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Homework Help: Please I need help on this problem, I can't figure out what to do

  1. Dec 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Let e(t), for t = 0,±1,±2, . . ., be a realization of an IID sequence of zero-mean random
    variables with variance σe^2, and let v(t) be the output of the filter:
    v(t) = (B(q)/A(q) )e(t)
    where B(q) and A(q) are polynomials in the forward shift operator q (i.e. qv(t) = v(t + 1)),
    given by:

    B(q) = 1 +Ʃ(from k=1 to m) bk*q^-k ---- bk →b subscript k
    A(q) = 1 +Ʃ(from k=1 to n ) ak*q^-k -----ak →a subscript k

    and the solutions to A(z) = 0 and B(z) are all inside the unit circle.
    (a) Let x(t) be the output of the following filter:
    x(t) = ((A(q) − B(q))/B(q)) v(t)
    (b) Write the relationship between x(t) and v(t) as a difference equation. What is the
    coefficient of v(t) (i.e. at zero delay)? Is this filter stable?
    (c) Compute the expected value E[x(t − k)e(t − k)], for k = 0, 1, 2, . . ..
    (d) Let y(t) be defined as:
    y(t) = (B(q)/A(q))x(t)
    Can this signal be used as a “one-step†ahead prediction of v(t)? Justify your answer.
    (e) Compute the expected value and autocorrelation of v(t) − y(t).


    2. Relevant equations



    3. The attempt at a solution[/b
     
    Last edited: Dec 21, 2013
  2. jcsd
  3. Dec 20, 2013 #2

    haruspex

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    You need to provide some attempt at a solution, or at the least an explanation of where you're stuck.
     
  4. Dec 20, 2013 #3
    Thank you Haruspex.
    I did substituted the A(q) and B(q) in to the x(t) equation in order to find the relationship between the x(t) and v(t) to get the coefficient of v(t). Although am not sure if that is right or not hence, i could not tell if the filter is stable or not. I could not compute the expected value either.

    I am actually got stock in b.
     
  5. Dec 20, 2013 #4

    haruspex

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    Then please post the working you have done.
     
  6. Dec 20, 2013 #5

    berkeman

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    Check your PMs. You *must* show your efforts toward the solution when you post questions here.
     
  7. Dec 21, 2013 #6
    x(t) = [1 + Æ©(from k= 1 to n) akq-k -(1 + Æ©(k = 1 to m)bkq-k] * v(t)]/ 1 + [k=1]\sum[m] bkq-k

    the relationship between x(t) and v(t) is therefore

    [q-1x(a-b)]\[1 + b*q-1] + [q-2x(a2-b2)]\[1 + b2*q-2] +.....+ [q-nx(an-bm)]\[1 + bm*q-m]

    Thank you.
     
  8. Dec 21, 2013 #7

    haruspex

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    There are several things I don't understand in that last line.
    First, there's no equals sign, so how does it express a relationship?
    Second, there's no mention of v.
    Third, what do the backslash symbols (\) represent?
    Fourth, how can you use ellipsis (...) between a term that has a2 and b2 and a term that has an and bm? What happens when m and n are different?
     
  9. Dec 21, 2013 #8

    x(t)/v(t) =

    [q-1x(a-b)]/[1 + b*q-1] + [q-2x(a2-b2)]/[1 + b2*q-2] +.....+ [q-nx(an-bm)]/[1 + bm*q-m]

    i used the ellipsis because the sup and sub are equal i.e k, and tend to m and n.
     
  10. Dec 21, 2013 #9

    haruspex

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    Yes, but if each goes up one at a time then they won't reach m and n together.
    To clarify the difficulty, try writing that out in full with m = 2, n = 1.
     
  11. Dec 21, 2013 #10
    This is what i am doing for this question :

    (b) Write the relationship between x(t) and v(t) as a difference equation. What is the
    coefficient of v(t) (i.e. at zero delay)? Is this filter stable?

    x(t) = ((A(q) − B(q))/B(q)) v(t)

    x(t) * B(q) = (A(q) - B(q) ) * v(t)

    bk *x(t-k) = ak*v(t-k) - bk*v(t-k)

    for zero delay ----- k=zero.. substitute into the equation
    b0 *x(t) = a0 * v(t) - b0 *v(t)
    therefore the coef. of v(t) be ..... ( a0 - b0 )

    b0 * x(t) = v(t) (a0 - b0)

    is that correct ?
     
  12. Dec 21, 2013 #11

    haruspex

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    I feel the LHS should be written B(q) * x(t). B(q) is an operator, so the order matters.
    But a0 = b0 = 1. Does that make things interesting?
    And I would have thought the coefficient would be of the form x(t) = (coefficient)*v(t), so it would be a0/b0 - 1.
     
  13. Dec 21, 2013 #12
    Thank you for the suggestions, i believe the coefficient of v(t) = 0 if a0 = b0 =1. i.e at zero delay. Any suggestion on how to go about with the c? thank you.
     
