# How to prepare observation table and plot frequency response of Butterworth Sallen-Key High Pass Filter

My assignment:

Design and simulate a Butterworth Sallen-Key High Pass Filter with the, fc = 10 kHz.

Prepare the observation table and plot the frequency response.

I want know how to prepare the observation table and plot the frequency response.

• I’m voting to close this question because it appears to be a homework or study exercise with no demonstration of attempt to do the work and no specific question asked. Apr 24 at 23:37
• this question for assignment i want see the different ideas about it , i have solved it but i want see more , because i want get more information , i found this topic hard to understand , i hope i can find anyone to help me Apr 24 at 23:42
• raees - Welcome :-) As commented, for assignment / test / homework-type questions like this, we need to see your own effort first. Therefore your question is now "on hold". You said you have solved it, but want "different ideas". Please click the Edit link below your question, add your solution and make it clear what sort of "different ideas" you are looking for. Remember, this is not a discussion forum, but a Q&A site, so please ask a specific question. (More site rules are in the tour and help center.) Then the question can be considered for re-opening. Thanks! Apr 24 at 23:49
• @raees - Thanks for adding the update. I have re-opened the question, after editing to make it clearer what the original assignment was, and the specific question you are asking. If I misunderstood that part, you can edit the text yourself. Thanks. Apr 25 at 0:13
• @raees This is a Butterworth! It's not hard to plot. You've not labeled the resistors, but the two values are right if they are in the right places. (I get 1125.396 k and 2250.789k.) I'm not sure what an observation table is, unless it's about building one of these and feeding it various frequencies from a generator and then measuring the output, writing out values into a table of some kind. You'll have to do that or else create it from thin air using the complex magnitude of the transfer function. The frequency response is close to trivial to plot by hand. Where are you stuck?
– jonk
Apr 25 at 2:16

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_3=\text{I}_1+\text{I}_2\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_3-\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2$$

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

$$\begin{cases} \frac{\text{V}_1-\text{V}_2}{\text{R}_3}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-\text{V}_1}{\text{R}_2}\\ \\ \frac{\text{V}_2}{\text{R}_4}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-\text{V}_1}{\text{R}_2} \end{cases}\tag3$$

Now, using an ideal opamp, we know that:

$$\text{V}_x:=\text{V}_+=\text{V}_-=\text{V}_2=\text{V}_3\tag4$$

So we can rewrite equation $$\(3)\$$ as follows:

$$\begin{cases} \frac{\text{V}_1-\text{V}_x}{\text{R}_3}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_x-\text{V}_1}{\text{R}_2}\\ \\ \frac{\text{V}_x}{\text{R}_4}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_x-\text{V}_1}{\text{R}_2} \end{cases}\tag5$$

Now, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_x}{\text{V}_\text{i}}=\frac{\text{R}_2\text{R}_4}{\text{R}_1\left(\text{R}_2+\text{R}_3\right)+\text{R}_2\left(\text{R}_3+\text{R}_4\right)}\tag6$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global*"];
V2 = Vx;
V3 = Vx;
FullSimplify[
Solve[{I3 == I1 + I2, I1 == (Vi - V1)/R1, I2 == (V3 - V1)/R2,
I3 == (V1 - V2)/R3, I3 == V2/R4}, {I1, I2, I3, V1, Vx}]]

Out[1]={{I1 -> ((R2 + R3) Vi)/(R1 (R2 + R3) + R2 (R3 + R4)),
I2 -> -((R3 Vi)/(R1 (R2 + R3) + R2 (R3 + R4))),
I3 -> (R2 Vi)/(R1 (R2 + R3) + R2 (R3 + R4)),
V1 -> (R2 (R3 + R4) Vi)/(R1 (R2 + R3) + R2 (R3 + R4)),
Vx -> (R2 R4 Vi)/(R1 (R2 + R3) + R2 (R3 + R4))}}


My equation was also confirmed using LTspice.

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so $$\\text{y}\left(\text{s}\right)\$$ is the Laplace transform of the function $$\\text{Y}\left(t\right)\$$):

• $$\text{R}_1=\frac{1}{\text{sC}_1}\tag7$$
• $$\text{R}_3=\frac{1}{\text{sC}_2}\tag8$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=\frac{\text{R}_2\text{R}_4}{\frac{1}{\text{sC}_1}\left(\text{R}_2+\frac{1}{\text{sC}_2}\right)+\text{R}_2\left(\frac{1}{\text{sC}_2}+\text{R}_4\right)}=$$ $$\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\text{s}^2}{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\text{s}^2+\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{s}+1}\tag9$$

