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Here you can see that the transfer function applied to a cosine input will give you a sinusoid and a transient term:

$$ x(t) = \underbrace{(x(0) + x'(0))(2 e^{-t} - e^{-2t}) + \frac{2}{5} e^{-t} - \frac{1}{2}e^{-t}}_{{\rm goes\ to\ } 0 {\rm\ as\ } t \rightarrow \infty }\ \ \ + \frac{1}{10} \cos(t) + \frac{3}{10} \sin(t) $$

However, I don't understand how this can be, aren't exponentials eigenfunctions of LTI systems? so how come it can give an extra (transient) term?

At the same time it makes sense that it has a transient term from the CF of the function. How do I reconcile this?

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    $\begingroup$ It would also be good to either post an extract of the document showing the bit you are referring to, or at least say where in the doc you are looking. $\endgroup$ – lxop Mar 19 '13 at 20:21
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There is a transient because the input isn't just a cosine, but a causal cosine; that is, the input isn't a cosine from $-\infty \rightarrow \infty$, only from 0 onwards. If the input were actually just an infinite length cosine, then you would not get any transients.

I notice that the notes you are working from don't show that the input is causal. One reason I maintain that it actually is, is that otherwise you have an output that is tending to $\infty$ as $t \rightarrow -\infty$. And also you don't get transients without a change to the system; the change in this case is that the system starts at time $t=0$.

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  • $\begingroup$ Thank you. Yeah, he says that y(n-1)=...=y(0)=0. Does this mean that to calculate the output for just a coisne, you'll need to put the boundary condition terms in the RHS of the subsidiary equation and then multiply by the transfer function? And if so, and from the convention that Output(s)=H(s)Input(s), does that mean that these boundary termas could be considered as part of the input? $\endgroup$ – Guillermo Valle Pérez Mar 19 '13 at 21:11
  • $\begingroup$ Regarding $Y(s) = H(s)X(s)$, to operate on a complete cosine, you'll need to use the bilateral Laplace transform, which integrates from $-\infty\rightarrow\infty$, rather than $0\rightarrow\infty$. Also, this convention only holds true if the initial conditions are zero. $\endgroup$ – lxop Mar 19 '13 at 23:03
  • $\begingroup$ Yes I just read about the initial conditions having to be zero for using the transfer function. But this is weird to me then. To me this means that when people say that the complex exponential is an eigenfunction of H, they have to mean a causal exponential, because the initial conditions have to be zero! But we saw that the causal one isn't an eigenfunction as it gives transient terms! Basically, how can a complete cosine be an eigenfunction of H if H only cares about what happens after 0? $\endgroup$ – Guillermo Valle Pérez Mar 19 '13 at 23:12
  • $\begingroup$ It looks to me as if the derivation assumes non-zero initial conditions, which is where the transient response comes from. $\endgroup$ – Peter K.♦ Mar 20 '13 at 0:18
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    $\begingroup$ 'Complete' complex exponentials are eigenfunctions of LTI systems (see en.wikipedia.org/wiki/…), while a causal exponential (like you have here) will give transients. $\endgroup$ – lxop Mar 20 '13 at 1:13

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