- 2070 Asar · 6 marks
Check linearity/TI: log(x(t)), t·x(t)
Short answer
i) y=log(x(t)): non-linear (log not additive) but time-invariant. ii) y=t·x(t): linear (amplification by t is linear) but time-variant because delay alters multiplier. Neither system is LTI.
Check whether the given system is linear and time invariant or not. i) y(t) = log(x(t)) ii) y(t) = t·x(t)
i) y(t) = log(x(t))
- Linearity: for x = a·x₁ + b·x₂, y = log(a·x₁ + b·x₂), while a·y₁ + b·y₂ = a·log x₁ + b·log x₂. They are not equal (log is not additive), and log(a·x) = log a + log x ≠ a·log x. Non-linear.
- Time invariance: input x(t - t₀) gives log x(t - t₀) = y(t - t₀). Time-invariant.
ii) y(t) = t·x(t)
- Linearity: response to a·x₁ + b·x₂ is t(a·x₁ + b·x₂) = a·t·x₁ + b·t·x₂ = a·y₁ + b·y₂. Linear.
- Time invariance: delay the input: y₁(t) = t·x(t - t₀). Delay the output: y(t - t₀) = (t - t₀)·x(t - t₀). These differ. Time-variant.
Answer: (i) non-linear and time-invariant; (ii) linear but time-variant. Neither system is LTI.
✓ Answer checked by Claude Opus 5.5 (high effort)
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