 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2024/paper-3/7/e/solution

The squared signal equals

1 V^{2}

except at irrelevant isolated jump times. Its mean-square signal power is therefore

P = \frac{1}{4 s} \int_{0}^{4 s} ∣ g (t) ∣^{2} d t = 1 V^{2} .

Here “power” has the signal-processing definition specified in the question. To convert this voltage mean square into electrical power in watts, a load electrical resistance would also be required:

P_{electrical} = ⟨ g^{2} ⟩ / R

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