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If the difference equation that defines the DT system is,

$$z[n] = (1-x[n]) + (1-y[n]) $$

How can we determine in the system is LTI or not when:

  1. $x[n]$ is the input
  2. $y[n]$ is the input
  3. both are inputs

Using the definitions is somewhat vague in this case, so I am lost how to explicitly explain that they are not linear and time invariant for 1 and 2 at least.

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Most likely it is a homework question - what did do so far? Testing for linearity and time-invariance of linear system is pretty straightforward. – jojek Jun 24 '14 at 8:29
@jojek It is not a homework question, in fact, I have the answer. But, the answer is vague so I am asking this question to have a solid way to show that a system is LTI or not. – user29568 Jun 24 '14 at 9:28
When you write $ x[n] $ is the input you mean $ y[n] $ is zero or what? – Drazick Jun 24 '14 at 9:40
up vote 2 down vote accepted


If $z_1[n]$ is the response to input $x_1[n]$ (and $y[n]$ is NOT an input signal), and $z_2[n]$ is the response to $x_2[n]$, then it is straightforward to show that the response to a linear combination $ax_1[n]+bx_2[n]$ is NOT equal to $az_1[n]+bz_2[n]$, which would be a requirement for linearity. And due to symmetry, the same is the case if $y[n]$ is the input (and $x[n]$ is NOT an input signal). The test for time-invariance also fails in both of these cases, because only the input signal is shifted (and not the other nuisance signal).

When both $x[n]$ and $y[n]$ are input signals, the linearity test above still fails, so the system is non-linear (because it is an affine mapping due to the constants in the system equation). It is, however, time-invariant because if both input signals are shifted by some constant, the output is shifted by the same constant (which is not the case if only one of the two signals is an input signal).

share|improve this answer
Perfectly clear. Thank you :) – user29568 Jun 24 '14 at 11:14

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