Question:medium
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).
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).
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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.
Updated On: Feb 15, 2026
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Solution and Explanation
Step 1: Understanding the Question
The task is to analyze a discrete-time system defined by a difference equation and determine if it possesses two fundamental properties: linearity and time-invariance.
Step 2: Key Formula or Approach
Linearity Check: A system is linear if it satisfies the superposition principle. A quick test for linearity is to check if zero input results in zero output (assuming zero initial conditions). If $x(n) = 0$ leads to $y(n) \neq 0$, the system is non-linear.
Time-Invariance Check: A system is time-invariant if its behavior does not change over time. In a difference equation, this means the coefficients should not be explicit functions of the time index 'n'.
Step 3: Detailed Explanation
Checking for Linearity:
The given difference equation is: \[ y(n) = \frac{5}{6}y(n-1) - \frac{1}{6}(4-n) + x(n) \] Let's check the zero-input response. If we set the input $x(n) = 0$, the equation becomes: \[ y(n) = \frac{5}{6}y(n-1) - \frac{1}{6}(4-n) \] The term $-\frac{1}{6}(4-n)$ is an extra term that is independent of the input $x(n)$. Because of this term, the output $y(n)$ is not zero even when the input $x(n)$ is zero. This violates the property of homogeneity, which is a necessary condition for linearity.
Therefore, the system is non-linear.
Checking for Time-Invariance:
Let's examine the coefficients in the difference equation. The term $-\frac{1}{6}(4-n)$ has a coefficient that explicitly depends on the time index 'n'.
Because the system's definition changes as 'n' changes, a time shift in the input will not result in an identical time shift in the output.
For instance, the term's value at time $n$ is $f(n) = -\frac{1}{6}(4-n)$. At time $n-n_0$, its value is $f(n-n_0) = -\frac{1}{6}(4-(n-n_0))$, which is not a simple shift of the original function.
Therefore, the system is time-varying.
Step 4: Final Answer
The system is both non-linear and time-varying.
\[ \boxed{\text{Non-linear and Time-Varying}} \]
The task is to analyze a discrete-time system defined by a difference equation and determine if it possesses two fundamental properties: linearity and time-invariance.
Step 2: Key Formula or Approach
Linearity Check: A system is linear if it satisfies the superposition principle. A quick test for linearity is to check if zero input results in zero output (assuming zero initial conditions). If $x(n) = 0$ leads to $y(n) \neq 0$, the system is non-linear.
Time-Invariance Check: A system is time-invariant if its behavior does not change over time. In a difference equation, this means the coefficients should not be explicit functions of the time index 'n'.
Step 3: Detailed Explanation
Checking for Linearity:
The given difference equation is: \[ y(n) = \frac{5}{6}y(n-1) - \frac{1}{6}(4-n) + x(n) \] Let's check the zero-input response. If we set the input $x(n) = 0$, the equation becomes: \[ y(n) = \frac{5}{6}y(n-1) - \frac{1}{6}(4-n) \] The term $-\frac{1}{6}(4-n)$ is an extra term that is independent of the input $x(n)$. Because of this term, the output $y(n)$ is not zero even when the input $x(n)$ is zero. This violates the property of homogeneity, which is a necessary condition for linearity.
Therefore, the system is non-linear.
Checking for Time-Invariance:
Let's examine the coefficients in the difference equation. The term $-\frac{1}{6}(4-n)$ has a coefficient that explicitly depends on the time index 'n'.
Because the system's definition changes as 'n' changes, a time shift in the input will not result in an identical time shift in the output.
For instance, the term's value at time $n$ is $f(n) = -\frac{1}{6}(4-n)$. At time $n-n_0$, its value is $f(n-n_0) = -\frac{1}{6}(4-(n-n_0))$, which is not a simple shift of the original function.
Therefore, the system is time-varying.
Step 4: Final Answer
The system is both non-linear and time-varying.
\[ \boxed{\text{Non-linear and Time-Varying}} \]
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