MrAl and Audioguru thank you for your helps
for MrAl in your post #49 you said the output is Vout=Vin*t/(C1*R3)+Vc0
i just want to know how you calculate it , is there any analysis ?
That expression comes from knowing beforehand how op amp integrators work, although it can also come from a knowledge of op amp amplifiers in general.
Those integrators very often appear as just a single capacitor in the feedback (output to inverting input) and a single resistor on the input (Vin to inverting terminal) with the non inverting terminal to ground. They appear so much that after a while you dont have to analyze it anymore because you remember that form.
If we look at it as just a general op amp amplifier, then we know that the 'gain' of the op amp is the feedback impedance divided by the input series impedance. So that would be zFB/zIN. Since zFB=1/(s*C) and zIN is just R, we get:
Gain=(1/(s*C))/R=1/(s*C*R)
and it is an inverter so the gain is -1/(s*C*R).
So the output is -Vin/(s*C*R) and for a step input the output is -Vin/(s^2*C*R) in the frequency domain. Using the Inverse Laplace Transform we get to the time domain (there are other ways however) and that gives us:
Vout=-Vin*t/(R*C)
and that is a ramp.
To account for the initial capacitor voltage Vc0 in this case we note that the capacitor voltage is equal to the output voltage so if there was any initial cap voltage it would equal Vout to start with at t=0, so we get finally:
Vout=-Vin*t/(R*C)+Vc0
Another way is to simply model the op ap as a voltage controlled voltage source with known gain A and just do a complete analysis that starts with:
Vout=A*(vp-vn)
where vp is the voltage at the non inverting terminal and vn is the voltage at the inverting terminal and A is the internal gain of the amplifier. The gain A is usually high in an op amp so what we do to simplify the equation a little is after we write the equation we let A go to infinity and that removes A from the equation and leaves us with the same result as above.
Still yet another way is to start in the time domain with a state vector differential equation and then solve it for all time.
So you see there are several approaches.
Also note that the time equation shown above is correct but we must apply it in a non linear way. We must manually limit the time period to 1/(2*f) because at that time the first op amp output polarity changes, so we have to stop and restart with a different Vin and previous Vc0 to get the output for the next half cycle. So for example if we calculated the output to be +11 volts at the end of the first half cycle with Vin=-13, then we have to make Vc0=+11 and Vin=+13 and then do the calculation again to get the next half cycle.