# If exponentials are eigenfunctions, where does the transient term come from?

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?

• 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. – lxop Mar 19 '13 at 20:21

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 $t=0$.
• 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. – lxop Mar 19 '13 at 23:03