ATOT: EEs, CEs, BMEs and otherwise: Simple Time Invariance question

[irishScott]

So for our exam tomorrow my professor gave us last year's exam and solution. I'm running through it now, and one question gives us a simple system and asks (among other things) if it's time invariant or not.

The System: y[n] = (x[n] + x[-n]) / 2

Now according to my logic, we first get the output for shifted input:
x[n-n0] --------> y2[n] = (x[n-n0] + x[-(n-n0)] / 2 = (x[n-n0] + x[-n+n0]) / 2

Then we shift the output: y[n-n0] = (x[n-n0] + x[-(n-n0)] / 2 = (x[n-n0] + x[-n+n0]) / 2

Since y[n-n0] = y2[n], the system is time invariant.

However according to the solution y[n-n0] actually = x[n-n0] + x[-n-n0], with != y2[n], therefore the system is not time invariant.

This makes no sense to me. When you write y[n-n0] you're simply substituting n-n0 for n, right? At least that's how every math class I've ever taken does it. I've solved other time invariance problems with no issue, but these are the prof-provided solutions to a major exam so I'm hesitant to question them. Just wondering what AT's opinion is.


Edit: Problem solved. Thanks for all your help!

 

[Matthiasa]

if y[n] = (x[n] + x[-n]) / 2
then
y[n-n0]= (x[n-n0] + x[-(n-n0)]) / 2
not
y[n-n0]=x[n-n0] + x[-n-n0]

Something was typed wrong.

 

[irishScott]

if y[n] = (x[n] + x[-n]) / 2
then
y[n-n0]= (x[n-n0] + x[-(n-n0)]) / 2
not
y[n-n0]=x[n-n0] + x[-n-n0]

Something was typed wrong.

Thing is, these are actual exam solutions for an exam given last year, and the class is over 100 people. Would think someone would have pointed this out if they were legitimately wrong.

Oh well, worse case scenario this part is only 3 points out of 100, hopefully if such a thing appears on tomorrow's exam it'll be of similar significance.

 

[TecHNooB]

it looks like its time varying. however, i do see your confusion. let me see if i can figure out how to show the work.

 

[irishScott]

Ahhh after consulting some of the textbook solutions (completely legitimately) I think I've got it.

To simplify things, let's just look at the signal y[n] = x[-n]

When stating x[n-n0], a sort of reverse-substitution takes place. It's not saying n = n-n0, it's saying (n = -n) - n0 = -n - n0. x[n - n0] is simply the notation for "shift the function by subtracting n0", it does not imply a literal interpretation of n = n - n0.

 

[TecHNooB]

Shifted output:
x(n) = f(n)
y(n - a) = [ x(n - a) + x(-(n - a)) ] / 2 = [ f(n - a) + f(-(n - a)) ] / 2

Shifted input:
x(n) = f(n - a)
y(n) = [ x(n) + x(-n) ] / 2 = [ f(n - a) + f(-n - a) ] / 2

 

[TuxDave]

Thing is, these are actual exam solutions for an exam given last year, and the class is over 100 people. Would think someone would have pointed this out if they were legitimately wrong.

Oh well, worse case scenario this part is only 3 points out of 100, hopefully if such a thing appears on tomorrow's exam it'll be of similar significance.

Well, the exam solutions are correct. That is a time variant.

If you shift the input (let's say by 1) and call that new input x2

x2[n] = x[n-1]
x2[-n] = x[-n-1] <-- where you made your mistake

You aren't substituting n for n-1, you're shifting the discrete time by one. The only way I can explain is just by a simple example.

For example if:

x[n] = ...-2,-1,0,1,2... w/ x[0] = 0
x2[n] SHOULD be ... -3,-2,-1,0,1 w/ x2[0] = -1

x2[-2] = x[-2-1] = x[-3] = -3 (correct!)
x2[-2] = x[-(2-1)] = x[-1] = -1 (incorrect!)

 

[TuxDave]

Ahhh after consulting some of the textbook solutions (completely legitimately) I think I've got it.

To simplify things, let's just look at the signal y[n] = x[-n]

When stating x[n-n0], a sort of reverse-substitution takes place. It's not saying n = n-n0, it's saying (n = -n) - n0 = -n - n0. x[n - n0] is simply the notation for "shift the function by subtracting n0", it does not imply a literal interpretation of n = n - n0.

Heh, I guess I have to give credit to my college professor who taught me linear systems about 12 years ago and haven't touched since. I hated that class so much too. 😛

 

[irishScott]

Got it, thanks for the help guys. :thumbsup:

 

[Matthiasa]

Ah right different maths. 🙁
Way to use to treating everything that goes into a function an entity.
I covered that junk so much to.

I'll just call it being to late to think.

 

[esun]

Just to add my 2 cents, this is a very simple problem without even thinking about the math behind it.

Really the major question is whether y[n] = x[-n] is time invariant (you know y[n] = x[n] is). So imagine a signal in your head. Something simple, maybe a delta function.

Now imagine shifting it right, then reversing it.

Now imagine reversing it, then shifting it right.

It should now be obvious that y[n] = x[-n] is not time invariant. By shifting first, you end up pushing the delta farther to the left (the impulse is at some negative n). By reversing first, you end up pushing it right.

 

[Matthiasa]

Well since it a time of day that I can sort of think I will say many real systems tend to be close enough to time invariant, within a certian operational range, and are treated as such due to it being a major pain in the @#$ to do otherwise.

Yes things due often change with temperature and breakdown with time, but that's what the datasheet of a product is suppose to help deal with.
To prevent that from being the case in cases where it is undesirable.