Question

A discrete-time system has an input-output relationship: \( y[n]=\begin{cases} x[n], & n \ge 1 \\ 0, & n=0 \\ x[n+1], & n \le -1 \end{cases} \). The given system may have the properties: A. Linearity, B. Time-invariance, C. Causality, D. Stability. Choose the correct properties.

• A. A and D only
• B. A and B only
• C. A, B and C only
• D. A, B, C, D

Hint:

To check system properties:

\textbf{Causality:} Does \(y[n]\) ever depend on \(x[k]\) for \(k>n\)? If yes, not causal.
\textbf{Time-Invariance:} Is the response to \(x[n-d]\) equal to \(y[n-d]\)? If no, not time-invariant.
\textbf{Stability:} If \(|x[n]|\) is always finite, is \(|y[n]|\) also always finite? If yes, it's stable.

Answer 1

The Correct Option is A

Step 1: Test for Linearity.
A system is linear if it satisfies superposition. Let \( y_1[n] \) be the output for input \( x_1[n] \) and \( y_2[n] \) for \( x_2[n] \). For an input \( x[n] = a x_1[n] + b x_2[n] \), the output is \( y[n] \).
For \( n \ge 1 \), \( y[n] = x[n] = a x_1[n] + b x_2[n] = a y_1[n] + b y_2[n] \).
For \( n \le -1 \), \( y[n] = x[n+1] = a x_1[n+1] + b x_2[n+1] = a y_1[n] + b y_2[n] \).
The system is composed of linear operations on the input, so it is linear.

Step 2: Test for Time-Invariance.
A system is time-invariant if a shift in the input causes an identical shift in the output. Let \( y'[n] \) be the response to \( x[n-d] \). We must check if \( y'[n] = y[n-d] \).
Let's find the output for \( x[n-1] \) (i.e., d=1): \( y'[n] = \begin{cases} x[n-1], & n \ge 1 \\ 0, & n=0 \\ x[n], & n \le -1 \end{cases} \).
Now let's find the shifted original output \( y[n-1] \): Replace \( n \) with \( n-1 \) in the definition of \( y[n] \).
\( y[n-1] = \begin{cases} x[n-1], & n-1 \ge 1 \implies n \ge 2 \\ 0, & n-1=0 \implies n=1 \\ x[n], & n-1 \le -1 \implies n \le 0 \end{cases} \).
Comparing \( y'[n] \) and \( y[n-1] \), they are not identical (e.g., at \( n=1 \), \( y'[1]=x[0] \) while \( y[1-1]=y[0]=0 \)). The system is not time-invariant.

Step 3: Test for Causality.
A system is causal if the output \( y[n] \) depends only on the current and past inputs \( x[k] \) where \( k \le n \).
For \( n \le -1 \), the output is \( y[n] = x[n+1] \). Since \( n+1 > n \), the output depends on a future value of the input. Therefore, the system is not causal.

Step 4: Test for Stability (BIBO).
A system is Bounded-Input, Bounded-Output (BIBO) stable if every bounded input produces a bounded output. Assume the input is bounded, i.e., \( |x[n]| \le M_x < \infty \) for all \( n \).
For \( n \ge 1 \), \( |y[n]| = |x[n]| \le M_x \).
For \( n=0 \), \( |y[0]| = 0 \).
For \( n \le -1 \), \( |y[n]| = |x[n+1]| \le M_x \).
In all cases, the output is bounded by \( M_x \). The system is stable.

Conclusion: The system is Linear (A) and Stable (D).