Z-transforms with initial conditions
To solve Nth order difference equation
with the non-zero initial conditions we need N initial conditions on output y(n) and M initial conditions on input x(n). Generally the input is applied suddenly (which means that it is stepped into the system) at n = 0, so that no initial conditions are required for it, i.e., x(n) = 0 for n < 0. The output y(n), however, generally will have non-zero initial conditions for n = -1 to -N.
Solving for y(n) for n ≥ 0, so that Y(z) = ?{y(n)} is one-sided z-transform. The difference equation contains other terms such as y(n-1), y(n-2), etc. which are delayed versions of y(n). Suppose N = 3, then we have y(n-1), y(n-2), and y(n-3) to deal with. The transform of y(n-1) is handled as shown below. 1st, for the sequence y(n) as shown below we define
Y+(z) = ?{y(n), n ≥ 0} = A + Bz -1 + Cz -2 + ...
Refer to this loosely as just Y(z) when there is no possibility of confusion.
We obtain y(n-1) by delaying the sequence by 2 unit, as shown below.
As it can be seen from the graph
In a similar way
and by extension ?{y(n-3), n ≥ 0} = z -3Y (z) + y(-3) + y(-2)z -1 + y(-1)z -2
For N = 3 this would be the last. But it can be generalize
As follows
In the case of the input x(n), as it is applied suddenly at n = 0, the initial conditions are zero, which means that, x(-1) = x(-2) = ...= x(-M) = 0, so that
?{x(n-k) u(n)} = z - k X+ (z)
We give below the mathematical derivation of z- transform of the delayed truncated sequence.
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