Now, when working with sinusoidal signals we can use $$\\text{s}:=\text{j}\omega\$$ (where $$\\text{j}^2=-1\$$ and $$\\omega=2\pi\text{f}\$$ with $$\\text{f}\$$ is the frequency of the input signal in Hertz). So we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\left(\text{j}\omega\right)^2}{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\left(\text{j}\omega\right)^2+\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{j}\omega+1}=$$ $$\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2}{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2-1-\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\text{j}}\tag{10}$$

Now, you want to find the absolute value:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2}{\sqrt{\left(\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2-1\right)^2+\left(\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\right)^2}}\tag{11}$$

Assuming that $$\\text{C}_1=\text{C}_2:=\text{C}\$$, we can simplify $$\(11)\$$ as follows:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{C}^2\text{R}_2\text{R}_4\omega^2}{\sqrt{\left(\text{C}^2\text{R}_2\text{R}_4\omega^2-1\right)^2+\left(2\text{C}\text{R}_2\omega\right)^2}}\tag{12}$$

Solving for the maximum gives:

$$\frac{\partial\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}{\partial\omega}=0\space\Longrightarrow\space\omega=\frac{1}{\text{C}\sqrt{\text{R}_2\left(\text{R}_4-2\text{R}_2\right)}}\tag{13}$$

With the condition that $$\2\text{R}_2<\text{R}_4\$$.

The code I used to find it is:

In[2]:=Clear["Global*"];
C1 = c;
C2 = c;
FullSimplify[
Solve[{D[(C1*C2*R2*R4*\[Omega]^2)/
Sqrt[(C1*C2*R2*R4*\[Omega]^2 -
1)^2 + (R2*(C1 + C2)*\[Omega])^2], \[Omega]] ==
0, \[Omega] > 0 && c > 0 && R2 > 0 && R4 > 0}, \[Omega]],
Assumptions -> c > 0 && R2 > 0 && R4 > 0]

Out[2]={{\[Omega] ->
ConditionalExpression[1/(c Sqrt[R2 (-2 R2 + R4)]), 2 R2 < R4]}}


Because we want one cut-off frequency, we must have that $$\2\text{R}_2\ge\text{R}_4\$$. Now, I used Mathematica to find that the following component values work:

$$\text{C}=\frac{\sqrt{\frac{1}{10}\left(\sqrt{3996002}+1999\right)}}{200000000 \pi }\approx3.1823\cdot10^{-8}\space\text{F}\space\wedge\space$$ $$\text{R}_2=1000000\space\Omega\space\wedge\space\text{R}_4=1000\space\Omega\tag{14}$$

Code:

In[3]:=Clear["Global*"];
C1 = c;
C2 = c;
FullSimplify[
Solve[{((C1*C2*R2*R4*(10*1000*2*Pi)^2)/
Sqrt[(C1*C2*R2*R4*(10*1000*2*Pi)^2 -
1)^2 + (R2*(C1 + C2)*(10*1000*2*Pi))^2]) == 1/Sqrt[2],
10^(-9) < c <= 10^(-3) && 10^3 <= R2 <= 10^6 &&
10^3 <= R4 <= 10^6 && 2 R2 >= R4}, {c, R2, R4}]]

Out[3]={{R4 -> ConditionalExpression[(-1 + Sqrt[
2 + 1600000000 c^2 \[Pi]^2 R2^2])/(400000000 c^2 \[Pi]^2 R2),
c > 0 &&
R2 <= 1000000 && ((20000000 c \[Pi] > Sqrt[1 + Sqrt[2]] &&
2000000000 c \[Pi] < Sqrt[10 (1999 + Sqrt[3996002])] &&
1/(400000000000 c^2 \[Pi]^2 -
40000 c \[Pi] Sqrt[-1 + 200000000000000 c^2 \[Pi]^2]) <=
R2) || (40000000 c \[Pi] > Sqrt[2] && R2 >= 1000 &&
20000000 c \[Pi] <= Sqrt[1 + Sqrt[2]]) || (1000000000 c > 1 &&
40000 c \[Pi] R2 >= Sqrt[2] &&
40000000 c \[Pi] <= Sqrt[2]))]}, {c -> Sqrt[
1/10 (1999 + Sqrt[3996002])]/(200000000 \[Pi]), R2 -> 1000000,
R4 -> 1000}}
`
• Can you show the Mathematica code for finding the omega?
– G36
Apr 26 at 20:04
• What equation do you mean? Equation $(13)$?
– Jan
Apr 26 at 20:09
• Yes, Equation 13
– G36
Apr 26 at 20:16
• @G36 I edited the code in the answer.
– Jan
Apr 26 at 20:20
• Well notice that when we multiply the numerator $\left(\text{j}\omega\right)^2=-\omega^2$ we can multiply the numerator and denominator by $-1$ and you will get my result.
– Jan
Apr 26 at 20:28