Final answer:
The given system y(t)=∫t₀x(τ)dτ is linear as it satisfies the principle of superposition and is time-invariant as the output is the same regardless of when the input is applied.
Step-by-step explanation:
The system given by y(t)=∫t₀x(τ)dτ will be linear if it obeys the principle of superposition, meaning if y(t) is the response to x(t), then for any two inputs x₁(t) and x₂(t) and constants a and b, the response to ax₁(t) + bx₂(t) should be ay₁(t) + by₂(t). It will be time-invariant if the output does not depend on when the input is applied, so for a time shift τ, the output y(t) for x(t - τ) should be the same as y(t - τ) for x(t).
To prove linearity, consider two arbitrary functions x₁(t) and x₂(t), and let Y₁(t) and Y₂(t) be their corresponding outputs. Then,
Y₁(t) = ∫t₀x₁(τ)dτ and
Y₂(t) = ∫t₀x₂(τ)dτ. For any two constants a and b, the response to ax₁(t) + bx₂(t) is:
∫t₀(a*x₁(τ) + b*x₂(τ))dτ = a*∫t₀x₁(τ)dτ + b*∫t₀x₂(τ)dτ = a*
Y₁(t) + b*
Y₂(t), proving the system is linear.
To prove time-invariance, consider a shifted input x(t - τ) and its output:
y(t) = ∫t₀x(t - τ)dτ. After changing the variable of integration, one can show that this output is the same as shifting the response of the unshifted input by τ, proving the system is time-invariant.