Why is the transfer function of this circuit first order, when it has two energy storage elements?

I have the following circuit (assuming an ideal opamp),

simulate this circuit – Schematic created using CircuitLab

which has a transfer function that looks something like $$\frac{As + 1}{Bs + 1}$$ when you take the output to be the output of the opamp, and the input to be the node before R1 and C1.

Trying to find the order of this circuit, my first impression was that it is a first order circuit. It's got a transfer function where the denominator has one for the highest power of s, and the differential equation that will describe it will be first order. However, a should be equivalent definition of the order of a system is the number of independant energy storage components, of which this circuit has two.

I know that the transfer function is first order, but there are two seemingly independant energy storage devices in this circuit. So what gives? Is it that the two capacitors are not "independant"?

• Welcome! Please mark input and output with names nodes. I can guess but I prefer not to. May 18, 2023 at 17:41
• Yes, I've edited to be more clear. May 18, 2023 at 19:07

there are two seemingly independent energy storage devices in this circuit. So what gives?

They are independent but only one stores the state of the system. The op-amp virtual ground allows the output to be determined by the voltage on $$\C_2\$$, which holds the system state.

$$\C_1\$$, on the input side along with $$\R_1\$$, determines how the state is modified by the input.

In order for independent storage elements to determine the order of the system, each one must hold a state variable.

The standard state equations demonstrate this.

$$\dot{x}=Ax+Bu$$and$$y=Cx$$

Here: $$\C=1\$$, $$\y=v_{out}\$$, $$\x=v_{C2}\$$ (the state variable), $$\Bu=\left[\begin{array}{c} b_{1}\dot{v}_{in}\\ b_{0}v_{in} \end{array}\right]\$$

• I think this is what I was looking for, thank you very much! So it's not true that the order of the system is the same as the number of energy storage elements in every case then. This also makes sense because the voltage across C1 is just the input voltage anyway so it can't also be a system state. May 18, 2023 at 23:27
• I did not write equations but if you insert a small resistance in series with $C_1$, like an ESR, then it becomes a 2nd-order system. May 19, 2023 at 8:12
• @JadonClugston: It is true for passive systems, but not necessarily true for active systems. May 19, 2023 at 15:25
• @VerbalKint: Yes it becomes 2nd Order while retaining a zero as well. May 19, 2023 at 15:26
• Oui, of course, the zero is still there : ) May 19, 2023 at 16:17

The transfer function can be found using the fast analytical circuits techniques or FACTs without writing a line of algebra, just going through a few simple sketches. In this analysis, you must realize that the (-) pin is at a 0-V potential considering the virtual ground:

First, determine the gain $$\H_0\$$ at dc by opening all the capacitors. Then reduce the stimulus to zero and determine the time constants of the circuit. Once done, find an impedance condition which would null the response at $$\V_{out}\$$. Assemble the final transfer function in a compact form.

A quick Mathcad sheet tells us if we are ok or not by comparing the brute-force expression with the low-entropy version:

If you want to learn about this technique, I recommend you look at my APEC 2016 seminar which smoothly introduces the method.

• I already know the transfer function, I derived it myself using nodal analysis :). Cool method though! May 18, 2023 at 23:01

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:

\begin{cases} \begin{alignat*}{1} \text{I}_0&=\text{I}_1+\text{I}_2\\ \\ 0&=\text{I}_1+\text{I}_2+\text{I}_3 \end{alignat*} \end{cases}\tag1

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

\begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2&=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3&=\frac{\text{V}_2-\text{V}_1}{\text{R}_3} \end{alignat*} \end{cases}\tag2

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

\begin{cases} \begin{alignat*}{1} \text{I}_0&=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_2}\\ \\ 0&=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_2}+\frac{\text{V}_2-\text{V}_1}{\text{R}_3} \end{alignat*} \end{cases}\tag3

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_1=0\space\text{V}\tag4$$

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

\begin{cases} \begin{alignat*}{1} \text{I}_0&=\frac{\text{V}_\text{i}-0}{\text{R}_1}+\frac{\text{V}_\text{i}-0}{\text{R}_2}\\ \\ 0&=\frac{\text{V}_\text{i}-0}{\text{R}_1}+\frac{\text{V}_\text{i}-0}{\text{R}_2}+\frac{\text{V}_2-0}{\text{R}_3} \end{alignat*} \end{cases}\tag5

Now, we can solve for the transfer function:

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

Where I used the following Mathematica-code:

In[1]:=Clear["Global*"];
V1 = 0;
FullSimplify[
Solve[{I0 == I1 + I2, 0 == I1 + I2 + I3, I1 == (Vi - V1)/R1,
I2 == (Vi - V1)/R2, I3 == (V2 - V1)/R3}, {V2, I0, I1, I2, I3}]]

Out[1]={{V2 -> -(((R1 + R2) R3 Vi)/(R1 R2)), I0 -> (1/R1 + 1/R2) Vi,
I1 -> Vi/R1, I2 -> Vi/R2, I3 -> -(((R1 + R2) Vi)/(R1 R2))}}
`

My equation was also confirmed using LTspice.

So, applying this to your case we can see that:

$$$$\begin{split} \mathscr{H}\left(\text{s}\right)&=\frac{\text{v}_2\left(\text{s}\right)}{\text{v}_\text{i}\left(\text{s}\right)}\\ \\ &=-\frac{\displaystyle\frac{\displaystyle\text{R}_1\cdot\frac{1}{\text{sC}_3}}{\displaystyle\text{R}_1+\frac{1}{\text{sC}_3}}\cdot\left(\frac{\displaystyle1}{\displaystyle\text{sC}_4}+\text{R}_2\right)}{\displaystyle\frac{\displaystyle1}{\displaystyle\text{sC}_4}\cdot\text{R}_2}\\ \\ &=-\frac{\displaystyle\frac{\displaystyle\text{R}_1}{\displaystyle1+\text{sC}_3\text{R}_1}\cdot\left(1+\text{sC}_4\text{R}_2\right)}{\text{R}_2}\\ \\ &=-\frac{\displaystyle\text{R}_1}{\displaystyle\text{R}_2}\cdot\frac{\displaystyle1+\text{sC}_4\text{R}_2}{\displaystyle1+\text{sC}_3\text{R}_1} \end{split}\tag7$$$$

Notice that when $$\\text{R}:=\text{R}_1=\text{R}_2\$$, this simplifies to:

$$\mathscr{H}\left(\text{s}\right)=-\frac{\displaystyle1+\text{sC}_4\text{R}}{\displaystyle1+\text{sC}_3\text{R}}\tag8$$

• The circuit in your answer does not match the circuit that I see in the question. Perhaps something is amiss somewhere? May 18, 2023 at 19:03
• @MathKeepsMeBusy look again. At the end of my answer I applied it to the question of the OP. First I derive the transfer function. May 18, 2023 at 20:04
• The schematic was edited by the OP after I gave my answer, at least according to my timeline. May 18, 2023 at 20:09