# Thread: Causal LTI system: homogeneous-, impulse- and step-response

1. ## Causal LTI system: homogeneous-, impulse- and step-response

1. The problem
A causal LTI system is described by the difference equation $y[n] - 5y[n-1] + 6y[n-2] = 2x[n-1]$.

Determine the
(a) homogeneous ( $x[n]=0$ for all $n$),
(b) impulse,
(c) step
response of the system.

2. Relevant equations and definitions
2.1 Causal are systems for which the output $y[n_0]$ depends only on the input samples $x[n]$, for $n \leq n_0$

2.2 For the (a) part of the problem:
$\sum_{k=0}^{N} a_k y_h[n-k] = 0 \text{ ................... (2.96)}$
Equation $(2.96)$ is called the homogeneous difference equation and $y_h[n]$ the homogeneous solution. The sequence $y_h[n]$ is in fact a member of a family of solutions of the form
$y_h[n] = \sum_{m=1}^{N} A_m z_m^{n} \text{ ................... (2.97)}$
where the coefficients $A_m$ can be chosen to satisfy a set of auxiliary conditions on $y[n]$.

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 ( $y[n_0]$ only depends on $x[n]$, $n \leq n_0$) is already in the given equation, correct?

(a) homogeneous response, y_h[n]
The official solution is given in the form of Eq. $(2.97)$: $y_h[n] = A_1(2)^n + A_2(3)^n$

I have no idea how one should arrive at this solution. Neither writing it out for particular $n$'s (and hoping to glimpse a pattern), nor writing the system in the general form of $\sum_{k=0}^N a_k y[n-k] = \sum_{m=0}^M b_m x[n-m]$ and then finding the relevant coefficients (N=2: $a_0=1, a_1=-5, a_2=6$; M=1: $b_0=0, b_1=2$), gives way toward the solution.

(b) impulse response, h[n]
I do get the official solution ( $h[n] = 2(3^n - 2^n)u[n]$), but I'm exploiting the Z-transform, which is introduced only after this chapter ... so there must be other way without Z-transform:

$H(z) = \frac{Y(z)}{X(z)} = \frac{2z^{-1}}{1 - 5z^{-1} + 6z^{-2}} = 2\left(\frac{1}{1-3z^{-1}} - \frac{1}{1-2z^{-1}}\right) \Rightarrow h[n] = 2(3^n - 2^n)u[n]$

(c) step response, s[n]
As in the (a) part, I'm completely stuck here. The solution should be $s[n] = \left( -8(2)^{n-1} + 9(3)^{n-1} + 1 \right) u[n]$, 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:
$s[n] = \sum_{k=-\infty}^{\infty} x[n-k] h[k]= \sum_{k=-\infty}^{\infty} u[n-k] 2(3^k - 2^k)u[k] = 2 \left( \sum_{k=-\infty}^{\infty} 3^k u[k] u[n-k] - \sum_{k=-\infty}^{\infty} 2^k u[k] u[n-k] \right) = \text{?}$

2. ## Re: Causal LTI system: homogeneous-, impulse- and step-response

Bumping in hope of an answer.

3. ## Re: Causal LTI system: homogeneous-, impulse- and step-response

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 $A e^{st}$, all that is needed for a homogeneous response, is to solve the characteristic polynomial $s^2 - 5s + 6 = (s-3)(s-2) = 0$. Hence, $y_h$ is of the form $A_1 (2)^n + A_2(3)^n$.

(c) Can somebody help me out? Is there an easy way to finding the step response?

4. ## Re: Causal LTI system: homogeneous-, impulse- and step-response

(c) The solution (the step response) in t-domain via evaluating the convolution sum:
$s[n] = \sum_{k=-\infty}^{\infty} x[n-k] h[k]= \sum_{k=-\infty}^{\infty} u[n-k] 2(3^k - 2^k)u[k] = 2 \sum_{k=0}^{n} (3^k - 2^k) = 2( \frac{1-3^{n+1}}{1 - 3} - \frac{1-2^{n+1}}{1-2} ) = 3^{n+1} - 2^{n+2} + 1$

Perhaps simpler to go directly in the z-domain: $S(z)=H(z)U(z)=...$

Thanks for the help.