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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.

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  • $\begingroup$ as @MBaz has already stated, you should specify which type of filter you are considering. The definition of a filter is very broad and indeed every signal processing system can be described as a filter. But for example a very important class is named as frequency selective filters, possibly which is the one you are referring to. Also linear and nonlinear filters will have completely different properties. So, please specify which type it is. $\endgroup$ – Fat32 Mar 6 '17 at 23:25
  • $\begingroup$ I edited my post $\endgroup$ – luis Mar 7 '17 at 8:19
  • $\begingroup$ Assuming you are referring to a continuous time Bessel filter. What you consider is posibly a transient response casued by the initialization of the filter that will cease out proportional to the bandwdith of the lowpass filter. In order to help you better, put your calculated output oscillation as a mathematical expression so that it can be checked to see whether you made a computation error or an interpretation mistake... $\endgroup$ – Fat32 Mar 7 '17 at 10:15
  • $\begingroup$ you can find the calculations here link $\endgroup$ – luis Mar 7 '17 at 14:08
  • $\begingroup$ Ok from the link, (whose latter part after the EDIT section is a mess btw) it's obvious that you are referrring to the damped oscillations that represent the zero input response of the second order bessel filter. This is a normal consequence and should not be confused with a steady state output of a pure sine wave, which by definition cannot exist at the output of an LTI (a Bessel for example) if the input does not contain any. Consult to a Circuit Theory book for the detailed treatment of finding transient responses of Electrical RLC circuits. I suggest Nilsson's classic text. $\endgroup$ – Fat32 Mar 7 '17 at 14:53
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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.

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A linear filter can increase the magnitude of frequency spectra already present in a signal. Note that the Fourier Transform of any (reasonable) waveform or signal (including ones that do not look at all like they are oscillating) will decompose it into sinusoids (which do oscillate). Amplify any one frequency, and an oscillation may start to stand out and make it look like it just appeared; but a lower amplitude form of that spectra was already hidden in the input signal to start with.

It wasn't added. It was amplified. Which is possible with an LTI filter.

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A linear, time-invariant system such as a filter is incapable of producing frequencies that are not already present at its input.

One simple way to see this is that if the system's frequency response is $H(f)$, then for an input $X(f)$ the output is $Y(f)=X(f)H(f)$. Cleary, if $X(f_0)=0$, then $Y(f_0)=0$ for any frequency $f_0$.

Note that this is not true for linear time-variant systems. The classic example is the system with output $y(t)=x(t)\cos(2\pi f_0t)$, which does produce new frequencies, but is time-variant.

If you provide more details of the system you're studying, it may be possible to expand this answer.

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An IIR filter can produce oscillations due to quantization error in implementation. For example Limit-Cycle oscillations can occur under very special conditions of input and output, caused by quantizing the data after a feedback in a recursive loop, and can be in significant excess of the maximum quantization error which is not at all desirable. Another example is oscillations that can occur due to overflow causing a gross non-linearity.

Both of these effects are non-linear, and therefore create other frequencies in what would otherwise be a linear, time-invariant system as MBaz has pointed out.

For more detailed information see this reference:

http://www.dsp-book.narod.ru/DSPMW/03.PDF

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