Bumping in hope of an answer.
1. The problem
A causal LTI system is described by the difference equation .
Determine the
(a) homogeneous ( for all ),
(b) impulse,
(c) step
response of the system.
2. Relevant equations and definitions
2.1 Causal are systems for which the output depends only on the input samples , for
2.2 For the (a) part of the problem:
Equation is called the homogeneous difference equation and the homogeneous solution. The sequence is in fact a member of a family of solutions of the form
where the coefficients can be chosen to satisfy a set of auxiliary conditions on .
3. The attempt at a solution
First of all, there is no additional information to be gained from knowing that the LTI system is also "causal", right? That is, being causal ( only depends on , ) is already in the given equation, correct?
(a) homogeneous response, y_h[n]
The official solution is given in the form of Eq. :
I have no idea how one should arrive at this solution. Neither writing it out for particular 's (and hoping to glimpse a pattern), nor writing the system in the general form of and then finding the relevant coefficients (N=2: ; M=1: ), gives way toward the solution.
(b) impulse response, h[n]
I do get the official solution ( ), but I'm exploiting the Z-transform, which is introduced only after this chapter ... so there must be other way without Z-transform:
(c) step response, s[n]
As in the (a) part, I'm completely stuck here. The solution should be , which smells of Z-transform, but I'm also interested if/how could one do without it?
All I can think of for (c) is using the result from (b) (the impulse response), but I get stuck:
Think I've found the solution to (a), but I'm still stuck with (c).
(a) Since every complex number can be represented as exponential , all that is needed for a homogeneous response, is to solve the characteristic polynomial . Hence, is of the form .
(c) Can somebody help me out? Is there an easy way to finding the step response?
(c) The solution (the step response) in t-domain via evaluating the convolution sum:
Perhaps simpler to go directly in the z-domain:
Thanks for the help.