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Homework Help: Input output relations in signals and systems

  1. Jul 12, 2011 #1
    1. The problem statement, all variables and given/known data

    I have a trouble understanding what would be the output when for example we say let's input the signal x(k*n) or x(n-n0) to the system...

    this has given me problems when having to solve systems for which I have to check the property of time invariance

    suppose we have a discrete system

    y[n] = x[2n]

    then, find what the system would be for the following inputs

    i. x[2n]
    ii. x[n-2]
    iii. x[n/3]
    iv. x[-n]
    v. x[-n+3]
    vi. x[-n/4 + 1]

    2. Relevant equations

    3. The attempt at a solution
    well the thing that makes it confusing is that x[n-3] or what ever is not y[n-3]... but rather something else..

    my solutions would be

    i. y[n] = x[4n]
    ii. y[n] = x[2n - 2]
    iii.y[n] = x[2/3 n]
    iv. y[n] = x[-n]
    v. y[n] = x[-2n + 3]
    vi. y[n] = x[-2*n/4 + 1]

    my questions:

    1. are my answers correct?
    2. is there any method on finding the answer using some formula?
    3. would these answers be the same for continuous time?

    thanks in advance
  2. jcsd
  3. Jul 12, 2011 #2


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    not 100% i understand the question but i would simplify rearranging with the x expression to find the related y index

    lets call
    y[n] = x[2n]

    first one is the same as the definition
    x[2n] = y[n]

    for the 2nd
    x[n-2] = x[2(n/2-1)]= y[(n/2-1)]

    what do you think?
  4. Jul 12, 2011 #3
    I'm not sure about it

    for the second one, my book has the same answer I gave.. but I can't understand why this is the answer..

    the y[n] = x[2n]

    is the definition of the system...

    now if I input a x[2n] won't it be y[n] = x[4n]? since, the system will take the input and make the x[2n] signal to be in a region which is 2 times less than the region of x[2n]
  5. Jul 12, 2011 #4


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    yeah, sorry i'm not totally getting it....

    can you explain exactly what the system, notation and indicies represent? also what other answers the book has, we may be able to work backwards

    otherwise i may have to pass this one onto someone else
    Last edited: Jul 12, 2011
  6. Jul 12, 2011 #5
    basically the system is

    y(t) = x(2t) where y(t) is the output and x(2t) is the input..

    and I'm trying to understand what the relation between the output would be for shifted inputs

    for example

    if the system is y(t) = x(t)

    then the relation in this system for input x(2t) would be y(t) = x(2t)

    while for x(t-1) it would be y(t) = x(t-1)

    now I saw an example of time invariance

    where the system y(t) = x(2t) is NOT time invariant...

    the explanation for this is that if we put an input of x(t-t0) we get an output of

    y(t) = x(2t - t0)

    the system would be time invariant if the same shifting for y would give us the same system

    but in this case

    y(t-t0) = x(2t - 2t0) <> x(2t - t0) hence the system is not time invariant...

    now what I don't get is, WHY, for input x(t-t0) we get the system y(t) = x(2t - t0) and not the system y(t) = x(2t - 2t0)...

    there is obviously a difference between "shifting the time for y" and "assuming a shifted time for input x" and this is the difference that I'm trying to understand..

    note, that for the examples above, I know the answer only for y[n] = x[2n] which is essentially what I just explained, and the same result is written in the book, the other examples are just what I thought of, when trying to figure out the result for different systems
  7. Jul 12, 2011 #6


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    Some of your statements appear to be contradictory - although I'm not necessarily sure I understand all of the notation.

    Do you have a few clear and fairly simple examples from your notes or book --- or a decent web site to look at?
  8. Jul 12, 2011 #7
    I asked a question some days ago


    and the user said

    hence it is time variant :P
    now, this is where it all started, and I figured out that maybe he was correct, but these gave me some more questions that had nothing to do with time invariance and that's why I posted this new question

    as you can see my question had to do with proving that the system y(t) = x(2t) is time variant

    I said

    we get an output of y2(t) = x2(t) = x1(2t - 2t0) = y1(t-t0)


    the output y2(t) is wrong, it should be y2(t) = x1(2t-t0)

    and this is what I don't understand, why it's wrong, why it is like that and not what I said before

    if you still have trouble understanding what I mean, it's ok, I will read my notes again and maybe I ll read something that will clear my thoughts
  9. Jul 12, 2011 #8
    ok I think after reading again the replies + notes I have this

    we have a system y(t) = x(2t)

    if we want to shift y by 1. for example we want y(t-1)

    it means that we want to shift it by 1 in the x axis..

    but in order to shift it by 1 in the x axis, we need to shift x(2t) too by 1 in the x axis..

    now, since the range of x(2t) in the x axis, is 2 time less than the range of x(t) in the x axis, it means that, if we want to shift the x(2t) by 1, we will need to shift x(2t) by 2..

    now this 2, is not the same range as in y, because y is just y(t) and x is x(2t)..

    it refers to the range of each step in x(2t).. which is obviously 2 times less than x(t) itself


    y(t-1) = x(2(t-1)) = x(2t - 2)


    having the system y(t) = x(2t)

    and if I say "let's input x(t-1) to this system

    we have that for input x2(t) = X(t-1)

    I get a y2(t) = X2(2t) = X(2t-1)

    and that's why it's not x(2t-2)...

    sorry for being stupid.. I think this is the correct explanation..

    hence for

    i. x[2n]
    ii. x[n-2]
    iii. x[n/3]
    iv. x[-n]
    v. x[-n+3]
    vi. x[-n/4 + 1]


    x1[n] = x[2n]

    hence y1[n] = x1[2n] = x[4n]


    x1[n] = x[n-2]

    hence y1[n] = x1[2n] = x[2n-2]


    x1[n] = x[n/3]

    hence y1[n] = x1[2n] = x[2n/3]


    x1[n] = x[-n]

    hence y1[n] = x1[2n] = x[-2n]


    x1[n] = x[-n+3]

    hence y1[n] = x1[2n] = x[-2n+3]


    x1[n] = x[-n/4 + 1]

    hence y1[n] = x1[2n] = x[-2n/4 + 1]
    Last edited: Jul 12, 2011
  10. Jul 12, 2011 #9


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    I think you can simplify that last one to x[-n/2 + 1] .

    I found a somewhat helpful example in http://en.wikipedia.org/wiki/Time_invariant#Simple_example".

    Basically it said, if you have the time variable outside of the argument, then it's time-variant -- as in: y(t) = t·x(2t).

    Then an input of x[2n] would give: y[n] = (2n)·x[4n] .

    I'll think about this spme more.
    Last edited by a moderator: Apr 26, 2017
  11. Jul 13, 2011 #10
    that was a good example thanks
    Last edited by a moderator: Apr 26, 2017
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