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I plotted the responses of two difference equation obtained from a Z transform transfer function. I used two methods and get two different results. Why? 

tf = y(output)/u(input) = z/ (z^2 - 3z + 3)

Method 1, using Matlab, taking the inverse Z transform

tf_difference = iztrans(tf, z, k);

yields: y = 2^k - 1, for timesteps 'k'. This is an exponential.

Method 2, algebraic rearranging by hand, turning z^n into (k-n):

y(k+2) = 3y(k+1) - 3y(k) + u(k+1)

or equivalently

y(k+1) = 3y(k) - 3y(k-1) + u(k)

Also, for the second method, i need to specify a 'u', whereas in the first method that's not needed, only k is needed. Why does the first method not need an input signal u, only a time step?

J B
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  • My best guess is that the first method determines the TF when the input is a unit impulse. That means that for the second method, `u(k) = 1` when `k = 0` and `u(k) = 0` otherwise. – rayryeng May 02 '14 at 19:09

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