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I'd like to understand when an input signal is time varying or time invariant. Specifically, I have this function:

$y(n) = x(n) + 0.1 + (0.2)^n + x(n-3)$

I thought that this is time varying because of the $n$ in the exponent. Would anyone mind to clear this up for me? Is this time invariant or time varying?

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Signals are (almost) always time variant. Otherwise they would be very boring :-). I believe your question is about whether "systems" are time invariant or not. A system basically is a "prescription" of how to create an output signal from an input signal. The equation you show is a system equation of how to create output y[n] from input x[n].

The idea of time invariance is the following: The behavior of the system doesn't depend on absolute time, i.e. the relationship between input and output is the same, regardless of whether it's Tuesday or Friday. You can find a more mathematical description here http://en.wikipedia.org/wiki/Time-invariant_system

The easiest invariance test is the following:

  1. pick any input signal, typically $\delta [n]$ will do fine.
  2. Calculate the output $y{_{0}}[n]$ for that intput
  3. now delay the input signal by some amount.
  4. Calculate the output $y{_{1}}[n]$ for the delayed input
  5. If $y{_{1}}[n]$ is NOT simply a delayed version of $y{_{0}}[n]$ then the system time variant

Another (equivalent) test is the following

  1. For time invariant system, the system equations only dependency on time must be as linear index into the input signal

Either one will work.

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  • $\begingroup$ And just to add on to your answer, the system in question is not time-invariant, due to the $(0.2)^n$ term. Notably, it is also not linear, for the same reason (the $0.1$ constant term would also make it nonlinear). $\endgroup$ – Jason R Mar 2 '13 at 14:39
  • $\begingroup$ @JasonR: That constant offset is always a killer --- many people think "affine" and "linear" (in this context) are the same, when they are not. :-) $\endgroup$ – Peter K. Mar 3 '13 at 17:21

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