Understanding Time Invariance in Signals

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

 

[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.

 

[kolycholy]

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.

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

 

[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

(i used unit step as an example)

 

[kolycholy]

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

(i used unit step as an example)

i am so sure it isn't time invariant, because the solution manual said so ...

 

[desA]

What is 'n'?

 

[kolycholy]

What is 'n'?

n is just time, but it assumes discrete value only

 

[desA]

So, you've answered your own question.

 

[kolycholy]

So, you've answered your own question.

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

 

[angel23]

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

 

[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.