The Z-transform of delayed truncated sequence The one-sided z-transform of

x(n) is given by

Given sequence x(n), we delay it by k units, and then truncate it to left of n = 0 to get x(n-k) u(n). We want to find the z-transform of x(n-k) u(n).

If we let n-k = r, then n = r+k, and summation limits n = 0 to ∞ become r = -k to ∞. Then

Refer to X+(z) as X(z) and write the result as

The result shown above is used to solve linear constant coefficient difference equations with inputs that are stepped into a system. Suppose that we want solution of

subject to initial conditions

{y(i), i = -1, -2, ..., -N}         and      {x(i), i = -1, -2, ..., -M}

We take z-transform of the equation using the result derived above for delayed-truncated sequences

here we have used Z to mean ? the z-transformation operation. The left hand side is given by

(In terms of the derivation earlier all the Y(z)'s are Y+(z)'s, that is , one-sided transforms). All the Y(z) terms are grouped together under a summation, and all the remaining terms, because of the initial conditions {y(i), i = -1, -2, ..., -N}, can be grouped together such that the above can be written as

By following a similar procedure the right hand side can also be written as follows (here again the X(z)'s are X+(z)'s, i.e. , one-sided transforms):

LHS = RHS becomes

To summarize: to solve for y(n) we take the z-transform of linear constant coefficient difference equation by using initial conditions, manipulate in the z-domain to get Y(z) and then take the inverse z-transform of Y(z) to obtain y(n).

Example Find the solution to

with initial conditions y(-1) = 4, y(-2) = 10.

Solution There are 3 methods of solution:

1.   Find iterative solution in discrete-time domain. Generally this will not give an analytical form of solution.

2.   Solve in discrete-time domain (homogeneous solution + particular solution).

3.   Solve in frequency domain as shown below.

For the input sequence x(n) which is stepped into a system, specified in words such as x(n) = 0 for n < 0, the initial conditions are zero and do not matter. But for the output sequence y(n) where the initial conditions y(-1), y(-2) are explicitly given to be non-zero we are required to use the above derived "z-transform for the delayed truncated sequence". Particularly we have

 
The time-domain solution was covered in HW. The solution is repeated below

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