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Finding the closed form of a recursive LTI system

  1. Jan 14, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the closed form of the impulse response of the system [itex]y[n] = 7y[n-1]-12y[n-2]+x[n][/itex] using the peel away and guess method. Ie, by using Python code to find the geometric ratios and amplitudes of the outputs as n grows large, then calculate residuals, and find the geometric ratios and amplitudes of the residuals, and so on.

    2. Relevant equations


    3. The attempt at a solution

    This is the code:

    Code (Text):
    memo = {}
    def f(n):
         if n <= 0: return 1.0
         if n == 1: return 7.0
         if n in memo: return memo[n]
         memo[n] = 7*f(n-1) - 12*f(n-2)
        return memo[n]
    # residual
    def g(n):
        return f(n) - 4*4**n

    for i in range(1,500):
        print i, ": ", f(i)
        # this closed form found from transfer function
        print i, ": ", 4*4**i - 3*3**i
    I was able to find an exact matching closed form for the system from performing partial fraction decomposition on the transfer function H(z) and then doing inverse Z-Transform on that...but I did this as a last resort.

    My problem was that when I check the ratio of the residuals, when n gets above 127, the ratios of the residuals don't stabilize, they oscillate.

    This code:

    Code (Text):
    for i in range(1,500):
        print i, ": ", g(i)/g(i-1)
    produces this output at large n:

    CTukJxO.png

    imgur link: https://i.imgur.com/CTukJxO.png

    I spent a long time trying to reconcile this in a closed form. As you can see, this pattern goes [itex]4*1.05^{-1}, 4*1.05^0, 4*1.05^1, 4*1.05^0,...[/itex]. This seemed very messy.

    Of course, the real closed form is simple enough when figured out from the transfer function, it's [itex]y[n]= 4*4^n - 3*3^n[/itex].

    And below n = 127, the ratio of the residuals seems to want to settle around 3, so guessing 3 at that point would be fine.

    Why are the residuals oscillating like that? Why doesn't it affect or appear in the actual closed form? How would one go about making a judgement on which value for the geometric ratio of the residuals to guess? If I had kept going with the later oscillating ratios of the residuals, wouldn't I have to introduce imaginary numbers?
     
  2. jcsd
  3. Jan 18, 2017 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    I'm unfamiliar with the "signals and systems" approach to difference equations. From glancing at the notes

    https://ocw.mit.edu/courses/electri...tems-fall-2011/readings/MIT6_003F11_chap2.pdf

    https://ocw.mit.edu/courses/electri...tems-fall-2011/readings/MIT6_003F11_chap4.pdf

    you are working exercise 23 of chapter 4.

    If I assume ##x[n] = 0## and consider the problem as a homogeneous linear recurrence relation with constant coefficients (https://en.wikipedia.org/wiki/Recur...currence_relations_with_constant_coefficients )

    Then we have the difference equation ## a[n] = 7 a[n-1] - 12 a[n-2] ##
    This has characteristic polynomial ## p(t) = t^2 - 7t + 12 = (t-4)(t-3) ##
    Which gives the general solution ## a[n] = k_14^n + k_23^n ##

    So I don't understand how the ## 4^i - 3^i## in your code reconciles with the your initial condition ## f[1] = 7 ##, which, in my notation would be ##a[1] = 7 ##. To get ##a[1] = 7##, I would use ## k_1 = k_2 = 1## instead of ##k_1 = 1, k_2 = -1 ##.
     
  4. Jan 18, 2017 #3

    Mark44

    Staff: Mentor

    Above, your equation is ##y[n] = 7y[n-1]-12y[n-2]+x[n]##, but your code below uses ##y[n] = 7y[n-1]-12y[n-2]##; i.e., no x[n] term. Why do you have this descrepancy?
    I believe the numbers you are seeing result from doing arithmetic on very large numbers, too large for the computer to represent exactly.
    For example, when i = 46, f(i) is approx. 1.98 X 10^28, and 4 * 4 ^ 46 is a 29-digit number. For larger i, the values get even larger. From about i = 121, the ratio of the residuals are probably at the limit of the computer's ability to do precise division, which, I believe, leads to the oscillation that you're seeing.
     
  5. Jan 18, 2017 #4

    Ibix

    User Avatar
    Science Advisor

    If Mark44 is correct, you might try setting your initial conditions to 1 and 7 rather than 1.0 and 7.0. Python has a built in big integer class that I think it will use if you do that.
     
  6. Jan 18, 2017 #5
    The x[n] is there as an impulse at time 0 and 1, at t = 0 the impulse is 1 and t = 1 the impulse is 7.

    Code (Text):
    if n <= 0: return 1.0
    if n == 1: return 7.0
     
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