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?

Difficulty: Easy

Correct Answer: 0

Explanation:


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

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