# dc offset in a miller integrator

I simulated a miller integrator circuit on multisim online. When I did so I made sure that dc offset in the op amp and the source is zero and also bias and offset currents are zero. But in output to a sine signal (ac) input, the output cosine signal has a negative dc offset. What is causing this offset? Here is the circuit simulated:

And here is the input and output waveform:

• Lack of ground reference? – MadHatter May 5 '17 at 19:29
• The answer is staring you in the face ! What are the voltages at t = 0, look at the graph. Both waves start at 0 V because that is the starting point the simulator uses. At this point the capacitor has no charge. Give it an initial condition of 10 mV and the offset will be gone. – Bimpelrekkie May 5 '17 at 19:32
• The choice to start the integration makes a lot of difference. The circuit is working like the math would. $\int^{t_2}_{t_1} A\cdot sin(x)\:\textrm{d}x=A\cdot\left[cos(t_1)-cos(t_2)\right]$. If you choose $t_1=0$ then the answer is always $A-A\cdot cos(t_2)$ and since $cos(t_2)$ can't be larger than 1, the integral is always positive or zero. Cool that an electronic circuit can obey the results expected from the math. – jonk May 5 '17 at 21:27