# How can we calculate the transfer function from this filter?

I'm trying to calculate the transfer function of this 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:

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

• Do you want the Laplace TF
– Chu
Commented Jun 2, 2016 at 17:15
• 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. Commented Jun 2, 2016 at 17:18
• In these simple circuit, you should derive the equation by yourself. Commented Jun 2, 2016 at 17:40
• @JohnD True ! I was trying to use Millman here but it's easier with the voltage divider, thank you ! Commented Jun 3, 2016 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

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

• 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!).
– LvW
Commented Jun 2, 2016 at 19:26
• @LvW good point, I will precise it in the answer. Commented Jun 2, 2016 at 19:28
• Thank you very much for your answer, but how do I know that $\omega_c = \frac{1}{CR}$ ? Commented Jun 3, 2016 at 11:46
• @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. Commented Jun 3, 2016 at 12:20
• @MathieuL Thank you I will then try to understand what LvW is saying ! I will come back if I do not fully understand. Commented Jun 3, 2016 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 $\omega_c$.

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

$$\omega=2\pi 1000$$

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

Either will give you about 0.9876.

• 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) Commented Jun 2, 2016 at 18:18 • Thanks ! But I can't seem to understand how you got the values for$ \omega_c \text{ and } \omega $? Commented Jun 3, 2016 at 11:47 •$\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\\$ Commented Jun 3, 2016 at 17:44