A voltage V(t) is a Gaussian, ergodic random process with mean 0 and variance 4 V^2. If measured with a direct-current (DC) meter (i.e., time average reading), what value will the meter indicate?

Question

Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:Understanding what various meters measure for random processes is fundamental in measurement and instrumentation. A DC (average-responding) meter indicates the time-average (mean) value of the input signal. For zero-mean random processes, this has a straightforward implication. Given Data / Assumptions: V(t) is Gaussian and ergodic. Mean μ = 0 V. Variance σ^2 = 4 V^2 (so rms = √σ^2 = 2 V, but rms is not what a DC meter reads). DC meter displays time average (mean) for an ergodic process. Concept / Approach:For an ergodic process, the time average equals the ensemble average. A DC meter, which essentially performs a long-time average, therefore indicates the process mean. Since μ = 0, the meter reading is 0 V. Note this differs from an rms-responding meter, which would show 2 V for the given variance. Step-by-Step Solution: DC meter → time average → equals ensemble mean for ergodic processes.Given mean μ = 0 → meter reading = 0 V. Verification / Alternative check: If an rms meter were used, the indicated value would be √(E[V^2]) = √(σ^2) = 2 V, highlighting the difference between average and rms measurements. Why Other Options Are Wrong: 4: This is the variance, not a DC average.2 or 2√2: These relate to rms or peak equivalents, not DC average.√2: No measurement mode here yields √2 V directly. Common Pitfalls: Confusing mean (average) with rms; instruments can be average-responding or true-rms—know which is used. Final Answer: 0

A voltage V(t) is a Gaussian, ergodic random process with mean 0 and variance 4 V^2. If measured with a direct-current (DC) meter (i.e., time average reading), what value will the meter indicate?

More Questions from Signals and Systems

Discussion & Comments