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Discrete-Time Function

  1. Oct 5, 2015 #1
    Consider that:

    f(x(n)) = n*x(n)

    and that x(n) is 1 for n = 0,1,2,3 and 0 everywhere else.

    Thus, f(x(n)) will be for n = -5 to 5 like this:

    0 0 0 0 0 0 1 2 3 0 0

    Am I right?
  2. jcsd
  3. Oct 5, 2015 #2


    Staff: Mentor

    Looks fine to me. It might be simpler to write this:
    f(0) = 0
    f(1) = 1
    f(2) = 2
    f(3) = 3
    f(n) = 0 elsewhere
  4. Oct 5, 2015 #3
    Thanks for the answer. So, ok.

    Now, can I write that:

    f(x(n-m)) = (n-m)x(n-m) ?

    For example,

    f(x(n-15)) = (n-15)x(n-15)
  5. Oct 5, 2015 #4


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    That is improper notation.

    Probably what you have in mind is:

    Let x(n) = 1 for x in {0, 1, 2, 3} and 0 otherwise.
    Let f(n) = n*x(n).
  6. Oct 5, 2015 #5
    But I want to view this function as something that takes as input a function and returns another function.
    So, is like a discrete-time system. You apply a discrete-time waveform at its input and at its output you see another discrete-time waveform which is the result of the system acting on the input.

    Can't I do this?
  7. Oct 5, 2015 #6


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    Ahh, that makes more sense then. I do not want to weigh in on the proper notation for defining functions whose domain and co-domain are both sets of functions.
  8. Oct 5, 2015 #7
    So, the real reason why I made this post and my confusion.

    I can write x(n) = u(n) - u(n-4), where u(n) is the unit step function.

    f(x(n)) = f(u(n)-u(n-4)) = f(u(n)) - f(u(n-4)) = nu(n) - (n-4)u(n-4).

    And this function, plotted with wolfram alpha (x*unitstep(x)-(x-4)*unitstep(x-4)) gives the following result:

    0 0 0 0 0 0 1 2 3 4 4 4 4 ....

    So, this is my confusion. What am I doing wrong? That is a linear function and superposition principle applies.

    If, on the other hand, I am writing:

    f(x(n)) = n*x(n) = n*(u(n)-u(n-4))

    it gives the same result as in the first post.
  9. Oct 5, 2015 #8
    The problem contradicts itself, as follows:

    We have f(x(n))=n x(n) and x(n)=1 for n=0,1,2,3 and zero otherwise.

    Plugging in n=0, we have f(x(0))=0, so f(1)=0
    Plugging in n=1, we have f(x(1))=1.x(1), so f(1)=1

  10. Oct 5, 2015 #9
    Ok, I see the contradiction.

    So, I cannot write f(x(n)) = n*x(n).

    But I can write y(n) = n*x(n). So, the system takes as input some discrete-time waveform x(n) and produces another discrete-time waveform y(n).

    But, can you explain to me why I cannot write f(x(n)) = y(n) = n*x(n) ?

    If someone could shed light on this, I would be very happy. I'm thinking about this for days now.
  11. Oct 5, 2015 #10


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    That was where I was coming from in post #4. However, since the formal parameter "x(n)" on the left hand side is being treated as a function and since the formula on the right hand side is being treated as a function definition for the resulting function, the notation is not unreasonable and not contradictory.

    The intent is that it be read as:

    [ Let f be a function from the set of one-parameter functions to the set of one-parameter functions such that ] f(x(n)) = [the function y defined by y(n) =] n * x(n)

    We could then meaningfully ask about f(x)(n): the function y obtained by applying f to x, applied to argument n. This would produce an integer. e.g. f(x)(3) = 3.
    Or we could ask about f(x(n)), treating n as a dummy variable. This result would be a function and could be presented as a list of values as was done in the original post.
  12. Oct 5, 2015 #11
    What about my confusion on post #7?
  13. Oct 5, 2015 #12


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    The relevant definition being...
    I am in deep water here. Fighting to interpret the notation and understand your difficulty both. So bear with me while I try to parse this thing out.

    f(x(n)) = f( u(n) - u(n-4) )​

    So we have this operator f. It takes in a function x and spits out a function y. This first part of your assertion looks solid. It is mere definition.

    f( u(n) - u(n-4) ) = f(u(n)) - f(u(n-4))​

    This part looks dodgy. It seems to assume that f is linear. But let's skip past that.

