Relationship between input and output voltage on a complex inverting integrator circuit

Question

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.

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.

Answer 1

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.