Question about Third order Butterworth Filter Design

I'm new to PSpice and filter design and I'm trying to design a third order Low pass Butterworth Filter with cut-off frequency = 10 Hz. Here is the circuit that I've designed:

I've applied a 5V sinusoid with 2V DC offset to check whether it is working or not. Here is the result that I'm obtaining:

At the output I should get 2V, which is exactly the DC component of the sinusoid that I want, correct? However, I have no idea why it is taking so long for the output to settle into 2V (~ 0.2 seconds). Is there any way I can improve this design in order to minimize this stabilization time? Furthermore, could someone provide me some recommendation on which OP AMP comercially available models it is best to use for filtering?

• Those circuit values don't produce a Butterworth filter - you should see a little overshoot when you apply the 2 volt step input. Commented Jul 14, 2020 at 22:23
• @Andyaka could you please tell me what I'm missing here in order to design a third-order butterworth filter with gain of 1? Commented Jul 14, 2020 at 23:38
• But looking at the normalized butterworth polynomials, for n = 3 we have (s + 1)(s.^2 + s + 1). For the third-order filter don't we have Q = 1? Commented Jul 14, 2020 at 23:44
• I agree that the opamp is a droopy 2nd-order Bessel filter, not a Butterworth. Then it is followed by another RC making it even droopier. Commented Jul 15, 2020 at 0:21
• Are you allowed any overshoot? Do you have a target settling time you need to meet? Settling time is normally defined at some percentage error : what is that value for your application? Fastest settling time is normally achieved with a slightly underdamped filter, that overshoots to just less than your permitted tolerance.
– user16324
Commented Jul 15, 2020 at 10:37

Your LPF with 10Hz cutoff has transient reponse that will take about 0.5s (about 5*1/fc) to fully settle. That's what you are seeing.

For faster step-response settling you need to pick a wider filter with higher Fc.

Have a look here for the relationship between rise time and filter Fc:

There are circuit tricks to get it to start at a desired known non-zero DC by initializing your filter to a desired "reset" state. Although this does not change the rise time, it does allow the circuit to start at a pre-fixed DC level immediately after a reset. (This requires additional analog switches and control circuitry)

Well, I am trying to analyze the following circuit (assuming an ideal opamp):

simulate this circuit – Schematic created using CircuitLab

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

$$\begin{cases} \text{I}_3=\text{I}_1+\text{I}_2\\ \\ \text{I}_3=\text{I}_4\\ \\ \text{I}_2=\text{I}_7\\ \\ \text{I}_5=\text{I}_6+\text{I}_7\\ \\ \text{I}_4+\text{I}_6=\text{I}_1+\text{I}_5 \end{cases}\tag1$$

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

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_3-\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_4}\\ \\ \text{I}_6=\frac{\text{V}_3-\text{V}_4}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_4}{\text{R}_6} \end{cases}\tag2$$

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

$$\begin{cases} \frac{\text{V}_1-\text{V}_2}{\text{R}_3}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-\text{V}_1}{\text{R}_2}\\ \\ \frac{\text{V}_1-\text{V}_2}{\text{R}_3}=\frac{\text{V}_2}{\text{R}_4}\\ \\ \frac{\text{V}_3-\text{V}_1}{\text{R}_2}=\text{I}_7\\ \\ \text{I}_5=\frac{\text{V}_3-\text{V}_4}{\text{R}_5}+\text{I}_7\\ \\ \frac{\text{V}_2}{\text{R}_4}+\frac{\text{V}_3-\text{V}_4}{\text{R}_5}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\text{I}_5 \end{cases}\tag3$$

Now, Using an ideal opamp, We know that $$\\text{V}_x:=\text{V}_+=\text{V}_-=\text{V}_2=\text{V}_3\$$. So we can rewrite equation $$\(3)\$$ as follows:

$$\begin{cases} \frac{\text{V}_1-\text{V}_x}{\text{R}_3}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_x-\text{V}_1}{\text{R}_2}\\ \\ \frac{\text{V}_1-\text{V}_x}{\text{R}_3}=\frac{\text{V}_x}{\text{R}_4}\\ \\ \frac{\text{V}_x-\text{V}_1}{\text{R}_2}=\text{I}_7\\ \\ \text{I}_5=\frac{\text{V}_x-\text{V}_4}{\text{R}_5}+\text{I}_7\\ \\ \frac{\text{V}_x}{\text{R}_4}+\frac{\text{V}_x-\text{V}_4}{\text{R}_5}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\text{I}_5 \end{cases}\tag4$$

Now, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_\text{o}}{\text{V}_\text{i}}=\frac{\text{R}_2\text{R}_4\text{R}_6}{\left(\text{R}_1\left(\text{R}_2+\text{R}_3\right)+\text{R}_2\left(\text{R}_3+\text{R}_4\right)\right)\left(\text{R}_5+\text{R}_6\right)}\tag5$$

