Question

Consider the system described by the difference equation 
\[ y(n) = \frac{5}{6}y(n-1) - \frac{1}{6}(4-n) + x(n). \] Determine whether the system is linear and time-invariant (LTI). 
 

Hint:

If a difference equation contains a standalone term involving $n$ (like $4-n$), the system is time-varying. If zero input does not give zero output, the system is not linear.

Answer 1

Step 1: Check Linearity.
A system is linear if it satisfies the principle of superposition.
That is, if for inputs $x_1(n)$ and $x_2(n)$ with outputs $y_1(n)$ and $y_2(n)$, then:
\[ a x_1(n) + b x_2(n) \Rightarrow a y_1(n) + b y_2(n). \]
Given equation:
\[ y(n) = \frac{5}{6}y(n-1) - \frac{1}{6}(4-n) + x(n). \]
The term \[ -\frac{1}{6}(4-n) \] is independent of input $x(n)$.
This is an additive term not scaled by the input.
If $x(n)=0$, output is still:
\[ y(n) = \frac{5}{6}y(n-1) - \frac{1}{6}(4-n). \] Thus zero input does not produce zero output.
Hence the system is non-linear (strictly speaking, non-homogeneous).
Step 2: Check Time Invariance.
A system is time-invariant if shifting input by $n_0$ shifts output by same amount.
Here the explicit term: \[ (4-n) \] depends directly on $n$.
If we replace $n$ by $(n-n_0)$, the term becomes: \[ 4-(n-n_0) = 4-n+n_0, \] which is not a simple shifted version of original.
Thus the system explicitly depends on time index $n$.
Hence the system is time-varying.
Final Conclusion:
The system is
\[ \boxed{\text{Non-linear and Time-Varying}}. \]