Question

Discuss whether LTI systems with the following impulse responses are causal and/or stable where u(t) is the unit-step function.

h(t)=e⁻²ᵗ⋅v(t−5)

Answer 1

The LTI system with impulse response h(t)=e^{-2t}⋅u(t-5) is causal because the output at time t depends only on the current and past inputs, as ensured by u(t-5). It is also stable because the impulse response is absolutely integrable, given the exponential decay starting from t=5.

Step-by-step explanation:

To determine whether the Linear Time-Invariant (LTI) system with impulse response h(t)=e^{-2t}⋅u(t-5) is causal and/or stable, we need to look at the definition of causality and stability in LTI systems.

A system is causal if the output of the system at any time depends only on the present and past inputs, not on future inputs. In mathematical terms, for an impulse response h(t) of the system, causality implies that h(t)=0 for t < 0. In this case, since the unit-step function u(t-5) is zero for t < 5 and one for t ≥ 5, the impulse response h(t) is zero for t < 5, which means the system is causal.

In terms of stability, an LTI system is stable if its impulse response is absolutely integrable, that is, the integral of the absolute value of h(t) over the entire time axis is finite. Mathematically, ∫|h(t)|dt < ∞. Given h(t) only starts at t=5 due to the unit-step function and the exponential term e^{-2t} decays as t → ∞, the impulse response will integrate to a finite value. Thus, the given system is stable.