Question

Which one of the following discrete-time systems is time-invariant?

• A. \( y(n) = nx(n) \)
• B. \( y(n) = x(3n) \)
• C. \( y(n) = x(-n)\pi \)
• D. \( y(n) = x(n-3) \)

Hint:

Test time invariance: shift input, check if output shifts identically.

Answer 1

The Correct Option is D

To determine which discrete-time system is time-invariant, we need to understand the concept of time-invariance. A system is time-invariant if a time shift in the input signal causes an identical time shift in the output signal. Mathematically, a system is time-invariant if for any input signal \( x(n) \), the output \( y(n) \) satisfies:

\( \text{If } x'(n) = x(n-n_0), \text{ then } y'(n) = y(n-n_0) \)

Let's analyze each option:

1. \( y(n) = nx(n) \):
   If the input is \( x(n-n_0) \), the output becomes \( y'(n) = n \cdot x(n-n_0) \). Comparing it with \( y(n-n_0) = (n-n_0) \cdot x(n-n_0) \), we see they are not the same. Hence, this system is not time-invariant.

2. \( y(n) = x(3n) \):
   If the input is \( x(n-n_0) \), the output becomes \( y'(n) = x(3n-n_0) \). Comparing it with \( y(n-n_0) = x(3(n-n_0)) = x(3n-3n_0) \), they are not equal. Thus, this system is not time-invariant.

3. \( y(n) = x(-n)\pi \):
   If the input is \( x(n-n_0) \), the output becomes \( y'(n) = x(-(n-n_0))\pi = x(-n+n_0)\pi \). Comparing it with \( y(n-n_0) = x(-(n-n_0))\pi \), these expressions are not equal, indicating this system is not time-invariant.

4. \( y(n) = x(n-3) \):
   If the input is \( x(n-n_0) \), the output becomes \( y'(n) = x((n-n_0)-3) = x(n-n_0-3) \). This is equivalent to \( y(n-n_0) = x((n-n_0)-3) \), satisfying the time-invariance condition. Hence, this system is time-invariant.

Therefore, the correct answer is \( y(n) = x(n-3) \), which indicates a time-invariant system.