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I am on my first semester of electrical engineering and I came upon an exercise where in the given circuit I have to calculate the relationship between the input and output voltages of the op-amp. enter image description here

After some searching I found that the given circuit is an inverting integrator. In wikipedia it states that the relationship between the input and output voltage is:$${\displaystyle V_{\text{out}}(t_{1})=V_{\text{out}}(t_{0})-{\frac {1}{R_{\text{i}}C_{\text{f}}}}\int _{t_{0}}^{t_{1}}V_{\text{in}}(t)\,dt.}$$

So I tried to solve it myself. I started with a 1st Kirchoff's law equation which came out as: $$I=I_1+I_2$$ then I substituted using Ohm's law: $$\frac{V_{in}}{R_1}=\frac{V_{out}}{R_2}+I_2 \space \space (1)$$ For the last current I used: $$Q=C_2V \rightarrow Q'=C_2 V'$$ and because $$Q'=I_2 \space and \space V'=V_{out}'$$ it resulted in $$I_2 = C_2 V_{out}$$ Then through substitution in the first equation I found: $$V_{out}'= -\frac{V_{out}}{R_2C_2}+\frac{V_{in}}{R_1C_2}$$ At this point I thought I would integrate and find the result given by wikipedia but I can't seem to work it out and I think I've made a mistake in my calculations.The last equation seems to me like a differential one which unfortunately I don't know how to solve. I also have my doubts about this circuit being an inverting integrator because of the existence of capacitor C1, that I did not take into consideration. Any help on what's wrong would be greatly appreciated. Please excuse my ignorance and/or any mistakes.

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    \$\begingroup\$ this is not a pure integrator. \$\endgroup\$ – analogsystemsrf Oct 21 '19 at 17:15
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    \$\begingroup\$ And there's a label "Vin", but it's shown as going to ground. \$\endgroup\$ – TimWescott Oct 21 '19 at 17:35
  • \$\begingroup\$ As is Vout... so, I think some imagination is required to understand it was probably not the intention of OP \$\endgroup\$ – Huisman Oct 21 '19 at 17:54
  • \$\begingroup\$ i just realized my whole circuit was wrong, thank you for the heads up. I have uploaded the correct circuit now, I am sorry I don't know a good tool to draw circuits. \$\endgroup\$ – Konstantinos Zafeiris Oct 21 '19 at 18:00
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    \$\begingroup\$ Your first node equation ignores current through C1 and R3. \$\endgroup\$ – Scott Seidman Oct 22 '19 at 14:44
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The formula I got (and I verified it using LTspice) is (in the 'complex' s-domain):

$$\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=-\frac{1}{\text{R}_1}\cdot\frac{\text{R}_2\cdot\frac{1}{\text{sC}_2}}{\text{R}_2+\frac{1}{\text{sC}_2}}=-\frac{1}{\text{R}_1}\cdot\frac{\text{R}_2}{1+\text{sC}_2\text{R}_2}\tag1$$

Now, we can look at a few cases:

  1. When \$\text{V}_\text{in}\left(t\right)=\hat{\text{v}}\$, where \$\hat{\text{v}}\$ is a constant DC-voltage. We get for the output voltage: $$\text{V}_\text{out}\left(t\right)=\hat{\text{v}}\cdot\frac{\text{R}_2}{\text{R}_1}\cdot\left(\exp\left(-\frac{t}{\text{C}_2\text{R}_2}\right)-1\right)\tag2$$
  2. When \$\text{V}_\text{in}\left(t\right)=\hat{\text{v}}\cdot\cos\left(\omega t\right)\$. We get for the output voltage: $$\text{V}_\text{out}\left(t\right)=\frac{\text{R}_2}{\text{R}_1}\cdot\frac{\hat{\text{v}}}{1+\left(\text{R}_2\text{C}_2\omega\right)^2}\cdot\left(\exp\left(-\frac{t}{\text{C}_2\text{R}_2}\right)-\cos\left(\omega t\right)-\text{C}_2\text{R}_2\omega\sin\left(\omega t\right)\right)\tag3$$
  3. When \$\text{V}_\text{in}\left(t\right)=\delta\left(t\right)\$. We get for the output voltage: $$\text{V}_\text{out}\left(t\right)=-\frac{\exp\left(-\frac{t}{\text{C}_2\text{R}_2}\right)}{\text{C}_2\text{R}_1}\tag4$$
  4. When \$\text{V}_\text{in}\left(t\right)=\theta\left(t\right)\$. We get for the output voltage: $$\text{V}_\text{out}\left(t\right)=\frac{\text{R}_2}{\text{R}_1}\cdot\left(\exp\left(-\frac{t}{\text{C}_2\text{R}_2}\right)-1\right)\tag5$$

Edit:

The 'complex' s-domain is also known by it's name: the Laplace transform. It plays a very important role in solving this kind of problems.

I solved the output voltage in terms of the input voltage, and it gave me:

$$\text{V}_\text{out}\left(t\right)=-\frac{1}{\text{C}_2\text{R}_1}\int_0^t\text{V}_\text{in}\left(\tau\right)\cdot\exp\left(\frac{\tau-t}{\text{C}_2\text{R}_2}\right)\space\text{d}\tau\tag6$$

So, no it is not a pure integrator.

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  • \$\begingroup\$ Thank you for your answer. As to my understanding, you solved the differential equation that results from my final answer, or did you approached the problem differently? Then you found 4 cases depending on the input voltage. Unfortunately I do not yet know, what the comlex s-domain is, could you please provide some material so that I could study it? In the end is this circuit still considered a complex inverting integrator or not? I thank you again in advance. \$\endgroup\$ – Konstantinos Zafeiris Oct 22 '19 at 13:03
  • \$\begingroup\$ @ΚωνσταντίνοςΖαφείρης I edited my answer a bit. \$\endgroup\$ – Jan Oct 22 '19 at 14:25
  • \$\begingroup\$ I feel like including the convolution integral as a time-domain method of solving the input-output problem might be a worthwhile addition to this answer. \$\endgroup\$ – Captainj2001 Oct 22 '19 at 15:27

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