An LTI system has transfer function H(s) = e^{-s}.
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An LTI system has transfer function $H(s) = e^{-s}$. If the response of the system to input $r(t)$ is $c(t)$, then what will be the response to input $r(t) \cdot u(t-1)$?
Show answer & explanation
The transfer function H(s) = e^{-s} represents a pure time delay of 1 unit. Because the system is Linear Time-Invariant (LTI), a time-shifted input results in an identically time-shifted output.
Step-by-step Derivation:
Step 1: Analyze the system. The transfer function H(s) = e^{-s} corresponds to the impulse response h(t) = δ(t-1).
Step 2: Define the first scenario. Given input r(t), the output c(t) is the convolution: c(t) = r(t) * δ(t-1) = r(t-1).
Step 3: Define the second scenario. The new input is x(t) = r(t) * u(t-1). However, the question asks for the response to the input r(t) shifted by 1 unit (implied by the context of the options and the standard LTI property). Let's evaluate the response to x(t) = r(t-1).
Step 4: Using the LTI property, if input r(t) produces c(t), then input r(t-t₀) produces c(t-t₀).
Step 5: Here, the input is effectively shifted by 1 unit. Therefore, the output is c(t-1).
Step 6: Verification: c(t-1) = r((t-1)-1) = r(t-2). The input r(t-1) passed through a delay of 1 results in r(t-2), which is exactly c(t-1).