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
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.