The filter is nonlinear and time invariant. The scaling property of linearity clearly fails since, scaling by gives the output signal , while . The filter is time invariant, however, because delaying by samples gives which is the same as .

The filter
is linear and *time varying*.
We can show linearity by setting the input to a linear combination of
two signals
, where
and
are constants:

Thus, scaling and superposition are verified. The filter is time-varying, however, since the time-shifted output is which is not the same as the filter applied to a time-shifted input ( ). Note that in applying the time-invariance test, we time-shift the input signal only, not the coefficients.

The filter
, where
is any constant, is *nonlinear*
and time-invariant, in general. The condition for time invariance is
satisfied (in a degenerate way) because a constant signal equals all
shifts of itself. The constant filter *is* technically linear,
however, for
, since
, even though the input
signal has no effect on the output signal at all.

Any filter of the form
is linear and
time-invariant. This is a special case of a *sliding linear
combination* (also called a *running weighted sum*, or
*moving average* when
).
All sliding linear combinations are linear,
and they are time-invariant as well when the coefficients (
) are constant with respect to time.

Sliding linear combinations may also include past *output*
samples as well (feedback terms). A simple example is any filter of
the form

Since linear combinations of linear combinations are linear combinations, we can use

If the input signal is now replaced by , which is delayed by samples, then the output is for , followed by

or for all and . This establishes that each output sample from the filter of Eq.(4.7) can be expressed as a time-invariant linear combination of present and past samples.

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Center for Computer Research in Music and Acoustics (CCRMA), Stanford University