In the Transfer Function A is the Open loop Gain and its equal to GB/s where GB is Gain Bandwidth Product
Hope i have given you all the details
No, you didn't.
In your posting#5 you have shown us a simulation result - but you didn`t give us the most important information: Which circuit configuration?
But - in this case, I could imagine (also from the magnitude response) that you have simulated the complete circuit as shown in your first posting.
But for future questions: Present simulation results in relation to the corresponding circuit only.
Now to your question: The magnitude and phase responses are OK. Up to approx. 10 kHz the circuit resembles the typical differentiator behaviour.
For larger frequencies the open loop opamp gain cannot be considered as ideal and causes deviations from the desired response.
Starting with approx. 15 kHz the loop gain approaches unity and the magnitude response is nearly identical to the open-loop response A(s) of the opamp.
Thus, the wanted GBW (transis frequency) is identical to the frequency where the gain is 0 dB. Either you perform simulations up to 10 MHz (by the way: m is milli and M is mega)
or you try to continue the gain slope up to the 0 dB crossing point (which will be approx. at 10 MHz).
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but i have another doubt, why A=GB/s? in low frequency, A is almost constant, above p1(dominant pole), it is about GB/s. right?
Yes, of course. The approximation A=GBW/s can be used only for frequencies far above the first pole frequency (and is valid only for unity-gain compensated opamps with one single pole in the active region).