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enter image description here

As you can see the gain is constant for the practical op-amp, even though gain is

enter image description here enter image description here

This shows that it clearly changes for every frequency. Is the one in the graph an approximation, maybe because it does not change much below cut-off frequency ?

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    \$\begingroup\$ Well, this isn't an integrator but a low-pass filter... \$\endgroup\$ – aschipfl Jun 6 at 17:16
  • \$\begingroup\$ Yes, a low pass filter is an integrator in this case because it is used with an op-amp \$\endgroup\$ – Allen Jun 6 at 17:19
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    \$\begingroup\$ If you want a pure integrator, you could remove Rf. But you need a way to be sure the capacitor is discharged at start up, and to recover if the output goes to the rails. \$\endgroup\$ – The Photon Jun 6 at 17:21
  • \$\begingroup\$ Okay, but why is the gain constant till cut-off frequency \$\endgroup\$ – Allen Jun 6 at 17:23
  • \$\begingroup\$ @Allen, how does a capacitor behave at low frequencies? What would this circuit be if you replaced the capacitor with its low-frequency equivalent? \$\endgroup\$ – The Photon Jun 6 at 17:29
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You have confused yourself by labeling your graph with frequency = 0 at the vertical axis.

The response of the integrator is a straight line only on a log-log graph, which means that 0 frequency (DC) is infinitely far away on the left side of the graph. What does this imply about the gain of an ideal integrator at DC?

But your circuit does have a definite value for DC gain. What is it, and at what frequency does it begin to deviate from that value?

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  • \$\begingroup\$ So, is the gain drawn constant because the it does not deviate that much, it deviates but not much, from the DC gain ? \$\endgroup\$ – Allen Jun 6 at 17:26
  • \$\begingroup\$ Right. This is a Bode plot, in which we don't draw the actual gain; instead we draw the asymptotes, which are easier to work with and visualize. \$\endgroup\$ – Dave Tweed Jun 6 at 17:31
  • \$\begingroup\$ Thanks a lot @Dave Tweed This was bugging me for a while. \$\endgroup\$ – Allen Jun 6 at 17:34

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