  14. Dec 22, 2013 #13
    That sounds about right Steff. Now lets move to part c:

    (c) Compute the expected value E[x(t − k)e(t − k)], for k = 0, 1, 2, . . ..

    x(t) = ((A(q) − B(q))/B(q)) v(t)
    where
    v(t) = (B(q)/A(q) )e(t)
    substituting v(t) into x(t) we get
    x(t) = ((A(q) − B(q))/B(q))*(B(q)/A(q) )e(t) ---> x(t) = (((A(q) − B(q))/A(q))*e(t)
    with a time shift to x(t) ---> x(t-k) the equation-> x(t-k) = (((A(q) − B(q))/A(q))*e(t-k)

    Now for: e(t) with a time shift to e(t)---> e(t-k)

    now we got both x(t-k) and e(t-k) ,,, multiply them together to get
    x(t-k) = (((A(q) − B(q))/A(q))*e(t-k)
    e(t-k)
    we obtain:
    x(t-k)e(t-k)= (((A(q) − B(q))/A(q))*e(t-k) *e(t-k) ----->(((A(q) − B(q))/A(q))*e^2(t-k)

    take the expected value :
    E[x(t-k)e(t-k)]=((A(q) − B(q))/A(q))*E[e^2(t-k)]
    and E[e^2(t-k)] = σe^2 since the mean is zero ----- it is equal to σe^2

    So the result is E[x(t-k)e(t-k)]=((A(q) − B(q))/A(q))* σe^2

    check this, this is what I am doing for this question.
    let me know if its correct
     
  15. Dec 22, 2013 #14

    haruspex

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    No, the A(q) etc. are operators. The '*' between (((A(q) − B(q))/A(q)) and e(t-k) does not represent ordinary multiplication, so you can't apply associativity like that.
    Let's start with something a bit simpler: can you evaluate E[e(t) q-1 e(t)]?
    Hint:
    e(t) and e(t-1) are independent
     
  16. Dec 22, 2013 #15
    Okay, can you evaluate E[e(t) q-1 e(t)]?
    with your hint : e(t) and e(t-1) are independent
    then E[e(t)]E[e(t-1)] ---> and that is zero since the mean is zero.
     
  17. Dec 22, 2013 #16
    I am not sure if I got that right or not. can you give me more hints if its not correct
     
  18. Dec 22, 2013 #17

    haruspex

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    That's right.
    When i first saw this question I was a bit non-plussed by the notion of dividing by an operator. So I did a bit of reading and it seems that x(t-k) = ((A(q) − B(q))/A(q))*e(t-k) means A(q)x(t-k) = (A(q) − B(q))*e(t-k), no more, no less. So I would favour working with that form.
    So consider E[e(t-k) A(q) e(t-k)]. Can you evaluate that now?
     
  19. Dec 22, 2013 #18
    Same it would equal to zero
    as
    E[e(t-k) A(q) e(t-k)] = A(q)E[e(t-k)]E[(e(t-k)] ----> and E[e(t-k)] is zero.
     
  20. Dec 22, 2013 #19

    haruspex

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    No. E[e(t-k) e(t-k)] is not zero. As you posted earlier it would be σe2.
    What in general will E[e(t-k)e(t-j)] evaluate to?
    In E[e(t-k) A(q) e(t-k)], expand A(q).
     
  21. Dec 22, 2013 #20
    Perfect I see it now after expanding A(q) I would get σe2

    What in general will E[e(t-k)e(t-j)] evaluate to?

    zero for all j not equal to k

    but if j=k then σe2
     
  22. Dec 23, 2013 #21
    so I get for part c ---> answer is zero
     
  23. Dec 23, 2013 #22

    haruspex

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    How do you get that? (I'm not saying it's wrong - I genuinely don't know.)

    Btw, it occurred to me that another approach is to consider that polynomials can be factorised. E.g. A(q) could be written as âˆ(1-ωiq-1). We're told all the roots ωi satisfy |ωi| < 1. As a result, A-1(q) = âˆiÆ©n(-ωiq-1)n.
     
  24. Dec 23, 2013 #23
    this is the final equation I got for :
    Need to compute E[x(t-k)e(t-k)]
    x(t) = (SUM(ak*e(t-k)) - (SUM(bk*e(t-k) - (SUM(ak*x(t-k))

    multiply x(t) by e(t):
    x(t)e(t)= (SUM(ak*e(t-k)e(t)) - (SUM(bk*e(t-k)e(t)) - (SUM(ak*x(t-k)x(t))
    then shift by j for both (x) and (e)
    x(t-j)e(t-j)= (SUM(ak*e(t-k-j)e(t-j)) - (SUM(bk*e(t-k-j)e(t-j)) - (SUM(ak*x(t-k-j)x(t-j))
    take the expected value:
    E[x(t-j)e(t-j)]= E[(SUM(ak*e(t-k-j)e(t-j))] - E[(SUM(bk*e(t-k-j)e(t-j))] - E[(SUM(ak*x(t-k-j)e(t-j))]
    lets look at each expectation by it self:
    E[(SUM(ak*e(t-k-j)e(t-j))] = zero since k is never zero ...the shift difference between e(t-k-j)e(t-j)) is never the same
    E[(SUM(bk*e(t-k-j)e(t-j))] similar to the above
    E[(SUM(ak*e(t-k-j)e(t-j))] Since x(t) doesn't contain current values of e(t) it only contains past values . therefore for example E[e(t-1-j)e(t-j)] is zero.

    so the result of the E[x(t-j)e(t-j)]= zero.
     
  25. Dec 23, 2013 #24
    (d) Let y(t) be defined as:
    y(t) = (B(q)/A(q))x(t)
    Can this signal be used as a ONE STEP ahead prediction of v(t)? Justify your answer

    I got that

    can you give me hints for

    part c
     
  26. Dec 23, 2013 #25

    haruspex

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    I.e. x(t) = (A-B)e(t-k) + Ax(t-k)?
    How do you arrive at that? I get:
    Av(t) = Be(t)
    Bx(t) = Av(t) - Bv(t) = Be(t) - (B/A)e(t)
    To get a mix of t and t-k I would need to introduce a qk somewhere.
     
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