Time-invariance of y[n]=y[n−4]+x[n−4]

A system is time-invariant if a shift of the input by n₀ shifts the output by the same n₀.

System: y[n] = y[n − 4] + x[n − 4]. It is a recursive (difference-equation) system, taken with initial rest conditions (output is zero before the input starts).

Step 1. Apply input x₁[n] = x[n − n₀]. Output y₁[n] satisfies the same equation:

y₁[n] = y₁[n − 4] + x₁[n − 4] = y₁[n − 4] + x[n − n₀ − 4]

Step 2. Delay the original output by n₀. From y[n] = y[n − 4] + x[n − 4], replacing n by n − n₀:

y[n − n₀] = y[n − n₀ − 4] + x[n − n₀ − 4]

Step 3. Compare. Let w[n] = y[n − n₀]. Then w[n] = w[n − 4] + x[n − n₀ − 4], which is exactly the equation satisfied by y₁[n]. With the same initial rest condition both have the same solution, so

y₁[n] = y[n − n₀].

Equivalently, unrolling the recursion: y[n] = Σ(k=0 to ∞) x[n − 4 − 4k], and a shift of x shifts every term, so y shifts equally.

Answer: The system is time-invariant (all coefficients are constants, and with initial rest it is also linear).