courteous
Junior Member level 1
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:
\[s[n] = \sum_{k=-\infty}^{\infty} x[n-k] h[k]= \sum_{k=-\infty}^{\infty} u[n-k] 2(3^k - 2^k)u[k] = 2 \left( \sum_{k=-\infty}^{\infty} 3^k u[k] u[n-k] - \sum_{k=-\infty}^{\infty} 2^k u[k] u[n-k] \right) = \text{?} \]
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:
\[s[n] = \sum_{k=-\infty}^{\infty} x[n-k] h[k]= \sum_{k=-\infty}^{\infty} u[n-k] 2(3^k - 2^k)u[k] = 2 \left( \sum_{k=-\infty}^{\infty} 3^k u[k] u[n-k] - \sum_{k=-\infty}^{\infty} 2^k u[k] u[n-k] \right) = \text{?} \]
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