How Does Changing Capacitor Value Affect OPAMP Output Voltage?

[etf]

Hi!
Here is my task:

OPAMP circuit and output voltag are shown in image below. Assuming that OPAMP is ideal, come up with a solution by hand.

View attachment 73553

Then change capacitor's value from 200nF to 33nF and output voltage will be:

View attachment 73554

Explain why output voltages are different ( first has triangle shape and gradually decreases and second has cut off bottom and doesn't decrease).

Here is what I have done:
Since there are two sources, one DC and one time dependent pulse waveform, I used superposition method. When time dependent source is active, our circuit is:
View attachment 73555
v2(t)=0\rightarrow v1(t)=0

i2(t)=0\rightarrow iin(t)=ic(t)

iin(t)=\frac{vin(t)}{R1}

v3(t)+\frac{1}{C}\int ic(t)dt+R1iin(t)-vin(t)=0

For positive input voltage vin(t)=1V we have:

v3(t)=-\frac{1}{RC1}\int vin(t)dt=-50\int 1dt=-50t,\,\, 0\leq t\leq 100*10^{-3}s

For negative input voltage vin(t)=-1V we have:
v3(t)=-\frac{1}{RC1}\int vin(t)dt=-50\int (-1)dt=50t,\,\,100*10^{-3}s\leq t\leq 200*10^{-3}s
But I don't know what to do next :(

 

[rude man]

1. you need a feedback resistor to keep from saturation if you're looking at a simulation or a real circuit. With an ideal op amp that's not necessary.
2. you have not described the inpt in words. I'm not familiar with LTSpice.

 

[lazyaditya]

The circuit you have is an integrator.The output for pulse waveform will be triangular wave and the output for the dc signal will be a straight line with a negative slope. When you add the equation of both of your output for ac and dc inputs you will have waveform as given in first figure. Second figure that you have shown has clipping of waveform at the bottom that is since the output waveform has gone below the lower threshold of OP-AMP and thus into saturation (-Vee).
I hope this would help.

 

[etf]

Solved it :)