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

• $$\text{R}_2=\frac{1}{\text{sC}_1}\tag6$$
• $$\text{R}_4=\frac{1}{\text{sC}_2}\tag7$$
• $$\text{R}_6=\frac{1}{\text{sC}_3}\tag8$$

So, the transfer function becomes:

$$\mathcal{h}\left(\text{s}\right)=\frac{\text{v}_\text{o}\left(\text{s}\right)}{\text{v}_\text{i}\left(\text{s}\right)}=\frac{\frac{1}{\text{sC}_1}\frac{1}{\text{sC}_2}\frac{1}{\text{sC}_3}}{\left(\text{R}_1\left(\frac{1}{\text{sC}_1}+\text{R}_3\right)+\frac{1}{\text{sC}_1}\left(\text{R}_3+\frac{1}{\text{sC}_2}\right)\right)\left(\text{R}_5+\frac{1}{\text{sC}_3}\right)}=$$ $$\frac{1}{\alpha_1\text{s}^3+\alpha_2\text{s}^2+\alpha_3\text{s}+1}\tag9$$

Where:

• $$\alpha_1=\text{C}_1\text{C}_2\text{C}_3\text{R}_1\text{R}_3\text{R}_5\tag{10}$$
• $$\alpha_2=\text{C}_2\left(\text{C}_1\text{R}_1\text{R}_3+\text{C}_3\text{R}_5\left(\text{R}_1+\text{R}_3\right)\right)\tag{11}$$
• $$\alpha_3=\text{C}_2\left(\text{R}_1+\text{R}_3\right)+\text{C}_3\text{R}_5\tag{12}$$

Because we are working with sinusodial signals, we can write $$\\text{s}=\text{j}\omega\$$ where $$\\text{j}^2=-1\$$ and $$\\omega=2\pi\text{f}\$$ with $$\\text{f}\$$ is the frequency in Hertz. So, we can write:

$$\underline{\mathcal{h}}\left(\text{j}\omega\right)=\frac{1}{\alpha_1\left(\text{j}\omega\right)^3+\alpha_2\left(\text{j}\omega\right)^2+\alpha_3\left(\text{j}\omega\right)+1}=\frac{1}{1-\alpha_2\omega^2+\omega\left(\alpha_3-\alpha_1\omega^2\right)\text{j}}\tag{13}$$

Now, we can find the amplitude by finding the absolute value of $$\(13)\$$:

$$\left|\underline{\mathcal{h}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{\left(1-\alpha_2\omega^2\right)^2+\left(\omega\left(\alpha_3-\alpha_1\omega^2\right)\right)^2}}\tag{14}$$

Now, when we know that (which is the case in your example) $$\\text{R}:=\text{R}_1=\text{R}_3=\text{R}_5\$$ and $$\\text{C}:=\text{C}_1=\text{C}_2=\text{C}_3\$$, we obtain the following transfer function:

$$\left|\underline{\mathcal{h}}\left(\text{j}\omega\right)\right|=\frac{1}{\left(1+\left(\omega\text{CR}\right)^2\right)^\frac{3}{2}}\tag{15}$$

We can find the cut-off frequency by solving:

$$\left|\underline{\mathcal{h}}\left(\text{j}\omega\right)\right|=\frac{1}{\left(1+\left(\omega\text{CR}\right)^2\right)^\frac{3}{2}}=\frac{1}{\sqrt{2}}\space\Longrightarrow\space\omega=\frac{\sqrt{2^\frac{1}{3}-1}}{\text{CR}}\tag{16}$$

So, for the frequency we get:

$$\omega=2\pi\text{f}=\frac{\sqrt{2^\frac{1}{3}-1}}{\text{CR}}\space\Longleftrightarrow\space\text{f}=\frac{\sqrt{2^\frac{1}{3}-1}}{2\pi\text{CR}}\tag{17}$$

So, when we know that the cut-off frequency must be $$\10\space\text{Hz}\$$ we need:

$$10=\frac{\sqrt{2^\frac{1}{3}-1}}{2\pi\text{CR}}\space\Longleftrightarrow\space\text{CR}=\frac{\sqrt{2^\frac{1}{3}-1}}{20\pi}\approx0.00811411\space\left[\Omega\text{F}\right]\tag{18}$$

And, so the transfer function becomes:

$$\left|\underline{\mathcal{h}}\left(\text{j}\omega\right)\right|=\frac{1}{\left(1+\left(\omega\cdot\frac{\sqrt{2^\frac{1}{3}-1}}{20\pi}\right)^2\right)^\frac{3}{2}}\tag{19}$$

Plotting that, gives: