filtering - Does a filter add oscillations to a signal?

this is a more general version of a question which i asked previously.

From what I understand, the purpose of a filter is to change the amplitude of a specific frequency band.

As I went through the analytic calculation of a filtered (2nd order) signal I observed that the output signal contains an oscillating component (a sinusoid, with the frequency being the imaginary part of the complex conjugate pole pair).

A complex pole (in the left half plane) does correspond to a damped oscillation. I really cannot understand how the addition of a specific frequency to the signal conforms with the job of a filter. Thanks for your insights!

EDIT:

The filter I am considering is a second order low pass Bessel Filter. The input signal is an exponential decay. I tried to calculate the output signal and observed there was a exponentially decaying cosine involved with its frequency being the imaginary part of the two poles. It really surprised me since I did not expect an oscillation in a filtered signal, which it did not express before.

I do understand, that this sinusoid was present in the signal already before filtering, it just struck me that it appears in the filtered signal as a single term.

Answer

There are two unrelated phenomena that need to be understood in this context. First of all, as pointed out in the answers by hotpaw2 and by MBaz, an LTI system cannot add any frequency components to an input signal. This is obvious from the input-output relation in the frequency domain:

$$Y(\omega)=X(\omega)H(\omega)\tag{1}$$

where $Y(\omega)$ is the output spectrum, $X(\omega)$ is the input spectrum, and $H(\omega)$ is the filter's frequency response. Clearly, frequencies not contained in the input signal (i.e., frequencies for which $X(\omega)=0$ holds), cannot appear at the output because if $X(\omega_0)=0$ it follows that $Y(\omega_0)=0$ (assuming finite $H(\omega)$). However, this statement must be understood correctly, as explained below.

As you've observed, there can be oscillations in the output signal that do not appear to be present in the input signal. These are caused by one of two phenomena. The first one can be observed for systems with rational transfer functions having complex conjugate poles. A stable system with poles at $s_{\infty}=-\alpha\pm j\omega_0$ ($\alpha>0$) will generally contain an exponentially damped output term with frequency $\omega_0$, even if $X(\omega_0)=0$! Note that this is no contradiction with $(1)$ because a damped oscillation at $\omega_0$ in the output signal can even occur if $Y(\omega_0)=0$. The equation $Y(\omega_0)=0$ only means that a sinusoidal signal with frequency $\omega_0$ extending from $t=-\infty$ to $t=\infty$ cannot occur in the output signal. It does not say that there cannot be a right-sided exponentially damped sinusoidal signal at that frequency. The part of the output signal related to the system's poles (even with zero initial conditions) is called natural response (in contrast to forced response, the shape of which is determined by the input signal). If you want to read more (than you might want to know) about natural and forced response, check out this answer.

The second phenomenon is the Gibbs phenomenon, which is most clearly observed with ideal (low pass) filters that completely suppress certain higher frequencies of the input signal, and in this way cause oscillations that were seemingly not present in the input signal. However, those oscillations actually don't occur because you add frequencies but because you remove frequencies from the input signal. Think about the ripples in the impulse response of an ideal low pass filter (a sinc function), which are often clearly visible in the output signal, especially if the input signal is wideband, such as an impulse or an ideal step. A nice figure of the step response of an ideal low pass filter (a sine integral) can be seen here.