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Problem 1 [LTI]. For each of the following systems determine whether or not the system is (1) linear, (2) time-invariant, and (3) causal. Explain your answers!

(a) y[n] = ∑[infinity]ₖ₌₀ x[n-k]

User Ppichier
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Final answer:

The system described by y[n] = sum from k=0 to infinity of x[n-k] is linear because it satisfies the superposition principle, time-invariant because a shift in the input signal results in an equivalent shift in the output, and causal because the output depends only on the current and past input values.

Step-by-step explanation:

The student has asked to determine whether a given system is linear, time-invariant, and causal based on its output equation y[n] = ∞∞k=0 x[n-k]. To assess these properties, we must apply certain tests to the system.

Linearity

A system is linear if it satisfies the properties of superposition (additivity and homogeneity). In the case of the given equation, if we consider two signals x1[n] and x2[n] and a scalar a, then the response to the combined input a*x1[n] + x2[n] will be a*y1[n] + y2[n], which satisfies the linearity criterion.

Time-Invariance

For time invariance, we check if a time-shift in the input signal results in an equivalent time-shift in the output signal. Here, shifting the input x[n] to x[n-m] results in an output shift, y[n-m], thus maintaining time invariance.

Causality

The system is causal if the output at any time n only depends on the input at the current and past times. Since the sum in the equation only goes up to the current time n and does not depend on future values of x, the system is indeed causal.

Therefore, the system described by y[n] = ∞∞k=0 x[n-k] is linear, time-invariant, and causal.

User David Rabinowitz
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