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I have a discrete-time system y[n] = x[n]*x[n-1]

I need to show that this system is non-linear by using a counter example rather than by disproving with algebra and the properties of additivity and homogeneity.

I think that the system is homogeneous although not additive, but I can't think of an example to show this. Would someone help me out please?

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  • $\begingroup$ Hint: try with constant signals. $\endgroup$ – MBaz Sep 30 '15 at 21:46
  • $\begingroup$ It's not even homogeneous. Really, nearly any two signals will work as a counter example. There's nothing complicated about constructing a counter example. Just take something and plug it in. $\endgroup$ – Jazzmaniac Sep 30 '15 at 22:03
  • $\begingroup$ Hmm when I tried to do the algebra to see if it was homogeneous I got a*x[n] -> a*y[n], I guess I screwed up somewhere. I'm really just not familiar with how this works. Say I take a unit impulse and put it in, then I get a signal of 0 out. Then I shift the unit impulse to the right and put it in and still get a signal of 0 out. I wish I had an example to get me started. $\endgroup$ – Austin Sep 30 '15 at 23:27
  • $\begingroup$ You are testing shift invariance with your example, not linearity. $\endgroup$ – Peter K. Sep 30 '15 at 23:43
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OK. Let's try homogeneity again: $$ y[n] = x[n]\cdot x[n-1] $$ for input $x[n]$.

For input $a\cdot x[n]$ we get $$ y'[n] = a\cdot x[n] \cdot a x[n-1] = a^2 \cdot x[n]\cdot x[n-1] \not = ay[n] $$ so homogeneity doesn't apply.

Let's try that with an example, as per MBaz's suggestion.

Let's try: $$ x[n] = 2 $$ so that $$ y[n] = 4 $$ Then let's try twice that: $$ x'[n] = 2\cdot x[n] = 4 $$ So that $$ y'[n] = 16 \not= 2\cdot y[n] $$

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  • $\begingroup$ Ah, I added instead of multiplying. Thank you I'm starting to get the examples now $\endgroup$ – Austin Oct 1 '15 at 1:27

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