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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?

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  • \$\begingroup\$ 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'? \$\endgroup\$ – laptop2d Apr 20 '18 at 15:27
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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.

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    \$\begingroup\$ You're welcome. I am happy to help. If I answered your question, then please mark my answer as such. Thanks! \$\endgroup\$ – PICyPICyPICy Apr 22 '18 at 18:33
  • \$\begingroup\$ @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! \$\endgroup\$ – Andy aka Apr 28 '18 at 17:55

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