 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/14d/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

y (0) = 0

gives

(α s + 1) Y (s) = F (s) .

Hence the transfer function is

G (s) = \frac{Y ( s )}{F ( s )} = \frac{1}{α s + 1} .

Its pole is at

- 1/ α

, and its causal impulse response is

g (t) = \frac{1}{α} e^{- t / α} H (t) .

It is absolutely integrable exactly when

α > 0 .

 New to topics? Read the docs here!