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The input is in sine form (AC input) but the output (in blue color) is the straight line

The input is in sine form (AC input) but the output (in blue color) is the straight line. So what is wrong with my schematic in LT Spice this time? I thought I should've got output in sine form if I have input in sine form

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  • \$\begingroup\$ It's probably computing a DC solution for the starting condition. Try turning that off in the simulation control panel. If you're going to do it by forcing initial conditions, you should force both the \$V_-\$ pin and the output pin of the op-amp to zero. \$\endgroup\$ – TimWescott Apr 23 at 18:39
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    \$\begingroup\$ There's a reason practical integrators usually have a resistor in parallel with the capacitor. \$\endgroup\$ – Hearth Apr 23 at 18:52
  • \$\begingroup\$ It's just the learning curve for a simulator with many options for trace display and intitial condition is one of them. \$\endgroup\$ – Sunnyskyguy EE75 Apr 23 at 19:49
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You have a 159.15Hz sine wave. The integral of \$A \sin(2\pi ft)\$ is \$ \frac{-A}{2 \pi f} \cos(2\pi ft)\$. You have an on transient which might be dying away slowly, and then your sine wave might be sitting on top of that, and is just too small to see on this y-scale. Try nuking your initial conditions, and try looking at the response 100s or so later to see if they print better.

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  • \$\begingroup\$ He's also got an RC constant of 10 seconds; I thought of that, but I don't think that's it. \$\endgroup\$ – TimWescott Apr 23 at 18:40
  • \$\begingroup\$ @TiimWescott, for some reason, I got it backwards. The response is small (integral), not big. \$\endgroup\$ – Scott Seidman Apr 23 at 19:08
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Try setting the initial condition within the capacitor:

enter image description here

The output of the integrator is very small (100uV peak or so ) so you'll have to display it without the input or it will look like a straight line at zero.

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The -1V initial transient comes from allowing the DC solution to establish the starting condition, the cap is charged up by the source being at +1V due to being at +90 degrees phase, once the simulation starts the op-amp forces the input to its virtual earth condition by stepping the output down. If you set the phase to 0 this won't happen, but equally, if you set "skip initial operating point solution" it will start up with the cap uncharged. The ".ic" directive becomes irrelevant then.

enter image description here

The output signal is still invisible, since it is scaled to the input amplitude.

enter image description here

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