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I know that a transfer function has the general form of $$ G(S) = Y(S)/U(S) $$ where \$Y(S)\$ is the output and \$U(S)\$ is the input.

However, if given the output \$u(t)\$ and the steady state response \$y_{ss}(t)\$, is is still possible to obtain your transfer function?

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2 Answers 2

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Is there a way to find the transfer function from only your input and the steady state response?

Clearly, no. Steady state response means assentially the 0 frequency response. Obviously systems can have the same 0 frequency (DC) response but various responses to other frequencies.

For example, consider a simple R-C low pass filter. The DC response is simply 1 (output = input), but that is not true of higher frequencies. Different values of R and C will have different frequency profiles, but all have the same response to DC.

Or consider a simple R-C high pass filter. The DC response is 0, but obviously that is not true of other frequencies.

Added:

As Roger C mentioned in a comment, perhaps you meant steady state response to a fixed input waveform, not necessarily a steady (DC) one. If that is the case, it is possible to get a good idea of the overall transfer function if you do the test at lots of frequencies. However, a good idea of the transfer function is not the same as the actual transfer function, so the answer is still no in theory.

One way to look at this is that by putting in a particular pure frequency (sine signal) and measuring the steady state output (amplitude and phase), you measure one point of the transfer function in frequency space. By point-sampling the transfer function in frequency space, you can infer the continuous frequency response. The inverse Fourier transform of that gives you the impulse response, which is what you are looking for.

The problem with this method is that it only point-samples the freqency response, which never guarantees what the response is between the samples. If you know something about your system, then point sampling it at strategic frequencies could be good enough in practice, but this does not work in the general case. For example, it is quite common to measure a audio amplifier this way, since you know there aren't supposed to be sharp resonant peaks or the like in the frequency response.

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  • \$\begingroup\$ Actually "steady state" is not really the DC response but the output when transients have gone. For example, if at time t=0 a sinusoidal signal enters into a filter, the steady state output is a sinusoidal signal of the same frequency, but before reaching this steady state there will be a transient. \$\endgroup\$
    – Roger C.
    Commented Dec 11, 2014 at 19:50
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    \$\begingroup\$ @Roger: Yes, I see the OP could have meant that. I have updated my answer accordingly. \$\endgroup\$ Commented Dec 11, 2014 at 20:06
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No. You need the whole output response (which includes the transient part) to your input stimulus to be able to calculate the transfer function of your linear time-invariant system.

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