Suppose I have a system:

$$y[n] = x[n] + x[n-1] + x[0]$$

To see if this system is time-variant, we can compare $H\{x[n-1]\}$ to $y[n-1]$ and check that they are different.

I think the delayed (by 1 unit) response to $x[n]$ is:

$$y[n-1] = x[n-1] + x[n-2] + x[-1]$$

Nevertheless, I'm having an issue understanding why. How does the $x[0]$ term (from $y[n]$) become $x[-1]$ while calculating $y[n-1]$?

Mathematically, to calculate $y[n-1]$, I thought it would be reasonable to replace $n$ by $(n-1)$ in the equation, which would result in $y[n-1] = x[n-1] + x[n-2] + x[0]$. I mean, if $y[n]$ response contains a constant tern $x[0]$, I expect this value to remain in $y[n]$ independent of the $n$ index. Why that's not the case ?

I also have a problem with a different system: $$y[n] = x[2n]$$

I don't understand why this system presents a delayed response: $$y[n-1] = x[2n - 1]$$

(instead of $y[n-1]=x[2n-2]$) and why its response to the shifted (by 1 unit) input is $H\{x[2n - 2]\}$ instead of $H\{x[2n -1]\}$.

Looks like I have doubts about how to calculate the delayed response. Can you point out to problems in my examples to help clarify it?

  • 2
    $\begingroup$ You may find this answer useful. $\endgroup$ – Juancho Feb 5 '13 at 2:06
  • $\begingroup$ I changed your question from "time-invariant systems" to "how to calculate a delayed response". A "how to verify systems time-invariance" question already exists and is linked by @Juancho (and we avoid duplicates), but if you do not understand some parts of that answer, feel free to ask about that in the comments section of that question/answer. $\endgroup$ – penelope Feb 5 '13 at 10:40

The answer @Juancho liked to is indeed very useful. To me it looks like you don't understand time invariance fully, and that answer sould clarify things.

But, concerning your questions of getting the delayed response of the systems:

With the first system, $y[n] = x[n] + x[n-1] + x[0]$, you are right. The constant term will not change, since it is constantly $x[0]$, and thus not dependent of $n$. The correct delayed response is:

$$y[n-1] = x[n-1] + x[n-2] + x[0]$$

You can calculate that by substituting $n$ by $m-1$ (or $n-1$, it's the same which letter you use, but it's simpler for me to use a different letter while substituting). Now, whit $n=m-1$, you have:

$$y[n] = y[m-1] = x[(m-1)] + x[(m-1)-1] + x[0] = x[m-1] + x[m-2] + x[0]$$

When you get the final result (when you can't simplify any more), you can change the variable name from $m$ to $n$ again and get the first formula above.

With the second system, you employ the same logic to get the delayed response. Starting from:

$$y[n] = x[2n]$$

To get a delayed response you must substitute $n$ with $m-1$ and you get:

$$n = m-1$$ $$y[n] = y[m-1] = x[2 \times (m-1)] = x[2m - 2]$$

and now you can change the variable name again to get:

$$y[n-1] = x[2n - 2]$$.

These are explanations on how to calculate a delayed response (by 1). Looking at a response of a system delayed by 1 actually means that at time $n=5$, you will be looking at a response that you would normally get for $n=4$. The system is delayed, i.e. it behaves as the original, just with some time-delay.

To put it very simply, if I mark the delayed response by $y'[n]$, and the original by $y[n]$, this would hold:

$$y'[1] = y[0]$$ $$y'[2] = y[1]$$ $$y'[100] = y[99]$$

and more generally:

$$y'[n] = y[n-1]$$

For understanding time-invariance in systems, consult the linked question since that's precisely its topic.

To comment on the incorrect answers you proposed as delayed responses: they are actually responses to a delayed input.

What that means is that you will be feeding your original system with a delayed input. If your original input signal was $x[n]$, to get a response to a delayed input, you actually want to "feed" your system to a signal $x'[n] = x[n-1]$. Let's mark systems response to a delayed input with $y''[n]$ and calculate that for your first example:

$$ y''[n] = x'[n] + x'[n-1] + x'[0]$$

The system didn't change: It's still calculating the same operation: summing up the input at time $n$ with the input at time $n-1$ with the value of the input at time $0$. But, the input changed!. Substituting $x'[n] = x[n-1]$, you get:

$$ y''[n] = x[(n)-1] + x[(n-1)-1] + x[(0)-1] = x[n-1] + x[n-2] + x[-1]$$

To summarize, in order for a system to be time-invariant, the response of the system delayed by $k$ should be the same as the response of the system to the input delayed by $k$.

In this example, we were just examining delays by $1$. What proves that the system is not time-invariant in you case is the fact that $y'[n] \neq y''[n]$ because you have:

$$y'[n] = y[n-1] = x[n-1] + x[n-2] + x[0]$$ $$y''[n] = x'[n] + x'[n-1] + x'[0] = x[n-1] + x[n-2] + x[-1]$$

  • $\begingroup$ Now, i'm even more confused because s3.11 ( pag 13 ) in MIT's signal and systems asignament solution shows how the term x[0] changes to x[-N] in the case of a shift by N units of the system response to x[n] (y[n]) : ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/… Considering y[n] = x[2n] It also shows how the delayed ( by N ) response to x[n] equals y[n - N] = x[2n - N] instead of x[2n - 2N] and the response to the delayed(by N units) input is T{x[2n - 2N]} rather than T{x[2n - N]} like you guys(penelope and Juancho) mentioned. $\endgroup$ – user1843665 Feb 5 '13 at 13:40
  • $\begingroup$ @hnl I edited my answer with more explanations, so I hope it is clearer now. $\endgroup$ – penelope Feb 5 '13 at 14:38
  • $\begingroup$ You are absolutely right, i find it amusing that the assignment solution of a MIT's course is really wrong, but that's the reality. How can i delete my answear ? $\endgroup$ – user1843665 Feb 5 '13 at 16:00
  • $\begingroup$ @hnl you should have a "delete" link under all of your own posts -- both answers and questions (and an 'x' sign next to your comments) $\endgroup$ – penelope Feb 5 '13 at 16:09
  • $\begingroup$ I had to setup a password to be able to delete.I also can't voteup your answear until 15+ reputation :/ .Anyways, thanks for the hints, the x',y' notation you used, clarified the topic for me. $\endgroup$ – user1843665 Feb 5 '13 at 17:14

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