • \$\begingroup\$ Thanks ! But I can't seem to understand how you got the values for \$ \omega_c \text{ and } \omega \$ ? \$\endgroup\$ – David Rigaux Jun 3 '16 at 11:47
  • \$\begingroup\$ \$\omega=2\pi f\$, so for 1000 Hz (V1 block states 1 kHz), \$\omega=2\pi 1000\$. If you want the transfer function in terms of \$\omega\$, then just leave it. \$\omega_c\$ for an RC filter is always \$\frac{1}{RC}\$, but if you didn't know that you could use a voltage divider as John suggested where \$Z_c=\frac{1}{j \omega C}\$ and \$Z_R=R\$ \$\endgroup\$ – MikeP Jun 3 '16 at 17:44

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