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Lead Lag Network
This can be simplified to:
The 2 poles can be calculated if we solve the denominator which can we written as:
If we substitute the following:
we can write (3) as:
We can solve for s as:
Now to simplify further we need to see the relative magnitudes to decide what terms can be neglected. Comparing A and B we can clearly see that if the circuit time constants are lower than even a milli second A ≪ B. That is because A is a product of 2 time constants while B is a time constant. Usually the network is designed to have 1 time constant much smaller than the other. So consider the case when one time constant is 100 times larger than the other. For example
So from this point on the assumption and approximation that we take is:
One time constant in the circuit is more dominant than the other one.
So now we can approximate the poles now as:
So substituting the values of A and B in we get the final poles of the circuit as: