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I wonder if the following acoustic system is i) considered to be linear time-invariant, and ii) if it is fully described by it's frequency response?

A speaker produces waves at the input location of a duct system with a branch and the output is measured after the branch point. The anechoic terminations completely absorb the sound waves, and it is assumed that only 1D low frequency waves exist.

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So the system, when subject to the input $$ x(t) = e^{j \omega t}, \quad t \ge 0 $$

experiences the output $$ y(t) = \left\{ \begin{array}{l l} 0 & \quad 0 \le t < t_0 \\ A e^{j \omega t} & \quad \text{$t_0 \le t<t_1$}\\ A e^{j \omega t} + B e^{j \omega t}& \quad \text{$t \ge t_1$} \end{array} \right., $$

where A and B are complex constants depending on the duct geometry, and $t_1$is the delay time due to the side branch geometry.

Yes the system itself is not changing with time, however due to the finite speed of sound there is a transient period before the reflection from a side branch has occurred.

The frequency response at $\omega$ (a complex number capturing only magnitude and phase relationship between input and output) would be related to the final part of the piecewise defined output (steady state part) and offer no information about the time $t_1$. So, for example, a second system which had a longer side branch may have the same frequency response yet a different value of $t_1$. Therefore my argument would be that the FR is inadequate to characterise the transient behaviour of the system. Is this correct?

xyz
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No, if your system can (with sufficient accuracy) be modeled as a linear time-invariant system, then it is fully characterized by its impulse response. Since the frequency response is the Fourier transform of the impulse response, the system is also fully characterized by its frequency response. The frequency response does not only describe steady-state behavior but also transients (because it simply is a full description of the system). See also this answer to a related question.

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Matt L.
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  • "No" meaning no it is not an LTI system? Or that my reasoning is not correct? I am aware that the FR describes transients, but not these specific transients. The previous question/answer does not deal with this – xyz Apr 07 '14 at 07:20
  • "No" to your last question ("is this correct?"). Why would the FR describe transients in general but "not these specific transients"? What's special about them? – Matt L. Apr 07 '14 at 07:22
  • Piece wise defined output, two separate systems have the same input/output phase and magnitude relation at steady state – xyz Apr 07 '14 at 07:23
  • That's just a kind of echo, isn't it? – Matt L. Apr 07 '14 at 07:24
  • Yes. My apologies that this is probably a very basic question that I am having trouble understanding. But for the said example then how could the FR capture an echo? – xyz Apr 07 '14 at 07:28
  • Do you know what a delta impulse $\delta(t)$ is? If so, consider the impulse response $h(t)=\delta(t)+a\delta(t-T)$. This is a system with output $x(t)+ax(t-T)$, where $x(t)$ is the input signal, $a$ is some real-valued constant, and $T$ is a (positive) delay time. This is a very simple LTI system, which can of course also be described by its frequency response. – Matt L. Apr 07 '14 at 07:31
  • Yep I know the Dirac delta, but can you point me towards where I can find the FR of the simple example you just gave? – xyz Apr 07 '14 at 07:37
  • Fourier transform table: $H(j\omega)=1+ae^{-j\omega T}$ – Matt L. Apr 07 '14 at 07:40
  • Does that not imply that the same $H(j \omega)$ is shared by multiple unique systems (i.e. delays $T$, $T+2 \pi/ \omega$, $T+4 \pi/ \omega$...) and therefore cannot be recreated when taken back to the time domain? – xyz Apr 07 '14 at 14:14
  • $\omega$ is not a constant but the independent frequency variable. There is only one impulse response corresponding to this frequency response. – Matt L. Apr 07 '14 at 14:33
  • Two systems can have the same complex value of their frequency responses for any specific frequency $\omega_0$, but still their impulse responses can uniquely be obtained from their respective frequency responses. No information is lost. You just take the inverse Fourier transform of both frequency responses to get the impulse responses. Look at it like this: if two functions share a finite number of values, are they necessarily the same? (The answer is "no".) – Matt L. Apr 07 '14 at 14:43
  • Ahhh ok I think that I am starting to grasp my misunderstanding - that the time delay cannot be a function of $\omega$. I still need to think about it more, but thanks heaps. – xyz Apr 07 '14 at 14:43
  • If you think that the answer (including the comments) helped you solve your problem, please accept it (by hitting the button next to it), so other users know that the question has been answered satisfactorily. If not, no problem, you might want to wait for other answers. – Matt L. Apr 07 '14 at 14:45