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Linear/TI/stable/memoryless: y(t), y[k]

Short answer

a) y(t) = (1 - e^{-4t})u(t) depends only on t, not input, so it is non-linear, time-variant, stable, and memoryless. b) y[k] = sin(x[k-4]) is non-linear (sin is not additive), time-invariant, stable (|y|≤1), and has memory (uses past sample).

Determine if the following systems are linear, time-invariant, stable and memoryless. a) y(t) = [1 - e^(-4t)][U(t)] where U(t) is the continuous-time unit step function. b) y[k] = sin(x[k-4])

a) y(t) = [1 - e^(-4t)] U(t)

The expression has no input x(t); it is the given output function. I take the system as one whose output is this fixed function of time, y(t) = (1 - e^(-4t))U(t) (independent of the input), and test it as such.

  • Linear? No. With zero input, the output is not zero (y ≠ 0 for t > 0), so superposition (zero in, zero out) fails.
  • Time-invariant? No. The output depends explicitly on t: shifting the input by t₀ does not shift the output, since the output does not follow the input at all.
  • Stable? Yes. 0 ≤ y(t) < 1 for all t, so the output is always bounded.
  • Memoryless? Yes. The output at t depends only on t, not on past or future values of any input.

(If the signal is meant to be the step response of a system, then h(t) = 4e^(-4t)U(t): that system is linear, time-invariant, stable, causal, and has memory.)

b) y[k] = sin(x[k - 4])

  • Linear? No. sin(a·x₁ + b·x₂) ≠ a·sin x₁ + b·sin x₂.
  • Time-invariant? Yes. A shift of n in the input gives sin(x[k - n - 4]) = y[k - n].
  • Stable? Yes. |y[k]| ≤ 1 for any input.
  • Memoryless? No. The output depends on the input four samples earlier, so the system has memory.
  • It is also causal (it uses a past value).

✓ Answer checked by Claude Opus 5.5 (high effort)

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