Question

Let \(f(t)\) be a periodic function with period \(\tau\). Explain clearly why the average of \(f\) over one period is not necessarily the same as the average over some other time interval. Explain why, on the other hand, the average over a long time \(T\) approaches the average over one period, as \(T \rightarrow \infty\)

Short answer

The average over one period may differ from other intervals, but averages converge as time interval grows longer.

Step-by-step solution

Understanding a Periodic Function

A periodic function, like \(f(t)\), repeats its values in regular intervals, known as its period \(\tau\). This means that the behavior of \(f(t)\) is consistent over each segment of length \(\tau\), making \(f(t + \tau) = f(t)\) for any \(t\).

Average Over One Period

The average value of \(f(t)\) over one period \(\tau\) is calculated as \(\frac{1}{\tau}\int_{0}^{\tau} f(t) \, dt\). This represents how the values of \(f(t)\) are distributed over a single cycle.

Difference with Average Over Other Intervals

If we take an average over a time interval different from \(\tau\), say from \(0\) to \(T\) (where \(T\) is not a multiple of \(\tau\)), the function \(f(t)\) may not complete its periods entirely within \(T\). Thus, some parts of \(f(t)\) might be overrepresented or underrepresented, leading to an average that differs from the one-period average.

Approaching the Period Average as Time Increases

When taking the average over a long time \(T\), the interval \(T\) can potentially include many full periods of \(f(t)\). As \(T \rightarrow \infty\), any discrepancies from incomplete periods at the beginning or end of the interval become negligible. Thus, the average value over \(T\) approaches the average over a complete period. This can be expressed as \(\frac{1}{T}\int_{0}^{T} f(t) \, dt \approx \frac{1}{\tau}\int_{0}^{\tau} f(t) \, dt \) as \(T \rightarrow \infty\).

Key concepts

Average Value Over a Period

In mathematics, the average value of a periodic function over one complete period is a foundational concept. When we talk about periodic functions like \( f(t) \), which consistently repeat their behavior every period \( \tau \), this periodicity ensures predictability over each cycle. The average is obtained by integrating the function over one entire period and then dividing by the period length, given by the formula: \[ \text{Average over one period} = \frac{1}{\tau}\int_{0}^{\tau} f(t) \, dt \]This calculation tells us how the function behaves over each individual cycle. It assumes that each possible value within a cycle contributes proportionally to the average. For instance, if \( f(t) \) represents a sine wave, the average over one full sine wave cycle would consider both the ups and downs balanced over a complete period.

Incomplete Periods

When a periodic function is averaged over an interval that does not line up perfectly with its period, such as \( T \) where \( T eq n\tau \) for any integer \( n \), we encounter the challenge of incomplete periods. Imagine slicing through a loaf of bread—the slices don’t complete a full loaf exactly, leading to uneven pieces. Similarly, in an incomplete period, parts of \( f(t) \) might not be represented fully. This results in some values being counted more than others, while others may be missed completely, skewing the average. For example, if our interval stops halfway through an oscillation, the average becomes biased towards that partially completed oscillation. This difference occurs because the function might not return to its starting point, creating an imbalance in representation.

Long-Term Averaging

Long-term averaging is like observing a constant drip turning into a steady stream. When you average \( f(t) \) over a very large interval \( T \), the partial periods at start and end become increasingly insignificant. This is because the ratio of partial periods to complete periods decreases as \( T \) becomes very large. Thus, expressed mathematically, the long-term average:\[ \lim_{T \to \infty} \frac{1}{T} \int_{0}^{T} f(t) \, dt = \frac{1}{\tau} \int_{0}^{\tau} f(t) \, dt \]ensures the function’s true periodic nature eventually dominates any initial irregularities. This situation reflects the principle of ergodicity, where time averages and ensemble averages yield the same result in the limit, assuming the function is reasonably well-behaved. By considering ever longer times, you "smooth out" any discrepancies caused by incomplete cycles, converging towards the true average of the period.