Inferring exponential decay from difference equations?I'm trying to justify, why the graph of the following system: v ( n ) = 0.6 ⋅ v ( n − 1 ) p ( n ) = 0.13 ⋅ v ( n ) + 0.87 ⋅ p ( n − 1 ) + 25 with initial values v ( 0 ) ≈ 1441.67, p(0)=3000 and p(1)=3500,seems to show exponential decay for both v(n) and p(n) (after some initial growth).Justifying this for v(n) is easy: v ( n ) = 0.6 n v ( 0 ) ⟹ v ( n ) decays exponentially decays exponentiallybut what about p(n)?I can write for example p(3): p ( 3 ) = a ∗ b 3 ∗ v ( 0 ) + c ∗ [ a ∗ b 2 ∗ v ( 0 ) + c ∗ [ a ∗ b ∗ v ( 0 ) + c ∗ p ( 0 ) + 25 ] ] = a ∗ b 3 ∗ v ( 0 ) + c ∗ [ a ∗ b 2 ∗ v ( 0 ) + c ∗ a ∗ b ∗ v ( 0 ) + c 2 ∗ p ( 0 ) + c ∗ 25 ] = a ∗ b 3 ∗ v ( 0 ) + c ∗ a ∗ b 2 ∗ v ( 0 ) + c 2 ∗ a ∗ b 2 ∗ v ( 0 ) + c 3 ∗ p ( 0 ) + c 2 ∗ 25 where a = 13 / 100 , b = 60 / 100 , c = 87 / 100and one can see exponential terms, but the whole expression is too complicated in order to infer whether the decaying is "clean" exponential decay or whether it would exhibit some sort of other more complex curves.

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Inferring exponential decay from difference equations?I&#039;m trying to justify, why the graph of the following system:                    v        (        n        )        =        0.6        ⋅        v        (        n        −        1        )                            p        (        n        )        =        0.13        ⋅        v        (        n        )        +        0.87        ⋅        p        (        n        −        1        )        +        25            with initial values   v  (  0  )  ≈  1441.67, p(0)=3000 and p(1)=3500,seems to show exponential decay for both v(n) and p(n) (after some initial growth).Justifying this for v(n) is easy:  v  (  n  )  =      0.6    n    v  (  0  )    ⟹    v  (  n  )   decays exponentially decays exponentiallybut what about p(n)?I can write for example p(3):                    p        (        3        )        =        a        ∗                  b          3                ∗        v        (        0        )        +        c        ∗                              [                          a        ∗                  b          2                ∗        v        (        0        )        +        c        ∗        [        a        ∗        b        ∗        v        (        0        )        +        c        ∗        p        (        0        )        +        25        ]                              ]                                              =        a        ∗                  b          3                ∗        v        (        0        )        +        c        ∗                              [                          a        ∗                  b          2                ∗        v        (        0        )        +        c        ∗        a        ∗        b        ∗        v        (        0        )        +                  c          2                ∗        p        (        0        )        +        c        ∗        25                              ]                                              =        a        ∗                  b          3                ∗        v        (        0        )        +        c        ∗        a        ∗                  b          2                ∗        v        (        0        )        +                  c          2                ∗        a        ∗                  b          2                ∗        v        (        0        )        +                  c          3                ∗        p        (        0        )        +                  c          2                ∗        25            where   a  =  13      /    100  ,  b  =  60      /    100  ,  c  =  87      /    100and one can see exponential terms, but the whole expression is too complicated in order to infer whether the decaying is &quot;clean&quot; exponential decay or whether it would exhibit some sort of other more complex curves.

Ashlee Ramos · Accepted Answer

Use z-transform as  V  (  z  )  =  0.6      z          −      1        V  (  z  )  +  v  (  0  )    P  (  z  )  =  0.13  V  (  z  )  +  0.87      z          −      1        P  (  z  )  +  p  (  0  )  +  25to get a linear set of equations. Solving this linear equation yields:                    V        (        z        )                            =                              v            (            0            )                                1            −            0.6                          z                              −                1                                                                        P        (        z        )                            =                              p            (            0            )            +            25                                1            −            0.87                          z                              −                1                                                    +                              0.13            v            (            0            )                                (            1            −            0.87                          z                              −                1                                      )            (            1            −            0.6                          z                              −                1                                      )                              Now use the inverse z-transform to get:                    v        (        n        )                            =                  0.6          n                        v        (        0        )                            p        (        n        )                            =                  0.87          n                                  (          25          +          p          (          0          )          )                +        v        (        0        )                  (          0.419          ×                      0.87            n                    −          0.289          ×                      0.6            n                    )                    Addendum:You can also use a discrete-time dynamical system model for this set of equations to better understand its behavior. Here I would define:      x    n    =      (                            v          (          n          )                                      p          (          n          )                      )  and obtain a model in the form of       x    n    =  A      x          n      −      1        +  B      u    n  , where:  A  =      (                            0                          0.6                                      0.078                          0.87                      )    ,    B  =      (                            0                                      25                      )  now it is sufficient to find the (unit)step-response of this system.

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