0
\$\begingroup\$

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:

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

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

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

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

\$\endgroup\$
  • \$\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
0
\$\begingroup\$

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\$.

$$\omega_c=\frac{1}{RC}=1000$$

$$\omega=2\pi 1000$$

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

Either will give you about 0.9876.

\$\endgroup\$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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