    f(u(n)) - f(u(n-4)) = nu(n) - (n-4)u(n-4)​

    This part looks wrong. Try evaluating f(u(n-4)). You should not get (n-4)u(n-4)
  14. Oct 5, 2015 #13
    Here 's another illustration of the fact that you can't use the notation f(x(n)) = n x(n) to map the function x(n) to y(n):

    Evaluate f(1). The problem is that you can't tell whether n is 0,1,2 or 3. Each of these gives a different result.
  15. Oct 5, 2015 #14


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    If we interpret "1" as the constant function which takes all inputs to 1 and take f() as defined in the original post then f(1) is function g(x) = x * 1(x) = x. That is f(1) is the identity mapping.

    f(1)(1) = 1. f(1)(2) = 2, f(1)(n) = n
  16. Oct 5, 2015 #15
    But surely:
    if y(n) = n*x(n) then y(n-4) = (n-4)*x(n-4)

    I mean, view the function as a black box. If I input x(n), then I will get n*x(n). If I input instead at a later time, so I input x(n-4), then I should get (n-4)x(n-4).
    So, the moment of time where I input the function should not matter.

    You can view the waveforms on wolfram alpha.
    If you write unitstep(x) then you will see a unit step function. And to this it corrresponds x*unitstep(x).
    If I write unitstep(x-4), then this function starts at x=4. but is the same as unitstep(x). And to this it corresponds (x-4)*unitstep(x-4) which has the same form as x*unitstep(x), only delayed in time.

    But x*unitstep(x-4) has a very different form.
  17. Oct 6, 2015 #16


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    f, as you have claimed to define it is a function that transforms one function into a different function.
    y as you have defined it above is a function that transforms values into different values.

    Let's evaluate f(u(n-4)).

    u() is the unit step function. u(x) = 1 for x >= 0 and u(x) = 0 for x < 0.
    u(n-4) is the unit step function shifted right by 4. so u(n-4)(x) = 1 for x >= 4 and u(n-4)(x) = 1 for x < 4.

    Note that because f is a function that takes functions as input we must interpret its argument as a function and not as an integer value. That is why I have chosen to use the above notation -- to stress that "u(n-4)" is a function.

    f(u(n-4)) transforms this right-shifted step function by multiplying each of its computed results by n.

    f(u(n-4))(1) = 1*u(n-4)(1) = 1*u(1-4) = 0
    f(u(n-4))(2) = 2*u(n-4)(2) = 2*u(2-4) = 0
    f(u(n-4))(3) = 3*u(n-4)(3) = 3*u(3-4) = 0
    f(u(n-4))(4) = 4*u(n-4)(4) = 4*u(4-4) = 4
    f(u(n-4))(n) = n*u(n-4)(n) = n*u(n-4) = [n for n >= 4 and 0 for n < 4]
  18. Oct 7, 2015 #17
    Ok, I agree. I asked other people and they all say is n*u(n-4).

    Now, lets take it a step further and talk about linear, time-invariant systems.

    Consider that the response of the system to u(n) is s(n) = (1-a^(n+1))*u(n).

    Basically, is the same as n*u(n). I just replaced n with the exponential (1-a^(n+1)), where a is given.

    So to u(n) it corresponds s(n).

    Now if I want to find the response to x(n) = u(n) - u(n-4), I apply the superposition principle and do the following thing:

    - I calculate the output of the system separately: so to u(n) it corresponds s(n), and to u(n-4) it corresponds s(n-4)
    - I add the outputs to obtain y(n) = s(n)-s(n-4) = (1-a^(n+1))*u(n) - (1-a^(n-4+1))*u(n-4).

    So, observe that to u(n-4) it corresponds (1-a^(n-4+1))*u(n-4) and not (1-a^(n+1))*u(n-4).

    This is what happens in an LTI system. I can't say that to u(n-4) it corresponds (1-a^(n+1))*u(n-4) because if I take not 4, but 100, then at n = 100, when the step down happens, the exponential would have died and I don't want this. I don't want this because when the step goes down, I want it to go down exponential and not suddenly.

    So, what happens? Aren't linear, time-invariant systems described by functions but by other mathematical objects?
  19. Oct 7, 2015 #18
    I figured it out. y(n) = nx(n) is linear but not time-invariant. That is why it does not work what I was trying to do.

    For example, the RC circuit is LTI but the relationship between the input and output is not time-invariant.

    Thanks for your time.
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