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Let's say that we have a LTI system (speaking strictly theoretical at the moment) and it has a transfer function H(s). Is this transfer function the same over all the operating points that the system can be in ?

For example a purely theoretical circuit, composed with ideal passive elements, has one input, one output and couple of nodes where we can set DC voltages. Will the output of that circuit be predictable with it's transfer function acquired with a certain set of DC voltages at it's nodes, and will that function be valid if I change the DC voltages on the nodes (implying a different operating point)?

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    \$\begingroup\$ An LTI system has a unique representation in terms of its transfer function. It is the same for all inputs. I am not sure what you mean by valid because validity is not a problem in theoretical setting. \$\endgroup\$ – Fraïssé Nov 21 '16 at 22:35
  • \$\begingroup\$ What do you meaning by 'setting' voltages at the nodes? \$\endgroup\$ – Chu Nov 22 '16 at 0:24
  • \$\begingroup\$ Yes, because it is linear. \$\endgroup\$ – mkeith Nov 22 '16 at 6:13
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If you're in purely theoretical circuit, than of course linear means linear. No matter what the actual value is. That is the definition of linear.

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Yes. This is the whole point of LTI systems. They utilize superposition. Consider this: $$V_{\mathrm{in}} H = (V_{\mathrm{DC}} + V) H = V_{\mathrm{DC}} H + V H$$

You can see how you can simply apply the superposition principle to get the final response of the system. The transfer function remains the same.

On the other hand, if the system is not linear, you cannot use the superposition principle. It is due to the fact that you can express the response of a linear system as: $$H = \frac{V_{\mathrm{out}}}{V_{\mathrm{in}}}$$ For a non-linear system you cannot reduce it as a fraction of this type. You get he input in the numerator for example.

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  • \$\begingroup\$ Thanks all for joining and answering me :-) I asked the question because I had a discussion with couple of colleagues at work and I found their arguments bothering me. At the end it turned out that we were talking about the same thing just using different words and terminology :-) Thanks a lot for the answers! \$\endgroup\$ – Aleks Dec 4 '16 at 21:13

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