Time invariant

1. Oct 9, 2006

kolycholy

i usually have such a hard time determining whether a signal is time invariant or not ...

for example, why would x[-n] not be time-invariant?

please dont just tell me why x[-n] would not be time invariant ...
tell me techniques that I can apply to other signals too

2. Oct 9, 2006

angel23

look at the parameters beside your function if they contain a t term then your signal is time varient while if the parameters are constants then the signal is time invarient.

3. Oct 9, 2006

kolycholy

that makes sense ... but then tell me why x[-n] is not time invariant?

4. Oct 9, 2006

angel23

do you see any t terms beside the function??
it is time invarient. why r u sure it isn't time invarient?
you can use this site to see the graph for check. http://www.jhu.edu/~signals/sys/resulta939.html [Broken]

(i used unit step as an example)

Last edited by a moderator: May 2, 2017
5. Oct 9, 2006

kolycholy

i am so sure it isnt time invariant, because the solution manual said so ...

Last edited by a moderator: May 2, 2017
6. Oct 9, 2006

What is 'n'?

7. Oct 9, 2006

kolycholy

n is just time, but it assumes discrete value only

8. Oct 9, 2006

9. Oct 9, 2006

kolycholy

no i did not ... please enlighten me ...

10. Oct 9, 2006

angel23

i am sure it is time invarient my mind says so.

11. Sep 10, 2009

SicSemper

The key is in understanding the test for time invariance.

To test: x[n] > DELAY > x[n-n0] > SYSTEM > w[n]
|
>>> SYSTEM > y[n] > DELAY > y[n-n0]

w[n] and y[n-n0] are equal if the system is time invariant

in the case of y[n]=x[-n], for the top approach, delaying the system results in n-n0 then we apply the system's effect of reversing JUST n, so w[n]=x[-n-n0]. With the second path, we apply the system and get y[n]=x[-n] and then apply the delay to get y[n-n0]=x[-(n-n0)]=x[-n+n0].

Since x[-n-n0] is not the same as x[-n+n0] the system is time VARIANT.