Here we construct an integrator :
In this case we need to add a 10MegaOhm resistor in parallel to the capacitor, other wise any small DC offset in the signal will build up in the capacitor and our Vout signal will start to go crazy, (think integral of a horizontal line).
what would I get?
Well, it takes 0.002 s to complete a cycle, if we look at how much V is integrated over the positive half of the cycle, that would be 2V * 0.001s = 0.002V times 1 / RC = 2V. Which is close to the peak to peak voltage you see in the picture (2.1V). There is a DC offset though, -0.5V, since the DC gain of the contraption is 100, that means the function generator is off by only 5mV.
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