The circuit shows a 50Hz sinewave input with a peak of 12V but shows no outputs. A simple calculation shows that the capacitor values are too small or the resistor values are too small for any effect at 50Hz.
If you are making a sinewave inverter then you need the RMS sinewave voltage not the peak voltage.
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Your LTspiceIV image also shows no arrow pointing to the output but it is probably at the resistor on the right side that is fed 50Hz from the capacitor with its value much too low for 50Hz which produces a low signal level.
Please learn about resistors and capacitors and how to calculate their values at a certain frequency. Try 3.3uF with the 1k resistor.
You do not know the basics of electronics. The peak voltage of a sinewave is "the root of 2" (1.414) times higher than its RMS voltage. Then the RMS voltage is the peak voltage times 0.707.
The calculation of the reactance of a capacitor is 1 divided by (2 x pi x R x f).
Please learn these basic formulas.
With the 3.3uF capacitor in the simulation do you see that the output is almost 0.707 times the input and do you see almost a 45 degrees phase shift?
You do not know the basics of electronics. The peak voltage of a sinewave is "the root of 2" (1.414) times higher than its RMS voltage. Then the RMS voltage is the peak voltage times 0.707.
The calculation of the reactance of a capacitor is 1 divided by (2 x pi x R x f).
Please learn these basic formulas.
There are comments about using Google in your previous posts, may I also suggest Wikipedia.org for your favourite list? https://en.wikipedia.org/wiki/Bode_plot
You have drawn the circuit in LTspice, why not switch to AC analysis and watch it's bode plot?