# Implication of s -> infinity in a transfer function

For a strictly proper transfer function with no poles/zeros at infinity, I understand that it gives G(inf) = 0. What does it for a proper TF?

• Please rephrase your question, it is confusing. I hope you mean $T(\infty)$ , evaluating the frequency space at infinity has little meaning (that I know of). What do you mean by 'proper tf'? – laptop2d Apr 20 '18 at 15:27

For a transfer function with more finite poles than finite zeros, the gain goes to zero as the frequency increases without bound.

For a transfer function with an equal number of finite poles and finite zeros, the gain will be non-zero as the frequency increases without bound. This is given by the "D" term in the state space representation of the transfer function.

• You're welcome. I am happy to help. If I answered your question, then please mark my answer as such. Thanks! – PICyPICyPICy Apr 22 '18 at 18:33
• @user551397 May I remind you what PICyPICyPICy requested. The currency on SE.EE is upvotes and answer acceptance although the word "thanks" does have some value! – Andy aka Apr 28 '18 at 17:55