# 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