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System properties; is y(t) = sin[x(t)] time-variant

Short answer

The system is time-invariant because a delay in input yields the same delay in output: sin[x(t - t0)] = y(t - t0). It is memoryless, causal, BIBO stable, nonlinear, and non-invertible.

What are the properties of system? Determine whether the given system is time-variant or not. y(t) = sin[x(t)]

Properties of systems

  1. Memory: memoryless if y(t)y(t) depends only on x(t)x(t) at the same instant (e.g. y=2x(t)y = 2x(t)); otherwise has memory (e.g. y=x(t−1)y = x(t-1)).
  2. Causality: causal if y(t)y(t) depends only on present and past inputs.
  3. Linearity: satisfies superposition: ax1+bx2→ay1+by2a x_1 + b x_2 \to a y_1 + b y_2.
  4. Time invariance: x(t−t0)→y(t−t0)x(t - t_0) \to y(t - t_0).
  5. Stability (BIBO): every bounded input gives a bounded output.
  6. Invertibility: distinct inputs give distinct outputs, so an inverse system exists.

Is y(t)=sin⁡[x(t)]y(t) = \sin[x(t)] time-variant?

Let the input be delayed: x1(t)=x(t−t0)x_1(t) = x(t - t_0). The output for this input is

y1(t)=sin⁡[x1(t)]=sin⁡[x(t−t0)]y_1(t) = \sin[x_1(t)] = \sin[x(t - t_0)]

The original output delayed by t0t_0 is

y(t−t0)=sin⁡[x(t−t0)]y(t - t_0) = \sin[x(t - t_0)]

Since y1(t)=y(t−t0)y_1(t) = y(t - t_0), the system is time-invariant (not time-variant). The operation sin⁡[⋅]\sin[\cdot] does not depend on tt explicitly.

Other properties, for completeness:

PropertyResultReason
Time invarianceTime-invariantshown above
LinearityNon-linearsin⁡(x1+x2)≠sin⁡x1+sin⁡x2\sin(x_1 + x_2) \ne \sin x_1 + \sin x_2
MemoryMemorylessuses only x(t)x(t)
CausalityCausalno future input
StabilityStable∣y(t)∣≤1\lvert y(t)\rvert \le 1 for any input
InvertibilityNot invertiblexx and x+2πx + 2\pi give same yy

✓ Answer checked by Claude Opus 5.5 (high effort)

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